AQA GCSE Chemistry Revision
Explore AQA GCSE Chemistry revision topics. Check the official specification for complete requirements for your exam year and route.
Course details
Assessment The assessment includes Paper 1 and Paper 2; 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.
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AQA GCSE Chemistry paper tier
- Foundation tier
- Higher tier
§2.2 and §6.1: AQA 8462 Foundation entry 8462F or Higher entry 8462H; §6.3 tier grade ranges
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Topic group
Exam and scientific method
- Core
Scope: Core · §2.2/§5.2/§5.3 written assessment
Open revision guidance
What to know
- AQA GCSE Chemistry 8462 is linear: all examinations are sat at the end of the course. Paper 1 and Paper 2 are each 1 hour 45 minutes, 100 marks and 50% of the qualification. The raw marks are scaled equally to a total of 200.
- Paper 1 assesses Topics 1–5: atomic structure and the periodic table; bonding, structure and properties of matter; quantitative chemistry; chemical changes; and energy changes. Paper 2 assesses Topics 6–10: rate and extent of chemical change; organic chemistry; chemical analysis; chemistry of the atmosphere; and using resources. Paper 2 may also draw on fundamentals in §§4.1–4.3.
- Both Foundation and Higher papers use multiple-choice, structured, closed short-answer and open-response questions. Practical knowledge is assessed in the written papers; questions drawing on required practical knowledge count for at least 15% of the qualification's overall marks. There is no separate practical examination.
- AO1 assesses knowledge and understanding of scientific ideas, techniques and procedures; AO2 assesses their application, including scientific enquiry; AO3 assesses analysis, interpretation, evaluation, judgements, conclusions and improvements to experimental procedures. The approximate overall weighting is 40% AO1, 40% AO2 and 20% AO3; the component weighting ranges are 37–43%, 37–43% and 17–23% respectively.
What to do in an assessment
- Read the paper context and identify whether the task is recall, application, calculation or evaluation. For longer responses, link a chemical idea to the evidence in the question and then to the requested conclusion.
- For a calculation, show the equation, substitution, units and a sensible significant-figure answer. For an extended response, organise the explanation as a connected sequence rather than a list of unlinked facts.
In plain language
The exam does not keep facts, maths and practical work in separate boxes. You may need to recognise a familiar chemical idea, use it with unfamiliar data and explain why your conclusion follows. Treat the paper's information as evidence to use, not decoration.
Self-check
A question gives a new reaction graph and asks you to explain a rate change in a six-mark response. What should your answer combine?
Show the answer
Use collision theory to explain the changed rate, identify the relevant variable from the graph, quote or compare the evidence, and build a linked explanation. The new setting does not make the underlying chemistry a different syllabus topic.
- Core
Scope: Core · §3.1–3.4 working scientifically, vocabulary and units
Open revision guidance
What to know
- Scientific models help represent, explain and predict observations but have limits; new evidence can support, challenge or replace them. Data may be uncertain, incomplete or unavailable. Scientific and technological applications can bring benefits and ethical, personal, social, economic and environmental consequences; solutions to human environmental impacts must be evaluated using evidence. Peer review helps detect false claims and establish consensus; popular-media reports are not peer reviewed and may be oversimplified, inaccurate or biased.
- Plan investigations from a testable hypothesis: identify independent, dependent and control variables; choose appropriate apparatus, techniques and materials; take representative samples; read and record measurements; manage hazards and reduce risk; and evaluate whether a method is valid and sufficiently precise. Use appropriate tables, diagrams, frequency tables, bar charts, histograms and graphs to present or translate observations and data; identify patterns, anomalies, predictions and conclusions linked to evidence or competing hypotheses.
- Measurements must be evaluated objectively. Accuracy is closeness to the true value; precision means repeated values cluster closely; repeatability means repetition under the same conditions by the same investigator gives similar results; reproducibility means different investigators using different equipment obtain similar results. Random error varies unpredictably and can be reduced by taking more measurements and reporting a mean; systematic error shifts readings consistently from the true value. Examine anomalous results to identify their cause; if an anomaly arose from a poor measurement, explain why it is excluded rather than silently discarding it. Represent distributions and estimate uncertainty with the range about the mean.
- Risk perception can differ from measured risk because a risk may be voluntary or imposed, familiar or unfamiliar, visible or invisible; judge likelihood and consequences using evidence. The area between a curve and the x-axis can have physical meaning and may be estimated by counting squares. Use scientific vocabulary, quantities, symbols and IUPAC chemical names correctly; recognise how mass, time, temperature, volume, concentration and rate are determined.
