Cambridge IGCSE Chemistry Revision
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Cambridge IGCSE Chemistry (0620)
Exam years · 2027Specification version · Version 2
Course details
CIE · IGCSE Chemistry0620Exam years · 2027
Assessment The assessment includes Paper 1, Paper 2, Paper 3, Paper 4, Paper 5 and Paper 6; see the official specification for the full structure.
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Reviewed course guidance
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Open a point to read what to know, how to use it in an assessment, a plain-language explanation and a self-check. Answers stay hidden until you choose to reveal them.
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Cambridge IGCSE Chemistry 0620 theory route
- Core route (Papers 1 and 3; Core content; grades C–G)
- Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G)
Cambridge IGCSE Chemistry 0620 practical assessment
- Paper 5 Practical Test (laboratory examination)
- Paper 6 Alternative to Practical (written examination)
§2 Assessment overview and §4 Practical assessment: choose one paper.
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Topic group
1 States of matter
- Course content
Scope: Course content · §3 1.1 Core 1–4; 1.2 Core 1
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What to know
- A solid has a fixed shape and volume: particles are very close in an ordered arrangement and vibrate about fixed positions. A liquid has a fixed volume but takes its container's shape: particles remain close but are disordered and slide past one another. A gas has neither fixed shape nor volume: widely separated particles move rapidly and randomly and fill the container.
- Melting changes solid to liquid, boiling and evaporation change liquid to gas, condensation changes gas to liquid, and freezing changes liquid to solid. These are physical changes: the substance's particles remain the same. Heating raises particle motion; cooling lowers it.
- At constant pressure, heating a gas increases its volume; at constant temperature, increasing pressure decreases its volume. The explanation is the balance between particle motion, collision frequency and the pressure exerted on the gas.
- Diffusion is the net movement of particles from higher to lower concentration due to random motion. It occurs in liquids and gases: for example, a coloured solute spreads through still water and a gas spreads through a room.
What to do in an assessment
- For each state, compare shape, volume, spacing, arrangement and motion; for a state change, name the direction and explain it without proposing a chemical change. State what is held constant when predicting a gas-volume change.
In plain language
The same substance behaves differently when its particles have different spacing and freedom to move.
Self-check
A fixed amount of gas is warmed at constant pressure. What happens to its volume, and why does perfume eventually spread across a room?
Show the answer
Its volume increases as heating raises particle motion. Perfume vapour diffuses because randomly moving gas particles spread from the higher-concentration region near the source into regions of lower concentration.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 1.1 Supplement 5–6; 1.2 Supplement 2
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What to know
- In a single phase, heating increases particles' average kinetic energy and temperature. During melting or boiling, the temperature remains constant while transferred energy separates particles against attractions (increases potential energy); cooling reverses these changes. On a heating/cooling curve, sloped regions are temperature changes and flat plateaux are state changes.
- At constant pressure, a temperature rise makes gas particles move faster and spread farther apart, so the gas volume increases; at constant temperature, a pressure rise forces particles closer together and decreases volume.
- At the same temperature, gas particles have the same average kinetic energy, but lighter molecules move faster on average and diffuse faster than heavier ones. For example, ammonia (Mᵣ 17) diffuses faster than hydrogen chloride (Mᵣ 36.5).
What to do in an assessment
- Read a heating/cooling curve from left to right and identify the state(s), plateaux and sloping regions; explain the temperature and energy change in each. When comparing gas volume or diffusion, state the controlled variable and compare the particle-level cause.
In plain language
Supplement particle theory connects energy transfer to state changes and explains gas expansion, compression and diffusion rates.
Self-check
At the same temperature, which diffuses faster: ammonia (Mᵣ 17) or hydrogen chloride (Mᵣ 36.5)? During boiling, why does added energy not raise the temperature?
Show the answer
Ammonia diffuses faster because its lighter molecules move faster on average at the same temperature. During boiling, transferred energy separates particles rather than increasing their average kinetic energy, so the temperature remains constant.
Topic group
2 Atoms, elements and compounds
- Course content
Scope: Course content · §3 2.1 Core 1; 2.2 Core 1–6; 2.3 Core 1–2
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What to know
- An element contains one type of atom. A compound contains two or more elements chemically bonded in a fixed ratio. A mixture contains substances together but not chemically bonded; components retain their own properties and may be separated physically.
- An atom has a central nucleus containing protons and neutrons, with electrons in shells. Relative charges are proton +1, neutron 0, electron −1; relative masses are approximately proton 1, neutron 1 and electron 1/1836. Proton number (atomic number) is the number of protons; mass number (nucleon number) is protons plus neutrons.
- For proton numbers 1–20, fill the first shell with 2 electrons, the second with 8, then the third with 8 before placing any remaining electrons in the fourth shell. Thus K (Z=19) is 2,8,8,1 and Ca (Z=20) is 2,8,8,2; chlorine (Z=17) is 2,8,7. Check each arrangement by summing its shell electrons and matching the atom's electron count. For ions, add electrons for a negative charge and remove them for a positive charge: Ca²⁺ has 20 protons but 18 electrons, so it is 2,8,8; Cl⁻ has 17 protons but 18 electrons, so it is 2,8,8. Group VIII noble gases have full outer shells; for Groups I–VII, outer-electron number equals group number, and occupied-shell number equals period number.
- Isotopes are atoms of the same element with the same proton number but different neutron numbers. In nuclide notation, the upper left number is mass number and the lower left is proton number; for example, ²³₁₁Na has 11 protons and 12 neutrons. For ³⁵₁₇Cl⁻, there are 17 protons, 18 neutrons and 18 electrons.
What to do in an assessment
- Calculate neutrons = mass number − proton number; check an electron arrangement by summing electrons and applying the ion charge; read both numbers in nuclide notation before naming protons, neutrons or electrons.
In plain language
Atomic numbers identify elements; neutron number and electron arrangement explain isotope and ion notation.
Self-check
A chloride ion is formed from chlorine-35 (proton number 17). Give its neutron count and electron arrangement.
Show the answer
Neutrons = 35 − 17 = 18. Neutral chlorine has 17 electrons; Cl⁻ has gained one, so it has 18 electrons arranged 2,8,8.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 2.3 Supplement 3–4
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What to know
- Isotopes of an element have the same proton number. Their neutral atoms therefore have the same number and arrangement of electrons, which gives them the same chemical properties.
- Relative atomic mass is the abundance-weighted mean of isotope masses compared with one-twelfth of a carbon-12 atom; it need not be a whole number.
What to do in an assessment
- Calculate relative atomic mass as Σ(isotope mass × fractional abundance); when using percentage abundances divide the sum of mass × percentage by 100. Check that the abundances total 100%.
In plain language
Relative atomic mass is an average that gives more abundant isotopes more influence.
Self-check
An element has isotopes of mass 35 (75%) and 37 (25%). Find its relative atomic mass.
Show the answer
Aᵣ = [(35 × 75) + (37 × 25)] ÷ 100 = 35.5; the result is an abundance-weighted mean.
- Course content
Scope: Course content · §3 2.4 Core 1–4; 2.5 Core 1–3
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What to know
- A cation forms when an atom loses electron(s); an anion forms when it gains electron(s). An ionic bond is the strong electrostatic attraction between oppositely charged ions. In sodium chloride, sodium transfers its one outer electron to chlorine, producing Na⁺ and Cl⁻; draw [Na]⁺ and [Cl]⁻ in brackets, with eight outer electrons around Cl (seven of its own plus the transferred electron shown as a cross).
- Ionic compounds have high melting and boiling points because strong electrostatic attractions between oppositely charged ions throughout the lattice require much energy to overcome. In a solid, ions are fixed and cannot carry current; when molten or dissolved in water, ions can move and conduct.
- A covalent bond is a shared pair of electrons; the shared electrons give each atom a noble-gas outer-shell arrangement. Draw dot-and-cross diagrams for H₂, Cl₂, H₂O, CH₄, NH₃ and HCl: each H forms one shared pair, O forms two bonds and has two lone pairs, N forms three bonds and has one lone pair, C forms four bonds, and each Cl forms one bond and has three lone pairs. These are small separate molecules with low melting/boiling points because only a small amount of energy is needed to separate the molecules; the covalent bonds within each molecule remain intact during melting or boiling. They contain no mobile ions/electrons and do not conduct electricity; naming specific intermolecular-force types is not required.
What to do in an assessment
- For NaCl, show Na's outer electron transferred to Cl and label the resulting charges; check total charge is zero. For a covalent molecule, draw one line or shared dot-cross pair per bond and add lone pairs so each atom reaches its outer-shell target.
In plain language
Ionic bonding transfers electrons between atoms; covalent bonding shares electron pairs.
Self-check
Why does sodium chloride conduct when molten but not when solid?
Show the answer
In the solid lattice, Na⁺ and Cl⁻ are fixed in place; when molten, these charged ions are mobile and carry current.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 2.4 Supplement 5–7; 2.5 Supplement 4–5
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What to know
- Ionic solids form a regular giant lattice of alternating positive and negative ions. Each ion is attracted to oppositely charged neighbours in all directions, so the lattice is strong and ionic compounds have high melting and boiling points. Ions carry current only when free to move in a melt or aqueous solution.
- Ionic bonding also forms between metals and non-metals beyond Group I/VII: magnesium transfers two outer electrons to oxygen, giving Mg²⁺ and O²⁻ in MgO. In a dot-and-cross diagram show the two transferred electrons and the full outer shells of both ions.
- Supplement dot-and-cross diagrams include CH₃OH, C₂H₄, O₂, CO₂ and N₂. Show one C=C double bond in ethene, O=O double bond in oxygen, two C=O double bonds in O=C=O, and an N≡N triple bond; include the appropriate non-bonding electron pairs on O and N. In methanol carbon forms four bonds, oxygen two bonds and two lone pairs, and hydrogen one bond.
- Small simple molecules have weak intermolecular attractions between molecules (specific force types are not required), so relatively little energy separates them and they have low melting and boiling points. The covalent bonds within each molecule are not broken during melting or boiling; the substances do not conduct because they have no mobile ions or electrons.
What to do in an assessment
- Explain a property by connecting structure, bonding and the particles or charge carriers that can move.
In plain language
A material's large-scale properties follow from its bonding and structure.
Self-check
Why does a simple covalent molecular substance have a low boiling point without breaking its covalent bonds?
Show the answer
Boiling separates molecules by overcoming their weak intermolecular attractions; the strong covalent bonds within each molecule remain intact.
- Course content
Scope: Course content · §3 2.6 Core 1–2
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What to know
- In diamond, each carbon atom forms four strong covalent bonds in a rigid three-dimensional giant structure. Many strong bonds make diamond hard and give it a high melting point; its hardness suits cutting tools.
- In graphite, each carbon forms three covalent bonds in flat layers. The layers can slide over one another, so graphite is useful as a lubricant. One electron per carbon is delocalised within the layers and can carry charge, so graphite is useful as an electrode.
What to do in an assessment
- Relate each stated use to a specific structural feature rather than merely naming a property.
In plain language
Different carbon structures create different useful properties.
Self-check
How can graphite be both a lubricant and an electrode?
Show the answer
Its carbon layers slide over each other, while delocalised electrons move through the layers and carry charge.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 2.6 Supplement 3–4; 2.7 Supplement 1–2
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What to know
- Silicon(IV) oxide, SiO₂, has a giant covalent structure: silicon and oxygen atoms are linked by strong covalent bonds throughout a three-dimensional network. Its hardness and high melting point resemble diamond's because both structures contain many strong covalent bonds.
- Metallic bonding is the electrostatic attraction between positive ions in a giant metallic lattice and a sea of delocalised electrons. The mobile electrons carry electrical charge. Layers of metal ions can slide while attraction to the electron sea remains, explaining malleability and ductility.
What to do in an assessment
- For a metal property, identify the role of the delocalised electrons or the ability of the lattice to deform.
In plain language
Metallic bonding combines fixed positive ions with mobile electrons.
Self-check
What moves when a metal conducts?
Show the answer
Delocalised electrons move through the metal lattice.
Topic group
3 Stoichiometry
- Course content
Scope: Course content · §3 3.1 Core 2–4; 3.2 Core 1–3
Open revision guidance
What to know
- A molecular formula gives the number and type of atoms in one molecule. From a model, count each atom: two hydrogen atoms and one oxygen atom give H₂O; one carbon and four hydrogens give CH₄.
- A word equation names reactants and products (magnesium + oxygen → magnesium oxide). A balanced symbol equation conserves the number of each type of atom: 2Mg(s) + O₂(g) → 2MgO(s). State symbols are (s), (l), (g) and (aq); use them when required.
- Relative atomic mass, Aᵣ, is the average mass of an element's isotopes relative to one-twelfth of a carbon-12 atom. Mᵣ is the sum of the Aᵣ values in a molecule (H₂O: 2(1) + 16 = 18); for an ionic compound use relative formula mass (MgO: 24 + 16 = 40).
- Core reacting-mass calculations use simple proportions, not moles. From 2Mg + O₂ → 2MgO, 48 g Mg reacts with 32 g O₂ to form 80 g MgO; 12 g Mg therefore forms 20 g MgO if oxygen is in excess.
What to do in an assessment
- Write correct formulae first, balance by changing coefficients (never subscripts), then check every element's atom count and the state symbols on both sides. For a reacting-mass proportion, use the balanced coefficients and relative formula masses.
In plain language
A chemical equation is a particle-counting statement: atoms are rearranged, not created or lost.
Self-check
Balance the equation and calculate the mass of carbon dioxide made from 12 g carbon in excess oxygen: C + O₂ → CO₂.
Show the answer
C + O₂ → CO₂ is already balanced. Aᵣ(C)=12 and Mᵣ(CO₂)=44, so 12 g C is in the 12:44 mass ratio and forms 44 g CO₂.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 3.1 Supplement 5–8
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What to know
- An empirical formula is the simplest whole-number ratio of atoms or ions. For an ionic formula, balance total charge and reduce to the simplest ratio: Mg²⁺ and Cl⁻ give MgCl₂; Al³⁺ and O²⁻ give Al₂O₃. A model gives the same ratios by counting ions: six Cl⁻ around three Mg²⁺ have a 2:1 ratio, so the formula unit is MgCl₂.
- Construct balanced symbol equations with state symbols; ionic equations show reacting ions and cancel spectator ions. For example, CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq) has the net ionic equation Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s). Check both atom numbers and total charge.
- When information describes a reaction, identify reactants and products, write their formulae, balance coefficients, and add states from the conditions given.
What to do in an assessment
- For empirical formulae, divide mole amounts by the smallest and scale to whole numbers. For ionic formulae, make the total charge zero. For symbol or ionic equations, balance atoms, then charge, and retain only reacting species in the net ionic equation.
In plain language
Formulae encode simplest composition; ionic equations show only the particles that change.
Self-check
Write the net ionic equation for aqueous barium nitrate reacting with aqueous sodium sulfate to form a precipitate.
Show the answer
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). It is balanced for atoms and total charge.
- Course content
Scope: Course content · §3 3.3 Core 1
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What to know
- Concentration states how much solute is in a stated solution volume. Mass concentration uses grams per dm³ (g/dm³); amount concentration uses moles per dm³ (mol/dm³). They have different numerators and are not interchangeable.
What to do in an assessment
- Convert cm³ to dm³ by dividing by 1000 before using a molar concentration; keep the specified g/dm³ or mol/dm³ unit with the answer.
In plain language
Concentration states how much solute is present in a stated volume of solution.
Self-check
A solution contains 2.0 g solute in 250 cm³. Find its mass concentration in g/dm³.
Show the answer
250 cm³ = 0.250 dm³, so concentration = 2.0 ÷ 0.250 = 8.0 g/dm³.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 3.3 Supplement 2–8
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What to know
- One mole is the unit for amount of substance and contains 6.02 × 10²³ particles (Avogadro constant). Use n=m/M, m=nM and M=m/n (mass in g, molar mass in g/mol); particle number N=nNₐ. Examples: 5.85 g NaCl with Mᵣ=58.5 g/mol is 0.100 mol; 0.20 mol of a substance with M=40 g/mol has mass 8.0 g; 4.0 g in 0.10 mol gives M=40 g/mol and relative mass 40; 0.25 mol contains 1.505 × 10²³ particles.
- At r.t.p., 1 mol gas occupies 24 dm³, so n=V/24. For 2H₂ + O₂ → 2H₂O, 16 g O₂ is 0.50 mol and is limiting if 4.0 g H₂ (2.0 mol) is present; it reacts with 1.0 mol H₂ and forms 1.0 mol H₂O = 18 g. A 48 dm³ sample of gas at r.t.p. is 2.0 mol.
- For solutions, n=cV with V in dm³; use the balanced equation's mole ratio. A 25.0 cm³ acid sample neutralised by 20.0 cm³ of 0.100 mol/dm³ NaOH (1:1 reaction) contains 0.00200 mol acid, so its concentration is 0.00200/0.0250 = 0.0800 mol/dm³. To prepare 0.050 mol at 0.250 mol/dm³, V=n/c=0.200 dm³; if Mᵣ=40, that amount has mass 2.0 g, giving 10 g/dm³. Convert cm³ to dm³ by dividing by 1000.
- For an empirical formula, convert masses (or percentages treated as grams in a 100 g sample) to moles and divide by the smallest; 40.0% C, 6.7% H and 53.3% O give a 1:2:1 ratio, CH₂O. If Mᵣ=180 and empirical formula mass=30, the molecular formula is six times larger: C₆H₁₂O₆.
- Percentage yield = actual yield/theoretical yield × 100%; percentage composition = element mass/formula mass × 100%; purity = pure-substance mass/impure-sample mass × 100%. Examples: 7.2 g product from a theoretical 8.0 g is 90% yield; hydrogen is 2/18 × 100 = 11.1% of water by mass; 9.0 g pure material in a 10.0 g sample is 90% pure.
What to do in an assessment
- Balance the equation; convert measured quantities to moles; identify the limiting reactant; apply the coefficient ratio; convert to the requested mass, gas volume, solution volume, concentration or particle number; include units and appropriate significant figures.
In plain language
The mole lets a balanced equation connect measurable mass, solution and gas quantities.
Self-check
A titration uses 20.0 cm³ of 0.100 mol/dm³ NaOH to neutralise 25.0 cm³ HCl. Find the HCl concentration, given HCl + NaOH → NaCl + H₂O.
Show the answer
n(NaOH)=0.100 × 0.0200=0.00200 mol. The 1:1 equation gives 0.00200 mol HCl in 0.0250 dm³; c=0.00200/0.0250=0.0800 mol/dm³.
- Course content
Scope: Course content · §3 3.1 Core 1
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What to know
- Element symbols/formulae used across the course: H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂ are diatomic; C, Si, S and P are written as element symbols in formulae. The named noble gases include He, Ne and Ar. Named metals include Li, Na, K, Mg, Ca, Al, Zn, Fe, Ni, Cu, Ag, Au, Cr, Co, Ba, Pb and Sn; a metal element's formula is its symbol. Use the periodic table to identify the symbol for any other element named.
- Common molecules and gases: water H₂O; hydrogen H₂; oxygen O₂; nitrogen N₂; chlorine Cl₂; hydrogen chloride HCl; ammonia NH₃; methane CH₄; carbon monoxide CO; carbon dioxide CO₂; sulfur dioxide SO₂; sulfur trioxide SO₃; nitrogen monoxide NO; nitrogen dioxide NO₂; glucose C₆H₁₂O₆.
- Common acids, bases and ions: hydrochloric acid HCl; nitric acid HNO₃; sulfuric acid H₂SO₄; ethanoic acid CH₃COOH; hydroxide OH⁻; ammonium NH₄⁺; nitrate NO₃⁻; carbonate CO₃²⁻; sulfate SO₄²⁻; sulfite SO₃²⁻; phosphate PO₄³⁻; chloride Cl⁻; bromide Br⁻; iodide I⁻; hydrogen ion H⁺. The aqueous cations named in the analysis syllabus are Al³⁺, NH₄⁺, Ca²⁺, Cr³⁺, Cu²⁺, Fe²⁺, Fe³⁺ and Zn²⁺; flame-test cations include Li⁺, Na⁺, K⁺, Ca²⁺, Ba²⁺ and Cu²⁺.
- Common oxides, hydroxides and carbonates: magnesium oxide MgO; aluminium oxide Al₂O₃; silicon dioxide SiO₂; iron(III) oxide Fe₂O₃; hydrated iron(III) oxide Fe₂O₃·xH₂O; copper(II) oxide CuO; calcium oxide CaO; calcium hydroxide Ca(OH)₂; sodium hydroxide NaOH; potassium hydroxide KOH; calcium carbonate CaCO₃; magnesium carbonate MgCO₃; carbon monoxide CO; carbon dioxide CO₂; sulfur dioxide SO₂; nitrogen monoxide NO; nitrogen dioxide NO₂.
- Common salts in reaction and analysis examples: sodium chloride NaCl; lithium chloride LiCl; potassium chloride KCl; aluminium chloride AlCl₃; magnesium chloride MgCl₂; zinc chloride ZnCl₂; iron(II) chloride FeCl₂; calcium chloride CaCl₂; copper(II) chloride CuCl₂; cobalt(II) chloride CoCl₂; ammonium chloride NH₄Cl; copper(II) sulfate CuSO₄; zinc sulfate ZnSO₄; iron(II) sulfate FeSO₄; magnesium sulfate MgSO₄; silver nitrate AgNO₃; barium nitrate Ba(NO₃)₂; calcium nitrate Ca(NO₃)₂; ammonium nitrate NH₄NO₃; ammonium sulfate (NH₄)₂SO₄; silver chloride AgCl; silver bromide AgBr; silver iodide AgI; barium sulfate BaSO₄; calcium sulfate CaSO₄; sodium carbonate Na₂CO₃; sodium ethanoate CH₃COONa; magnesium ethanoate Mg(CH₃COO)₂; calcium silicate CaSiO₃; cryolite Na₃AlF₆; potassium manganate(VII) KMnO₄. Hydrated salts include blue copper(II) sulfate CuSO₄·5H₂O and hydrated cobalt(II) chloride CoCl₂·6H₂O.
- Organic formulas include the series CₙH₂ₙ₊₂ (alkanes), CₙH₂ₙ (alkenes), CₙH₂ₙ₊₁OH (alcohols) and CₙH₂ₙ₊₁COOH (carboxylic acids); examples include methane CH₄, ethane C₂H₆, propane C₃H₈, butane C₄H₁₀, ethene C₂H₄, propene C₃H₆, butene C₄H₈, methanol CH₃OH, ethanol C₂H₅OH, propanol C₃H₇OH, butanol C₄H₉OH, methanoic acid HCOOH, ethanoic acid CH₃COOH, propanoic acid C₂H₅COOH and butanoic acid C₃H₇COOH. Also recall ethyl ethanoate CH₃COOCH₂CH₃, poly(ethene) repeat unit [–CH₂–CH₂–]ₙ and glucose C₆H₁₂O₆. Use displayed/structural formula rules to write the specified unbranched isomers, ester combinations and polymer repeat units.
What to do in an assessment
- Recall the symbol or formula from the chemical name; use subscripts for atom counts, brackets for a repeated polymer unit, and charges for ions. For any other named ionic compound, use the ions' charges to determine the simplest neutral formula and retain any stated water of crystallisation. Check that ionic formulae have zero total charge and molecular formulae preserve each atom count.
In plain language
A chemical name identifies a fixed symbol or atom ratio; this bank gathers the formulae needed throughout the course.
Self-check
Give the formula for calcium hydroxide, the nitrate ion, and a typical alkane with four carbon atoms.
Show the answer
Calcium hydroxide is Ca(OH)₂, nitrate is NO₃⁻, and the alkane formula CₙH₂ₙ₊₂ gives C₄H₁₀.
Topic group
4 Electrochemistry
- Course content
Scope: Course content · §3 4.1 Core 1–7
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What to know
- Electrolysis is decomposition of an ionic compound by an electric current when it is molten or in aqueous solution. The anode is positive, the cathode negative, and the electrolyte is the molten or aqueous ionic substance that undergoes electrolysis.
- With inert electrodes, molten PbBr₂ gives lead at the cathode and bromine at the anode (silvery lead forms; reddish-brown bromine vapour appears). Concentrated aqueous NaCl gives hydrogen at the cathode and chlorine at the anode (bubbles form; chlorine is a greenish-yellow gas that bleaches damp indicator paper). Dilute H₂SO₄ gives hydrogen at the cathode and oxygen at the anode (bubbles; hydrogen gives a squeaky pop and oxygen relights a glowing splint).
- At the cathode, metals or hydrogen form; at the anode, non-metals other than hydrogen form. To predict products from a molten binary compound, send the metal cation to the cathode and the non-metal anion to the anode, then identify the elements formed.
- Electroplating improves appearance and resistance to corrosion. Connect the object to be coated as the cathode, the coating metal as the anode, and use an electrolyte containing ions of the coating metal; metal deposits on the object as the anode supplies ions.
What to do in an assessment
- For each electrode, identify its sign, arriving ion, product and observation. Use the stated physical state and concentration of the electrolyte; label anode/cathode correctly before predicting products.
In plain language
An electric current drives oppositely charged ions to electrodes where products form.
Self-check
Predict the products in molten lead(II) bromide and name the electrode signs.
Show the answer
Pb²⁺ goes to the negative cathode and forms lead; Br⁻ goes to the positive anode and forms reddish-brown bromine. The electrolyte must be molten so its ions can move.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 4.1 Supplement 8–11
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What to know
- Electrons move through the external circuit, while ions carry charge through the electrolyte: cations move to the cathode and anions to the anode. At the cathode cations gain electrons (reduction); at the anode anions lose electrons (oxidation). Electrons are not the charge carriers travelling through the solution.
- In aqueous CuSO₄ with inert carbon/graphite electrodes, reddish-brown copper deposits at the cathode and oxygen bubbles at the anode; Cu²⁺ is removed, so the blue solution becomes paler. With copper electrodes, copper deposits at the cathode while the anode dissolves; its mass decreases and the blue colour/concentration is maintained.
- For an aqueous halide, the cathode product depends on the cation: a less reactive metal than hydrogen deposits as metal; otherwise hydrogen forms. At the anode, a concentrated chloride/bromide/iodide solution forms the halogen; a dilute halide solution forms oxygen instead. Thus concentrated NaCl(aq) gives H₂ and Cl₂; dilute NaCl(aq) gives H₂ and O₂.
- Half-equations show electron transfer and conserve atoms and charge. Reduction at the cathode: Cu²⁺ + 2e⁻ → Cu; hydrogen formation: 2H⁺ + 2e⁻ → H₂. Oxidation at the anode: 2Cl⁻ → Cl₂ + 2e⁻; oxygen formation in dilute acid: 2H₂O → O₂ + 4H⁺ + 4e⁻.
What to do in an assessment
- Write the cathode and anode processes separately; balance atoms first and electrons/charge next. Electrons appear as reactants for reduction and products for oxidation; multiply half-equations only when combining them so electron numbers cancel.