- Use SI units unless context makes another unit appropriate; recognise prefixes and powers of ten (tera, giga, mega, kilo, centi, milli, micro, nano) and interconvert units such as 1000 cm³ = 1 dm³ and 1 nm = 10⁻⁹ m. Numerical work can require decimals, standard form, ratios, fractions, percentages, estimates, an appropriate number of significant figures, means, frequency diagrams and order-of-magnitude calculations. Use equality, inequality and proportionality symbols; change the subject of an equation; substitute values with appropriate units; translate data between graph and number; interpret y = mx + c, plot two variables, find a linear gradient/intercept, and use a tangent gradient as a rate. Geometry includes triangle and rectangle area and cube surface area and volume.
What to do in an assessment
- Plan a fair investigation by stating a testable hypothesis, independent/dependent/control variables, apparatus, a representative sampling method, measurements, repeats and safety controls. Explain why variables are manipulated or held constant and how the method serves its purpose. Evaluate hazard and risk, method validity and measurement sufficiency; distinguish accuracy, precision, repeatability and reproducibility; identify random/systematic error; investigate anomalies and only exclude one when evidence shows a poor measurement caused it.
- Choose a suitable data display, construct/interpret frequency tables and diagrams, bar charts or histograms, plot two variables, translate between graphical and numeric forms, and draw a conclusion consistent with observations and the hypothesis. Calculate a mean and range, select sensible significant figures and units, estimate uncertainty using the range, and report patterns or anomalies. Rearrange and substitute into equations, make an order-of-magnitude estimate, interpret y = mx + c, find a gradient/intercept, draw a tangent for an instantaneous rate and estimate area under a curve by counting squares when relevant. Give a reasoned environmental or technology evaluation using evidence, benefits, consequences and uncertainty.
In plain language
A good science answer shows how you know. A result is an observation; a model explains it. Repeating a measurement checks whether a pattern is dependable, while a fair test changes one factor at a time. Scientists revise explanations when good evidence no longer fits them.
Self-check
A student says one result proves a catalyst always doubles reaction rate. Name two weaknesses in that claim.
Show the answer
One result provides no check on repeatability, and the statement generalises beyond the tested conditions. The student should repeat measurements, compare a suitable control, calculate rates and limit the conclusion to the evidence and conditions.
- Core
Scope: Core · §4.11 key ideas
Open revision guidance
What to know
- Matter is made from tiny particles called atoms; about 100 naturally occurring types of atoms are elements. Elements show periodic relationships in chemical and physical properties, explained by atomic structure. Atoms bond by transferring or sharing electrons; molecular shapes and the arrangement of giant structures are important to how substances behave.
- Reaction barriers explain why reactions occur at different rates. Chemical reactions can be understood through three fundamental processes: proton transfer, electron transfer or electron sharing. Energy is conserved in chemical reactions and cannot be created or destroyed. These universal key ideas are embedded through the course and assessed across all papers.
What to do in an assessment
- When a question crosses topics, identify the level needed: particles and electrons for a mechanism, structure for a material property, or energy/rate for a process. Link the levels explicitly rather than naming a key idea without using it.
In plain language
Many apparently different chemistry questions use the same few tools: identify the particles, work out how they are bonded, explain what changes in a reaction, and track matter and energy. These ideas let you apply known chemistry in a new setting.
Self-check
Why can a question about a metal's conductivity be answered using both bonding and particle structure?
Show the answer
Metallic bonding describes positive ions surrounded by delocalised electrons; the mobile electrons carry charge through the structure, explaining conductivity.
- Core
Scope: Core · §7 mathematical requirements
Open revision guidance
What to know
- AQA §7 arithmetic includes decimal/standard form, ratios/fractions/percentages and estimates. Data handling includes suitable significant figures, arithmetic means, frequency tables/diagrams/bar charts/histograms and order-of-magnitude calculations. Algebra includes the symbols =, <, <<, >>, >, ∝ and ~; changing equation subject; and substituting values with appropriate physical units.
- Graph skills include translating graphical/numerical data, understanding y = mx + c, plotting two variables, determining gradient and intercept, and drawing/using a tangent gradient as rate of change. Geometry requires visualising and representing 2D/3D forms (including 2D representations of 3D objects), calculating triangle and rectangle areas, and calculating surface area and volume of cubes. Foundation maths questions are not below Key Stage 3 level; Higher maths questions are not below the demand of Foundation-tier GCSE Mathematics.
What to do in an assessment
- Write the relationship, rearrange before substitution, convert units, substitute values with units, then round appropriately. Use all repeat values to calculate a mean; represent data in a suitable frequency table/chart/histogram; estimate magnitude before calculating.
- For y = mx + c, identify gradient m and intercept c; plot labelled axes with a sensible scale and determine a straight-line gradient/intercept. Use a tangent for instantaneous rate. Calculate triangle area = ½bh, rectangle area = lw, cube surface area = 6s² and cube volume = s³; check units become squared or cubed.
In plain language
Marks are easier to earn when your method can be followed. A correct number with no conversion or unit can be hard to verify; showing each step also helps you spot whether you accidentally used cm³ where dm³ was required.
Self-check
Convert 250 cm³ to dm³ before using it in a concentration calculation.
Show the answer
250 ÷ 1000 = 0.250 dm³.
Topic group
Atomic structure and periodic table
- Core
Scope: Core · §4.1.1.1–4.1.1.7 atoms, mixtures and electronic structure
Open revision guidance
What to know
- An element contains one type of atom; a compound contains elements chemically combined in fixed proportions and is separated only by chemical reactions. A mixture contains uncombined substances that retain their properties. Reactions form new substances and often have a detectable energy change. Chemical symbols identify elements; formula subscripts show atom ratios, while balanced-equation coefficients show reacting ratios.
- A proton has relative charge +1 and mass 1; a neutron has charge 0 and mass 1; an electron has charge −1 and relative mass about 1/1840. A neutral atom has equal numbers of protons and electrons. Almost all atomic mass is in the nucleus. An atom's radius is about 0.1 nm (1 × 10⁻¹⁰ m); a nucleus radius is about 1 × 10⁻¹⁴ m, less than 1/10,000 of the atom's radius. A 1 mm grain of sand is about five million atom diameters across (using a 0.2 nm atom diameter). Proton number identifies the element; mass number is protons plus neutrons. Isotopes have the same proton number and different neutron numbers.
- Atomic models changed as evidence arrived. Before the electron was discovered, atoms were considered indivisible; the plum-pudding model then represented a ball of positive charge with embedded negative electrons. In alpha scattering, most positively charged alpha particles passed straight through thin metal foil, showing that atoms are mostly empty space; some were deflected, and very few rebounded. Those rare large deflections indicated a very small, dense, positively charged nucleus containing most of the mass, rather than positive charge spread throughout the atom. Bohr proposed electrons occupy specific shells around the nucleus; later evidence identified protons and neutrons. Relative atomic mass is a weighted mean of isotope masses. Electrons occupy the lowest available shells: for the first twenty elements, represent arrangements with a maximum of two in the first shell, then eight in the second and eight in the third at this level. Use the supplied periodic table for element names and symbols. The names and symbols of the first 20 elements, Group 1 and Group 7 elements, and other elements named in this specification are examinable.
What to do in an assessment
- For any specified reaction, name the reactants and products, write its word equation, translate names to correct formulae, then balance with coefficients (never alter subscripts). For example: magnesium + oxygen → magnesium oxide; 2Mg + O₂ → 2MgO. Name compounds from formulas or symbol equations (NaCl is sodium chloride).
- From an atom/ion symbol, read proton and mass numbers, calculate neutrons = mass number − proton number, then use charge to determine electrons. Represent each of the first twenty atoms as shell numbers and a shell diagram. For relative atomic mass, multiply each isotope mass by its percentage abundance, sum and divide by 100.
- Choose separation by the physical property and carry out the sequence: filter an insoluble solid from liquid; crystallise a dissolved solid from a concentrated solution; use simple distillation to recover solvent from solution; use fractional distillation for miscible liquids with different boiling points; use chromatography when components travel differently in a solvent. For example, separate sand from salt solution by filtration, then crystallise the salt; separate the water from salt water by distillation.
In plain language
Think of proton number as an element's identity card: changing it makes a different element. Isotopes keep that identity but have different numbers of neutrons. Scientific models are useful pictures that change when evidence improves. A mixture is like ingredients that keep their identities; a compound is a new substance with a fixed recipe.
Self-check
An ion has 17 protons, 18 neutrons and charge 1−. How many electrons does it have, and what is its mass number?
Show the answer
It has 18 electrons: a neutral atom with 17 protons would have 17, and a 1− ion has gained one. Its mass number is 17 + 18 = 35.