In plain language
The electrodes exchange electrons while ions carry charge inside the solution.
Self-check
Write the half-equations for concentrated aqueous sodium chloride and identify oxidation/reduction.
Show the answer
Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ (reduction). Anode: 2Cl⁻ → Cl₂ + 2e⁻ (oxidation). At the cathode electrons are gained; at the anode electrons are lost.
- Course content
Scope: Course content · §3 4.2 Core 1
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What to know
- A hydrogen–oxygen fuel cell uses hydrogen and oxygen to produce electricity, with water as its only chemical product. The cell converts chemical energy into electrical energy; the reactants are supplied continuously.
What to do in an assessment
- Distinguish the cell's chemical product from electricity, its useful output.
In plain language
A hydrogen–oxygen fuel cell produces electricity from supplied hydrogen and oxygen, with water as its only chemical product.
Self-check
What are the reactants and only chemical product of a hydrogen–oxygen fuel cell?
Show the answer
Hydrogen and oxygen are used to produce electricity; water is the only chemical product.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 4.2 Supplement 2
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What to know
- At the vehicle, a hydrogen–oxygen fuel cell produces electricity and water and emits no carbon dioxide from the cell; a petrol engine releases energy by combustion and produces carbon dioxide and other exhaust gases.
- Fuel cells can be more energy-efficient and hydrogen can be made using renewable electricity, but hydrogen must be produced, compressed or liquefied, transported and stored; these steps use energy and add cost. Petrol has an established storage/refuelling network, while hydrogen infrastructure and fuel-cell systems are expensive and less widespread.
- The environmental comparison depends on how hydrogen and petrol are produced and delivered, not only on the vehicle's tailpipe.
What to do in an assessment
- Compare at least one advantage and one disadvantage on both point-of-use emissions and fuel supply; support a conclusion with the stated criteria.
In plain language
A fuel cell is cleaner at the point of use, but its total merits depend on how hydrogen is made and supplied.
Self-check
Give one advantage and one disadvantage of a hydrogen fuel-cell vehicle compared with a petrol vehicle.
Show the answer
An advantage is that the cell's only chemical product is water and it emits no CO₂ at the vehicle. A disadvantage is that hydrogen production, storage and distribution require energy and costly infrastructure.
Topic group
5 Chemical energetics
- Course content
Scope: Course content · §3 5.1 Core 1–3
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What to know
- An exothermic reaction transfers thermal energy from the reacting system to its surroundings, so the surroundings warm. Combustion is a familiar exothermic example.
- An endothermic reaction takes thermal energy from the surroundings into the reacting system, so the surroundings cool. Thermal decomposition of calcium carbonate is endothermic.
- On a reaction pathway diagram, energy is the vertical axis and reaction progress the horizontal axis. The reactants and products are horizontal energy levels joined by a pathway: products lower than reactants mean a net energy release (exothermic); products higher than reactants mean energy is taken in (endothermic).
What to do in an assessment
- Read the vertical energy levels of reactants and products, compare them, and state the direction of thermal-energy transfer and the resulting surroundings-temperature change.
In plain language
The temperature change around a reaction reveals the direction of thermal-energy transfer.
Self-check
A pathway diagram shows products below reactants. Classify the reaction and predict the change in the surroundings.
Show the answer
It is exothermic: energy is transferred to the surroundings, so their temperature rises.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 5.1 Supplement 4–8
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What to know
- The transfer of thermal energy in a reaction is its enthalpy change, ΔH. ΔH is negative for an exothermic reaction and positive for an endothermic reaction.
- Activation energy, Eₐ, is the minimum energy colliding particles must have to react. On a pathway diagram, Eₐ is the vertical energy difference from the reactant level to the peak; ΔH is the signed vertical difference from the reactant level to the product level.
- To draw either pathway from information, label the vertical axis energy and horizontal axis reaction progress, put reactants and products at their stated relative energies, draw the pathway peak, and add arrows for ΔH and Eₐ. For an exothermic profile the products are lower; for an endothermic profile they are higher.
- Breaking bonds is endothermic because energy is absorbed; making bonds is exothermic because energy is released. ΔH = total energy needed to break bonds − total energy released when new bonds form.
- Worked bond-energy example: H₂ + Cl₂ → 2HCl. With H–H = 436, Cl–Cl = 243 and H–Cl = 431 kJ/mol, ΔH = (436 + 243) − 2(431) = −183 kJ/mol, so the reaction is exothermic.
What to do in an assessment
- Label reactants, products, ΔH and Eₐ on both types of pathway. For a bond-energy calculation, count every bond broken and formed using the balanced equation, multiply each bond energy by its coefficient, then subtract total energy released from total energy absorbed; include kJ/mol and interpret the sign.
In plain language
A reaction releases energy only when forming its new bonds releases more than breaking the old bonds requires.
Self-check
For H₂ + Br₂ → 2HBr, use H–H = 436, Br–Br = 193 and H–Br = 366 kJ/mol to calculate ΔH.
Show the answer
ΔH = (436 + 193) − 2(366) = −103 kJ/mol. The negative value means the reaction is exothermic.
Topic group
6 Chemical reactions
- Course content
Scope: Course content · §3 6.1 Core 1; 6.2 Core 1–4
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What to know
- A physical change changes a substance's state or form without making a new substance; melting ice is physical. A chemical change forms one or more new substances because atoms are rearranged; burning magnesium is chemical.
- Reaction rate increases when solution concentration or gas pressure is increased, when a solid's surface area is increased, when temperature is raised, or when a suitable catalyst (including an enzyme) is added. Removing a catalyst or reversing these changes reduces the rate. A catalyst increases rate and is chemically unchanged at the end.
- Investigate rate by measuring a reactant's mass decrease or product's mass increase over time, or by measuring gas volume formed over time. For example, place a flask containing acid and marble chips on a balance and record mass at fixed time intervals as CO₂ escapes; alternatively connect the flask to a gas syringe and record gas volume at the same intervals.
- Calculate an average rate as change in mass or gas volume divided by elapsed time. On a quantity-versus-time graph, the gradient shows rate: a steeper curve means faster reaction; the curve levels when the reaction stops. Compare initial rates using the initial gradient.
What to do in an assessment
- For a fair rate test, change one factor only, keep other conditions and quantities the same, start timing consistently, record mass or gas volume at regular intervals, and plot the measured quantity against time. Read whether the chosen measurement increases or decreases as the reaction proceeds.
In plain language
Rate is how quickly reactants are used or products form.
Self-check
In a marble-chip/acid experiment, mass falls from 105.0 g to 101.0 g in 40 s. What is the average mass-loss rate, and what does a plateau on a mass–time graph mean?
Show the answer
Average rate = 4.0 g ÷ 40 s = 0.10 g/s. A plateau means the mass is no longer changing, so the reaction has stopped producing escaping gas.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 6.2 Supplement 5–8
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What to know
- Collision theory describes reacting particles as colliding. Increasing the number of particles per unit volume (by increasing solution concentration or gas pressure) raises collision frequency. A larger solid surface area exposes more particles for collisions.
- Particles have a distribution of kinetic energies. Raising temperature increases their average kinetic energy, so collisions occur more frequently and a greater fraction have energy at least the activation energy, Eₐ. Only collisions with sufficient energy can react.
- A catalyst supplies an alternative pathway with lower Eₐ, increasing the fraction of successful collisions; it is not consumed and does not change the overall enthalpy change of the reaction.
- Use collision theory to explain each rate factor: concentration/pressure changes particles per unit volume and collision frequency; surface area changes exposed particles; temperature changes particle kinetic energy and the fraction above Eₐ; catalyst changes Eₐ.
- Evaluate a practical rate method against the measurement needed. A gas syringe gives a direct gas-volume/time record but leaks, sticking or a delayed start can reduce accuracy; improve by checking airtight connections and collecting repeats. Measuring mass loss on a balance is simple for a gas that escapes, but drafts, balance resolution and the start-time delay affect readings; shield the balance and repeat. Keep reactant amounts, temperature and apparatus constant when comparing runs.
What to do in an assessment
- Build an explanation as condition change → particle-level change → change in frequency or proportion of successful collisions → rate change. Distinguish collision frequency from collision energy.
In plain language
Faster reactions have more frequent successful collisions.
Self-check
Why does raising temperature usually increase reaction rate more than merely increasing collision frequency?
Show the answer
Higher temperature raises average kinetic energy and greatly increases the fraction of collisions with energy at least Eₐ, so more collisions are successful as well as more frequent.
- Course content
Scope: Course content · §3 6.3 Core 1–2
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What to know
- The symbol ⇌ shows that a chemical reaction is reversible: products can react to form the original reactants as well as reactants forming products.
- Heating hydrated copper(II) sulfate drives off water and changes blue crystals to white anhydrous CuSO₄; adding water reverses the change and restores blue hydrated crystals. Hydrated cobalt(II) chloride is pink; heating removes water and forms blue anhydrous CoCl₂, while adding water restores the pink hydrate.
What to do in an assessment
- Read the ⇌ symbol as a two-way reaction. For the two named salts, connect heating to loss of water and the anhydrous colour, and addition of water to formation of the hydrate and its colour.
In plain language
In a reversible reaction, products can react to remake reactants.
Self-check
A blue hydrated copper(II) sulfate sample is heated and becomes white. What has happened, and what change reverses it?
Show the answer
It has lost water of crystallisation to form white anhydrous CuSO₄. Adding water reforms the blue hydrated salt.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 6.3 Supplement 3–11
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What to know
- In a closed system, dynamic equilibrium is reached when the forward and reverse reaction rates are equal. Both reactions continue, but reactant and product concentrations remain constant.
- A change in temperature shifts equilibrium toward the endothermic direction (and away from the exothermic direction). Increasing pressure shifts a gaseous equilibrium toward the side with fewer gas molecules; lowering pressure favours the side with more. Increasing a reactant concentration or removing a product shifts the position to use up that change; adding product or removing reactant shifts it the opposite way. A catalyst speeds both directions equally, so equilibrium is reached sooner but its position and yield do not change.
- Apply the rules using the equation and conditions provided: compare gas-molecule counts for pressure, identify the endothermic direction for temperature, and see which direction consumes an added substance or replaces a removed one. For N₂ + 3H₂ ⇌ 2NH₃, higher pressure favours the right (2 gas molecules rather than 4); the forward reaction is exothermic, so lower temperature favours ammonia; adding N₂ or removing NH₃ also shifts right. For 2SO₂ + O₂ ⇌ 2SO₃, higher pressure favours the right (2 rather than 3 gas molecules), and lower temperature favours the exothermic product direction.
- Haber process: N₂ + 3H₂ ⇌ 2NH₃. Nitrogen is obtained from air and hydrogen from methane. The typical conditions are 450 °C, 20 000 kPa (200 atm) and an iron catalyst. Higher pressure improves ammonia yield and rate but requires costly, robust equipment; 450 °C is a compromise between equilibrium yield and a useful rate; the catalyst speeds attainment of equilibrium without changing yield.
- Contact process: 2SO₂ + O₂ ⇌ 2SO₃. Sulfur dioxide is made by burning sulfur or roasting sulfide ores; oxygen comes from air. The typical conditions are 450 °C, 200 kPa (2 atm) and vanadium(V) oxide catalyst. The temperature balances exothermic yield against rate; although higher pressure favours SO₃, the modest extra yield does not justify the extra pressure cost and safety demands. The catalyst increases rate, not equilibrium yield.
- Industrial operating conditions are selected by balancing reaction rate, equilibrium position/yield, safety and economics rather than choosing only the condition that maximises equilibrium yield.
What to do in an assessment
- For a changed condition, predict the direction of shift and justify it with the endothermic direction, gas count or concentration change. For each process, state sources, balanced reversible equation and typical conditions; explain how those conditions balance rate, equilibrium, safety and cost.
In plain language
Industry chooses conditions that make product quickly and economically without ignoring equilibrium, safety or cost.
Self-check
For N₂ + 3H₂ ⇌ 2NH₃ (forward exothermic), predict the yield effect of raising pressure and temperature. Why is iron added?
Show the answer
Raising pressure favours ammonia because the product side has fewer gas molecules; raising temperature shifts equilibrium left, away from exothermic ammonia formation. Iron increases the rates of both directions so equilibrium is reached faster, without changing its position or yield.
- Course content
Scope: Course content · §3 6.4 Core 1–5
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What to know
- A redox reaction involves oxidation and reduction simultaneously. At Core, oxidation is gain of oxygen and reduction is loss of oxygen. For example, in Fe₂O₃ + 3CO → 2Fe + 3CO₂, iron(III) oxide loses oxygen to become iron, while carbon monoxide gains oxygen to become carbon dioxide.
- Roman numerals in names show an element's oxidation number in a compound: FeO is iron(II) oxide, while Fe₂O₃ is iron(III) oxide. Identify the oxygen gained and lost to decide which reactant is oxidised and which is reduced.
What to do in an assessment
- Compare the reactant and product formulas for oxygen transfer; name the species gaining oxygen as oxidised and the species losing oxygen as reduced. Check that both processes occur in the same redox reaction.
In plain language
When oxygen transfers, one substance gains it as another loses it.
Self-check
In Fe₂O₃ + 3CO → 2Fe + 3CO₂, identify what is oxidised and reduced; what does iron(III) mean?
Show the answer
CO gains oxygen to form CO₂, so CO is oxidised; Fe₂O₃ loses oxygen to form Fe, so it is reduced. Iron(III) indicates oxidation number +3 for iron in that compound.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 6.4 Supplement 6–13
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What to know
- Oxidation is loss of electrons and an increase in oxidation number; reduction is gain of electrons and a decrease in oxidation number. Thus in 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, Fe²⁺ is oxidised (loses electrons; +2 to +3) and Cl₂ is reduced (gains electrons; 0 to −1). Redox involves both at once.
- Use these oxidation-number rules: an uncombined element has oxidation number 0 (O₂); a monatomic ion's oxidation number equals its charge (Fe³⁺ is +3); oxidation numbers sum to 0 in a neutral compound; and in a polyatomic ion they sum to the ion's charge. For example, in H₂SO₄, 2(+1) + S + 4(−2) = 0, so S = +6; in SO₄²⁻, S + 4(−2) = −2, so S = +6.
- Acidified potassium manganate(VII) is purple and becomes colourless as Mn(VII) is reduced. Aqueous iodide is colourless; when I⁻ is oxidised to iodine, the mixture becomes brown (starch, if present, gives a blue-black colour). These observations can identify a redox change.
- An oxidising agent causes another substance to be oxidised and is itself reduced; a reducing agent causes another substance to be reduced and is itself oxidised. In 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, Cl₂ is the oxidising agent and Fe²⁺ is the reducing agent.
What to do in an assessment
- Write oxidation numbers for relevant atoms/ions using the rules, compare reactants with products, and identify increases as oxidation and decreases as reduction. Track electron gain/loss when charges are shown; identify the oxidising/reducing agent by the change it causes and the change it undergoes.
In plain language
Redox is electron transfer; oxidation numbers provide a systematic way to track it.
Self-check
In 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, identify oxidation, reduction, oxidising agent and reducing agent.
Show the answer
Fe²⁺ loses electrons and rises from +2 to +3, so it is oxidised and is the reducing agent. Cl₂ gains electrons and falls from 0 to −1, so it is reduced and is the oxidising agent.
Topic group
7 Acids, bases and salts
- Course content
Scope: Course content · §3 7.1 Core 1–8
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What to know
- Acids react with suitable reactive metals to make a salt and hydrogen (e.g. magnesium + hydrochloric acid → magnesium chloride + hydrogen), with bases to make a salt and water, and with carbonates to make a salt, water and carbon dioxide.
- Acids turn blue litmus red, leave thymolphthalein colourless, and turn methyl orange red. Alkalis turn red litmus blue, turn thymolphthalein blue, and turn methyl orange yellow; methyl orange is orange near neutral. Use the indicator's colour change to distinguish acidic, neutral and alkaline conditions.
- A base is a metal oxide or metal hydroxide; an alkali is a base that dissolves in water. Bases neutralise acids to give a salt and water. When a base is warmed with an ammonium salt, ammonia is released; for example, NaOH + NH₄Cl → NaCl + H₂O + NH₃.
- Aqueous acids contain H⁺(aq) ions and aqueous alkalis contain OH⁻(aq) ions. Neutralisation is H⁺(aq) + OH⁻(aq) → H₂O(l). Universal indicator paper compares hydrogen-ion concentration: acidic solutions are below pH 7, neutral solution is pH 7, and alkaline solutions are above pH 7; the colour scale progresses red/orange/yellow through green to blue/purple as pH rises.
What to do in an assessment
- For acid reactions, identify the reactant class and predict the salt and any gas/water. Distinguish a base from a soluble alkali; use the stated indicator colours or pH key to classify acidity/alkalinity, and use H⁺ + OH⁻ → H₂O for neutralisation.
In plain language
Acids and bases react to make salts; acid and alkali neutralise when H⁺ and OH⁻ form water.
Self-check
What happens when hydrochloric acid reacts with calcium carbonate, and what colour does methyl orange show in the acid before reaction?
Show the answer
Calcium chloride, water and carbon dioxide form; acid turns methyl orange red.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 7.1 Supplement 9–12
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What to know
- A Brønsted acid is a proton (H⁺) donor; a Brønsted base is a proton acceptor.
- A strong acid is completely dissociated in aqueous solution; a weak acid is only partially dissociated. Strength describes degree of dissociation, not the acid's concentration.
- Hydrochloric acid is strong and is shown by a one-way arrow: HCl(aq) → H⁺(aq) + Cl⁻(aq). Ethanoic acid is weak and is shown with an equilibrium arrow: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq).
What to do in an assessment
- Use a one-way arrow for complete dissociation and an equilibrium arrow for partial dissociation; distinguish acid strength from solution concentration.
In plain language
Strong and weak describe how completely an acid ionises, not how much acid was added.
Self-check
Why is HCl(aq) shown with → but CH₃COOH(aq) with ⇌?
Show the answer
HCl is completely dissociated in water (strong acid), while ethanoic acid is only partially dissociated (weak acid), so both forward and reverse changes are represented.
- Course content
Scope: Course content · §3 7.2 Core 1; 7.3 Core 1–3
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What to know
- Classify non-metal oxides such as SO₂ and CO₂ as acidic oxides; metal oxides such as CuO and CaO are basic oxides. This pattern is related to the metallic or non-metallic character of the element.
- Prepare a soluble salt from an acid and a soluble alkali by titration: measure a fixed volume of alkali, add indicator, titrate with acid to the end-point, repeat using the exact volumes without indicator, gently evaporate some water to concentrate the neutral solution, cool to crystallise the salt, then filter and dry the crystals.
- For a soluble salt made from acid and an insoluble solid, warm the acid and add excess metal, insoluble base or insoluble carbonate until no more reacts. Filter off the excess solid, evaporate the filtrate to concentrate it, allow crystals to form, then filter and dry the crystals. Example: warm sulfuric acid with excess CuO, filter, then crystallise CuSO₄ from the filtrate.
- Solubility rules: all sodium, potassium and ammonium salts are soluble; all nitrates are soluble; chlorides are soluble except lead and silver chlorides; sulfates are soluble except barium, calcium and lead sulfates; carbonates are insoluble except sodium, potassium and ammonium carbonates; hydroxides are insoluble except sodium, potassium and ammonium hydroxides, while calcium hydroxide is partially soluble.
- A hydrated substance is chemically combined with water; an anhydrous substance contains no water. Heating a hydrated salt can produce the anhydrous form.
What to do in an assessment
- Choose titration when both starting solutions are soluble, so no reactant can be removed by filtration. For a metal, insoluble base or insoluble carbonate, add excess solid, filter, then crystallise the dissolved salt. For each route, state how the salt is separated and purified.
In plain language
The salt-preparation method depends on whether the reactants and product dissolve.
Self-check
How would you prepare pure hydrated copper(II) sulfate crystals from sulfuric acid and copper(II) oxide?
Show the answer
Warm dilute sulfuric acid, add CuO in excess until no more reacts, filter off the excess CuO, gently evaporate the filtrate to concentrate it, allow it to cool and crystallise, then filter and dry the crystals.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 7.2 Supplement 2–3; 7.3 Supplement 4–5
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What to know
- An amphoteric oxide reacts with both acids and bases to form a salt and water. Aluminium oxide, Al₂O₃, and zinc oxide, ZnO, are amphoteric; they are neither simply acidic nor simply basic.
- Prepare an insoluble salt by precipitation: mix solutions of two soluble salts so their ions form the required insoluble product, then filter, wash the precipitate with distilled water and dry it. For example, Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) produces yellow lead(II) iodide.
- Water of crystallisation is the water molecules present in hydrated crystals; the formula shows their fixed number, as in CuSO₄·5H₂O and CoCl₂·6H₂O.
What to do in an assessment
- Use the solubility rules to select two soluble starting solutions that supply the ions for the required precipitate.
In plain language
A precipitation reaction makes an insoluble salt directly from dissolved ions.
Self-check
Describe how to make and isolate an insoluble salt from two soluble salt solutions.
Show the answer
Mix the solutions so the required insoluble salt precipitates, filter it off, wash it with distilled water, and dry it. This is precipitation.
Topic group
8 The Periodic Table
- Course content
Scope: Course content · §3 8.1 Core 1–5; 8.2 Core 1–2; 8.3 Core 1–4; 8.4 Core 1; 8.5 Core 1
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What to know
- The Periodic Table arranges elements in horizontal periods and vertical groups in increasing proton (atomic) number. Metallic character generally decreases from left to right across a period. For main-group elements, group position helps predict common ion charge: Groups I, II and III typically form +1, +2 and +3 ions, while Groups V, VI and VII typically form −3, −2 and −1 ions; Group IV elements generally share electrons rather than form simple ions.
- Elements in the same group have similar chemical properties because their atoms have the same number of outer-shell electrons. An element's position can therefore predict its outer-shell arrangement, common ion charge, metallic/non-metallic character and how it behaves relative to other members of its group.
- Group I lithium, sodium and potassium are relatively soft metals. Down the group, melting point decreases, while density and reactivity increase. When information about another Group I element is supplied, identify the trend and use it to predict the stated property rather than assuming every property changes in the same direction.
- Group VII chlorine, bromine and iodine are diatomic non-metals. Down the group, density increases and reactivity decreases. At r.t.p., chlorine is a pale yellow-green gas, bromine a red-brown liquid, and iodine a grey-black solid. A more reactive halogen displaces a less reactive halide from solution: Cl₂ + 2KBr → 2KCl + Br₂; Br₂ + 2KI → 2KBr + I₂.
- Transition elements are metals with high densities and high melting points; they form coloured compounds and often act as catalysts both as elements and in compounds (for example, iron and vanadium(V) oxide in industrial processes).
- Group VIII (Group 18) noble gases are unreactive, monatomic gases. Their outer electron shell is full (helium's first shell contains two electrons), so they have little tendency to gain, lose or share electrons.
What to do in an assessment
- Read a group's direction and stated trend carefully; use supplied data to predict an unknown member only for that property. For halogen displacement, compare reactivity first, then write the displaced halogen and new halide salt. Connect similar group chemistry to outer-shell electron number.
In plain language
The table's position and repeated outer-electron patterns help predict chemical behaviour.
Self-check
Will bromine displace chloride ions from potassium chloride solution? Explain, and state chlorine's appearance at r.t.p.
Show the answer
No. Bromine is less reactive than chlorine, so it cannot displace chloride; chlorine is a pale yellow-green gas at r.t.p.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 8.1 Supplement 6; 8.4 Supplement 2
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What to know
- A question may supply observations or data for elements in a group. Compare values down the group, state the trend using the correct direction and property, then apply that trend to predict a value or property for another member.
- Transition elements can form ions with variable oxidation numbers. Iron(II) is Fe²⁺ and iron(III) is Fe³⁺; the Roman numeral gives the ion's charge/oxidation number.
What to do in an assessment
- State the pattern in the given data and use it to predict the next property; identify the iron ion charge from its Roman numeral.
In plain language
Group patterns support predictions, while transition metals can form ions with different charges.
Self-check
What does a supplied Group I data table let you do, and what ions are named by iron(II) and iron(III)?
Show the answer
State the observed down-group trend and use it to predict a property for another member. Iron(II) is Fe²⁺; iron(III) is Fe³⁺.
Topic group
9 Metals
- Course content
Scope: Course content · §3 9.1 Core 1–2; 9.2 Core 1; 9.3 Core 1–4
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What to know
- Metals generally conduct heat and electricity well, are malleable (can be hammered into shape) and ductile (can be drawn into wires), and have relatively high melting and boiling points. Non-metals generally have poorer thermal/electrical conductivity, are brittle when solid, and often have lower melting and boiling points. These are general comparisons, not rules without exceptions.
- The general chemical reactions of metals are with dilute acids, cold water or steam, and oxygen. A sufficiently reactive metal plus dilute acid forms a salt and hydrogen; reactive metals plus cold water form a metal hydroxide and hydrogen, while some metals react with steam to form a metal oxide and hydrogen; reaction with oxygen forms a metal oxide. Whether a particular metal reacts depends on its position in the reactivity series.
- Aluminium's low density suits aircraft; its low density and good electrical conductivity suit overhead cables; its corrosion resistance suits food containers. Copper's good electrical conductivity and ductility make it suitable for electrical wiring.
- An alloy is a mixture of a metal with other elements: brass is copper and zinc; stainless steel is iron mixed with elements such as chromium, nickel and carbon. Alloys can be harder and stronger than pure metals. Stainless steel is used for cutlery because of its hardness and resistance to rusting. In a particle diagram, an alloy has atoms of more than one element mixed in the metal structure.
What to do in an assessment
- Explain a material choice by linking the required function to a stated physical property.
In plain language
Metal applications depend on a match between material properties and the job.
Self-check
Why is copper suitable for electrical wiring?
Show the answer
It conducts electricity well and can be drawn into ductile wires.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 9.3 Supplement 5
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What to know
- Different-sized atoms in an alloy distort the regular metal layers, so those layers cannot slide over one another as easily as in a pure metal. This explains increased hardness and strength.
What to do in an assessment
- Use an atom-layer diagram to show how unlike atom sizes interrupt sliding.
In plain language
Alloy atoms make it harder for metal layers to move past each other.
Self-check
Why can an alloy be harder than a pure metal?
Show the answer
Different-sized atoms disrupt the regular layers and hinder sliding.
- Course content
Scope: Course content · §3 9.4 Core 1–3; 9.5 Core 1–3
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What to know
- The reactivity series, most to least reactive, is potassium, sodium, calcium, magnesium, aluminium, carbon, zinc, iron, hydrogen, copper, silver, gold. Potassium, sodium and calcium react with cold water to form the metal hydroxide and hydrogen; magnesium reacts with steam to form magnesium oxide and hydrogen. Magnesium, zinc and iron react with dilute hydrochloric acid to form the metal chloride and hydrogen, while copper, silver and gold do not react with dilute hydrochloric acid.