- Core
Scope: Core · §4.1.2.1–4.1.2.6 periodic table and groups
Open revision guidance
What to know
- The modern periodic table is ordered by proton number; elements in a group share outer-electron counts and often have similar chemistry. Mendeleev arranged elements by atomic weight, left gaps for undiscovered elements and changed some positions; predicted elements were later found, and isotopes explained why atomic-weight order sometimes differed. Metals usually form positive ions and lie left/below; they are typically shiny, malleable and good conductors. Non-metals lie right/above, do not form positive ions, and are generally poor conductors; many form covalent molecules or acidic oxides.
- Group 0 atoms have stable full outer shells (helium has two), so they are very unreactive and do not readily form molecules; boiling points increase with relative atomic mass down the group. Group 1 atoms have one outer electron and form +1 ions; reactivity increases down the group. Lithium, sodium and potassium react with oxygen to form oxides, with chlorine to form ionic chlorides, and with room-temperature water to form alkaline hydroxides plus hydrogen; reaction vigour increases down the group. Group 7 halogens are diatomic non-metals with seven outer electrons; relative molecular mass and melting/boiling points increase down the group while reactivity decreases.
- Chlorine, bromine and iodine form ionic compounds (metal halides) with metals and covalent compounds with non-metals; for example, 2Na + Cl₂ → 2NaCl, while H₂ + Cl₂ → 2HCl. A more reactive halogen displaces a less reactive halogen from its aqueous salt: chlorine displaces bromine and iodine, and bromine displaces iodine.
What to do in an assessment
- Infer an atom's outer-electron arrangement from its group and period, then link the outer electrons to likely ion charge, bonding and reactivity. When comparing metals and non-metals, state physical and chemical contrasts rather than relying on location alone.
- Describe each Group 1 reaction with its named reactant and product: metal + oxygen → metal oxide; metal + chlorine → metal chloride; metal + water → metal hydroxide + hydrogen. Explain increasing vigour down the group from the one outer electron being lost more readily. Use the supplied observation to distinguish an ionic metal halide from a covalent non-metal halide, and predict displacement from the reactivity order.
In plain language
The table is a pattern map. Elements in the same column have a similar outer-electron setup, so their atoms tend to react in related ways. Moving down a group changes how strongly the outer electrons are held, which helps explain the trend in reactivity.
Self-check
A halogen solution displaces bromide ions but does not displace chloride ions. What can be concluded about its reactivity?
Show the answer
It is more reactive than bromine, because it displaced bromide, but less reactive than chlorine, because it could not displace chloride.
- Core
Scope: Core · §4.1.3.1–4.1.3.2 transition metals
Open revision guidance
What to know
- Compared with Group 1, transition metals are generally stronger, harder, denser and have higher melting points, and react less readily with oxygen, water and halogens.
- Chromium, manganese, iron, cobalt, nickel and copper are the named examples. Their ions may have different charges; many compounds are coloured, and transition metals or their compounds can act as catalysts.
What to do in an assessment
- Use a supplied comparison table to select evidence for the stated property; do not treat every member as identical when a question supplies an exception or data.
- When asked for a transition-metal application, link the property to the use: a catalyst speeds a reaction without being used up, while hardness or strength supports structural uses.
In plain language
Transition metals are not simply heavier Group 1 metals. They tend to be tougher and less reactive, and their chemistry often gives coloured compounds or useful catalysts.
Self-check
A reaction mixture contains a metal compound that speeds a reaction, remains after the reaction and has a coloured solution. Which transition-metal features fit?
Show the answer
Catalytic action and coloured compounds are characteristic transition-metal features; the catalyst is not consumed in the reaction.
- Higher
Scope: Higher · §4.1.1.1 Higher-only ionic and half-equations
Open revision guidance
What to know
- A balanced ionic equation contains only the reacting ions or molecules; spectator ions that remain unchanged can be cancelled. Both atom count and overall charge must balance.
- A half-equation shows a single oxidation or reduction process. Add electrons on the side needed to conserve charge, then check that the number of atoms and the net charge match on both sides.
What to do in an assessment
- Start with the full balanced equation, write each aqueous ionic substance as its ions, and cancel unchanged spectator ions. For a half-equation, balance atoms first and then charge with electrons.
- Use state symbols when the question asks for them and do not cancel an ion that changes during the reaction.
In plain language
An ionic equation removes the spectators so the actual chemical change is visible. A half-equation accounts for electrons moving in or out of one substance; charge is part of the balancing, not an optional decoration.