- A more reactive metal displaces a less reactive metal from its aqueous ions; no displacement occurs in the opposite order. For example, zinc displaces copper from copper(II) sulfate: Zn + CuSO₄ → ZnSO₄ + Cu. Use experimental results (including which metal reacts or displaces another) to deduce an order of reactivity.
- Iron and steel rust in the presence of both oxygen and water, forming hydrated iron(III) oxide, Fe₂O₃·xH₂O. Painting, greasing and coating with plastic are barrier methods: they exclude oxygen or water from the iron and so prevent rusting.
What to do in an assessment
- Use the reactivity series to predict whether a reaction occurs and identify which metal is oxidised in a displacement.
In plain language
A more reactive metal loses electrons more readily and can displace a less reactive metal.
Self-check
Will zinc displace copper from copper(II) sulfate?
Show the answer
Yes. Zinc is above copper in the reactivity series.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 9.4 Supplement 4–5; 9.5 Supplement 4–5
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What to know
- A more reactive metal has a greater tendency to form positive ions by losing electrons. In displacement tests it displaces the ions of a less reactive metal: for example, Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s), but Cu does not displace Mg²⁺. Compare the metal with Mg²⁺, Zn²⁺, Fe²⁺, Cu²⁺ and Ag⁺ solutions; a metal displaces ions of metals below it in the series.
- Aluminium appears relatively unreactive because a thin, protective aluminium oxide layer covers its surface and prevents further reaction with the underlying metal.
- Galvanising coats iron/steel with zinc. The zinc coating is a barrier to oxygen and water; zinc also provides sacrificial protection if the coating is scratched because it is more reactive than iron and oxidises first. Zinc loses electrons, Zn → Zn²⁺ + 2e⁻, while the iron is protected.
What to do in an assessment
- Distinguish a physical barrier from sacrificial protection and relate the latter to relative reactivity/electron loss.
In plain language
A sacrificial metal corrodes in preference to the iron it protects.
Self-check
Why can zinc still protect scratched galvanised iron?
Show the answer
Zinc is more reactive and oxidises preferentially, supplying sacrificial protection.
- Course content
Scope: Course content · §3 9.6 Core 1–3
Open revision guidance
What to know
- The ease of extracting a metal from its ore depends on its position in the reactivity series. Metals below carbon can be extracted by reduction with carbon or carbon monoxide; metals above carbon are more difficult to extract and are commonly extracted by electrolysis.
- In a blast furnace, coke (carbon) burns in oxygen, releasing heat and making carbon dioxide. Carbon dioxide reacts with more hot carbon to produce carbon monoxide. Carbon monoxide reduces iron(III) oxide in hematite to iron. Limestone (calcium carbonate) thermally decomposes to calcium oxide; this reacts with silica impurities to form calcium silicate slag, which can be separated from molten iron. Core candidates describe these stages without symbol equations.
- Bauxite is the main ore of aluminium, and aluminium is extracted from aluminium oxide by electrolysis.
What to do in an assessment
- Describe the function of coke, limestone, carbon monoxide and slag separately in the iron-extraction process.
In plain language
Extraction method depends on how strongly the metal holds onto its compounds.
Self-check
What is the role of limestone in the blast furnace?
Show the answer
It decomposes to calcium oxide, which reacts with silica impurities to form slag.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 9.6 Supplement 4–5
Open revision guidance
What to know
- The balanced blast-furnace equations are: C + O₂ → CO₂; C + CO₂ → 2CO; Fe₂O₃ + 3CO → 2Fe + 3CO₂; CaCO₃ → CaO + CO₂; CaO + SiO₂ → CaSiO₃. They show coke combustion, carbon-monoxide formation, iron-oxide reduction, limestone decomposition and slag formation.
- Electrolyse aluminium oxide from purified bauxite dissolved in molten cryolite. Cryolite lowers the melting/operating temperature and reduces the energy required. Al³⁺ ions move to the negative cathode and gain electrons: Al³⁺ + 3e⁻ → Al. O²⁻ ions move to the positive carbon anode and lose electrons: 2O²⁻ → O₂ + 4e⁻. The oxygen reacts with the carbon anodes to form carbon dioxide, so the anodes are gradually consumed and must be replaced.
What to do in an assessment
- Balance each symbol equation and half-equation; explain cryolite and anode replacement in terms of the process.
In plain language
Aluminium is extracted by electrolysis because it is more reactive than carbon.
Self-check
Why must the carbon anodes be replaced?
Show the answer
Oxygen formed at the anode reacts with the carbon, consuming it.
Topic group
10 Chemistry of the environment
- Course content
Scope: Course content · §3 10.1 Core 1–7; 10.2 Core 1–2
Open revision guidance
What to know
- Test for water with anhydrous cobalt(II) chloride paper: blue changes to pink. Anhydrous copper(II) sulfate is white and turns blue when water is present. To test purity, measure melting/freezing point and boiling point at a stated pressure: pure water has fixed characteristic values (0 °C and 100 °C at standard atmospheric pressure), while dissolved impurities change these values and broaden the melting/freezing range. Distilled water is used in practical chemistry rather than tap water because it contains fewer chemical impurities.
- Water from natural sources may contain dissolved oxygen, metal compounds, plastics, sewage, harmful microbes, nitrates from fertilisers, and phosphates from fertilisers and detergents. Dissolved oxygen supports aquatic life and some metal compounds provide essential minerals. Some metal compounds are toxic; some plastics harm aquatic life; sewage can contain disease-causing microbes; nitrates and phosphates can cause deoxygenation and damage aquatic life. Detailed stages of eutrophication are not required.
- Treat domestic water by sedimentation and filtration to remove solids, use carbon to remove tastes and odours, and chlorinate to kill microbes. Ammonium salts and nitrates are used as fertilisers. NPK fertilisers supply nitrogen, phosphorus and potassium to improve plant growth.
What to do in an assessment
- Separate beneficial dissolved substances from contaminants and describe the treatment step that removes or controls each.
In plain language
Water treatment removes particles, improves taste and controls microbes; fertilisers supply plant nutrients.
Self-check
Why is carbon used in domestic water treatment?
Show the answer
It removes unwanted tastes and odours.
- Course content
Scope: Course content · §3 10.3 Core 1–6
Open revision guidance
What to know
- Clean, dry air is approximately 78% nitrogen (N₂), 21% oxygen (O₂), and the remainder is a mixture of noble gases and carbon dioxide (CO₂).
- Sources of the named air pollutants: complete combustion of carbon-containing fuels produces CO₂; incomplete combustion produces carbon monoxide and particulates; methane comes from decomposition of vegetation and waste gases from animal digestion; nitrogen and oxygen from air react at high temperatures in car engines to form nitrogen oxides; sulfur dioxide forms when fossil fuels containing sulfur compounds are burned.
- Adverse effects: increased CO₂ and methane levels contribute to global warming and climate change; carbon monoxide is toxic; particulates increase respiratory-problem and cancer risk; nitrogen oxides contribute to acid rain and photochemical smog and cause respiratory problems; sulfur dioxide causes acid rain.
- Strategies: planting trees removes CO₂ by photosynthesis; reducing livestock farming can reduce methane emissions; decreasing fossil-fuel use and increasing hydrogen and renewable energy (e.g. wind and solar) reduce greenhouse-gas emissions. Catalytic converters reduce vehicle emissions of nitrogen oxides; low-sulfur fuels and flue-gas desulfurisation with calcium oxide reduce sulfur dioxide and therefore acid rain. Photosynthesis uses light energy and chlorophyll to react carbon dioxide with water and produce glucose and oxygen: carbon dioxide + water → glucose + oxygen.
What to do in an assessment
- Link each pollutant to its source and adverse effect, then choose a mitigation strategy that addresses that source or effect.
In plain language
Air-quality explanations connect what is emitted, how it affects people or climate, and how emissions can be reduced.
Self-check
Which pollutant from incomplete combustion is a toxic gas?
Show the answer
Carbon monoxide.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 10.3 Supplement 7–9
Open revision guidance
What to know
- The Earth's surface absorbs energy from sunlight and emits thermal (infrared) radiation. Greenhouse gases, including carbon dioxide and methane, absorb some of this outgoing thermal energy and emit it in all directions; some is returned toward Earth's surface. This reduces thermal-energy loss to space and raises global temperatures.
- At the high temperatures in car engines, nitrogen and oxygen from air react to form nitrogen oxides. In a catalytic converter, carbon monoxide reduces nitrogen oxide to nitrogen while itself being oxidised to carbon dioxide; for example, 2CO + 2NO → 2CO₂ + N₂.
- The balanced symbol equation for photosynthesis is 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
What to do in an assessment
- Explain greenhouse warming as an energy-transfer sequence, and check atom counts when using the converter or photosynthesis equation.
In plain language
Greenhouse gases slow infrared energy escaping to space; converters change harmful exhaust gases into less harmful products.
Self-check
What is the role of NO in the converter equation?
Show the answer
It is reduced to nitrogen gas while carbon monoxide is oxidised to carbon dioxide.
Topic group
11 Organic chemistry
- Course content
Scope: Course content · §3 11.1 Core 1–6; 11.2 Core 1–2
Open revision guidance
What to know
- A displayed formula shows every atom and every covalent bond. To draw one, show each carbon with four bonds and each oxygen with two; add hydrogen atoms and bonds rather than hiding them. The syllabus general formulae are alkanes CₙH₂ₙ₊₂, alkenes CₙH₂ₙ, alcohols CₙH₂ₙ₊₁OH and carboxylic acids CₙH₂ₙ₊₁COOH.
- A functional group is an atom or group of atoms that determines the characteristic chemical properties of a homologous series. Alkenes contain C=C, alcohols contain –OH and carboxylic acids contain –COOH. A saturated compound has only C–C single bonds; an unsaturated compound has one or more non-single C–C bonds.
- A homologous series is a family of similar compounds with similar chemical properties because of the same functional group. Identify the series from its functional group or formula, and recognise names ending in -ane, -ene, -ol and -oic acid as alkane, alkene, alcohol and carboxylic acid respectively.
- The named Core molecules include methane CH₄, ethane CH₃–CH₃, ethene CH₂=CH₂, ethanol CH₃–CH₂–OH and ethanoic acid CH₃–C(=O)–OH. Convert these structural shortcuts to displayed formulae by drawing every atom and bond; check valency and include displayed formulae for products of the syllabus reactions.
What to do in an assessment
- Count carbon atoms, identify the functional group and check that each carbon has four covalent bonds.
In plain language
Organic names and formula patterns reveal the carbon skeleton and its reactive group.
Self-check
Which functional group identifies an alcohol?
Show the answer
The hydroxyl group, –OH.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 11.1 Supplement 7–9; 11.2 Supplement 3–4
Open revision guidance
What to know
- A structural formula is an unambiguous description of how atoms are arranged: CH₂=CH₂, CH₃CH₂OH and CH₃COOCH₃ are examples. Structural isomers have the same molecular formula but different structural formulae: C₄H₁₀ can be CH₃CH₂CH₂CH₃ (butane) or CH₃CH(CH₃)CH₃ (2-methylpropane); C₄H₈ can be CH₃CH₂CH=CH₂ (but-1-ene) or CH₃CH=CHCH₃ (but-2-ene).
- A homologous series has the same functional group and general formula; consecutive members differ by one –CH₂– unit, show a trend in physical properties and have similar chemical properties. In these series, increasing chain length generally raises boiling point and viscosity and lowers volatility.
- Use the carbon-number roots meth-, eth-, prop- and but- and the suffix for the functional group. Unbranched alkanes through four carbons are methane, ethane, propane, butane; alkenes are ethene, propene, but-1-ene and but-2-ene; alcohols are methanol, ethanol, propan-1-ol, propan-2-ol, butan-1-ol and butan-2-ol; carboxylic acids up to four carbons are methanoic, ethanoic, propanoic and butanoic acid. Number the chain from the end nearest the functional group or double bond.
- An ester has the structural pattern R–C(=O)–O–R′. Name it as the alkyl group from the alcohol followed by the carboxylate name from the acid, ending in -oate. The full unbranched combinations are methyl, ethyl, propyl or butyl methanoate; methyl, ethyl, propyl or butyl ethanoate; methyl, ethyl, propyl or butyl propanoate; and methyl, ethyl, propyl or butyl butanoate. To draw an ester, join the acid's C(=O)–O part to the alcohol-derived alkyl group and show every atom/bond in the displayed formula.
What to do in an assessment
- Check that a displayed/structural formula preserves the molecular formula when identifying isomers.
In plain language
Structural formulae distinguish compounds that contain the same numbers of atoms arranged differently.
Self-check
Are but-1-ene and but-2-ene structural isomers?
Show the answer
Yes. They share C₄H₈ but the double bond is in a different position.
- Course content
Scope: Course content · §3 11.3 Core 1–7
Open revision guidance
What to know
- Coal, natural gas and petroleum are fossil fuels. Methane is the main constituent of natural gas. A hydrocarbon contains hydrogen and carbon only; petroleum is a mixture of hydrocarbons.
- Fractional distillation separates petroleum into useful fractions because hydrocarbons have different boiling points. Heat the petroleum so much of it vaporises; vapours rise through a fractionating column that is hot at the bottom and cooler at the top, condense at different heights according to their boiling ranges, and are collected as fractions.
- From the bottom to the top of the fractionating column, fractions have decreasing chain length and boiling point, and increasing volatility with decreasing viscosity.
- Fraction uses: refinery gas for heating and cooking; gasoline/petrol for car fuel; naphtha as a chemical feedstock; kerosene/paraffin for jet fuel; diesel/gas oil for diesel engines; fuel oil for ships and home heating; lubricating oil for lubricants, waxes and polishes; bitumen for making roads.
What to do in an assessment
- Explain separation by differences in boiling point and match each fraction's stated use to its source-column position.
In plain language
Petroleum is separated into useful groups because its hydrocarbons boil at different temperatures.
Self-check
How do chain length and boiling point change up the fractionating column?
Show the answer
Chain length and boiling point decrease.
- Course content
Scope: Course content · §3 11.4 Core 1–2; 11.5 Core 1–4
Open revision guidance
What to know
- Alkanes are saturated hydrocarbons with only single covalent bonds between carbon atoms. They are generally unreactive, except for combustion and substitution reactions with chlorine. Complete combustion in excess oxygen produces carbon dioxide and water (CH₄ + 2O₂ → CO₂ + 2H₂O); inadequate oxygen can cause incomplete combustion.
- Alkenes are unsaturated hydrocarbons because they contain a carbon–carbon double bond, C=C. Cracking larger alkane molecules at high temperature with a catalyst makes smaller alkane and alkene molecules, and can also make hydrogen. Cracking supplies smaller, more useful fuels and alkene feedstocks for polymer manufacture.
- Aqueous bromine distinguishes unsaturated from saturated hydrocarbons: an alkene decolourises orange/brown bromine water, while an alkane does not under these test conditions.
What to do in an assessment
- Use the bromine-water observation as evidence for a C=C bond, not as a general test for every functional group.
In plain language
The carbon double bond makes an alkene react differently from a saturated alkane.
Self-check
What observation indicates an alkene in the aqueous bromine test?
Show the answer
The orange/brown bromine water is decolourised.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 11.4 Supplement 3–4; 11.5 Supplement 5–6
Open revision guidance
What to know
- In a substitution reaction, one atom or group of atoms is replaced by another. Chlorination of an alkane is a photochemical substitution: ultraviolet light supplies activation energy, Eₐ. In the syllabus monosubstitution scope, one hydrogen is replaced by one chlorine; for example, CH₄ + Cl₂ → CH₃Cl + HCl.
- An addition reaction forms only one product. At an alkene C=C bond, bromine adds to form a dibromoalkane (ethene + Br₂ → 1,2-dibromoethane), hydrogen adds with a nickel catalyst to form an alkane (ethene + H₂ → ethane), and steam adds with an acid catalyst to form an alcohol (ethene + H₂O → ethanol). Draw the product with correct valency and all atoms from both reactants.
What to do in an assessment
- Identify whether bonds are replaced or added across C=C and draw the product with correct valency.
In plain language
Addition uses the double bond to join new atoms to an alkene; substitution replaces an atom in an alkane.
Self-check
What is the product when ethene adds hydrogen?
Show the answer
Ethane, C₂H₆.
- Course content
Scope: Course content · §3 11.6 Core 1–3; 11.7 Core 1
Open revision guidance
What to know
- Fermentation makes ethanol from aqueous glucose using yeast at 25–35 °C in the absence of oxygen: glucose → ethanol + carbon dioxide (C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂). Ethanol is also made by catalytic addition of steam to ethene at 300 °C and 6000 kPa (60 atm) with an acid catalyst.
- Ethanol combusts in oxygen to form carbon dioxide and water. Its uses include solvent and fuel.
- Ethanoic acid reacts with metals to form an ethanoate salt and hydrogen, with bases to form an ethanoate salt and water, and with carbonates to form an ethanoate salt, water and carbon dioxide. Examples: 2CH₃COOH + Mg → (CH₃COO)₂Mg + H₂; CH₃COOH + NaOH → CH₃COONa + H₂O; 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂.
What to do in an assessment
- Compare fermentation and catalytic hydration using their conditions; write products and salt formulae for acid reactions.
In plain language
Ethanol can be produced biologically or from ethene, and ethanoic acid behaves as a carboxylic acid.
Self-check
What condition must be absent during yeast fermentation?
Show the answer
Oxygen.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 11.6 Supplement 4; 11.7 Supplement 2–3
Open revision guidance
What to know
- Fermentation uses renewable glucose from plant material at a relatively low temperature and pressure, but is slow and dilute ethanol must be separated from water; growing the crop uses land. Catalytic addition uses ethene (usually obtained from fossil-fuel feedstocks), operates rapidly and continuously at 300 °C and 6000 kPa with an acid catalyst, but requires high temperature/pressure and a non-renewable feedstock. Evaluate the routes by comparing rate, conditions, feedstock and separation.
- Ethanol is oxidised to ethanoic acid by acidified aqueous potassium manganate(VII); the purple oxidising solution becomes colourless as it is reduced. Ethanoic acid is also produced by bacterial oxidation of ethanol during vinegar production.
- A carboxylic acid reacts with an alcohol using an acid catalyst to form an ester and water. For example, ethanoic acid + ethanol ⇌ ethyl ethanoate + water.
What to do in an assessment
- Use evidence about conditions and resources to justify a production method; identify the alcohol and acid parts of an ester.
In plain language
Ethanol can be oxidised to an acid, and acids plus alcohols make esters.
Self-check
What two classes of reactant make an ester?
Show the answer
A carboxylic acid and an alcohol.
- Course content
Scope: Course content · §3 11.8 Core 1–5
Open revision guidance
What to know
- A polymer is a large molecule built up from many smaller molecules called monomers. Poly(ethene) forms by addition polymerisation of ethene: each C=C opens and forms new C–C bonds, giving the repeat unit [–CH₂–CH₂–]ₙ.
- Plastics are made from polymers. Disposal in landfill uses space and many plastics persist; plastic accumulation in oceans harms aquatic life; burning some plastics can form toxic gases. Consider these environmental challenges when assessing disposal.
What to do in an assessment
- Draw the repeat unit in brackets with continuation bonds, preserving the atoms in the alkene monomer.
In plain language
A polymer is a long molecule made by linking many smaller molecules.
Self-check
What happens to ethene's double bond during addition polymerisation?
Show the answer
The double bond opens so each carbon can form new single bonds to neighbouring units.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 11.8 Supplement 6–13
Open revision guidance
What to know
- In an addition polymer, identify the repeat unit by changing the alkene C=C to C–C and adding continuation bonds; conversely, remove those continuation bonds and restore C=C to infer the alkene monomer. Condensation polymers contain linkages formed with elimination of a small molecule.
- A dicarboxylic acid plus a diamine forms a polyamide; a dicarboxylic acid plus a diol forms a polyester. To deduce a polymer from monomers, join the acid and amine/alcohol end groups, form the –CONH– or –COO– linkage and remove H₂O at each link; reverse these steps to find the monomers.
- Nylon is a polyamide with repeating amide linkages, –C(=O)–NH–; one syllabus structure is nylon-6,6, [–NH–(CH₂)₆–NH–C(=O)–(CH₂)₄–C(=O)–]ₙ. PET is a polyester with repeating ester linkages, –C(=O)–O–; its repeat unit is [–O–CH₂–CH₂–O–C(=O)–C₆H₄–C(=O)–]ₙ. PET can be converted back into its monomers and repolymerised; the full name polyethylene terephthalate is not required.
- Proteins are natural polyamides formed from amino-acid monomers, each with the general structure H₂N–CH(R)–COOH, where R is a variable side chain. The protein chain has repeated –NH–CH(R)–C(=O)– units joined by amide links; identify and draw the linkage in the given protein structure.
What to do in an assessment
- Use the end groups of monomers to build a repeat unit and preserve the correct linkage when reversing the polymerisation.
In plain language
Polymer structures can be read backwards to identify the monomers that joined together.
Self-check
What functional linkage identifies a polyamide?
Show the answer
The amide linkage, –CONH–.
Topic group
12 Experimental techniques and chemical analysis
- Course content
Scope: Course content · §3 12.1 Core 1–3; 12.2 Core 1–2
Open revision guidance
What to know
- Choose apparatus for the quantity: a stopwatch measures time, a thermometer measures temperature, a balance measures mass, and a burette, volumetric pipette, measuring cylinder or gas syringe measures volumes. A volumetric pipette delivers one accurate fixed volume; a burette delivers and measures a variable volume more precisely than a measuring cylinder; a cylinder is quicker but less precise; a gas syringe collects a gas volume directly. Explain the advantages/disadvantages in relation to the precision, range and task.
- A solvent is a substance that dissolves a solute; a solute is the substance dissolved; a solution is a mixture of one or more solutes dissolved in a solvent. A saturated solution contains the maximum concentration of solute dissolved at a specified temperature. Residue is material remaining after evaporation, distillation, filtration or a similar separation; filtrate is liquid/solution that has passed through a filter.
- In an acid–base titration, use a volumetric pipette to measure a fixed aliquot of one solution into a flask, add a suitable indicator, and add the other solution from a burette while swirling. Record initial and final burette readings; the titre is the volume delivered. Add titrant dropwise near the end-point, identified by the indicator's permanent colour change. Repeat titrations and use concordant titres when appropriate.
What to do in an assessment
- Measure a fixed aliquot with the pipette, add indicator, run titrant from the burette while swirling, approach the end-point dropwise and record initial/final readings.
In plain language
Good experimental design matches each instrument to the quantity and records enough detail to judge the result.
Self-check
Why use a pipette rather than a measuring cylinder for a precise aliquot?
Show the answer
A volumetric pipette delivers a more precise fixed volume.
- Course content
Scope: Course content · §3 12.3 Core 1–2; 12.4 Core 1–3
Open revision guidance
What to know
- In paper chromatography, draw a pencil baseline near the bottom of the paper, place small spots of soluble coloured samples on it, and suspend the paper in a suitable solvent with the spots above the solvent. The solvent rises and carries components at different rates according to their solubility and attraction to the paper; remove the paper before the solvent reaches the top and mark the solvent front. Compare unknown spots with known substances under the same conditions. One spot suggests a pure substance in this test; multiple spots indicate a mixture.
- Choose a separation method from the mixture's properties. A suitable solvent dissolves the wanted substance but not unwanted insoluble material: stir the sample with the solvent, filter off insoluble residue, then recover the dissolved substance by evaporation/crystallisation or distillation. Filtration separates an insoluble solid from a liquid. For crystallisation, concentrate the solution by gentle evaporation, cool it to form crystals, then filter and dry them. Simple distillation heats a solution and condenses the vapour to collect a solvent, or separates liquids with widely different boiling points. Fractional distillation uses a fractionating column to separate miscible liquids with different/nearby boiling ranges; vapours condense at different heights according to boiling point.
- Use melting or boiling points to assess identity and purity at the stated pressure. A pure substance has a characteristic sharp melting point and fixed boiling point; impurities change the characteristic value and generally broaden the melting range.
What to do in an assessment
- Keep the sample spot above the solvent, mark the solvent front, and choose a separation method whose physical-property difference fits the mixture.
In plain language
Separation works because mixture components differ in solubility, particle size or boiling point.
Self-check
Which technique separates an insoluble solid from a liquid?
Show the answer
Filtration.
- Course content
Scope: Course content · Cambridge IGCSE Chemistry 0620 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades A*–G) · §3 12.3 Supplement 3–4
Open revision guidance
What to know
- Paper chromatography also separates soluble colourless substances. Run the chromatogram with a suitable solvent, remove it before the solvent reaches the top, mark the solvent front, then use a locating agent to make colourless spots visible; a specific locating agent need not be memorised.
- Calculate retention factor, Rf = distance travelled by substance ÷ distance travelled by solvent front. Measure both from the same baseline in the same units; Rf has no units and is compared meaningfully only when the chromatographic conditions are the same.
What to do in an assessment
- Measure both distances in the same units and report Rf without units; compare values only under matching conditions.
In plain language
Rf describes how far a substance travelled compared with the solvent front.
Self-check
A spot moves 3 cm and the solvent front 6 cm. Find Rf.
Show the answer
3 ÷ 6 = 0.50.
- Course content
Scope: Course content · §3 12.5 Core 1–4 and §4 Notes for use in qualitative analysis
Open revision guidance
What to know
- Anion tests: carbonate, CO₃²⁻, reacts with dilute acid with effervescence; the gas turns limewater milky (CO₂). Acidify chloride, bromide or iodide solution with dilute nitric acid, then add aqueous silver nitrate: white AgCl, cream AgBr or yellow AgI precipitate respectively. For nitrate, add aqueous sodium hydroxide and aluminium foil, then warm carefully: ammonia is evolved and confirmed by holding damp red litmus at the mouth of the tube until it turns blue. Acidify sulfate with dilute nitric acid and add aqueous barium nitrate: a white precipitate forms. Sulfite reacts with acidified aqueous potassium manganate(VII), changing it from purple to colourless.
- Aqueous-cation tests (precipitate observations with excess reagent): Al³⁺ gives a white precipitate soluble in excess NaOH (colourless solution), but insoluble in excess aqueous ammonia; NH₄⁺ produces ammonia on warming with NaOH. Ca²⁺ gives a white precipitate insoluble in excess NaOH and no/slight white precipitate with ammonia. Cr³⁺ gives a green precipitate soluble in excess NaOH but insoluble in excess ammonia. Cu²⁺ gives a light-blue precipitate insoluble in excess NaOH; with ammonia it dissolves in excess to a dark-blue solution. Fe²⁺ gives a green precipitate insoluble in excess of either reagent, turning brown near the surface on standing. Fe³⁺ gives a red-brown precipitate insoluble in excess of either reagent. Zn²⁺ gives a white precipitate soluble in excess of either reagent to a colourless solution.