Self-check
In Ag⁺ + e⁻ → Ag, why is one electron needed?
Show the answer
The silver ion has a +1 charge and the silver atom is neutral. Gaining one negatively charged electron makes the total charge zero.
Topic group
Bonding, structure and properties
- Core
Scope: Core · §4.2.1.1–4.2.1.5 bonding
Open revision guidance
What to know
- Ionic bonding forms when outer electrons transfer from a metal atom to a non-metal atom. Group 1 metals lose one electron and form +1 ions; Group 2 metals lose two and form +2 ions. Group 6 non-metals gain two and form 2− ions; Group 7 non-metals gain one and form 1− ions. The resulting ions have noble-gas electron arrangements and attract in all directions. Covalent bonding joins non-metal atoms by shared pairs of electrons; covalent substances may be small molecules, very large polymer molecules or giant covalent structures. Metallic bonding is the strong attraction between positive metal ions in a regular giant structure and delocalised electrons; it occurs in metals and alloys.
- For example, Na transfers one electron to Cl, making Na⁺ and Cl⁻ (both noble-gas arrangements); Mg transfers two to O, making Mg²⁺ and O²⁻. Ionic formulae have the simplest whole-number ion ratio and total charge zero: MgCl₂ needs two Cl⁻ for each Mg²⁺, while Na₂O needs two Na⁺ for each O²⁻. Ionic compounds form giant lattices rather than discrete molecule pairs.
- A dot-and-cross diagram shows valence-electron transfer or sharing and which atom supplied each electron. For example, show Na's single outer electron as × and chlorine's seven as dots: Na× + Cl(•••••••) → [Na]⁺ [Cl(•••••••×)]⁻. For magnesium oxide show transfer of both Mg outer electrons: Mg×× + O(••••••) → [Mg]²⁺ [O(••••••××)]²⁻. For covalent diagrams, H₂, Cl₂ and HCl each have one shared pair; O₂ has two and N₂ has three; H₂O has two bonding pairs and two lone pairs around O; NH₃ has three bonding pairs and one lone pair around N; CH₄ has four bonding pairs around C. Show the three lone pairs on each chlorine atom, the two lone pairs on oxygen in water, and the one lone pair on nitrogen in ammonia. Brackets and ion charges show ionic particles; the dot/cross distinction tracks electron origin. A dot-and-cross formula unit does not show a complete ionic lattice. Dot-and-cross, ball-and-stick and 2D/3D diagrams also have limitations: they simplify relative sizes, spacing, shapes or continuous giant structures.
What to do in an assessment
- Draw ionic dot-and-cross diagrams for a Group 1 or 2 metal with a Group 6 or 7 non-metal: show outer electrons before and after transfer, use brackets and charges around ions, and check that each ion has a noble-gas arrangement. Practise NaCl, MgO, Na₂O and MgCl₂.
- Draw dot-and-cross diagrams for H₂, Cl₂, O₂, N₂, HCl, H₂O, NH₃ and CH₄; show each bonding pair and any lone pairs, then count atoms to deduce the molecular formula. Represent a covalent bond as one line in a small molecule, part of a giant covalent structure or a polymer repeat unit; use diagrams to distinguish a molecule, polymer and giant structure.
- From a model of a giant ionic structure, count each ion type, reduce the counts to the simplest whole-number ratio, and write the empirical formula; check net charge is zero. Explain why a single dot-and-cross formula unit or a spaced ball-and-stick diagram does not represent the full lattice, and identify a metallic giant structure from positive ions surrounded by mobile electrons.
In plain language
Bonding is about what holds particles together. Ions attract because their charges are opposite; covalent atoms share electrons; metals have electrons that can move through the whole structure. The structure matters because it predicts how a material behaves.
Self-check
A diagram shows positive metal ions in a regular array surrounded by mobile electrons. Identify the bonding and give one property it helps explain.
Show the answer
It shows metallic bonding. Delocalised electrons can carry charge, explaining why metals conduct electricity.
- Core
Scope: Core · §4.2.2.1–4.2.2.8 bonding, structure and properties
Open revision guidance
What to know
- In the simple particle model, solids have closely packed particles in fixed positions, liquids have particles close together that move past each other, and gases have widely spaced particles moving freely. Melting/freezing occur at the melting point; boiling/condensing occur at the boiling point. Melting, boiling, freezing and condensing are physical state changes. Energy is transferred to overcome attractions or change particle motion; the stronger the forces between particles, the more energy is needed to change state and generally the higher the melting/boiling temperature. Atoms themselves do not have the bulk properties of a material. Equations use (s), (l), (g) and (aq) for solid, liquid, gas and aqueous solution; include appropriate symbols for reactions.