- Gas tests: ammonia, NH₃, turns damp red litmus blue; carbon dioxide, CO₂, turns limewater milky; chlorine, Cl₂, bleaches damp litmus; hydrogen, H₂, gives a squeaky pop with a lighted splint; oxygen, O₂, relights a glowing splint; sulfur dioxide, SO₂, changes acidified aqueous potassium manganate(VII) from purple to colourless.
- Flame-test colours: Li⁺ red; Na⁺ yellow; K⁺ lilac; Ca²⁺ orange-red; Ba²⁺ light green; Cu²⁺ blue-green. Follow the reagent order in the anion tests and record observations (including precipitate colour and solubility in excess) before identifying the substance.
What to do in an assessment
- For nitrate, use a small sample in a test-tube, add aqueous sodium hydroxide then aluminium foil, warm gently with the tube held in a holder and pointed away from people, and test gas at the mouth with damp red litmus; do not stopper the tube, sniff the gas, or allow litmus to touch the liquid. Wear eye protection and work under teacher/centre risk controls. Record both the reagent sequence and the red-to-blue litmus result before identifying nitrate. For all tests, add reagents in the specified order and record observations before inferring an identity.
In plain language
Qualitative analysis identifies an unknown from a characteristic combination of colour, precipitate and gas-test results.
Self-check
What gas turns limewater milky?
Show the answer
Carbon dioxide.
Topic group
Assessment and practical skills
- Course content
Scope: Course content · §2 assessment overview; §4 Practical assessment and experimental contexts
Open revision guidance
What to know
- All candidates take three components. Candidates studying Core, or expected to achieve grade D or below, enter for Paper 1 (Core multiple choice: 45 minutes, 40 marks, 30%) and Paper 3 (Core theory: 1 hour 15 minutes, 80 marks, 50%), plus Paper 5 or 6; they are eligible for grades C–G. Candidates studying Extended and expected to achieve grade C or above enter for Paper 2 (Extended multiple choice: 45 minutes, 40 marks, 30%) and Paper 4 (Extended theory: 1 hour 15 minutes, 80 marks, 50%), plus Paper 5 or 6; they are eligible for grades A*–G. Papers 1–4 are externally assessed.
- Choose one externally assessed practical component: Paper 5 is a 1 hour 15 minute, 40-mark laboratory Practical Test (20%); Paper 6 is a 1-hour, 40-mark written Alternative to Practical (20%). Both assess the same Section 4 experimental skills and AO3; candidates do not perform experiments in Paper 6. Component weightings are fixed: Papers 1 and 2 each assess AO1 at 63%, AO2 at 37%, AO3 at 0%; Papers 3 and 4 have the same AO1/AO2/AO3 weighting; Papers 5 and 6 each assess AO1 at 0%, AO2 at 0%, AO3 at 100%.
- Practical contexts include quantitative volumes, masses, temperatures, times and lengths; reaction rates; salt preparation; filtration, crystallisation, solvent extraction, simple/fractional distillation and chromatography; electrolysis; identifying ions and gases; water tests/treatment; dilute-acid test-tube reactions; redox tests; heating/cooling curves; titration; solubility; melting/boiling points; metal/halogen displacement; reaction temperature changes; corrosion; and unfamiliar procedures with simple apparatus.
- Worked measurement/titration routine: choose the instrument whose scale/precision suits the quantity, set it up level and read at eye level. For a titration rinse a volumetric pipette with the analyte, transfer its fixed volume to a conical flask, add only the stated indicator, rinse/fill the burette with titrant and clear its jet bubble, record the initial reading, swirl while adding titrant, approach the endpoint dropwise, then record the final reading and calculate titre = final − initial. Repeat until concordant; do not infer precision beyond the instrument.
- Salt preparation: for a soluble salt from an acid and insoluble base, warm the acid gently, add the base in small portions with stirring until some remains, filter off excess solid, evaporate the filtrate only until concentrated, allow it to cool and crystallise, then filter and dry the crystals. For a soluble salt from two solutions, use a titration to find neutral volumes, repeat without indicator, then evaporate/cool/crystallise. Wear goggles and never evaporate to dryness unless specifically instructed.
- Separation operations: filtration retains an insoluble solid on wetted filter paper while liquid passes into a labelled vessel. To crystallise a dissolved solid, gently evaporate some solvent, stop before dryness, cool undisturbed, collect and dry crystals. For solvent extraction, shake a small sample with the specified immiscible solvent in a stoppered funnel, vent away from people, allow layers to settle, then separate the named layer. In simple distillation collect a lower-boiling solvent from a solution; in fractional distillation use a fractionating column and collect the fraction at its boiling range. For paper chromatography draw a pencil baseline, spot small concentrated samples, place solvent below the baseline, cover the chamber, stop before the solvent front reaches the top, mark the front and dry the chromatogram.
- Electrolysis: use the specified electrolyte and inert electrodes, connect to a low-voltage direct-current supply with power off while assembling, keep electrodes apart, then switch on and record observations at each electrode. Do not touch the solution/electrodes while powered; switch off before disconnecting. Identify products only by the required source-listed tests and apply centre safety rules.
- For water treatment/testing follow the supplied sequence: allow suspended solids to settle, filter, use carbon for taste/odour where specified and chlorinate only under the stated supervised conditions. For water tests or ion/gas identification, use small labelled samples, clean droppers, the exact reagent order and the source's positive observation; never mix stock reagents or smell gases directly.
- For dilute-acid, redox and displacement test-tube work, place a small measured sample in a labelled tube, add the specified dilute reagent dropwise, keep the opening pointed away, and record colour, precipitate, bubbles or temperature before inference. Compare metal/halogen displacement only with the supplied safe dilute solutions; use separate clean wells/tubes and dispose of residues under local rules.
- For a heating/cooling curve, put the thermometer bulb in the substance without touching the vessel, stir gently if directed, record temperature at equal time intervals during steady heating/cooling, and plot temperature / °C against time / min. For solubility, keep solvent volume constant, add weighed solute in small portions to a measured water sample held at a stated temperature, stir until the last portion just dissolves or a little remains undissolved, record the saturated mass/concentration, and repeat at other temperatures with fresh equal-volume samples; keep the temperature stable and compare solubility at the same endpoint. For melting point, place a small dry solid sample in a capillary tube with a thermometer in a heating bath/apparatus, heat slowly near the expected point and record the temperature at which melting starts and finishes; report a range and do not touch hot glass. For boiling point, heat a small sample in the specified open/vented school apparatus with thermometer bulb at the vapour level, record the steady boiling temperature, use a water bath for flammable liquids, and never heat a sealed system. Compare corrosion using labelled, otherwise identical iron samples, vary one condition only (for example dry air versus water, or air-free water versus aerated water), keep time, temperature, surface area and container the same, and record rusting/appearance at fixed intervals.
- Water and ion tests: use separate small, labelled portions, clean droppers and the source-specified order; acidify halide samples with dilute nitric acid before adding silver nitrate, and acidify sulfate samples with dilute nitric acid before adding barium nitrate. Observe precipitate colour and any change in excess reagent before identifying the ion. For water treatment, allow suspended solids to settle, decant and filter through wetted paper, use activated carbon only where specified for dissolved colour/taste/odour, then use the prescribed supervised disinfection step; do not taste samples or prepare/use chlorine outside the stated controlled context. In redox tests use small volumes and dilute, specified reagents: add acidified potassium manganate(VII) dropwise to the sample, note purple-to-colourless change where a reducing substance is present, wear goggles and keep oxidiser away from combustible material; never mix unknown stock reagents.
- For simple distillation of a solution, put the liquid in a flask with anti-bumping granules, fit a thermometer with its bulb at the side-arm/vapour path, connect the side-arm to a Liebig condenser, run cooling water in at the condenser's lower inlet and out at the upper outlet, and collect distillate from the outlet; heat gently, keep the apparatus vented, and never distil to dryness. For fractional distillation, place the fractionating column vertically between flask and still head, pack/use it as specified, position the thermometer bulb at the vapour inlet to the condenser, allow the column to establish a temperature gradient, and collect separate fractions as the head temperature reaches each boiling range. Support/clamp glassware, check water flow and joints before heating, keep the receiver open to the atmosphere, and use a water bath for flammable liquids.
- For an unfamiliar simple-apparatus procedure, select apparatus that can measure the expected value at suitable precision, draw and label a simple usable setup (vessels, connections, supports, liquid levels and flow direction), and state why a plausible alternative is less suitable. Estimate an expected value/range before starting; translate instructions into ordered actions; identify the measurement, endpoint and hazard controls; record numerical readings with units and qualitative observations in separate labelled table columns; repeat/blank where appropriate; and evaluate a second context by naming a specific limitation, its effect on results and a workable improvement. If the source supplies a method, follow its sequence rather than substituting a memorised procedure.
- Use the supplied qualitative-analysis notes when required. A paper choice changes the assessment mode, not the experimental-skills or practical-context scope.
- Chemical tests for water: use a clean dropper to put one small drop of the sample onto teacher-provided blue cobalt(II) chloride paper held with forceps; water changes it from blue to pink. In a separate test, put a few teacher-provided grains of anhydrous copper(II) sulfate on a spotting tile and add a small drop of sample; water changes the solid from white to blue. Compare with the unused reagent and record colour before and after. Wear eye protection, avoid skin contact and inhaling powders, and follow the centre’s risk assessment and disposal instructions; staff prepare/handle restricted reagents, or demonstrate the test where learner use is not permitted. A positive chemical test detects water, not its purity.
What to do in an assessment
- For Paper 5, perform the requested method safely and record actual observations; for Paper 6, interpret diagrams, methods and data without claiming to have carried out an experiment.
In plain language
Paper 5 is hands-on; Paper 6 tests practical understanding in writing, with the same AO3 focus.
Self-check
Does Paper 6 require the candidate to perform experiments?
Show the answer
No. It is a written Alternative to Practical.
- Course content
Scope: Course content · §2 AO3; §4 Practical assessment, required experimental skills
Open revision guidance
What to know
- Assessment-objective weightings are component-specific: Papers 1–4 each assess AO1 at 63%, AO2 at 37%, AO3 at 0%; Papers 5 and 6 each assess AO1 and AO2 at 0%, AO3 at 100%. AO1 covers facts, laws, definitions, vocabulary, symbols, quantities/units, apparatus and safety, plus applications and their social, economic and environmental implications. AO2 covers selecting/organising/presenting information, translating between words/symbols/graphs/numbers, manipulating data, identifying patterns/trends, explaining relationships, predicting and solving quantitative or unfamiliar-context problems. AO3 covers selecting and safely using techniques, apparatus and materials, planning, observing/measuring/estimating, interpreting/evaluating evidence and improving methods.
- Select and safely use apparatus/materials and follow a procedure; identify equipment from diagrams/descriptions, draw/complete/label apparatus diagrams, explain technique, justify method/apparatus choice, describe qualitative/gas/other tests, hazards/precautions and techniques that improve accuracy.
- Plan an investigation by identifying independent and dependent variables, controlling other variables (and explaining why), choosing an appropriate range and number of values, selecting apparatus, writing a sequence, identifying risks/precautions, specifying how results will be recorded and processed, predicting expected results, then using the evidence to form a conclusion or evaluate a prediction.
- Read analogue and digital scales or diagrams, using suitable precision (including the nearest half-scale division where required); make expected-value estimates, collect sufficient readings and repeats, record qualitative test observations, and enter measurements systematically in a table with units and appropriate precision.
- Process data for calculations or graphing; draw/read suitable plots and best-fit lines; analyse patterns; interpolate/extrapolate to find values, gradients or intercepts; justify conclusions with observations/data and explanation; judge data quality and handle anomalies appropriately. Evaluate variable control, identify error sources and suggest improvements to apparatus, arrangement, method or technique.
- Worked method — effect of acid concentration on magnesium reaction rate: wear splash goggles, use 0.5 mol/dm³ hydrochloric acid in 25.0 cm³ portions, keep ignition sources away because hydrogen forms, and use a gas syringe attached to a clamped side-arm flask. Change only acid concentration (for example 0.20, 0.40, 0.60, 0.80 mol/dm³); keep acid volume, temperature, same cleaned 2.0 cm magnesium strip and apparatus constant. Add magnesium through the side-arm arrangement, seal without obstructing the outlet, start the timer at contact, and record gas volume every 10 s until it stops increasing. Rinse safely, repeat each concentration at least three times, and record actual readings rather than predicted ones.
- Second worked context — solubility versus temperature: use five clean test-tubes or small beakers, each with 10.0 cm³ water; hold them at measured temperatures such as 20, 30, 40, 50 and 60 °C in a water bath. Add the same dry solute (for example potassium nitrate) in weighed 0.5 g portions, stir each portion until it dissolves, and stop when a small amount remains after stirring. Record total mass dissolved at each temperature, keep solvent volume and endpoint rule the same, and repeat one temperature to check reproducibility. A labelled table separates temperature / °C from solute mass / g and observations; estimate a likely range before selecting balance and sample sizes. Draw a simple setup showing the water bath, sample vessel, thermometer, stirrer and balance; explain that a hotplate is an alternative to a Bunsen near water but a water bath gives steadier temperatures. Evaluate heat loss and subjective saturation endpoint; use a lid/insulation and smaller addition increments as improvements.
- Worked data and graph analysis: for one trial, record time / s = 0, 10, 20, 30, 40 and H₂ volume / cm³ = 0, 19, 34, 44, 50. Plot time on x and volume on y, label both axes with quantity / unit, choose scales using >half the grid, mark each point to half a small square, then draw a smooth best-fit curve. The curve becomes less steep as magnesium is used up. To estimate initial rate, draw a tangent at t=0; if two widely spaced points on that tangent are (0 s, 0 cm³) and (20 s, 36 cm³), gradient = Δvolume/Δtime = 36/20 = 1.8 cm³/s. Mark the gradient triangle on the graph and use points on the tangent, not raw readings.
- Repeat/anomaly/evaluation example: times for the same fixed gas volume are 18.2 s, 18.8 s and 27.0 s. The third is anomalous relative to the close pair; inspect for a loose bung, gas leak or timing error and repeat that trial before deciding whether to exclude it. Mean of the agreeing pair is (18.2 + 18.8)/2 = 18.5 s. A leak makes the measured rate too low; improve the seal and check the syringe moves freely before each run. Temperature drift changes rate, so use a water bath and check temperature. A reasoned conclusion compares repeated initial rates across concentrations and cites the plotted slopes; greater concentration is expected to increase collision frequency and rate.
- Planning/recording worked structure: write the changed variable and its range, measured variable and instrument, each control and how it is held constant, repeat count, hazard and precaution, ordered steps, result table headings with units, processing (mean/rate/graph), predicted trend and what evidence would support the conclusion. Estimate a likely volume/time range first so the syringe or stopwatch is neither overloaded nor too coarse.
What to do in an assessment
- Use the worked method to write a numbered safe procedure, draw and label the apparatus arrangement, fill a results table with units, calculate a mean/rate, plot and interpret the data, mark a large gradient triangle, identify an anomaly, and link one specific error to a practical improvement. State changed/measured/controlled variables and justify apparatus choice.
In plain language
A good investigation is controlled, safe, measurable and evaluated using evidence rather than guesswork.
Self-check
Why repeat a measurement?
Show the answer
Repeats help reveal random variation, identify anomalies and calculate a more reliable mean.
- Course content
Scope: Course content · §4 Mathematical requirements; Presentation of data
Open revision guidance
What to know
- Number work includes addition, subtraction, multiplication and division; decimals, fractions, percentages, ratios (including x:y), reciprocals, positive whole-number indices, standard form, rounding and significant figures. Calculate a mean when asked and use a calculator appropriately.
- Algebra includes substituting measured values into equations with consistent units, simple rearrangement and direct or inverse proportion. Show a calculation in a traceable sequence and carry the units through it.
- Use appropriate geometrical/measurement terms for lines, angles, curves, diagonals, squares, rectangles and circles (including radius, diameter and circumference). A curve changes direction continuously; a diagonal joins non-adjacent corners of a polygon. Select suitable units and convert cm³↔dm³, mg↔g↔kg, J↔kJ and Pa↔kPa; use SI or approved units.
- Read and plot graphs/charts; interpolate/extrapolate; determine a straight-line gradient and intercept; recognise direct proportionality; find a tangent to a curve where required. Read instrument scales to suitable precision, including halfway between divisions when required.
- In tables, give each heading as quantity / unit (e.g. time / s), keep units out of data cells, record repeats and preserve measured precision. Give a calculated result to the same number of significant figures as the least precise raw datum when the calculation rule applies: 12.0 ÷ 0.75 = 16 (two significant figures, limited by 0.75), not 16.0; keep guard digits until rounding the final answer. For ratios, state the required form such as x:y.
- For graphs use a sharp pencil, label axes with quantity and unit, put the independent variable on x and dependent on y unless instructed otherwise, and choose sensible scales that use more than half of the grid (e.g. 1, 2 or 5 units per 2 cm). Mark points with crosses or encircled dots accurately to half a small square; draw one thin, smooth best-fit straight line or curve through the trend, with roughly equal numbers of plotted points on either side; do not force the line through the origin unless the data support it, and consider an apparent anomaly rather than silently discarding it. Read/interpolate/extrapolate to half a small square and use a marked gradient triangle spanning at least half the best-fit line.
- Worked arithmetic: convert 18.0 cm³ to dm³ by dividing by 1000: 0.0180 dm³. Convert 450 mg to grams: 0.450 g; 2.4 kJ = 2400 J; 180 kPa = 180 000 Pa. Write 0.00045 as 4.5 × 10⁻⁴. For a 1:2 ratio, keep the requested x:y order. A reciprocal rate for 25.0 s is 1/25.0 = 0.0400 s⁻¹. Keep guard digits during working and round only the final result; e.g. 12.746 to 3 significant figures is 12.7.
- Worked mean and anomaly: repeated titres of 24.60, 24.75 and 27.10 cm³ include a distant value. Check the burette record/apparatus and repeat before excluding it; if the first two are concordant, their mean is (24.60 + 24.75)/2 = 24.675 cm³, reported to a precision justified by the readings (24.68 cm³). Do not average an unexplained anomalous value blindly.
- Worked equation/proportion: rate = gas volume ÷ time. At 36 cm³ in 20 s, rate = 1.8 cm³/s. Rearrange V = rt to t = V/r; for V = 50 cm³ and r = 2.0 cm³/s, t = 25 s. If doubling concentration halves the time to a fixed gas volume, time is inversely proportional to concentration; if equal time intervals give equal volume changes, volume is directly proportional to time over that interval.
- Worked graph reading: between plotted points (20 s, 34 cm³) and (30 s, 44 cm³), a straight-line interpolation gives the 25 s volume as 39 cm³. For a line through (0, 2) and (20, 42), gradient = (42 − 2)/(20 − 0) = 2 cm³/s and y-intercept = 2 cm³. On a curved rate graph, draw a tangent at the required time and calculate its gradient from two widely spaced points on that tangent; tangent work is Extended-only.
- A scale with 1 cm³ divisions should be read to the nearest 0.5 cm³ where required; between 13 and 14 cm³, an estimate halfway is 13.5 cm³. A graph point plotted at (25 s, 39 cm³) must be marked within half a small square. In a table write time / s and volume / cm³ in headings, put numbers only in cells, preserve instrument precision, and report calculated precision no finer than the least precise measurement.
- Use the geometry terms accurately: if a circle has radius 2.0 cm, its diameter is 4.0 cm and circumference is 2πr ≈ 12.6 cm. Select units before calculation and convert every term in an equation to the same unit.
What to do in an assessment
- Keep intermediate values unrounded, show formula/substitution, write units and round only the final answer appropriately.
In plain language
Good quantitative chemistry preserves units and instrument precision from measurement through calculation and graph.
Self-check
How should a table heading show the unit for volume?
Show the answer
For example, volume / cm³.
- Course content
Scope: Course content · §4 Apparatus and reagents; Safety in the laboratory; Conventions; Command words
Open revision guidance
What to know
- Be familiar with aluminium foil; balances to 500 g with 0.1 g precision; insulating cups/beakers; 1 dm³, 250 cm³ and 100 cm³ beakers; boiling tubes; Bunsen burners; 50 cm³ burettes; conical flasks; delivery tubes; filter funnels/papers; flame-test wires; measuring cylinders (100, 50, 25 and 10 cm³); glassware labelling pens; pipette fillers; test-tube/boiling-tube racks; red/blue litmus; retort stands, bosses and clamps; droppers; small burette-filling funnels; spatulas; stirring rods; thermometers (−10 to +110 °C, 1 °C graduations); stoppers; stopwatches (1 s); test-tube holders and hard-glass test-tubes; tripods; universal-indicator paper; 25 cm³ volumetric pipettes; wash bottles and white tiles.
- Key listed sizes also include insulating cups/beakers about 150 cm³, squat lip beakers of 1 dm³/250 cm³/100 cm³, boiling tubes about 150 mm × 25 mm, conical flasks 50–250 cm³ and hard-glass test-tubes about 125 mm × 16 mm. Use specified graduations/precision for balances, thermometers, stopwatches and volumetric equipment.
- Operate measurement apparatus deliberately: place a balance on a level bench, zero it, weigh a container then reweigh with the sample and subtract; do not put chemicals directly on the pan. Use a measuring cylinder for approximate volumes, reading the bottom of a liquid meniscus at eye level. For a fixed 25.0 cm³ aliquot rinse a volumetric pipette with the solution, use a pipette filler (never mouth-pipette), set the meniscus to the mark at eye level, let it drain into the receiving flask without blowing out the tip. For titration rinse/fill a burette with the titrant, run liquid through the tap to remove the jet bubble, record the initial reading at eye level, add while swirling the conical flask, then add dropwise near the endpoint and record final reading; titre = final − initial. Clamp the burette vertically using a retort stand/boss/clamp and use a small funnel only to fill, removing it before titrating.
- Read thermometers with the bulb immersed in the sample but not touching the vessel; wait for a stable reading and report the scale precision. Start a stopwatch at the defined reaction event and stop at the stated endpoint; repeat timings. Before a gas-syringe run check that the plunger moves freely and the bung/delivery tube connections are airtight; keep the outlet unblocked. A delivery tube directs gas into the syringe or through liquid only when that test is intended; never heat a sealed vessel.
- For heating, support a beaker or boiling tube on a tripod as appropriate, keep it stable and never use a stoppered vessel over a flame. Light a Bunsen safely, use the visible yellow safety flame when not heating, and use the air hole only when directed; tie hair back and keep flammable solvents away. A test-tube holder grips the upper part of a heated tube, pointed away from people. Clamp apparatus securely before heating or collecting gas.
- For filtration, seat filter paper in a funnel, wet it to seal, place the funnel over a labelled receiving beaker/flask, and pour down a glass stirring rod without filling above the paper rim; rinse residue with a wash bottle if directed. Use spatulas for solids, droppers/teat pipettes for small additions, a glass rod for mixing, and a rack/holder for test-tubes. Put a white tile beneath a flask to see a faint indicator endpoint. Label every sample and never return used reagent to its stock bottle.
- For flame tests use a clean flame-test wire (or specified alternative): clean it between samples, dip only the tip into a small solid sample, introduce it at the edge of the blue flame and observe against the stated background. Litmus/universal-indicator paper is touched with a clean dropper or glass rod; do not dip used paper into a reagent bottle. In a labelled apparatus drawing show each vessel, connection, clamp/support, liquid level and direction of gas flow.
- Recognise the standard bench reagents/indicators: dilute HCl (1.0 mol/dm³), dilute HNO₃ (1.0 mol/dm³), dilute H₂SO₄ (0.5 mol/dm³), aqueous NH₃ (1.0 mol/dm³), NaOH (1.0 mol/dm³), Ba(NO₃)₂ (0.1 mol/dm³), AgNO₃ (0.05 mol/dm³), saturated Ca(OH)₂ limewater, KI (0.1 mol/dm³), KMnO₄ (0.02 mol/dm³), acidified KMnO₄ (0.01 mol/dm³ in 0.5 mol/dm³ H₂SO₄), methyl-orange/screened methyl-orange, thymolphthalein and freshly made starch indicator. Reagent preparation is not expected.
- Apparatus-table qualifications: the stated laboratory specifications include ISO 385 or Grade B burettes, ISO 6706 / ISO 4788 or Grade B measuring cylinders, and ISO 648 or Grade B volumetric pipettes. Flame-test wire or the stated alternative apparatus may be used. These procurement/specification details identify the source's equipment; they do not replace learning how to operate it. Indicator ranges are methyl orange pH 3.1–4.4, screened methyl orange pH 3.2–4.2 and thymolphthalein pH 9.3–10.5. Candidates are not expected to learn how to prepare the listed reagents or indicators.
- Follow centre/local safety and disposal rules, use risk-specific precautions and hazard data, and wear appropriate safety equipment including eye protection. Know the hazard-code terms corrosive (C), health hazard (HH), moderate hazard (MH), acutely toxic (T), flammable (F), oxidising (O) and aquatic-environment hazard (N); the codes themselves are not assessed.
- Use syllabus nomenclature and symbols. Traditional sulfate, sulfite, nitrate, nitrite, sulfuric acid and nitric acid names are used; traditional and systematic names receive credit unless oxidation-number naming is specifically required. Use a line decimal point; print 1000–9999 without separators, and use spaces between digit groups at 10 000 and above. Identify independent (changed) and dependent (measured) variables correctly. Use dm³ rather than litre/l, SI or approved units, no imperial/customary units, and keep units consistent with supplied data.
- Command words: analyse—examine details, elements and relationships; calculate—work out from information; compare—comment on similarities and/or differences; consider—review/respond to information; contrast—comment on differences; deduce—conclude from information; define—give precise meaning; demonstrate—show how/give an example; describe—state characteristics/main features; determine—establish an answer from available information; discuss—write about issues in a structured, in-depth way; evaluate—judge quality/importance/amount/value; examine—investigate closely; explain—give purposes/reasons and make relationships clear, supporting with evidence; give—recall or use a source; identify—name/select/recognise; justify—support with evidence/argument; predict—say what may happen from information; show—that—give structured evidence leading to a result; sketch—draw key features with care over proportions; state—express clearly; suggest—apply knowledge to propose a valid response.
What to do in an assessment
- Identify the command word, then answer in the requested form: for a definition be precise, for an explanation link cause and effect, and for evaluation make a judgement supported by evidence.
In plain language
Correct apparatus, safe method, notation and command-word response make chemistry results interpretable.
Self-check
What does 'evaluate' require?
Show the answer
Judge quality, importance, amount or value using relevant evidence or criteria.
Verified qualification
Cambridge IGCSE (9–1) Chemistry (0971)
Exam years · 2027Specification version · Version 2
Course details
CIE · IGCSE Chemistry0971Exam years · 2027
Assessment The assessment includes Paper 1, Paper 2, Paper 3, Paper 4, Paper 5 and Paper 6; see the official specification for the full structure.
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Reviewed course guidance
Revision points, with the explanation beside them
Open a point to read what to know, how to use it in an assessment, a plain-language explanation and a self-check. Answers stay hidden until you choose to reveal them.
Options and source
Cambridge IGCSE (9–1) Chemistry 0971 theory route
- Core route (Papers 1 and 3; Core content; grades 5–1)
- Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1)
Cambridge IGCSE (9–1) Chemistry 0971 practical assessment
- Paper 5 Practical Test (laboratory examination)
- Paper 6 Alternative to Practical (written examination)
§2 Assessment overview and §4 Practical assessment: choose one paper.
Choose each applicable course selection before counting reviewed requirements.
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Topic group
1 States of matter
- Course content
Scope: Course content · §3 1.1 Core 1–4; 1.2 Core 1
Open revision guidance
What to know
- A solid has a fixed shape and volume: particles are very close in an ordered arrangement and vibrate about fixed positions. A liquid has a fixed volume but takes its container's shape: particles remain close but are disordered and slide past one another. A gas has neither fixed shape nor volume: widely separated particles move rapidly and randomly and fill the container.
- Melting changes solid to liquid, boiling and evaporation change liquid to gas, condensation changes gas to liquid, and freezing changes liquid to solid. These are physical changes: the substance's particles remain the same. Heating raises particle motion; cooling lowers it.
- At constant pressure, heating a gas increases its volume; at constant temperature, increasing pressure decreases its volume. The explanation is the balance between particle motion, collision frequency and the pressure exerted on the gas.
- Diffusion is the net movement of particles from higher to lower concentration due to random motion. It occurs in liquids and gases: for example, a coloured solute spreads through still water and a gas spreads through a room.
What to do in an assessment
- For each state, compare shape, volume, spacing, arrangement and motion; for a state change, name the direction and explain it without proposing a chemical change. State what is held constant when predicting a gas-volume change.
In plain language
The same substance behaves differently when its particles have different spacing and freedom to move.
Self-check
A fixed amount of gas is warmed at constant pressure. What happens to its volume, and why does perfume eventually spread across a room?
Show the answer
Its volume increases as heating raises particle motion. Perfume vapour diffuses because randomly moving gas particles spread from the higher-concentration region near the source into regions of lower concentration.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 1.1 Supplement 5–6; 1.2 Supplement 2
Open revision guidance
What to know
- In a single phase, heating increases particles' average kinetic energy and temperature. During melting or boiling, the temperature remains constant while transferred energy separates particles against attractions (increases potential energy); cooling reverses these changes. On a heating/cooling curve, sloped regions are temperature changes and flat plateaux are state changes.
- At constant pressure, a temperature rise makes gas particles move faster and spread farther apart, so the gas volume increases; at constant temperature, a pressure rise forces particles closer together and decreases volume.
- At the same temperature, gas particles have the same average kinetic energy, but lighter molecules move faster on average and diffuse faster than heavier ones. For example, ammonia (Mᵣ 17) diffuses faster than hydrogen chloride (Mᵣ 36.5).
What to do in an assessment
- Read a heating/cooling curve from left to right and identify the state(s), plateaux and sloping regions; explain the temperature and energy change in each. When comparing gas volume or diffusion, state the controlled variable and compare the particle-level cause.
In plain language
Supplement particle theory connects energy transfer to state changes and explains gas expansion, compression and diffusion rates.
Self-check
At the same temperature, which diffuses faster: ammonia (Mᵣ 17) or hydrogen chloride (Mᵣ 36.5)? During boiling, why does added energy not raise the temperature?
Show the answer
Ammonia diffuses faster because its lighter molecules move faster on average at the same temperature. During boiling, transferred energy separates particles rather than increasing their average kinetic energy, so the temperature remains constant.
Topic group
2 Atoms, elements and compounds
- Course content
Scope: Course content · §3 2.1 Core 1; 2.2 Core 1–6; 2.3 Core 1–2
Open revision guidance
What to know
- An element contains one type of atom. A compound contains two or more elements chemically bonded in a fixed ratio. A mixture contains substances together but not chemically bonded; components retain their own properties and may be separated physically.
- An atom has a central nucleus containing protons and neutrons, with electrons in shells. Relative charges are proton +1, neutron 0, electron −1; relative masses are approximately proton 1, neutron 1 and electron 1/1836. Proton number (atomic number) is the number of protons; mass number (nucleon number) is protons plus neutrons.
- For proton numbers 1–20, fill the first shell with 2 electrons, the second with 8, then the third with 8 before placing any remaining electrons in the fourth shell. Thus K (Z=19) is 2,8,8,1 and Ca (Z=20) is 2,8,8,2; chlorine (Z=17) is 2,8,7. Check each arrangement by summing its shell electrons and matching the atom's electron count. For ions, add electrons for a negative charge and remove them for a positive charge: Ca²⁺ has 20 protons but 18 electrons, so it is 2,8,8; Cl⁻ has 17 protons but 18 electrons, so it is 2,8,8. Group VIII noble gases have full outer shells; for Groups I–VII, outer-electron number equals group number, and occupied-shell number equals period number.
- Isotopes are atoms of the same element with the same proton number but different neutron numbers. In nuclide notation, the upper left number is mass number and the lower left is proton number; for example, ²³₁₁Na has 11 protons and 12 neutrons. For ³⁵₁₇Cl⁻, there are 17 protons, 18 neutrons and 18 electrons.
What to do in an assessment
- Calculate neutrons = mass number − proton number; check an electron arrangement by summing electrons and applying the ion charge; read both numbers in nuclide notation before naming protons, neutrons or electrons.
In plain language
Atomic numbers identify elements; neutron number and electron arrangement explain isotope and ion notation.
Self-check
A chloride ion is formed from chlorine-35 (proton number 17). Give its neutron count and electron arrangement.
Show the answer
Neutrons = 35 − 17 = 18. Neutral chlorine has 17 electrons; Cl⁻ has gained one, so it has 18 electrons arranged 2,8,8.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 2.3 Supplement 3–4
Open revision guidance
What to know
- Isotopes of an element have the same proton number. Their neutral atoms therefore have the same number and arrangement of electrons, which gives them the same chemical properties.
- Relative atomic mass is the abundance-weighted mean of isotope masses compared with one-twelfth of a carbon-12 atom; it need not be a whole number.
What to do in an assessment
- Calculate relative atomic mass as Σ(isotope mass × fractional abundance); when using percentage abundances divide the sum of mass × percentage by 100. Check that the abundances total 100%.
In plain language
Relative atomic mass is an average that gives more abundant isotopes more influence.
Self-check
An element has isotopes of mass 35 (75%) and 37 (25%). Find its relative atomic mass.
Show the answer
Aᵣ = [(35 × 75) + (37 × 25)] ÷ 100 = 35.5; the result is an abundance-weighted mean.
- Course content
Scope: Course content · §3 2.4 Core 1–4; 2.5 Core 1–3
Open revision guidance
What to know
- A cation forms when an atom loses electron(s); an anion forms when it gains electron(s). An ionic bond is the strong electrostatic attraction between oppositely charged ions. In sodium chloride, sodium transfers its one outer electron to chlorine, producing Na⁺ and Cl⁻; draw [Na]⁺ and [Cl]⁻ in brackets, with eight outer electrons around Cl (seven of its own plus the transferred electron shown as a cross).
- Ionic compounds have high melting and boiling points because strong electrostatic attractions between oppositely charged ions throughout the lattice require much energy to overcome. In a solid, ions are fixed and cannot carry current; when molten or dissolved in water, ions can move and conduct.
- A covalent bond is a shared pair of electrons; the shared electrons give each atom a noble-gas outer-shell arrangement. Draw dot-and-cross diagrams for H₂, Cl₂, H₂O, CH₄, NH₃ and HCl: each H forms one shared pair, O forms two bonds and has two lone pairs, N forms three bonds and has one lone pair, C forms four bonds, and each Cl forms one bond and has three lone pairs. These are small separate molecules with low melting/boiling points because only a small amount of energy is needed to separate the molecules; the covalent bonds within each molecule remain intact during melting or boiling. They contain no mobile ions/electrons and do not conduct electricity; naming specific intermolecular-force types is not required.
What to do in an assessment
- For NaCl, show Na's outer electron transferred to Cl and label the resulting charges; check total charge is zero. For a covalent molecule, draw one line or shared dot-cross pair per bond and add lone pairs so each atom reaches its outer-shell target.
In plain language
Ionic bonding transfers electrons between atoms; covalent bonding shares electron pairs.
Self-check
Why does sodium chloride conduct when molten but not when solid?
Show the answer
In the solid lattice, Na⁺ and Cl⁻ are fixed in place; when molten, these charged ions are mobile and carry current.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 2.4 Supplement 5–7; 2.5 Supplement 4–5
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What to know
- Ionic solids form a regular giant lattice of alternating positive and negative ions. Each ion is attracted to oppositely charged neighbours in all directions, so the lattice is strong and ionic compounds have high melting and boiling points. Ions carry current only when free to move in a melt or aqueous solution.
- Ionic bonding also forms between metals and non-metals beyond Group I/VII: magnesium transfers two outer electrons to oxygen, giving Mg²⁺ and O²⁻ in MgO. In a dot-and-cross diagram show the two transferred electrons and the full outer shells of both ions.
- Supplement dot-and-cross diagrams include CH₃OH, C₂H₄, O₂, CO₂ and N₂. Show one C=C double bond in ethene, O=O double bond in oxygen, two C=O double bonds in O=C=O, and an N≡N triple bond; include the appropriate non-bonding electron pairs on O and N. In methanol carbon forms four bonds, oxygen two bonds and two lone pairs, and hydrogen one bond.
- Small simple molecules have weak intermolecular attractions between molecules (specific force types are not required), so relatively little energy separates them and they have low melting and boiling points. The covalent bonds within each molecule are not broken during melting or boiling; the substances do not conduct because they have no mobile ions or electrons.
What to do in an assessment
- Explain a property by connecting structure, bonding and the particles or charge carriers that can move.
In plain language
A material's large-scale properties follow from its bonding and structure.
Self-check
Why does a simple covalent molecular substance have a low boiling point without breaking its covalent bonds?
Show the answer
Boiling separates molecules by overcoming their weak intermolecular attractions; the strong covalent bonds within each molecule remain intact.
- Course content
Scope: Course content · §3 2.6 Core 1–2
Open revision guidance
What to know
- In diamond, each carbon atom forms four strong covalent bonds in a rigid three-dimensional giant structure. Many strong bonds make diamond hard and give it a high melting point; its hardness suits cutting tools.
- In graphite, each carbon forms three covalent bonds in flat layers. The layers can slide over one another, so graphite is useful as a lubricant. One electron per carbon is delocalised within the layers and can carry charge, so graphite is useful as an electrode.
What to do in an assessment
- Relate each stated use to a specific structural feature rather than merely naming a property.
In plain language
Different carbon structures create different useful properties.
Self-check
How can graphite be both a lubricant and an electrode?
Show the answer
Its carbon layers slide over each other, while delocalised electrons move through the layers and carry charge.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 2.6 Supplement 3–4; 2.7 Supplement 1–2
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What to know
- Silicon(IV) oxide, SiO₂, has a giant covalent structure: silicon and oxygen atoms are linked by strong covalent bonds throughout a three-dimensional network. Its hardness and high melting point resemble diamond's because both structures contain many strong covalent bonds.
- Metallic bonding is the electrostatic attraction between positive ions in a giant metallic lattice and a sea of delocalised electrons. The mobile electrons carry electrical charge. Layers of metal ions can slide while attraction to the electron sea remains, explaining malleability and ductility.
What to do in an assessment
- For a metal property, identify the role of the delocalised electrons or the ability of the lattice to deform.
In plain language
Metallic bonding combines fixed positive ions with mobile electrons.
Self-check
What moves when a metal conducts?
Show the answer
Delocalised electrons move through the metal lattice.
Topic group
3 Stoichiometry
- Course content
Scope: Course content · §3 3.1 Core 2–4; 3.2 Core 1–3
Open revision guidance
What to know
- A molecular formula gives the number and type of atoms in one molecule. From a model, count each atom: two hydrogen atoms and one oxygen atom give H₂O; one carbon and four hydrogens give CH₄.
- A word equation names reactants and products (magnesium + oxygen → magnesium oxide). A balanced symbol equation conserves the number of each type of atom: 2Mg(s) + O₂(g) → 2MgO(s). State symbols are (s), (l), (g) and (aq); use them when required.
- Relative atomic mass, Aᵣ, is the average mass of an element's isotopes relative to one-twelfth of a carbon-12 atom. Mᵣ is the sum of the Aᵣ values in a molecule (H₂O: 2(1) + 16 = 18); for an ionic compound use relative formula mass (MgO: 24 + 16 = 40).
- Core reacting-mass calculations use simple proportions, not moles. From 2Mg + O₂ → 2MgO, 48 g Mg reacts with 32 g O₂ to form 80 g MgO; 12 g Mg therefore forms 20 g MgO if oxygen is in excess.
What to do in an assessment
- Write correct formulae first, balance by changing coefficients (never subscripts), then check every element's atom count and the state symbols on both sides. For a reacting-mass proportion, use the balanced coefficients and relative formula masses.
In plain language
A chemical equation is a particle-counting statement: atoms are rearranged, not created or lost.
Self-check
Balance the equation and calculate the mass of carbon dioxide made from 12 g carbon in excess oxygen: C + O₂ → CO₂.
Show the answer
C + O₂ → CO₂ is already balanced. Aᵣ(C)=12 and Mᵣ(CO₂)=44, so 12 g C is in the 12:44 mass ratio and forms 44 g CO₂.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 3.1 Supplement 5–8
Open revision guidance
What to know
- An empirical formula is the simplest whole-number ratio of atoms or ions. For an ionic formula, balance total charge and reduce to the simplest ratio: Mg²⁺ and Cl⁻ give MgCl₂; Al³⁺ and O²⁻ give Al₂O₃. A model gives the same ratios by counting ions: six Cl⁻ around three Mg²⁺ have a 2:1 ratio, so the formula unit is MgCl₂.
- Construct balanced symbol equations with state symbols; ionic equations show reacting ions and cancel spectator ions. For example, CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq) has the net ionic equation Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s). Check both atom numbers and total charge.
- When information describes a reaction, identify reactants and products, write their formulae, balance coefficients, and add states from the conditions given.
What to do in an assessment
- For empirical formulae, divide mole amounts by the smallest and scale to whole numbers. For ionic formulae, make the total charge zero. For symbol or ionic equations, balance atoms, then charge, and retain only reacting species in the net ionic equation.
In plain language
Formulae encode simplest composition; ionic equations show only the particles that change.
Self-check
Write the net ionic equation for aqueous barium nitrate reacting with aqueous sodium sulfate to form a precipitate.
Show the answer
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). It is balanced for atoms and total charge.
- Course content
Scope: Course content · §3 3.3 Core 1
Open revision guidance
What to know
- Concentration states how much solute is in a stated solution volume. Mass concentration uses grams per dm³ (g/dm³); amount concentration uses moles per dm³ (mol/dm³). They have different numerators and are not interchangeable.
What to do in an assessment
- Convert cm³ to dm³ by dividing by 1000 before using a molar concentration; keep the specified g/dm³ or mol/dm³ unit with the answer.
In plain language
Concentration states how much solute is present in a stated volume of solution.
Self-check
A solution contains 2.0 g solute in 250 cm³. Find its mass concentration in g/dm³.
Show the answer
250 cm³ = 0.250 dm³, so concentration = 2.0 ÷ 0.250 = 8.0 g/dm³.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 3.3 Supplement 2–8
Open revision guidance
What to know
- One mole is the unit for amount of substance and contains 6.02 × 10²³ particles (Avogadro constant). Use n=m/M, m=nM and M=m/n (mass in g, molar mass in g/mol); particle number N=nNₐ. Examples: 5.85 g NaCl with Mᵣ=58.5 g/mol is 0.100 mol; 0.20 mol of a substance with M=40 g/mol has mass 8.0 g; 4.0 g in 0.10 mol gives M=40 g/mol and relative mass 40; 0.25 mol contains 1.505 × 10²³ particles.
- At r.t.p., 1 mol gas occupies 24 dm³, so n=V/24. For 2H₂ + O₂ → 2H₂O, 16 g O₂ is 0.50 mol and is limiting if 4.0 g H₂ (2.0 mol) is present; it reacts with 1.0 mol H₂ and forms 1.0 mol H₂O = 18 g. A 48 dm³ sample of gas at r.t.p. is 2.0 mol.
- For solutions, n=cV with V in dm³; use the balanced equation's mole ratio. A 25.0 cm³ acid sample neutralised by 20.0 cm³ of 0.100 mol/dm³ NaOH (1:1 reaction) contains 0.00200 mol acid, so its concentration is 0.00200/0.0250 = 0.0800 mol/dm³. To prepare 0.050 mol at 0.250 mol/dm³, V=n/c=0.200 dm³; if Mᵣ=40, that amount has mass 2.0 g, giving 10 g/dm³. Convert cm³ to dm³ by dividing by 1000.
- For an empirical formula, convert masses (or percentages treated as grams in a 100 g sample) to moles and divide by the smallest; 40.0% C, 6.7% H and 53.3% O give a 1:2:1 ratio, CH₂O. If Mᵣ=180 and empirical formula mass=30, the molecular formula is six times larger: C₆H₁₂O₆.
- Percentage yield = actual yield/theoretical yield × 100%; percentage composition = element mass/formula mass × 100%; purity = pure-substance mass/impure-sample mass × 100%. Examples: 7.2 g product from a theoretical 8.0 g is 90% yield; hydrogen is 2/18 × 100 = 11.1% of water by mass; 9.0 g pure material in a 10.0 g sample is 90% pure.
What to do in an assessment
- Balance the equation; convert measured quantities to moles; identify the limiting reactant; apply the coefficient ratio; convert to the requested mass, gas volume, solution volume, concentration or particle number; include units and appropriate significant figures.
In plain language
The mole lets a balanced equation connect measurable mass, solution and gas quantities.
Self-check
A titration uses 20.0 cm³ of 0.100 mol/dm³ NaOH to neutralise 25.0 cm³ HCl. Find the HCl concentration, given HCl + NaOH → NaCl + H₂O.
Show the answer
n(NaOH)=0.100 × 0.0200=0.00200 mol. The 1:1 equation gives 0.00200 mol HCl in 0.0250 dm³; c=0.00200/0.0250=0.0800 mol/dm³.
- Course content
Scope: Course content · §3 3.1 Core 1
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What to know
- Element symbols/formulae used across the course: H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂ are diatomic; C, Si, S and P are written as element symbols in formulae. The named noble gases include He, Ne and Ar. Named metals include Li, Na, K, Mg, Ca, Al, Zn, Fe, Ni, Cu, Ag, Au, Cr, Co, Ba, Pb and Sn; a metal element's formula is its symbol. Use the periodic table to identify the symbol for any other element named.
- Common molecules and gases: water H₂O; hydrogen H₂; oxygen O₂; nitrogen N₂; chlorine Cl₂; hydrogen chloride HCl; ammonia NH₃; methane CH₄; carbon monoxide CO; carbon dioxide CO₂; sulfur dioxide SO₂; sulfur trioxide SO₃; nitrogen monoxide NO; nitrogen dioxide NO₂; glucose C₆H₁₂O₆.
- Common acids, bases and ions: hydrochloric acid HCl; nitric acid HNO₃; sulfuric acid H₂SO₄; ethanoic acid CH₃COOH; hydroxide OH⁻; ammonium NH₄⁺; nitrate NO₃⁻; carbonate CO₃²⁻; sulfate SO₄²⁻; sulfite SO₃²⁻; phosphate PO₄³⁻; chloride Cl⁻; bromide Br⁻; iodide I⁻; hydrogen ion H⁺. The aqueous cations named in the analysis syllabus are Al³⁺, NH₄⁺, Ca²⁺, Cr³⁺, Cu²⁺, Fe²⁺, Fe³⁺ and Zn²⁺; flame-test cations include Li⁺, Na⁺, K⁺, Ca²⁺, Ba²⁺ and Cu²⁺.
- Common oxides, hydroxides and carbonates: magnesium oxide MgO; aluminium oxide Al₂O₃; silicon dioxide SiO₂; iron(III) oxide Fe₂O₃; hydrated iron(III) oxide Fe₂O₃·xH₂O; copper(II) oxide CuO; calcium oxide CaO; calcium hydroxide Ca(OH)₂; sodium hydroxide NaOH; potassium hydroxide KOH; calcium carbonate CaCO₃; magnesium carbonate MgCO₃; carbon monoxide CO; carbon dioxide CO₂; sulfur dioxide SO₂; nitrogen monoxide NO; nitrogen dioxide NO₂.
- Common salts in reaction and analysis examples: sodium chloride NaCl; lithium chloride LiCl; potassium chloride KCl; aluminium chloride AlCl₃; magnesium chloride MgCl₂; zinc chloride ZnCl₂; iron(II) chloride FeCl₂; calcium chloride CaCl₂; copper(II) chloride CuCl₂; cobalt(II) chloride CoCl₂; ammonium chloride NH₄Cl; copper(II) sulfate CuSO₄; zinc sulfate ZnSO₄; iron(II) sulfate FeSO₄; magnesium sulfate MgSO₄; silver nitrate AgNO₃; barium nitrate Ba(NO₃)₂; calcium nitrate Ca(NO₃)₂; ammonium nitrate NH₄NO₃; ammonium sulfate (NH₄)₂SO₄; silver chloride AgCl; silver bromide AgBr; silver iodide AgI; barium sulfate BaSO₄; calcium sulfate CaSO₄; sodium carbonate Na₂CO₃; sodium ethanoate CH₃COONa; magnesium ethanoate Mg(CH₃COO)₂; calcium silicate CaSiO₃; cryolite Na₃AlF₆; potassium manganate(VII) KMnO₄. Hydrated salts include blue copper(II) sulfate CuSO₄·5H₂O and hydrated cobalt(II) chloride CoCl₂·6H₂O.
- Organic formulas include the series CₙH₂ₙ₊₂ (alkanes), CₙH₂ₙ (alkenes), CₙH₂ₙ₊₁OH (alcohols) and CₙH₂ₙ₊₁COOH (carboxylic acids); examples include methane CH₄, ethane C₂H₆, propane C₃H₈, butane C₄H₁₀, ethene C₂H₄, propene C₃H₆, butene C₄H₈, methanol CH₃OH, ethanol C₂H₅OH, propanol C₃H₇OH, butanol C₄H₉OH, methanoic acid HCOOH, ethanoic acid CH₃COOH, propanoic acid C₂H₅COOH and butanoic acid C₃H₇COOH. Also recall ethyl ethanoate CH₃COOCH₂CH₃, poly(ethene) repeat unit [–CH₂–CH₂–]ₙ and glucose C₆H₁₂O₆. Use displayed/structural formula rules to write the specified unbranched isomers, ester combinations and polymer repeat units.
What to do in an assessment
- Recall the symbol or formula from the chemical name; use subscripts for atom counts, brackets for a repeated polymer unit, and charges for ions. For any other named ionic compound, use the ions' charges to determine the simplest neutral formula and retain any stated water of crystallisation. Check that ionic formulae have zero total charge and molecular formulae preserve each atom count.
In plain language
A chemical name identifies a fixed symbol or atom ratio; this bank gathers the formulae needed throughout the course.
Self-check
Give the formula for calcium hydroxide, the nitrate ion, and a typical alkane with four carbon atoms.
Show the answer
Calcium hydroxide is Ca(OH)₂, nitrate is NO₃⁻, and the alkane formula CₙH₂ₙ₊₂ gives C₄H₁₀.
Topic group
4 Electrochemistry
- Course content
Scope: Course content · §3 4.1 Core 1–7
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What to know
- Electrolysis is decomposition of an ionic compound by an electric current when it is molten or in aqueous solution. The anode is positive, the cathode negative, and the electrolyte is the molten or aqueous ionic substance that undergoes electrolysis.
- With inert electrodes, molten PbBr₂ gives lead at the cathode and bromine at the anode (silvery lead forms; reddish-brown bromine vapour appears). Concentrated aqueous NaCl gives hydrogen at the cathode and chlorine at the anode (bubbles form; chlorine is a greenish-yellow gas that bleaches damp indicator paper). Dilute H₂SO₄ gives hydrogen at the cathode and oxygen at the anode (bubbles; hydrogen gives a squeaky pop and oxygen relights a glowing splint).
- At the cathode, metals or hydrogen form; at the anode, non-metals other than hydrogen form. To predict products from a molten binary compound, send the metal cation to the cathode and the non-metal anion to the anode, then identify the elements formed.
- Electroplating improves appearance and resistance to corrosion. Connect the object to be coated as the cathode, the coating metal as the anode, and use an electrolyte containing ions of the coating metal; metal deposits on the object as the anode supplies ions.
What to do in an assessment
- For each electrode, identify its sign, arriving ion, product and observation. Use the stated physical state and concentration of the electrolyte; label anode/cathode correctly before predicting products.
In plain language
An electric current drives oppositely charged ions to electrodes where products form.
Self-check
Predict the products in molten lead(II) bromide and name the electrode signs.
Show the answer
Pb²⁺ goes to the negative cathode and forms lead; Br⁻ goes to the positive anode and forms reddish-brown bromine. The electrolyte must be molten so its ions can move.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 4.1 Supplement 8–11
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What to know
- Electrons move through the external circuit, while ions carry charge through the electrolyte: cations move to the cathode and anions to the anode. At the cathode cations gain electrons (reduction); at the anode anions lose electrons (oxidation). Electrons are not the charge carriers travelling through the solution.
- In aqueous CuSO₄ with inert carbon/graphite electrodes, reddish-brown copper deposits at the cathode and oxygen bubbles at the anode; Cu²⁺ is removed, so the blue solution becomes paler. With copper electrodes, copper deposits at the cathode while the anode dissolves; its mass decreases and the blue colour/concentration is maintained.
- For an aqueous halide, the cathode product depends on the cation: a less reactive metal than hydrogen deposits as metal; otherwise hydrogen forms. At the anode, a concentrated chloride/bromide/iodide solution forms the halogen; a dilute halide solution forms oxygen instead. Thus concentrated NaCl(aq) gives H₂ and Cl₂; dilute NaCl(aq) gives H₂ and O₂.
- Half-equations show electron transfer and conserve atoms and charge. Reduction at the cathode: Cu²⁺ + 2e⁻ → Cu; hydrogen formation: 2H⁺ + 2e⁻ → H₂. Oxidation at the anode: 2Cl⁻ → Cl₂ + 2e⁻; oxygen formation in dilute acid: 2H₂O → O₂ + 4H⁺ + 4e⁻.
What to do in an assessment
- Write the cathode and anode processes separately; balance atoms first and electrons/charge next. Electrons appear as reactants for reduction and products for oxidation; multiply half-equations only when combining them so electron numbers cancel.