- Ionic lattices have strong electrostatic attractions between oppositely charged ions in all directions, so they need much energy to melt or boil; ions conduct only when molten or dissolved and able to move. Small molecular substances are usually gases or liquids with relatively low melting/boiling points. Their weak intermolecular forces are overcome at melting/boiling; covalent bonds inside molecules remain intact. Intermolecular forces become stronger as molecule size increases, so larger molecules generally have higher melting and boiling points. Small molecules have no overall charge and do not conduct electricity.
- Polymers have very large molecules joined by strong covalent bonds within each chain and relatively strong intermolecular attractions between chains, so they are solids at room temperature. Giant covalent solids have strong covalent bonds throughout and very high melting points; diamond, graphite and silicon dioxide (silica) are examples. Metals have giant structures with strong metallic bonding, generally high melting/boiling points and layers that let pure metal bend or be shaped. Adding different-sized atoms forms harder alloys because the layers are distorted. Metals conduct electricity and transfer thermal energy because delocalised electrons move.
What to do in an assessment
- Use given melting/boiling-point data and a stated temperature to predict whether a substance is solid, liquid or gas; explain the answer through the energy transfer and strength of attractions between its particles. Build each property explanation as structure → forces or mobile charge carriers → property, and recognise that the bulk properties belong to the material's structure, not isolated atoms.
- For molecular substances, explain low melting/boiling points through weak intermolecular forces and use increasing molecule size → stronger intermolecular forces → higher melting/boiling point; distinguish these attractions from covalent bonds and explain why uncharged molecules do not conduct. Recognise polymers, giant covalent structures and metallic structures from diagrams; explain why mobile ions conduct only in molten/aqueous ionic material and why delocalised electrons carry electrical and thermal energy in metals.
- Explain why giant ionic and covalent structures need much energy to change state; identify silicon dioxide as giant covalent. Explain why pure metal layers can slide and how unlike-sized atoms distort layers in harder alloys. Use the correct state symbols (s), (l), (g) and (aq) in equations for reactions in the specification.
In plain language
A material's properties come from its whole arrangement, not just the atoms it contains. A solid with a strong attraction repeated throughout needs lots of energy to come apart; a molecule can boil when weak attractions between molecules are overcome while its atoms stay bonded.
Self-check
Why does solid sodium chloride not conduct, while molten sodium chloride does?
Show the answer
In the solid lattice the ions are fixed, so charge cannot flow. When molten, the ions are free to move and carry charge through the liquid.
- Higher
Scope: Higher · §4.2.2.1 Higher-only limitations of simple particle model
Open revision guidance
What to know
- The simple particle model represents every particle as a solid, inelastic sphere. It can help show arrangement and movement, but it omits forces between particles, treats every particle as spherical and solid, and does not show the particles' internal structure.
- When explaining state changes, distinguish energy used to overcome attractions between particles from energy that changes particle motion. The simple model's missing forces mean it cannot explain every detail of melting or boiling by itself.
What to do in an assessment
- Use the model for the question it can answer (such as particle arrangement), then name a relevant limitation when asked to evaluate it.
- In a change-of-state explanation, connect energy transfer to overcoming attractions or changing particle motion, and avoid saying that particles themselves melt or boil.
In plain language
A model is a useful map, not the real thing. Solid balls are a helpful picture for particle spacing, but real particles are not rigid billiard balls and they attract one another.
Self-check
A solid-sphere model shows gas particles far apart but has no forces. Why can it not fully explain boiling?
Show the answer
Boiling requires energy to overcome attractions between particles. The model has no forces, so it cannot represent that energy requirement even though it shows that gas particles are spread out.
- Core
Scope: Core · §4.2.3.1–4.2.3.3 carbon structures
Open revision guidance
What to know
- Diamond has a giant covalent network in which every carbon makes four bonds; this makes it hard and gives a high melting point, but it has no mobile electrons to conduct electricity.
- In graphite, each carbon bonds to three neighbours in hexagonal layers, leaving one delocalised electron per atom; layers can slide and the electrons conduct. Graphene is a single graphite layer: its strong carbon bonds form a very thin sheet, while its delocalised electrons conduct, making it useful in electronics and composites. Fullerenes are hollow carbon molecules based on hexagonal rings that may also contain five- or seven-membered rings. Buckminsterfullerene C₆₀ is spherical. Fullerene applications include carrying drug molecules in drug delivery, catalysts and lubricants; their hollow shapes and large surface-area-to-volume ratios suit these different uses. Carbon nanotubes are cylindrical fullerenes with a very high length-to-diameter ratio. Their high tensile strength suits reinforcing composite materials, while their electrical conductivity suits nanoscale electronics.