In plain language
The electrodes exchange electrons while ions carry charge inside the solution.
Self-check
Write the half-equations for concentrated aqueous sodium chloride and identify oxidation/reduction.
Show the answer
Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ (reduction). Anode: 2Cl⁻ → Cl₂ + 2e⁻ (oxidation). At the cathode electrons are gained; at the anode electrons are lost.
- Course content
Scope: Course content · §3 4.2 Core 1
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What to know
- A hydrogen–oxygen fuel cell uses hydrogen and oxygen to produce electricity, with water as its only chemical product. The cell converts chemical energy into electrical energy; the reactants are supplied continuously.
What to do in an assessment
- Distinguish the cell's chemical product from electricity, its useful output.
In plain language
A hydrogen–oxygen fuel cell produces electricity from supplied hydrogen and oxygen, with water as its only chemical product.
Self-check
What are the reactants and only chemical product of a hydrogen–oxygen fuel cell?
Show the answer
Hydrogen and oxygen are used to produce electricity; water is the only chemical product.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 4.2 Supplement 2
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What to know
- At the vehicle, a hydrogen–oxygen fuel cell produces electricity and water and emits no carbon dioxide from the cell; a petrol engine releases energy by combustion and produces carbon dioxide and other exhaust gases.
- Fuel cells can be more energy-efficient and hydrogen can be made using renewable electricity, but hydrogen must be produced, compressed or liquefied, transported and stored; these steps use energy and add cost. Petrol has an established storage/refuelling network, while hydrogen infrastructure and fuel-cell systems are expensive and less widespread.
- The environmental comparison depends on how hydrogen and petrol are produced and delivered, not only on the vehicle's tailpipe.
What to do in an assessment
- Compare at least one advantage and one disadvantage on both point-of-use emissions and fuel supply; support a conclusion with the stated criteria.
In plain language
A fuel cell is cleaner at the point of use, but its total merits depend on how hydrogen is made and supplied.
Self-check
Give one advantage and one disadvantage of a hydrogen fuel-cell vehicle compared with a petrol vehicle.
Show the answer
An advantage is that the cell's only chemical product is water and it emits no CO₂ at the vehicle. A disadvantage is that hydrogen production, storage and distribution require energy and costly infrastructure.
Topic group
5 Chemical energetics
- Course content
Scope: Course content · §3 5.1 Core 1–3
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What to know
- An exothermic reaction transfers thermal energy from the reacting system to its surroundings, so the surroundings warm. Combustion is a familiar exothermic example.
- An endothermic reaction takes thermal energy from the surroundings into the reacting system, so the surroundings cool. Thermal decomposition of calcium carbonate is endothermic.
- On a reaction pathway diagram, energy is the vertical axis and reaction progress the horizontal axis. The reactants and products are horizontal energy levels joined by a pathway: products lower than reactants mean a net energy release (exothermic); products higher than reactants mean energy is taken in (endothermic).
What to do in an assessment
- Read the vertical energy levels of reactants and products, compare them, and state the direction of thermal-energy transfer and the resulting surroundings-temperature change.
In plain language
The temperature change around a reaction reveals the direction of thermal-energy transfer.
Self-check
A pathway diagram shows products below reactants. Classify the reaction and predict the change in the surroundings.
Show the answer
It is exothermic: energy is transferred to the surroundings, so their temperature rises.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 5.1 Supplement 4–8
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What to know
- The transfer of thermal energy in a reaction is its enthalpy change, ΔH. ΔH is negative for an exothermic reaction and positive for an endothermic reaction.
- Activation energy, Eₐ, is the minimum energy colliding particles must have to react. On a pathway diagram, Eₐ is the vertical energy difference from the reactant level to the peak; ΔH is the signed vertical difference from the reactant level to the product level.
- To draw either pathway from information, label the vertical axis energy and horizontal axis reaction progress, put reactants and products at their stated relative energies, draw the pathway peak, and add arrows for ΔH and Eₐ. For an exothermic profile the products are lower; for an endothermic profile they are higher.
- Breaking bonds is endothermic because energy is absorbed; making bonds is exothermic because energy is released. ΔH = total energy needed to break bonds − total energy released when new bonds form.
- Worked bond-energy example: H₂ + Cl₂ → 2HCl. With H–H = 436, Cl–Cl = 243 and H–Cl = 431 kJ/mol, ΔH = (436 + 243) − 2(431) = −183 kJ/mol, so the reaction is exothermic.
What to do in an assessment
- Label reactants, products, ΔH and Eₐ on both types of pathway. For a bond-energy calculation, count every bond broken and formed using the balanced equation, multiply each bond energy by its coefficient, then subtract total energy released from total energy absorbed; include kJ/mol and interpret the sign.
In plain language
A reaction releases energy only when forming its new bonds releases more than breaking the old bonds requires.
Self-check
For H₂ + Br₂ → 2HBr, use H–H = 436, Br–Br = 193 and H–Br = 366 kJ/mol to calculate ΔH.
Show the answer
ΔH = (436 + 193) − 2(366) = −103 kJ/mol. The negative value means the reaction is exothermic.
Topic group
6 Chemical reactions
- Course content
Scope: Course content · §3 6.1 Core 1; 6.2 Core 1–4
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What to know
- A physical change changes a substance's state or form without making a new substance; melting ice is physical. A chemical change forms one or more new substances because atoms are rearranged; burning magnesium is chemical.
- Reaction rate increases when solution concentration or gas pressure is increased, when a solid's surface area is increased, when temperature is raised, or when a suitable catalyst (including an enzyme) is added. Removing a catalyst or reversing these changes reduces the rate. A catalyst increases rate and is chemically unchanged at the end.
- Investigate rate by measuring a reactant's mass decrease or product's mass increase over time, or by measuring gas volume formed over time. For example, place a flask containing acid and marble chips on a balance and record mass at fixed time intervals as CO₂ escapes; alternatively connect the flask to a gas syringe and record gas volume at the same intervals.
- Calculate an average rate as change in mass or gas volume divided by elapsed time. On a quantity-versus-time graph, the gradient shows rate: a steeper curve means faster reaction; the curve levels when the reaction stops. Compare initial rates using the initial gradient.
What to do in an assessment
- For a fair rate test, change one factor only, keep other conditions and quantities the same, start timing consistently, record mass or gas volume at regular intervals, and plot the measured quantity against time. Read whether the chosen measurement increases or decreases as the reaction proceeds.
In plain language
Rate is how quickly reactants are used or products form.
Self-check
In a marble-chip/acid experiment, mass falls from 105.0 g to 101.0 g in 40 s. What is the average mass-loss rate, and what does a plateau on a mass–time graph mean?
Show the answer
Average rate = 4.0 g ÷ 40 s = 0.10 g/s. A plateau means the mass is no longer changing, so the reaction has stopped producing escaping gas.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 6.2 Supplement 5–8
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What to know
- Collision theory describes reacting particles as colliding. Increasing the number of particles per unit volume (by increasing solution concentration or gas pressure) raises collision frequency. A larger solid surface area exposes more particles for collisions.
- Particles have a distribution of kinetic energies. Raising temperature increases their average kinetic energy, so collisions occur more frequently and a greater fraction have energy at least the activation energy, Eₐ. Only collisions with sufficient energy can react.
- A catalyst supplies an alternative pathway with lower Eₐ, increasing the fraction of successful collisions; it is not consumed and does not change the overall enthalpy change of the reaction.
- Use collision theory to explain each rate factor: concentration/pressure changes particles per unit volume and collision frequency; surface area changes exposed particles; temperature changes particle kinetic energy and the fraction above Eₐ; catalyst changes Eₐ.
- Evaluate a practical rate method against the measurement needed. A gas syringe gives a direct gas-volume/time record but leaks, sticking or a delayed start can reduce accuracy; improve by checking airtight connections and collecting repeats. Measuring mass loss on a balance is simple for a gas that escapes, but drafts, balance resolution and the start-time delay affect readings; shield the balance and repeat. Keep reactant amounts, temperature and apparatus constant when comparing runs.
What to do in an assessment
- Build an explanation as condition change → particle-level change → change in frequency or proportion of successful collisions → rate change. Distinguish collision frequency from collision energy.
In plain language
Faster reactions have more frequent successful collisions.
Self-check
Why does raising temperature usually increase reaction rate more than merely increasing collision frequency?
Show the answer
Higher temperature raises average kinetic energy and greatly increases the fraction of collisions with energy at least Eₐ, so more collisions are successful as well as more frequent.
- Course content
Scope: Course content · §3 6.3 Core 1–2
Open revision guidance
What to know
- The symbol ⇌ shows that a chemical reaction is reversible: products can react to form the original reactants as well as reactants forming products.
- Heating hydrated copper(II) sulfate drives off water and changes blue crystals to white anhydrous CuSO₄; adding water reverses the change and restores blue hydrated crystals. Hydrated cobalt(II) chloride is pink; heating removes water and forms blue anhydrous CoCl₂, while adding water restores the pink hydrate.
What to do in an assessment
- Read the ⇌ symbol as a two-way reaction. For the two named salts, connect heating to loss of water and the anhydrous colour, and addition of water to formation of the hydrate and its colour.
In plain language
In a reversible reaction, products can react to remake reactants.
Self-check
A blue hydrated copper(II) sulfate sample is heated and becomes white. What has happened, and what change reverses it?
Show the answer
It has lost water of crystallisation to form white anhydrous CuSO₄. Adding water reforms the blue hydrated salt.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 6.3 Supplement 3–11
Open revision guidance
What to know
- In a closed system, dynamic equilibrium is reached when the forward and reverse reaction rates are equal. Both reactions continue, but reactant and product concentrations remain constant.
- A change in temperature shifts equilibrium toward the endothermic direction (and away from the exothermic direction). Increasing pressure shifts a gaseous equilibrium toward the side with fewer gas molecules; lowering pressure favours the side with more. Increasing a reactant concentration or removing a product shifts the position to use up that change; adding product or removing reactant shifts it the opposite way. A catalyst speeds both directions equally, so equilibrium is reached sooner but its position and yield do not change.
- Apply the rules using the equation and conditions provided: compare gas-molecule counts for pressure, identify the endothermic direction for temperature, and see which direction consumes an added substance or replaces a removed one. For N₂ + 3H₂ ⇌ 2NH₃, higher pressure favours the right (2 gas molecules rather than 4); the forward reaction is exothermic, so lower temperature favours ammonia; adding N₂ or removing NH₃ also shifts right. For 2SO₂ + O₂ ⇌ 2SO₃, higher pressure favours the right (2 rather than 3 gas molecules), and lower temperature favours the exothermic product direction.
- Haber process: N₂ + 3H₂ ⇌ 2NH₃. Nitrogen is obtained from air and hydrogen from methane. The typical conditions are 450 °C, 20 000 kPa (200 atm) and an iron catalyst. Higher pressure improves ammonia yield and rate but requires costly, robust equipment; 450 °C is a compromise between equilibrium yield and a useful rate; the catalyst speeds attainment of equilibrium without changing yield.
- Contact process: 2SO₂ + O₂ ⇌ 2SO₃. Sulfur dioxide is made by burning sulfur or roasting sulfide ores; oxygen comes from air. The typical conditions are 450 °C, 200 kPa (2 atm) and vanadium(V) oxide catalyst. The temperature balances exothermic yield against rate; although higher pressure favours SO₃, the modest extra yield does not justify the extra pressure cost and safety demands. The catalyst increases rate, not equilibrium yield.
- Industrial operating conditions are selected by balancing reaction rate, equilibrium position/yield, safety and economics rather than choosing only the condition that maximises equilibrium yield.
What to do in an assessment
- For a changed condition, predict the direction of shift and justify it with the endothermic direction, gas count or concentration change. For each process, state sources, balanced reversible equation and typical conditions; explain how those conditions balance rate, equilibrium, safety and cost.
In plain language
Industry chooses conditions that make product quickly and economically without ignoring equilibrium, safety or cost.
Self-check
For N₂ + 3H₂ ⇌ 2NH₃ (forward exothermic), predict the yield effect of raising pressure and temperature. Why is iron added?
Show the answer
Raising pressure favours ammonia because the product side has fewer gas molecules; raising temperature shifts equilibrium left, away from exothermic ammonia formation. Iron increases the rates of both directions so equilibrium is reached faster, without changing its position or yield.
- Course content
Scope: Course content · §3 6.4 Core 1–5
Open revision guidance
What to know
- A redox reaction involves oxidation and reduction simultaneously. At Core, oxidation is gain of oxygen and reduction is loss of oxygen. For example, in Fe₂O₃ + 3CO → 2Fe + 3CO₂, iron(III) oxide loses oxygen to become iron, while carbon monoxide gains oxygen to become carbon dioxide.
- Roman numerals in names show an element's oxidation number in a compound: FeO is iron(II) oxide, while Fe₂O₃ is iron(III) oxide. Identify the oxygen gained and lost to decide which reactant is oxidised and which is reduced.
What to do in an assessment
- Compare the reactant and product formulas for oxygen transfer; name the species gaining oxygen as oxidised and the species losing oxygen as reduced. Check that both processes occur in the same redox reaction.
In plain language
When oxygen transfers, one substance gains it as another loses it.
Self-check
In Fe₂O₃ + 3CO → 2Fe + 3CO₂, identify what is oxidised and reduced; what does iron(III) mean?
Show the answer
CO gains oxygen to form CO₂, so CO is oxidised; Fe₂O₃ loses oxygen to form Fe, so it is reduced. Iron(III) indicates oxidation number +3 for iron in that compound.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 6.4 Supplement 6–13
Open revision guidance
What to know
- Oxidation is loss of electrons and an increase in oxidation number; reduction is gain of electrons and a decrease in oxidation number. Thus in 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, Fe²⁺ is oxidised (loses electrons; +2 to +3) and Cl₂ is reduced (gains electrons; 0 to −1). Redox involves both at once.
- Use these oxidation-number rules: an uncombined element has oxidation number 0 (O₂); a monatomic ion's oxidation number equals its charge (Fe³⁺ is +3); oxidation numbers sum to 0 in a neutral compound; and in a polyatomic ion they sum to the ion's charge. For example, in H₂SO₄, 2(+1) + S + 4(−2) = 0, so S = +6; in SO₄²⁻, S + 4(−2) = −2, so S = +6.
- Acidified potassium manganate(VII) is purple and becomes colourless as Mn(VII) is reduced. Aqueous iodide is colourless; when I⁻ is oxidised to iodine, the mixture becomes brown (starch, if present, gives a blue-black colour). These observations can identify a redox change.
- An oxidising agent causes another substance to be oxidised and is itself reduced; a reducing agent causes another substance to be reduced and is itself oxidised. In 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, Cl₂ is the oxidising agent and Fe²⁺ is the reducing agent.
What to do in an assessment
- Write oxidation numbers for relevant atoms/ions using the rules, compare reactants with products, and identify increases as oxidation and decreases as reduction. Track electron gain/loss when charges are shown; identify the oxidising/reducing agent by the change it causes and the change it undergoes.
In plain language
Redox is electron transfer; oxidation numbers provide a systematic way to track it.
Self-check
In 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, identify oxidation, reduction, oxidising agent and reducing agent.
Show the answer
Fe²⁺ loses electrons and rises from +2 to +3, so it is oxidised and is the reducing agent. Cl₂ gains electrons and falls from 0 to −1, so it is reduced and is the oxidising agent.
Topic group
7 Acids, bases and salts
- Course content
Scope: Course content · §3 7.1 Core 1–8
Open revision guidance
What to know
- Acids react with suitable reactive metals to make a salt and hydrogen (e.g. magnesium + hydrochloric acid → magnesium chloride + hydrogen), with bases to make a salt and water, and with carbonates to make a salt, water and carbon dioxide.
- Acids turn blue litmus red, leave thymolphthalein colourless, and turn methyl orange red. Alkalis turn red litmus blue, turn thymolphthalein blue, and turn methyl orange yellow; methyl orange is orange near neutral. Use the indicator's colour change to distinguish acidic, neutral and alkaline conditions.
- A base is a metal oxide or metal hydroxide; an alkali is a base that dissolves in water. Bases neutralise acids to give a salt and water. When a base is warmed with an ammonium salt, ammonia is released; for example, NaOH + NH₄Cl → NaCl + H₂O + NH₃.
- Aqueous acids contain H⁺(aq) ions and aqueous alkalis contain OH⁻(aq) ions. Neutralisation is H⁺(aq) + OH⁻(aq) → H₂O(l). Universal indicator paper compares hydrogen-ion concentration: acidic solutions are below pH 7, neutral solution is pH 7, and alkaline solutions are above pH 7; the colour scale progresses red/orange/yellow through green to blue/purple as pH rises.
What to do in an assessment
- For acid reactions, identify the reactant class and predict the salt and any gas/water. Distinguish a base from a soluble alkali; use the stated indicator colours or pH key to classify acidity/alkalinity, and use H⁺ + OH⁻ → H₂O for neutralisation.
In plain language
Acids and bases react to make salts; acid and alkali neutralise when H⁺ and OH⁻ form water.
Self-check
What happens when hydrochloric acid reacts with calcium carbonate, and what colour does methyl orange show in the acid before reaction?
Show the answer
Calcium chloride, water and carbon dioxide form; acid turns methyl orange red.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 7.1 Supplement 9–12
Open revision guidance
What to know
- A Brønsted acid is a proton (H⁺) donor; a Brønsted base is a proton acceptor.
- A strong acid is completely dissociated in aqueous solution; a weak acid is only partially dissociated. Strength describes degree of dissociation, not the acid's concentration.
- Hydrochloric acid is strong and is shown by a one-way arrow: HCl(aq) → H⁺(aq) + Cl⁻(aq). Ethanoic acid is weak and is shown with an equilibrium arrow: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq).
What to do in an assessment
- Use a one-way arrow for complete dissociation and an equilibrium arrow for partial dissociation; distinguish acid strength from solution concentration.
In plain language
Strong and weak describe how completely an acid ionises, not how much acid was added.
Self-check
Why is HCl(aq) shown with → but CH₃COOH(aq) with ⇌?
Show the answer
HCl is completely dissociated in water (strong acid), while ethanoic acid is only partially dissociated (weak acid), so both forward and reverse changes are represented.
- Course content
Scope: Course content · §3 7.2 Core 1; 7.3 Core 1–3
Open revision guidance
What to know
- Classify non-metal oxides such as SO₂ and CO₂ as acidic oxides; metal oxides such as CuO and CaO are basic oxides. This pattern is related to the metallic or non-metallic character of the element.
- Prepare a soluble salt from an acid and a soluble alkali by titration: measure a fixed volume of alkali, add indicator, titrate with acid to the end-point, repeat using the exact volumes without indicator, gently evaporate some water to concentrate the neutral solution, cool to crystallise the salt, then filter and dry the crystals.
- For a soluble salt made from acid and an insoluble solid, warm the acid and add excess metal, insoluble base or insoluble carbonate until no more reacts. Filter off the excess solid, evaporate the filtrate to concentrate it, allow crystals to form, then filter and dry the crystals. Example: warm sulfuric acid with excess CuO, filter, then crystallise CuSO₄ from the filtrate.
- Solubility rules: all sodium, potassium and ammonium salts are soluble; all nitrates are soluble; chlorides are soluble except lead and silver chlorides; sulfates are soluble except barium, calcium and lead sulfates; carbonates are insoluble except sodium, potassium and ammonium carbonates; hydroxides are insoluble except sodium, potassium and ammonium hydroxides, while calcium hydroxide is partially soluble.
- A hydrated substance is chemically combined with water; an anhydrous substance contains no water. Heating a hydrated salt can produce the anhydrous form.
What to do in an assessment
- Choose titration when both starting solutions are soluble, so no reactant can be removed by filtration. For a metal, insoluble base or insoluble carbonate, add excess solid, filter, then crystallise the dissolved salt. For each route, state how the salt is separated and purified.
In plain language
The salt-preparation method depends on whether the reactants and product dissolve.
Self-check
How would you prepare pure hydrated copper(II) sulfate crystals from sulfuric acid and copper(II) oxide?
Show the answer
Warm dilute sulfuric acid, add CuO in excess until no more reacts, filter off the excess CuO, gently evaporate the filtrate to concentrate it, allow it to cool and crystallise, then filter and dry the crystals.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 7.2 Supplement 2–3; 7.3 Supplement 4–5
Open revision guidance
What to know
- An amphoteric oxide reacts with both acids and bases to form a salt and water. Aluminium oxide, Al₂O₃, and zinc oxide, ZnO, are amphoteric; they are neither simply acidic nor simply basic.
- Prepare an insoluble salt by precipitation: mix solutions of two soluble salts so their ions form the required insoluble product, then filter, wash the precipitate with distilled water and dry it. For example, Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) produces yellow lead(II) iodide.
- Water of crystallisation is the water molecules present in hydrated crystals; the formula shows their fixed number, as in CuSO₄·5H₂O and CoCl₂·6H₂O.
What to do in an assessment
- Use the solubility rules to select two soluble starting solutions that supply the ions for the required precipitate.
In plain language
A precipitation reaction makes an insoluble salt directly from dissolved ions.
Self-check
Describe how to make and isolate an insoluble salt from two soluble salt solutions.
Show the answer
Mix the solutions so the required insoluble salt precipitates, filter it off, wash it with distilled water, and dry it. This is precipitation.
Topic group
8 The Periodic Table
- Course content
Scope: Course content · §3 8.1 Core 1–5; 8.2 Core 1–2; 8.3 Core 1–4; 8.4 Core 1; 8.5 Core 1
Open revision guidance
What to know
- The Periodic Table arranges elements in horizontal periods and vertical groups in increasing proton (atomic) number. Metallic character generally decreases from left to right across a period. For main-group elements, group position helps predict common ion charge: Groups I, II and III typically form +1, +2 and +3 ions, while Groups V, VI and VII typically form −3, −2 and −1 ions; Group IV elements generally share electrons rather than form simple ions.
- Elements in the same group have similar chemical properties because their atoms have the same number of outer-shell electrons. An element's position can therefore predict its outer-shell arrangement, common ion charge, metallic/non-metallic character and how it behaves relative to other members of its group.
- Group I lithium, sodium and potassium are relatively soft metals. Down the group, melting point decreases, while density and reactivity increase. When information about another Group I element is supplied, identify the trend and use it to predict the stated property rather than assuming every property changes in the same direction.
- Group VII chlorine, bromine and iodine are diatomic non-metals. Down the group, density increases and reactivity decreases. At r.t.p., chlorine is a pale yellow-green gas, bromine a red-brown liquid, and iodine a grey-black solid. A more reactive halogen displaces a less reactive halide from solution: Cl₂ + 2KBr → 2KCl + Br₂; Br₂ + 2KI → 2KBr + I₂.
- Transition elements are metals with high densities and high melting points; they form coloured compounds and often act as catalysts both as elements and in compounds (for example, iron and vanadium(V) oxide in industrial processes).
- Group VIII (Group 18) noble gases are unreactive, monatomic gases. Their outer electron shell is full (helium's first shell contains two electrons), so they have little tendency to gain, lose or share electrons.
What to do in an assessment
- Read a group's direction and stated trend carefully; use supplied data to predict an unknown member only for that property. For halogen displacement, compare reactivity first, then write the displaced halogen and new halide salt. Connect similar group chemistry to outer-shell electron number.
In plain language
The table's position and repeated outer-electron patterns help predict chemical behaviour.
Self-check
Will bromine displace chloride ions from potassium chloride solution? Explain, and state chlorine's appearance at r.t.p.
Show the answer
No. Bromine is less reactive than chlorine, so it cannot displace chloride; chlorine is a pale yellow-green gas at r.t.p.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 8.1 Supplement 6; 8.4 Supplement 2
Open revision guidance
What to know
- A question may supply observations or data for elements in a group. Compare values down the group, state the trend using the correct direction and property, then apply that trend to predict a value or property for another member.
- Transition elements can form ions with variable oxidation numbers. Iron(II) is Fe²⁺ and iron(III) is Fe³⁺; the Roman numeral gives the ion's charge/oxidation number.
What to do in an assessment
- State the pattern in the given data and use it to predict the next property; identify the iron ion charge from its Roman numeral.
In plain language
Group patterns support predictions, while transition metals can form ions with different charges.
Self-check
What does a supplied Group I data table let you do, and what ions are named by iron(II) and iron(III)?
Show the answer
State the observed down-group trend and use it to predict a property for another member. Iron(II) is Fe²⁺; iron(III) is Fe³⁺.
Topic group
9 Metals
- Course content
Scope: Course content · §3 9.1 Core 1–2; 9.2 Core 1; 9.3 Core 1–4
Open revision guidance
What to know
- Metals generally conduct heat and electricity well, are malleable (can be hammered into shape) and ductile (can be drawn into wires), and have relatively high melting and boiling points. Non-metals generally have poorer thermal/electrical conductivity, are brittle when solid, and often have lower melting and boiling points. These are general comparisons, not rules without exceptions.
- The general chemical reactions of metals are with dilute acids, cold water or steam, and oxygen. A sufficiently reactive metal plus dilute acid forms a salt and hydrogen; reactive metals plus cold water form a metal hydroxide and hydrogen, while some metals react with steam to form a metal oxide and hydrogen; reaction with oxygen forms a metal oxide. Whether a particular metal reacts depends on its position in the reactivity series.
- Aluminium's low density suits aircraft; its low density and good electrical conductivity suit overhead cables; its corrosion resistance suits food containers. Copper's good electrical conductivity and ductility make it suitable for electrical wiring.
- An alloy is a mixture of a metal with other elements: brass is copper and zinc; stainless steel is iron mixed with elements such as chromium, nickel and carbon. Alloys can be harder and stronger than pure metals. Stainless steel is used for cutlery because of its hardness and resistance to rusting. In a particle diagram, an alloy has atoms of more than one element mixed in the metal structure.
What to do in an assessment
- Explain a material choice by linking the required function to a stated physical property.
In plain language
Metal applications depend on a match between material properties and the job.
Self-check
Why is copper suitable for electrical wiring?
Show the answer
It conducts electricity well and can be drawn into ductile wires.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 9.3 Supplement 5
Open revision guidance
What to know
- Different-sized atoms in an alloy distort the regular metal layers, so those layers cannot slide over one another as easily as in a pure metal. This explains increased hardness and strength.
What to do in an assessment
- Use an atom-layer diagram to show how unlike atom sizes interrupt sliding.
In plain language
Alloy atoms make it harder for metal layers to move past each other.
Self-check
Why can an alloy be harder than a pure metal?
Show the answer
Different-sized atoms disrupt the regular layers and hinder sliding.
- Course content
Scope: Course content · §3 9.4 Core 1–3; 9.5 Core 1–3
Open revision guidance
What to know
- The reactivity series, most to least reactive, is potassium, sodium, calcium, magnesium, aluminium, carbon, zinc, iron, hydrogen, copper, silver, gold. Potassium, sodium and calcium react with cold water to form the metal hydroxide and hydrogen; magnesium reacts with steam to form magnesium oxide and hydrogen. Magnesium, zinc and iron react with dilute hydrochloric acid to form the metal chloride and hydrogen, while copper, silver and gold do not react with dilute hydrochloric acid.
- A more reactive metal displaces a less reactive metal from its aqueous ions; no displacement occurs in the opposite order. For example, zinc displaces copper from copper(II) sulfate: Zn + CuSO₄ → ZnSO₄ + Cu. Use experimental results (including which metal reacts or displaces another) to deduce an order of reactivity.
- Iron and steel rust in the presence of both oxygen and water, forming hydrated iron(III) oxide, Fe₂O₃·xH₂O. Painting, greasing and coating with plastic are barrier methods: they exclude oxygen or water from the iron and so prevent rusting.
What to do in an assessment
- Use the reactivity series to predict whether a reaction occurs and identify which metal is oxidised in a displacement.
In plain language
A more reactive metal loses electrons more readily and can displace a less reactive metal.
Self-check
Will zinc displace copper from copper(II) sulfate?
Show the answer
Yes. Zinc is above copper in the reactivity series.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 9.4 Supplement 4–5; 9.5 Supplement 4–5
Open revision guidance
What to know
- A more reactive metal has a greater tendency to form positive ions by losing electrons. In displacement tests it displaces the ions of a less reactive metal: for example, Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s), but Cu does not displace Mg²⁺. Compare the metal with Mg²⁺, Zn²⁺, Fe²⁺, Cu²⁺ and Ag⁺ solutions; a metal displaces ions of metals below it in the series.
- Aluminium appears relatively unreactive because a thin, protective aluminium oxide layer covers its surface and prevents further reaction with the underlying metal.
- Galvanising coats iron/steel with zinc. The zinc coating is a barrier to oxygen and water; zinc also provides sacrificial protection if the coating is scratched because it is more reactive than iron and oxidises first. Zinc loses electrons, Zn → Zn²⁺ + 2e⁻, while the iron is protected.
What to do in an assessment
- Distinguish a physical barrier from sacrificial protection and relate the latter to relative reactivity/electron loss.
In plain language
A sacrificial metal corrodes in preference to the iron it protects.
Self-check
Why can zinc still protect scratched galvanised iron?
Show the answer
Zinc is more reactive and oxidises preferentially, supplying sacrificial protection.
- Course content
Scope: Course content · §3 9.6 Core 1–3
Open revision guidance
What to know
- The ease of extracting a metal from its ore depends on its position in the reactivity series. Metals below carbon can be extracted by reduction with carbon or carbon monoxide; metals above carbon are more difficult to extract and are commonly extracted by electrolysis.
- In a blast furnace, coke (carbon) burns in oxygen, releasing heat and making carbon dioxide. Carbon dioxide reacts with more hot carbon to produce carbon monoxide. Carbon monoxide reduces iron(III) oxide in hematite to iron. Limestone (calcium carbonate) thermally decomposes to calcium oxide; this reacts with silica impurities to form calcium silicate slag, which can be separated from molten iron. Core candidates describe these stages without symbol equations.
- Bauxite is the main ore of aluminium, and aluminium is extracted from aluminium oxide by electrolysis.
What to do in an assessment
- Describe the function of coke, limestone, carbon monoxide and slag separately in the iron-extraction process.
In plain language
Extraction method depends on how strongly the metal holds onto its compounds.
Self-check
What is the role of limestone in the blast furnace?
Show the answer
It decomposes to calcium oxide, which reacts with silica impurities to form slag.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 9.6 Supplement 4–5
Open revision guidance
What to know
- The balanced blast-furnace equations are: C + O₂ → CO₂; C + CO₂ → 2CO; Fe₂O₃ + 3CO → 2Fe + 3CO₂; CaCO₃ → CaO + CO₂; CaO + SiO₂ → CaSiO₃. They show coke combustion, carbon-monoxide formation, iron-oxide reduction, limestone decomposition and slag formation.
- Electrolyse aluminium oxide from purified bauxite dissolved in molten cryolite. Cryolite lowers the melting/operating temperature and reduces the energy required. Al³⁺ ions move to the negative cathode and gain electrons: Al³⁺ + 3e⁻ → Al. O²⁻ ions move to the positive carbon anode and lose electrons: 2O²⁻ → O₂ + 4e⁻. The oxygen reacts with the carbon anodes to form carbon dioxide, so the anodes are gradually consumed and must be replaced.
What to do in an assessment
- Balance each symbol equation and half-equation; explain cryolite and anode replacement in terms of the process.
In plain language
Aluminium is extracted by electrolysis because it is more reactive than carbon.
Self-check
Why must the carbon anodes be replaced?
Show the answer
Oxygen formed at the anode reacts with the carbon, consuming it.
Topic group
10 Chemistry of the environment
- Course content
Scope: Course content · §3 10.1 Core 1–7; 10.2 Core 1–2
Open revision guidance
What to know
- Test for water with anhydrous cobalt(II) chloride paper: blue changes to pink. Anhydrous copper(II) sulfate is white and turns blue when water is present. To test purity, measure melting/freezing point and boiling point at a stated pressure: pure water has fixed characteristic values (0 °C and 100 °C at standard atmospheric pressure), while dissolved impurities change these values and broaden the melting/freezing range. Distilled water is used in practical chemistry rather than tap water because it contains fewer chemical impurities.
- Water from natural sources may contain dissolved oxygen, metal compounds, plastics, sewage, harmful microbes, nitrates from fertilisers, and phosphates from fertilisers and detergents. Dissolved oxygen supports aquatic life and some metal compounds provide essential minerals. Some metal compounds are toxic; some plastics harm aquatic life; sewage can contain disease-causing microbes; nitrates and phosphates can cause deoxygenation and damage aquatic life. Detailed stages of eutrophication are not required.
- Treat domestic water by sedimentation and filtration to remove solids, use carbon to remove tastes and odours, and chlorinate to kill microbes. Ammonium salts and nitrates are used as fertilisers. NPK fertilisers supply nitrogen, phosphorus and potassium to improve plant growth.
What to do in an assessment
- Separate beneficial dissolved substances from contaminants and describe the treatment step that removes or controls each.
In plain language
Water treatment removes particles, improves taste and controls microbes; fertilisers supply plant nutrients.
Self-check
Why is carbon used in domestic water treatment?
Show the answer
It removes unwanted tastes and odours.
- Course content
Scope: Course content · §3 10.3 Core 1–6
Open revision guidance
What to know
- Clean, dry air is approximately 78% nitrogen (N₂), 21% oxygen (O₂), and the remainder is a mixture of noble gases and carbon dioxide (CO₂).
- Sources of the named air pollutants: complete combustion of carbon-containing fuels produces CO₂; incomplete combustion produces carbon monoxide and particulates; methane comes from decomposition of vegetation and waste gases from animal digestion; nitrogen and oxygen from air react at high temperatures in car engines to form nitrogen oxides; sulfur dioxide forms when fossil fuels containing sulfur compounds are burned.
- Adverse effects: increased CO₂ and methane levels contribute to global warming and climate change; carbon monoxide is toxic; particulates increase respiratory-problem and cancer risk; nitrogen oxides contribute to acid rain and photochemical smog and cause respiratory problems; sulfur dioxide causes acid rain.
- Strategies: planting trees removes CO₂ by photosynthesis; reducing livestock farming can reduce methane emissions; decreasing fossil-fuel use and increasing hydrogen and renewable energy (e.g. wind and solar) reduce greenhouse-gas emissions. Catalytic converters reduce vehicle emissions of nitrogen oxides; low-sulfur fuels and flue-gas desulfurisation with calcium oxide reduce sulfur dioxide and therefore acid rain. Photosynthesis uses light energy and chlorophyll to react carbon dioxide with water and produce glucose and oxygen: carbon dioxide + water → glucose + oxygen.
What to do in an assessment
- Link each pollutant to its source and adverse effect, then choose a mitigation strategy that addresses that source or effect.
In plain language
Air-quality explanations connect what is emitted, how it affects people or climate, and how emissions can be reduced.
Self-check
Which pollutant from incomplete combustion is a toxic gas?
Show the answer
Carbon monoxide.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 10.3 Supplement 7–9
Open revision guidance
What to know
- The Earth's surface absorbs energy from sunlight and emits thermal (infrared) radiation. Greenhouse gases, including carbon dioxide and methane, absorb some of this outgoing thermal energy and emit it in all directions; some is returned toward Earth's surface. This reduces thermal-energy loss to space and raises global temperatures.
- At the high temperatures in car engines, nitrogen and oxygen from air react to form nitrogen oxides. In a catalytic converter, carbon monoxide reduces nitrogen oxide to nitrogen while itself being oxidised to carbon dioxide; for example, 2CO + 2NO → 2CO₂ + N₂.
- The balanced symbol equation for photosynthesis is 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
What to do in an assessment
- Explain greenhouse warming as an energy-transfer sequence, and check atom counts when using the converter or photosynthesis equation.
In plain language
Greenhouse gases slow infrared energy escaping to space; converters change harmful exhaust gases into less harmful products.
Self-check
What is the role of NO in the converter equation?
Show the answer
It is reduced to nitrogen gas while carbon monoxide is oxidised to carbon dioxide.
Topic group
11 Organic chemistry
- Course content
Scope: Course content · §3 11.1 Core 1–6; 11.2 Core 1–2
Open revision guidance
What to know
- A displayed formula shows every atom and every covalent bond. To draw one, show each carbon with four bonds and each oxygen with two; add hydrogen atoms and bonds rather than hiding them. The syllabus general formulae are alkanes CₙH₂ₙ₊₂, alkenes CₙH₂ₙ, alcohols CₙH₂ₙ₊₁OH and carboxylic acids CₙH₂ₙ₊₁COOH.
- A functional group is an atom or group of atoms that determines the characteristic chemical properties of a homologous series. Alkenes contain C=C, alcohols contain –OH and carboxylic acids contain –COOH. A saturated compound has only C–C single bonds; an unsaturated compound has one or more non-single C–C bonds.
- A homologous series is a family of similar compounds with similar chemical properties because of the same functional group. Identify the series from its functional group or formula, and recognise names ending in -ane, -ene, -ol and -oic acid as alkane, alkene, alcohol and carboxylic acid respectively.
- The named Core molecules include methane CH₄, ethane CH₃–CH₃, ethene CH₂=CH₂, ethanol CH₃–CH₂–OH and ethanoic acid CH₃–C(=O)–OH. Convert these structural shortcuts to displayed formulae by drawing every atom and bond; check valency and include displayed formulae for products of the syllabus reactions.
What to do in an assessment
- Count carbon atoms, identify the functional group and check that each carbon has four covalent bonds.
In plain language
Organic names and formula patterns reveal the carbon skeleton and its reactive group.
Self-check
Which functional group identifies an alcohol?
Show the answer
The hydroxyl group, –OH.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 11.1 Supplement 7–9; 11.2 Supplement 3–4
Open revision guidance
What to know
- A structural formula is an unambiguous description of how atoms are arranged: CH₂=CH₂, CH₃CH₂OH and CH₃COOCH₃ are examples. Structural isomers have the same molecular formula but different structural formulae: C₄H₁₀ can be CH₃CH₂CH₂CH₃ (butane) or CH₃CH(CH₃)CH₃ (2-methylpropane); C₄H₈ can be CH₃CH₂CH=CH₂ (but-1-ene) or CH₃CH=CHCH₃ (but-2-ene).
- A homologous series has the same functional group and general formula; consecutive members differ by one –CH₂– unit, show a trend in physical properties and have similar chemical properties. In these series, increasing chain length generally raises boiling point and viscosity and lowers volatility.
- Use the carbon-number roots meth-, eth-, prop- and but- and the suffix for the functional group. Unbranched alkanes through four carbons are methane, ethane, propane, butane; alkenes are ethene, propene, but-1-ene and but-2-ene; alcohols are methanol, ethanol, propan-1-ol, propan-2-ol, butan-1-ol and butan-2-ol; carboxylic acids up to four carbons are methanoic, ethanoic, propanoic and butanoic acid. Number the chain from the end nearest the functional group or double bond.
- An ester has the structural pattern R–C(=O)–O–R′. Name it as the alkyl group from the alcohol followed by the carboxylate name from the acid, ending in -oate. The full unbranched combinations are methyl, ethyl, propyl or butyl methanoate; methyl, ethyl, propyl or butyl ethanoate; methyl, ethyl, propyl or butyl propanoate; and methyl, ethyl, propyl or butyl butanoate. To draw an ester, join the acid's C(=O)–O part to the alcohol-derived alkyl group and show every atom/bond in the displayed formula.
What to do in an assessment
- Check that a displayed/structural formula preserves the molecular formula when identifying isomers.
In plain language
Structural formulae distinguish compounds that contain the same numbers of atoms arranged differently.
Self-check
Are but-1-ene and but-2-ene structural isomers?
Show the answer
Yes. They share C₄H₈ but the double bond is in a different position.
- Course content
Scope: Course content · §3 11.3 Core 1–7
Open revision guidance
What to know
- Coal, natural gas and petroleum are fossil fuels. Methane is the main constituent of natural gas. A hydrocarbon contains hydrogen and carbon only; petroleum is a mixture of hydrocarbons.
- Fractional distillation separates petroleum into useful fractions because hydrocarbons have different boiling points. Heat the petroleum so much of it vaporises; vapours rise through a fractionating column that is hot at the bottom and cooler at the top, condense at different heights according to their boiling ranges, and are collected as fractions.
- From the bottom to the top of the fractionating column, fractions have decreasing chain length and boiling point, and increasing volatility with decreasing viscosity.
- Fraction uses: refinery gas for heating and cooking; gasoline/petrol for car fuel; naphtha as a chemical feedstock; kerosene/paraffin for jet fuel; diesel/gas oil for diesel engines; fuel oil for ships and home heating; lubricating oil for lubricants, waxes and polishes; bitumen for making roads.
What to do in an assessment
- Explain separation by differences in boiling point and match each fraction's stated use to its source-column position.
In plain language
Petroleum is separated into useful groups because its hydrocarbons boil at different temperatures.
Self-check
How do chain length and boiling point change up the fractionating column?
Show the answer
Chain length and boiling point decrease.
- Course content
Scope: Course content · §3 11.4 Core 1–2; 11.5 Core 1–4
Open revision guidance
What to know
- Alkanes are saturated hydrocarbons with only single covalent bonds between carbon atoms. They are generally unreactive, except for combustion and substitution reactions with chlorine. Complete combustion in excess oxygen produces carbon dioxide and water (CH₄ + 2O₂ → CO₂ + 2H₂O); inadequate oxygen can cause incomplete combustion.
- Alkenes are unsaturated hydrocarbons because they contain a carbon–carbon double bond, C=C. Cracking larger alkane molecules at high temperature with a catalyst makes smaller alkane and alkene molecules, and can also make hydrogen. Cracking supplies smaller, more useful fuels and alkene feedstocks for polymer manufacture.
- Aqueous bromine distinguishes unsaturated from saturated hydrocarbons: an alkene decolourises orange/brown bromine water, while an alkane does not under these test conditions.
What to do in an assessment
- Use the bromine-water observation as evidence for a C=C bond, not as a general test for every functional group.
In plain language
The carbon double bond makes an alkene react differently from a saturated alkane.
Self-check
What observation indicates an alkene in the aqueous bromine test?
Show the answer
The orange/brown bromine water is decolourised.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 11.4 Supplement 3–4; 11.5 Supplement 5–6
Open revision guidance
What to know
- In a substitution reaction, one atom or group of atoms is replaced by another. Chlorination of an alkane is a photochemical substitution: ultraviolet light supplies activation energy, Eₐ. In the syllabus monosubstitution scope, one hydrogen is replaced by one chlorine; for example, CH₄ + Cl₂ → CH₃Cl + HCl.
- An addition reaction forms only one product. At an alkene C=C bond, bromine adds to form a dibromoalkane (ethene + Br₂ → 1,2-dibromoethane), hydrogen adds with a nickel catalyst to form an alkane (ethene + H₂ → ethane), and steam adds with an acid catalyst to form an alcohol (ethene + H₂O → ethanol). Draw the product with correct valency and all atoms from both reactants.
What to do in an assessment
- Identify whether bonds are replaced or added across C=C and draw the product with correct valency.
In plain language
Addition uses the double bond to join new atoms to an alkene; substitution replaces an atom in an alkane.
Self-check
What is the product when ethene adds hydrogen?
Show the answer
Ethane, C₂H₆.
- Course content
Scope: Course content · §3 11.6 Core 1–3; 11.7 Core 1
Open revision guidance
What to know
- Fermentation makes ethanol from aqueous glucose using yeast at 25–35 °C in the absence of oxygen: glucose → ethanol + carbon dioxide (C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂). Ethanol is also made by catalytic addition of steam to ethene at 300 °C and 6000 kPa (60 atm) with an acid catalyst.
- Ethanol combusts in oxygen to form carbon dioxide and water. Its uses include solvent and fuel.
- Ethanoic acid reacts with metals to form an ethanoate salt and hydrogen, with bases to form an ethanoate salt and water, and with carbonates to form an ethanoate salt, water and carbon dioxide. Examples: 2CH₃COOH + Mg → (CH₃COO)₂Mg + H₂; CH₃COOH + NaOH → CH₃COONa + H₂O; 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂.
What to do in an assessment
- Compare fermentation and catalytic hydration using their conditions; write products and salt formulae for acid reactions.
In plain language
Ethanol can be produced biologically or from ethene, and ethanoic acid behaves as a carboxylic acid.
Self-check
What condition must be absent during yeast fermentation?
Show the answer
Oxygen.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 11.6 Supplement 4; 11.7 Supplement 2–3
Open revision guidance
What to know
- Fermentation uses renewable glucose from plant material at a relatively low temperature and pressure, but is slow and dilute ethanol must be separated from water; growing the crop uses land. Catalytic addition uses ethene (usually obtained from fossil-fuel feedstocks), operates rapidly and continuously at 300 °C and 6000 kPa with an acid catalyst, but requires high temperature/pressure and a non-renewable feedstock. Evaluate the routes by comparing rate, conditions, feedstock and separation.
- Ethanol is oxidised to ethanoic acid by acidified aqueous potassium manganate(VII); the purple oxidising solution becomes colourless as it is reduced. Ethanoic acid is also produced by bacterial oxidation of ethanol during vinegar production.
- A carboxylic acid reacts with an alcohol using an acid catalyst to form an ester and water. For example, ethanoic acid + ethanol ⇌ ethyl ethanoate + water.
What to do in an assessment
- Use evidence about conditions and resources to justify a production method; identify the alcohol and acid parts of an ester.
In plain language
Ethanol can be oxidised to an acid, and acids plus alcohols make esters.
Self-check
What two classes of reactant make an ester?
Show the answer
A carboxylic acid and an alcohol.
- Course content
Scope: Course content · §3 11.8 Core 1–5
Open revision guidance
What to know
- A polymer is a large molecule built up from many smaller molecules called monomers. Poly(ethene) forms by addition polymerisation of ethene: each C=C opens and forms new C–C bonds, giving the repeat unit [–CH₂–CH₂–]ₙ.
- Plastics are made from polymers. Disposal in landfill uses space and many plastics persist; plastic accumulation in oceans harms aquatic life; burning some plastics can form toxic gases. Consider these environmental challenges when assessing disposal.
What to do in an assessment
- Draw the repeat unit in brackets with continuation bonds, preserving the atoms in the alkene monomer.
In plain language
A polymer is a long molecule made by linking many smaller molecules.
Self-check
What happens to ethene's double bond during addition polymerisation?
Show the answer
The double bond opens so each carbon can form new single bonds to neighbouring units.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 11.8 Supplement 6–13
Open revision guidance
What to know
- In an addition polymer, identify the repeat unit by changing the alkene C=C to C–C and adding continuation bonds; conversely, remove those continuation bonds and restore C=C to infer the alkene monomer. Condensation polymers contain linkages formed with elimination of a small molecule.
- A dicarboxylic acid plus a diamine forms a polyamide; a dicarboxylic acid plus a diol forms a polyester. To deduce a polymer from monomers, join the acid and amine/alcohol end groups, form the –CONH– or –COO– linkage and remove H₂O at each link; reverse these steps to find the monomers.
- Nylon is a polyamide with repeating amide linkages, –C(=O)–NH–; one syllabus structure is nylon-6,6, [–NH–(CH₂)₆–NH–C(=O)–(CH₂)₄–C(=O)–]ₙ. PET is a polyester with repeating ester linkages, –C(=O)–O–; its repeat unit is [–O–CH₂–CH₂–O–C(=O)–C₆H₄–C(=O)–]ₙ. PET can be converted back into its monomers and repolymerised; the full name polyethylene terephthalate is not required.
- Proteins are natural polyamides formed from amino-acid monomers, each with the general structure H₂N–CH(R)–COOH, where R is a variable side chain. The protein chain has repeated –NH–CH(R)–C(=O)– units joined by amide links; identify and draw the linkage in the given protein structure.
What to do in an assessment
- Use the end groups of monomers to build a repeat unit and preserve the correct linkage when reversing the polymerisation.
In plain language
Polymer structures can be read backwards to identify the monomers that joined together.
Self-check
What functional linkage identifies a polyamide?
Show the answer
The amide linkage, –CONH–.
Topic group
12 Experimental techniques and chemical analysis
- Course content
Scope: Course content · §3 12.1 Core 1–3; 12.2 Core 1–2
Open revision guidance
What to know
- Choose apparatus for the quantity: a stopwatch measures time, a thermometer measures temperature, a balance measures mass, and a burette, volumetric pipette, measuring cylinder or gas syringe measures volumes. A volumetric pipette delivers one accurate fixed volume; a burette delivers and measures a variable volume more precisely than a measuring cylinder; a cylinder is quicker but less precise; a gas syringe collects a gas volume directly. Explain the advantages/disadvantages in relation to the precision, range and task.
- A solvent is a substance that dissolves a solute; a solute is the substance dissolved; a solution is a mixture of one or more solutes dissolved in a solvent. A saturated solution contains the maximum concentration of solute dissolved at a specified temperature. Residue is material remaining after evaporation, distillation, filtration or a similar separation; filtrate is liquid/solution that has passed through a filter.
- In an acid–base titration, use a volumetric pipette to measure a fixed aliquot of one solution into a flask, add a suitable indicator, and add the other solution from a burette while swirling. Record initial and final burette readings; the titre is the volume delivered. Add titrant dropwise near the end-point, identified by the indicator's permanent colour change. Repeat titrations and use concordant titres when appropriate.
What to do in an assessment
- Measure a fixed aliquot with the pipette, add indicator, run titrant from the burette while swirling, approach the end-point dropwise and record initial/final readings.
In plain language
Good experimental design matches each instrument to the quantity and records enough detail to judge the result.
Self-check
Why use a pipette rather than a measuring cylinder for a precise aliquot?
Show the answer
A volumetric pipette delivers a more precise fixed volume.
- Course content
Scope: Course content · §3 12.3 Core 1–2; 12.4 Core 1–3
Open revision guidance
What to know
- In paper chromatography, draw a pencil baseline near the bottom of the paper, place small spots of soluble coloured samples on it, and suspend the paper in a suitable solvent with the spots above the solvent. The solvent rises and carries components at different rates according to their solubility and attraction to the paper; remove the paper before the solvent reaches the top and mark the solvent front. Compare unknown spots with known substances under the same conditions. One spot suggests a pure substance in this test; multiple spots indicate a mixture.
- Choose a separation method from the mixture's properties. A suitable solvent dissolves the wanted substance but not unwanted insoluble material: stir the sample with the solvent, filter off insoluble residue, then recover the dissolved substance by evaporation/crystallisation or distillation. Filtration separates an insoluble solid from a liquid. For crystallisation, concentrate the solution by gentle evaporation, cool it to form crystals, then filter and dry them. Simple distillation heats a solution and condenses the vapour to collect a solvent, or separates liquids with widely different boiling points. Fractional distillation uses a fractionating column to separate miscible liquids with different/nearby boiling ranges; vapours condense at different heights according to boiling point.
- Use melting or boiling points to assess identity and purity at the stated pressure. A pure substance has a characteristic sharp melting point and fixed boiling point; impurities change the characteristic value and generally broaden the melting range.
What to do in an assessment
- Keep the sample spot above the solvent, mark the solvent front, and choose a separation method whose physical-property difference fits the mixture.
In plain language
Separation works because mixture components differ in solubility, particle size or boiling point.
Self-check
Which technique separates an insoluble solid from a liquid?
Show the answer
Filtration.
- Course content
Scope: Course content · Cambridge IGCSE (9–1) Chemistry 0971 theory route: Extended route (Papers 2 and 4; Core plus Supplement; grades 9–1) · §3 12.3 Supplement 3–4
Open revision guidance
What to know
- Paper chromatography also separates soluble colourless substances. Run the chromatogram with a suitable solvent, remove it before the solvent reaches the top, mark the solvent front, then use a locating agent to make colourless spots visible; a specific locating agent need not be memorised.
- Calculate retention factor, Rf = distance travelled by substance ÷ distance travelled by solvent front. Measure both from the same baseline in the same units; Rf has no units and is compared meaningfully only when the chromatographic conditions are the same.
What to do in an assessment
- Measure both distances in the same units and report Rf without units; compare values only under matching conditions.
In plain language
Rf describes how far a substance travelled compared with the solvent front.
Self-check
A spot moves 3 cm and the solvent front 6 cm. Find Rf.
Show the answer
3 ÷ 6 = 0.50.
- Course content
Scope: Course content · §3 12.5 Core 1–4 and §4 Notes for use in qualitative analysis
Open revision guidance
What to know
- Anion tests: carbonate, CO₃²⁻, reacts with dilute acid with effervescence; the gas turns limewater milky (CO₂). Acidify chloride, bromide or iodide solution with dilute nitric acid, then add aqueous silver nitrate: white AgCl, cream AgBr or yellow AgI precipitate respectively. For nitrate, add aqueous sodium hydroxide and aluminium foil, then warm carefully: ammonia is evolved and confirmed by holding damp red litmus at the mouth of the tube until it turns blue. Acidify sulfate with dilute nitric acid and add aqueous barium nitrate: a white precipitate forms. Sulfite reacts with acidified aqueous potassium manganate(VII), changing it from purple to colourless.
- Aqueous-cation tests (precipitate observations with excess reagent): Al³⁺ gives a white precipitate soluble in excess NaOH (colourless solution), but insoluble in excess aqueous ammonia; NH₄⁺ produces ammonia on warming with NaOH. Ca²⁺ gives a white precipitate insoluble in excess NaOH and no/slight white precipitate with ammonia. Cr³⁺ gives a green precipitate soluble in excess NaOH but insoluble in excess ammonia. Cu²⁺ gives a light-blue precipitate insoluble in excess NaOH; with ammonia it dissolves in excess to a dark-blue solution. Fe²⁺ gives a green precipitate insoluble in excess of either reagent, turning brown near the surface on standing. Fe³⁺ gives a red-brown precipitate insoluble in excess of either reagent. Zn²⁺ gives a white precipitate soluble in excess of either reagent to a colourless solution.