What to do in an assessment
- In a structure comparison, state the number of bonds per carbon and connect that arrangement to hardness, layer movement, electron mobility or an identified use. Recognise a graphene sheet and fullerene diagrams from their structures; identify fullerene hexagonal and five-/seven-membered rings.
- Explain why nanotubes' structure and properties suit nanotechnology, electronics or materials; distinguish this discrete molecular form from a giant covalent carbon network.
In plain language
Diamond and graphite are both carbon, but their atoms are connected differently. Diamond locks each atom into a 3-D network; graphite has sliding sheets and spare electrons. Graphene is one of those sheets, so a tiny thickness can still provide useful strength and electrical behaviour.
Self-check
Which carbon structure conducts electricity and can be used as a lubricant, and why?
Show the answer
Graphite: each carbon leaves one delocalised electron that can carry charge, and weak attractions between its layers let them slide over one another.
- Core
Scope: Core · §4.2.4.1–4.2.4.2 nanoparticles
Open revision guidance
What to know
- Nanoscience refers to structures 1–100 nm in size, of the order of a few hundred atoms. A nanoparticle is smaller than PM2.5 fine particles, whose diameters are 100–2500 nm; PM10 coarse-particle diameters are 2500–10,000 nm. A typical atom is about 0.2 nm in diameter and a small molecule is of order 1 nm, so a nanoparticle is far larger than atoms/molecules but still nanoscale. (1 nm = 10⁻⁹ m.)
- As a cube's side becomes ten times smaller, its surface-area-to-volume ratio becomes ten times larger. A high surface-area-to-volume ratio may give nanoparticles different properties from the same material in bulk and may let a smaller quantity be effective. Uses include medicine, electronics, cosmetics and sun creams, deodorants and catalysts. Their applications are researched and possible risks must be considered.
What to do in an assessment
- Compare a 1–100 nm nano dimension with typical atom and molecule dimensions using the same units and an order-of-magnitude estimate. For a cube, calculate surface area = 6s² and volume = s³, then explain how the ratio changes when its side becomes ten times smaller.
- Evaluate a specified application by weighing its benefit against supplied evidence about exposure, safety or environmental effects; distinguish a demonstrated risk from an unresolved possibility.
In plain language
A tiny cube has much more surface compared with the amount of material inside. Reactions happen at surfaces, so shrinking a material can change how well it works. The same high exposure is why safety questions need evidence.
Self-check
Why might less nanoparticle material be needed than bulk material to provide the same catalytic effect?
Show the answer
The nanoparticle has a greater surface-area-to-volume ratio, so more of its atoms are at or near the surface where reactions occur.
Topic group
Quantitative chemistry
- Core
Scope: Core · §4.3.1.1–4.3.1.4 equations, mass and measurement
Open revision guidance
What to know
- A balanced symbol equation has the same number of each type of atom on both sides. A coefficient multiplies every atom in a formula; a subscript is part of the substance's formula and must never be changed to balance an equation. The relative formula mass, Mr, is the sum of relative atomic masses in the formula. In a balanced equation, total coefficient-weighted Mr is conserved. Percentage by mass of an element = (its total Ar contribution in the formula ÷ Mr) × 100.
- Mass is conserved because atoms are rearranged, not gained or lost. In a non-enclosed system, measured mass can appear to rise when oxygen particles from the air react with a metal, or fall when gas particles such as carbon dioxide escape during a carbonate's thermal decomposition. The balanced equation identifies which atoms and particles enter or leave the measured system; they have not vanished or been made.
- Every measurement has uncertainty. Represent the distribution of repeated results and estimate uncertainty using the range of values about their mean (highest minus lowest repeat); a mean does not remove measurement uncertainty.
What to do in an assessment
- Balance first by changing coefficients only. For a percentage-composition question, multiply the element's Ar by its subscript, divide by compound Mr and multiply by 100. If mass changes in an open system, use the balanced equation and particle model to explain which gas particles crossed the system boundary and why the measured solid's mass changed.
- Record repeat measurements in a table, calculate their mean and range, represent the spread/distribution (for example on a suitable graph), and use the range as an uncertainty estimate. Keep units and suitable significant figures.