- Gas tests: ammonia, NH₃, turns damp red litmus blue; carbon dioxide, CO₂, turns limewater milky; chlorine, Cl₂, bleaches damp litmus; hydrogen, H₂, gives a squeaky pop with a lighted splint; oxygen, O₂, relights a glowing splint; sulfur dioxide, SO₂, changes acidified aqueous potassium manganate(VII) from purple to colourless.
- Flame-test colours: Li⁺ red; Na⁺ yellow; K⁺ lilac; Ca²⁺ orange-red; Ba²⁺ light green; Cu²⁺ blue-green. Follow the reagent order in the anion tests and record observations (including precipitate colour and solubility in excess) before identifying the substance.
What to do in an assessment
- For nitrate, use a small sample in a test-tube, add aqueous sodium hydroxide then aluminium foil, warm gently with the tube held in a holder and pointed away from people, and test gas at the mouth with damp red litmus; do not stopper the tube, sniff the gas, or allow litmus to touch the liquid. Wear eye protection and work under teacher/centre risk controls. Record both the reagent sequence and the red-to-blue litmus result before identifying nitrate. For all tests, add reagents in the specified order and record observations before inferring an identity.
In plain language
Qualitative analysis identifies an unknown from a characteristic combination of colour, precipitate and gas-test results.
Self-check
What gas turns limewater milky?
Show the answer
Carbon dioxide.
Topic group
Assessment and practical skills
- Course content
Scope: Course content · §2 assessment overview; §4 Practical assessment and experimental contexts
Open revision guidance
What to know
- All candidates take three components. Candidates studying Core, or expected to achieve grade 3 or below, enter for Paper 1 (Core multiple choice: 45 minutes, 40 marks, 30%) and Paper 3 (Core theory: 1 hour 15 minutes, 80 marks, 50%), plus Paper 5 or 6; they are eligible for grades 5–1. Candidates studying Extended and expected to achieve grade 4 or above enter for Paper 2 (Extended multiple choice: 45 minutes, 40 marks, 30%) and Paper 4 (Extended theory: 1 hour 15 minutes, 80 marks, 50%), plus Paper 5 or 6; they are eligible for grades 9–1. Papers 1–4 are externally assessed.
- Choose one externally assessed practical component: Paper 5 is a 1 hour 15 minute, 40-mark laboratory Practical Test (20%); Paper 6 is a 1-hour, 40-mark written Alternative to Practical (20%). Both assess the same Section 4 experimental skills and AO3; candidates do not perform experiments in Paper 6. Component weightings are fixed: Papers 1 and 2 each assess AO1 at 63%, AO2 at 37%, AO3 at 0%; Papers 3 and 4 have the same AO1/AO2/AO3 weighting; Papers 5 and 6 each assess AO1 at 0%, AO2 at 0%, AO3 at 100%.
- Practical contexts include quantitative volumes, masses, temperatures, times and lengths; reaction rates; salt preparation; filtration, crystallisation, solvent extraction, simple/fractional distillation and chromatography; electrolysis; identifying ions and gases; water tests/treatment; dilute-acid test-tube reactions; redox tests; heating/cooling curves; titration; solubility; melting/boiling points; metal/halogen displacement; reaction temperature changes; corrosion; and unfamiliar procedures with simple apparatus.
- Worked measurement/titration routine: choose the instrument whose scale/precision suits the quantity, set it up level and read at eye level. For a titration rinse a volumetric pipette with the analyte, transfer its fixed volume to a conical flask, add only the stated indicator, rinse/fill the burette with titrant and clear its jet bubble, record the initial reading, swirl while adding titrant, approach the endpoint dropwise, then record the final reading and calculate titre = final − initial. Repeat until concordant; do not infer precision beyond the instrument.
- Salt preparation: for a soluble salt from an acid and insoluble base, warm the acid gently, add the base in small portions with stirring until some remains, filter off excess solid, evaporate the filtrate only until concentrated, allow it to cool and crystallise, then filter and dry the crystals. For a soluble salt from two solutions, use a titration to find neutral volumes, repeat without indicator, then evaporate/cool/crystallise. Wear goggles and never evaporate to dryness unless specifically instructed.
- Separation operations: filtration retains an insoluble solid on wetted filter paper while liquid passes into a labelled vessel. To crystallise a dissolved solid, gently evaporate some solvent, stop before dryness, cool undisturbed, collect and dry crystals. For solvent extraction, shake a small sample with the specified immiscible solvent in a stoppered funnel, vent away from people, allow layers to settle, then separate the named layer. In simple distillation collect a lower-boiling solvent from a solution; in fractional distillation use a fractionating column and collect the fraction at its boiling range. For paper chromatography draw a pencil baseline, spot small concentrated samples, place solvent below the baseline, cover the chamber, stop before the solvent front reaches the top, mark the front and dry the chromatogram.
- Electrolysis: use the specified electrolyte and inert electrodes, connect to a low-voltage direct-current supply with power off while assembling, keep electrodes apart, then switch on and record observations at each electrode. Do not touch the solution/electrodes while powered; switch off before disconnecting. Identify products only by the required source-listed tests and apply centre safety rules.
- For water treatment/testing follow the supplied sequence: allow suspended solids to settle, filter, use carbon for taste/odour where specified and chlorinate only under the stated supervised conditions. For water tests or ion/gas identification, use small labelled samples, clean droppers, the exact reagent order and the source's positive observation; never mix stock reagents or smell gases directly.
- For dilute-acid, redox and displacement test-tube work, place a small measured sample in a labelled tube, add the specified dilute reagent dropwise, keep the opening pointed away, and record colour, precipitate, bubbles or temperature before inference. Compare metal/halogen displacement only with the supplied safe dilute solutions; use separate clean wells/tubes and dispose of residues under local rules.
- For a heating/cooling curve, put the thermometer bulb in the substance without touching the vessel, stir gently if directed, record temperature at equal time intervals during steady heating/cooling, and plot temperature / °C against time / min. For solubility, keep solvent volume constant, add weighed solute in small portions to a measured water sample held at a stated temperature, stir until the last portion just dissolves or a little remains undissolved, record the saturated mass/concentration, and repeat at other temperatures with fresh equal-volume samples; keep temperature stable and compare solubility at the same endpoint. For melting point, place a small dry solid sample in a capillary tube with a thermometer in a heating bath/apparatus, heat slowly near the expected point and record the temperature at which melting starts and finishes; report a range and do not touch hot glass. For boiling point, heat a small sample in the specified open/vented school apparatus with thermometer bulb at the vapour level, record the steady boiling temperature, use a water bath for flammable liquids, and never heat a sealed system. Compare corrosion using labelled, otherwise identical iron samples, vary one condition only (for example dry air versus water, or air-free water versus aerated water), keep time, temperature, surface area and container the same, and record rusting/appearance at fixed intervals.
- Water and ion tests: use separate small, labelled portions, clean droppers and the source-specified order; acidify halide samples with dilute nitric acid before adding silver nitrate, and acidify sulfate samples with dilute nitric acid before adding barium nitrate. Observe precipitate colour and any change in excess reagent before identifying the ion. For water treatment, allow suspended solids to settle, decant and filter through wetted paper, use activated carbon only where specified for dissolved colour/taste/odour, then use the prescribed supervised disinfection step; do not taste samples or prepare/use chlorine outside the stated controlled context. In redox tests use small volumes and dilute, specified reagents: add acidified potassium manganate(VII) dropwise to the sample, note purple-to-colourless change where a reducing substance is present, wear goggles and keep oxidiser away from combustible material; never mix unknown stock reagents.
- For simple distillation of a solution, put the liquid in a flask with anti-bumping granules, fit a thermometer with its bulb at the side-arm/vapour path, connect the side-arm to a Liebig condenser, run cooling water in at the condenser's lower inlet and out at the upper outlet, and collect distillate from the outlet; heat gently, keep the apparatus vented, and never distil to dryness. For fractional distillation, place the fractionating column vertically between flask and still head, pack/use it as specified, position the thermometer bulb at the vapour inlet to the condenser, allow the column to establish a temperature gradient, and collect separate fractions as the head temperature reaches each boiling range. Support/clamp glassware, check water flow and joints before heating, keep the receiver open to the atmosphere, and use a water bath for flammable liquids.
- For an unfamiliar simple-apparatus procedure, select apparatus that can measure the expected value at suitable precision, draw and label a simple usable setup (vessels, connections, supports, liquid levels and flow direction), and state why a plausible alternative is less suitable. Estimate an expected value/range before starting; translate instructions into ordered actions; identify the measurement, endpoint and hazard controls; record numerical readings with units and qualitative observations in separate labelled table columns; repeat/blank where appropriate; and evaluate a second context by naming a specific limitation, its effect on results and a workable improvement. If the source supplies a method, follow its sequence rather than substituting a memorised procedure.
- Use the supplied qualitative-analysis notes when required. A paper choice changes the assessment mode, not the experimental-skills or practical-context scope.
- Chemical tests for water: use a clean dropper to put one small drop of the sample onto teacher-provided blue cobalt(II) chloride paper held with forceps; water changes it from blue to pink. In a separate test, put a few teacher-provided grains of anhydrous copper(II) sulfate on a spotting tile and add a small drop of sample; water changes the solid from white to blue. Compare with the unused reagent and record colour before and after. Wear eye protection, avoid skin contact and inhaling powders, and follow the centre’s risk assessment and disposal instructions; staff prepare/handle restricted reagents, or demonstrate the test where learner use is not permitted. A positive chemical test detects water, not its purity.
What to do in an assessment
- For Paper 5, perform the requested method safely and record actual observations; for Paper 6, interpret diagrams, methods and data without claiming to have carried out an experiment.
In plain language
Paper 5 is hands-on; Paper 6 tests practical understanding in writing, with the same AO3 focus.
Self-check
Does Paper 6 require the candidate to perform experiments?
Show the answer
No. It is a written Alternative to Practical.
- Course content
Scope: Course content · §2 AO3; §4 Practical assessment, required experimental skills
Open revision guidance
What to know
- Assessment-objective weightings are component-specific: Papers 1–4 each assess AO1 at 63%, AO2 at 37%, AO3 at 0%; Papers 5 and 6 each assess AO1 and AO2 at 0%, AO3 at 100%. AO1 covers facts, laws, definitions, vocabulary, symbols, quantities/units, apparatus and safety, plus applications and their social, economic and environmental implications. AO2 covers selecting/organising/presenting information, translating between words/symbols/graphs/numbers, manipulating data, identifying patterns/trends, explaining relationships, predicting and solving quantitative or unfamiliar-context problems. AO3 covers selecting and safely using techniques, apparatus and materials, planning, observing/measuring/estimating, interpreting/evaluating evidence and improving methods.
- Select and safely use apparatus/materials and follow a procedure; identify equipment from diagrams/descriptions, draw/complete/label apparatus diagrams, explain technique, justify method/apparatus choice, describe qualitative/gas/other tests, hazards/precautions and techniques that improve accuracy.
- Plan an investigation by identifying independent and dependent variables, controlling other variables (and explaining why), choosing an appropriate range and number of values, selecting apparatus, writing a sequence, identifying risks/precautions, specifying how results will be recorded and processed, predicting expected results, then using the evidence to form a conclusion or evaluate a prediction.
- Read analogue and digital scales or diagrams, using suitable precision (including the nearest half-scale division where required); make expected-value estimates, collect sufficient readings and repeats, record qualitative test observations, and enter measurements systematically in a table with units and appropriate precision.
- Process data for calculations or graphing; draw/read suitable plots and best-fit lines; analyse patterns; interpolate/extrapolate to find values, gradients or intercepts; justify conclusions with observations/data and explanation; judge data quality and handle anomalies appropriately. Evaluate variable control, identify error sources and suggest improvements to apparatus, arrangement, method or technique.
- Worked method — effect of acid concentration on magnesium reaction rate: wear splash goggles, use 0.5 mol/dm³ hydrochloric acid in 25.0 cm³ portions, keep ignition sources away because hydrogen forms, and use a gas syringe attached to a clamped side-arm flask. Change only acid concentration (for example 0.20, 0.40, 0.60, 0.80 mol/dm³); keep acid volume, temperature, same cleaned 2.0 cm magnesium strip and apparatus constant. Add magnesium through the side-arm arrangement, seal without obstructing the outlet, start the timer at contact, and record gas volume every 10 s until it stops increasing. Rinse safely, repeat each concentration at least three times, and record actual readings rather than predicted ones.
- Second worked context — solubility versus temperature: use five clean test-tubes or small beakers, each with 10.0 cm³ water; hold them at measured temperatures such as 20, 30, 40, 50 and 60 °C in a water bath. Add the same dry solute (for example potassium nitrate) in weighed 0.5 g portions, stir each portion until it dissolves, and stop when a small amount remains after stirring. Record total mass dissolved at each temperature, keep solvent volume and endpoint rule the same, and repeat one temperature to check reproducibility. A labelled table separates temperature / °C from solute mass / g and observations; estimate a likely range before selecting balance and sample sizes. Draw a simple setup showing the water bath, sample vessel, thermometer, stirrer and balance; explain that a hotplate is an alternative to a Bunsen near water but a water bath gives steadier temperatures. Evaluate heat loss and subjective saturation endpoint; use a lid/insulation and smaller addition increments as improvements.
- Worked data and graph analysis: for one trial, record time / s = 0, 10, 20, 30, 40 and H₂ volume / cm³ = 0, 19, 34, 44, 50. Plot time on x and volume on y, label both axes with quantity / unit, choose scales using >half the grid, mark each point to half a small square, then draw a smooth best-fit curve. The curve becomes less steep as magnesium is used up. To estimate initial rate, draw a tangent at t=0; if two widely spaced points on that tangent are (0 s, 0 cm³) and (20 s, 36 cm³), gradient = Δvolume/Δtime = 36/20 = 1.8 cm³/s. Mark the gradient triangle on the graph and use points on the tangent, not raw readings.
- Repeat/anomaly/evaluation example: times for the same fixed gas volume are 18.2 s, 18.8 s and 27.0 s. The third is anomalous relative to the close pair; inspect for a loose bung, gas leak or timing error and repeat that trial before deciding whether to exclude it. Mean of the agreeing pair is (18.2 + 18.8)/2 = 18.5 s. A leak makes the measured rate too low; improve the seal and check the syringe moves freely before each run. Temperature drift changes rate, so use a water bath and check temperature. A reasoned conclusion compares repeated initial rates across concentrations and cites the plotted slopes; greater concentration is expected to increase collision frequency and rate.
- Planning/recording worked structure: write the changed variable and its range, measured variable and instrument, each control and how it is held constant, repeat count, hazard and precaution, ordered steps, result table headings with units, processing (mean/rate/graph), predicted trend and what evidence would support the conclusion. Estimate a likely volume/time range first so the syringe or stopwatch is neither overloaded nor too coarse.
What to do in an assessment
- Use the worked method to write a numbered safe procedure, draw and label the apparatus arrangement, fill a results table with units, calculate a mean/rate, plot and interpret the data, mark a large gradient triangle, identify an anomaly, and link one specific error to a practical improvement. State changed/measured/controlled variables and justify apparatus choice.
In plain language
A good investigation is controlled, safe, measurable and evaluated using evidence rather than guesswork.
Self-check
Why repeat a measurement?
Show the answer
Repeats help reveal random variation, identify anomalies and calculate a more reliable mean.
- Course content
Scope: Course content · §4 Mathematical requirements; Presentation of data
Open revision guidance
What to know
- Number work includes addition, subtraction, multiplication and division; decimals, fractions, percentages, ratios (including x:y), reciprocals, positive whole-number indices, standard form, rounding and significant figures. Calculate a mean when asked and use a calculator appropriately.
- Algebra includes substituting measured values into equations with consistent units, simple rearrangement and direct or inverse proportion. Show a calculation in a traceable sequence and carry the units through it.
- Use appropriate geometrical/measurement terms for lines, angles, curves, diagonals, squares, rectangles and circles (including radius, diameter and circumference). A curve changes direction continuously; a diagonal joins non-adjacent corners of a polygon. Select suitable units and convert cm³↔dm³, mg↔g↔kg, J↔kJ and Pa↔kPa; use SI or approved units.
- Read and plot graphs/charts; interpolate/extrapolate; determine a straight-line gradient and intercept; recognise direct proportionality; find a tangent to a curve where required. Read instrument scales to suitable precision, including halfway between divisions when required.
- In tables, give each heading as quantity / unit (e.g. time / s), keep units out of data cells, record repeats and preserve measured precision. Give a calculated result to the same number of significant figures as the least precise raw datum when the calculation rule applies: 12.0 ÷ 0.75 = 16 (two significant figures, limited by 0.75), not 16.0; keep guard digits until rounding the final answer. For ratios, state the required form such as x:y.
- For graphs use a sharp pencil, label axes with quantity and unit, put the independent variable on x and dependent on y unless instructed otherwise, and choose sensible scales that use more than half of the grid (e.g. 1, 2 or 5 units per 2 cm). Mark points with crosses or encircled dots accurately to half a small square; draw one thin, smooth best-fit straight line or curve through the trend, with roughly equal numbers of plotted points on either side; do not force the line through the origin unless the data support it, and consider an apparent anomaly rather than silently discarding it. Read/interpolate/extrapolate to half a small square and use a marked gradient triangle spanning at least half the best-fit line.
- Worked arithmetic: convert 18.0 cm³ to dm³ by dividing by 1000: 0.0180 dm³. Convert 450 mg to grams: 0.450 g; 2.4 kJ = 2400 J; 180 kPa = 180 000 Pa. Write 0.00045 as 4.5 × 10⁻⁴. For a 1:2 ratio, keep the requested x:y order. A reciprocal rate for 25.0 s is 1/25.0 = 0.0400 s⁻¹. Keep guard digits during working and round only the final result; e.g. 12.746 to 3 significant figures is 12.7.
- Worked mean and anomaly: repeated titres of 24.60, 24.75 and 27.10 cm³ include a distant value. Check the burette record/apparatus and repeat before excluding it; if the first two are concordant, their mean is (24.60 + 24.75)/2 = 24.675 cm³, reported to a precision justified by the readings (24.68 cm³). Do not average an unexplained anomalous value blindly.
- Worked equation/proportion: rate = gas volume ÷ time. At 36 cm³ in 20 s, rate = 1.8 cm³/s. Rearrange V = rt to t = V/r; for V = 50 cm³ and r = 2.0 cm³/s, t = 25 s. If doubling concentration halves the time to a fixed gas volume, time is inversely proportional to concentration; if equal time intervals give equal volume changes, volume is directly proportional to time over that interval.
- Worked graph reading: between plotted points (20 s, 34 cm³) and (30 s, 44 cm³), a straight-line interpolation gives the 25 s volume as 39 cm³. For a line through (0, 2) and (20, 42), gradient = (42 − 2)/(20 − 0) = 2 cm³/s and y-intercept = 2 cm³. On a curved rate graph, draw a tangent at the required time and calculate its gradient from two widely spaced points on that tangent; tangent work is Extended-only.
- A scale with 1 cm³ divisions should be read to the nearest 0.5 cm³ where required; between 13 and 14 cm³, an estimate halfway is 13.5 cm³. A graph point plotted at (25 s, 39 cm³) must be marked within half a small square. In a table write time / s and volume / cm³ in headings, put numbers only in cells, preserve instrument precision, and report calculated precision no finer than the least precise measurement.
- Use the geometry terms accurately: if a circle has radius 2.0 cm, its diameter is 4.0 cm and circumference is 2πr ≈ 12.6 cm. Select units before calculation and convert every term in an equation to the same unit.
What to do in an assessment
- Keep intermediate values unrounded, show formula/substitution, write units and round only the final answer appropriately.
In plain language
Good quantitative chemistry preserves units and instrument precision from measurement through calculation and graph.
Self-check
How should a table heading show the unit for volume?
Show the answer
For example, volume / cm³.
- Course content
Scope: Course content · §4 Apparatus and reagents; Safety in the laboratory; Conventions; Command words
Open revision guidance
What to know
- Be familiar with aluminium foil; balances to 500 g with 0.1 g precision; insulating cups/beakers; 1 dm³, 250 cm³ and 100 cm³ beakers; boiling tubes; Bunsen burners; 50 cm³ burettes; conical flasks; delivery tubes; filter funnels/papers; flame-test wires; measuring cylinders (100, 50, 25 and 10 cm³); glassware labelling pens; pipette fillers; test-tube/boiling-tube racks; red/blue litmus; retort stands, bosses and clamps; droppers; small burette-filling funnels; spatulas; stirring rods; thermometers (−10 to +110 °C, 1 °C graduations); stoppers; stopwatches (1 s); test-tube holders and hard-glass test-tubes; tripods; universal-indicator paper; 25 cm³ volumetric pipettes; wash bottles and white tiles.
- Key listed sizes also include insulating cups/beakers about 150 cm³, squat lip beakers of 1 dm³/250 cm³/100 cm³, boiling tubes about 150 mm × 25 mm, conical flasks 50–250 cm³ and hard-glass test-tubes about 125 mm × 16 mm. Use specified graduations/precision for balances, thermometers, stopwatches and volumetric equipment.
- Operate measurement apparatus deliberately: place a balance on a level bench, zero it, weigh a container then reweigh with the sample and subtract; do not put chemicals directly on the pan. Use a measuring cylinder for approximate volumes, reading the bottom of a liquid meniscus at eye level. For a fixed 25.0 cm³ aliquot rinse a volumetric pipette with the solution, use a pipette filler (never mouth-pipette), set the meniscus to the mark at eye level, let it drain into the receiving flask without blowing out the tip. For titration rinse/fill a burette with the titrant, run liquid through the tap to remove the jet bubble, record the initial reading at eye level, add while swirling the conical flask, then add dropwise near the endpoint and record final reading; titre = final − initial. Clamp the burette vertically using a retort stand/boss/clamp and use a small funnel only to fill, removing it before titrating.
- Read thermometers with the bulb immersed in the sample but not touching the vessel; wait for a stable reading and report the scale precision. Start a stopwatch at the defined reaction event and stop at the stated endpoint; repeat timings. Before a gas-syringe run check that the plunger moves freely and the bung/delivery tube connections are airtight; keep the outlet unblocked. A delivery tube directs gas into the syringe or through liquid only when that test is intended; never heat a sealed vessel.
- For heating, support a beaker or boiling tube on a tripod as appropriate, keep it stable and never use a stoppered vessel over a flame. Light a Bunsen safely, use the visible yellow safety flame when not heating, and use the air hole only when directed; tie hair back and keep flammable solvents away. A test-tube holder grips the upper part of a heated tube, pointed away from people. Clamp apparatus securely before heating or collecting gas.
- For filtration, seat filter paper in a funnel, wet it to seal, place the funnel over a labelled receiving beaker/flask, and pour down a glass stirring rod without filling above the paper rim; rinse residue with a wash bottle if directed. Use spatulas for solids, droppers/teat pipettes for small additions, a glass rod for mixing, and a rack/holder for test-tubes. Put a white tile beneath a flask to see a faint indicator endpoint. Label every sample and never return used reagent to its stock bottle.
- For flame tests use a clean flame-test wire (or specified alternative): clean it between samples, dip only the tip into a small solid sample, introduce it at the edge of the blue flame and observe against the stated background. Litmus/universal-indicator paper is touched with a clean dropper or glass rod; do not dip used paper into a reagent bottle. In a labelled apparatus drawing show each vessel, connection, clamp/support, liquid level and direction of gas flow.
- Recognise the standard bench reagents/indicators: dilute HCl (1.0 mol/dm³), dilute HNO₃ (1.0 mol/dm³), dilute H₂SO₄ (0.5 mol/dm³), aqueous NH₃ (1.0 mol/dm³), NaOH (1.0 mol/dm³), Ba(NO₃)₂ (0.1 mol/dm³), AgNO₃ (0.05 mol/dm³), saturated Ca(OH)₂ limewater, KI (0.1 mol/dm³), KMnO₄ (0.02 mol/dm³), acidified KMnO₄ (0.01 mol/dm³ in 0.5 mol/dm³ H₂SO₄), methyl-orange/screened methyl-orange, thymolphthalein and freshly made starch indicator. Reagent preparation is not expected.
- Apparatus-table qualifications: the stated laboratory specifications include ISO 385 or Grade B burettes, ISO 6706 / ISO 4788 or Grade B measuring cylinders, and ISO 648 or Grade B volumetric pipettes. Flame-test wire or the stated alternative apparatus may be used. These procurement/specification details identify the source's equipment; they do not replace learning how to operate it. Indicator ranges are methyl orange pH 3.1–4.4, screened methyl orange pH 3.2–4.2 and thymolphthalein pH 9.3–10.5. Candidates are not expected to learn how to prepare the listed reagents or indicators.
- Follow centre/local safety and disposal rules, use risk-specific precautions and hazard data, and wear appropriate safety equipment including eye protection. Know the hazard-code terms corrosive (C), health hazard (HH), moderate hazard (MH), acutely toxic (T), flammable (F), oxidising (O) and aquatic-environment hazard (N); the codes themselves are not assessed.
- Use syllabus nomenclature and symbols. Traditional sulfate, sulfite, nitrate, nitrite, sulfuric acid and nitric acid names are used; traditional and systematic names receive credit unless oxidation-number naming is specifically required. Use a line decimal point; print 1000–9999 without separators, and use spaces between digit groups at 10 000 and above. Identify independent (changed) and dependent (measured) variables correctly. Use dm³ rather than litre/l, SI or approved units, no imperial/customary units, and keep units consistent with supplied data.
- Command words: analyse—examine details, elements and relationships; calculate—work out from information; compare—comment on similarities and/or differences; consider—review/respond to information; contrast—comment on differences; deduce—conclude from information; define—give precise meaning; demonstrate—show how/give an example; describe—state characteristics/main features; determine—establish an answer from available information; discuss—write about issues in a structured, in-depth way; evaluate—judge quality/importance/amount/value; examine—investigate closely; explain—give purposes/reasons and make relationships clear, supporting with evidence; give—recall or use a source; identify—name/select/recognise; justify—support with evidence/argument; predict—say what may happen from information; show—that—give structured evidence leading to a result; sketch—draw key features with care over proportions; state—express clearly; suggest—apply knowledge to propose a valid response.
What to do in an assessment
- Identify the command word, then answer in the requested form: for a definition be precise, for an explanation link cause and effect, and for evaluation make a judgement supported by evidence.
In plain language
Correct apparatus, safe method, notation and command-word response make chemistry results interpretable.
Self-check
What does 'evaluate' require?
Show the answer
Judge quality, importance, amount or value using relevant evidence or criteria.
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