In plain language
Atoms are rearranged, not created or destroyed. A balance will only appear to lose or gain mass if material can enter or leave, often as a gas. The equation is the map that tells you which particles and amounts to compare.
Self-check
A metal is heated in air and its mass increases. Does this break conservation of mass?
Show the answer
No. Oxygen from the air has joined the metal to form an oxide. The measured solid gained mass because oxygen entered the open system; include oxygen in the whole-system mass balance.
- Core
Scope: Core · §4.3.2.5 and §4.3.3.1–4.3.3.2 shared concentration, yield and atom economy
Open revision guidance
What to know
- For a mass-concentration question in g/dm³, calculate solute mass as concentration × final solution volume; convert 1000 cm³ = 1 dm³ before substituting. The full explanation of the concentration–mass–volume relationship is Higher tier.
- Percentage yield = (actual mass of product ÷ maximum theoretical mass of product) × 100. Yield may be low because a reversible reaction does not go to completion, product is lost in separation or reactants form other products. Percentage atom economy = (coefficient × Mr of desired product from the balanced equation ÷ sum of coefficient × Mr for all reactants from the balanced equation) × 100. Apply each equation coefficient when calculating formula masses. High atom economy is important for sustainable development and economic reasons because more starting-material mass becomes the desired product.
What to do in an assessment
- For a mass-concentration calculation, convert solution volume to dm³, multiply its stated concentration by that volume to find mass, and include units. For percentage yield, identify actual and theoretical product masses before dividing. For atom economy, use the balanced equation and coefficient-weighted formula masses for desired product and all reactants.
In plain language
Quantitative chemistry is bookkeeping with particles. A balanced equation gives the recipe, and a formula mass helps compare amounts. If a solution's concentration and volume are given, their product gives the solute mass. Write down each conversion so a small unit change does not derail the calculation.
Self-check
For CaCO₃ → CaO + CO₂, the theoretical CO₂ mass is 44 g and the actual mass is 22 g. Find the percentage yield and percentage atom economy for CO₂ (Mr: CaCO₃ = 100, CO₂ = 44).
Show the answer
Percentage yield = (22 ÷ 44) × 100 = 50%. Percentage atom economy = (Mr of desired CO₂ ÷ Mr of reactant CaCO₃) × 100 = (44 ÷ 100) × 100 = 44%. The first compares actual with theoretical product; the second compares desired-product formula mass with all reactant formula mass in the balanced equation.
- Higher
Scope: Higher · §4.3.2.1–4.3.2.4 Higher-only moles, equation calculations and limiting reactants
Open revision guidance
What to know
- One mole is 6.02 × 10²³ stated particles (the Avogadro constant per mole); its mass in grams is numerically equal to Mr. The mole can count atoms, molecules, ions, electrons or formula units, as specified; for example, one mole of carbon atoms and one mole of CO₂ molecules contain the same number of their stated particles. n (mol) = mass (g) ÷ Mr; mass = n × Mr.
- The balanced equation gives mole ratios and can be interpreted as ratios of particles and amounts in moles. A worked mass-to-mass route is: balance the equation → calculate Mr → convert known mass to moles → use coefficient ratio → convert wanted moles back to mass. To balance a reaction from measured masses, convert each measured reactant/product mass to moles, divide by the smallest amount and scale the ratio to simple whole numbers. In a reaction with two reactants, the limiting reactant is completely used first and caps the amount of product; any excess reactant remains.
What to do in an assessment
- Example: 12 g Mg is 12 ÷ 24 = 0.50 mol; 8 g O₂ is 8 ÷ 32 = 0.25 mol. Their 2:1 mole ratio gives 2Mg + O₂ → 2MgO. For a product-mass calculation, write the balanced equation, calculate Mr, find moles of the known mass, apply coefficients, then multiply product moles by product Mr; carry units through each line.
- For two reactants, compare available mole amounts with the coefficient ratio; identify the limiting reactant before calculating the maximum product. Explain that any excess reactant remains after the limiting reactant is used up.
In plain language
A mole is a counting unit too large for individual atoms. Convert everything into that common count, use the balanced equation as a recipe, and only then return to grams or gas volume. The ingredient that runs out first is the limit on the batch.
Self-check
For 2H₂ + O₂ → 2H₂O, 3 mol H₂ is mixed with 2 mol O₂. Which reactant limits water, and how many moles of water can form?
Show the answer
Three moles of H₂ require 1.5 mol O₂, and 2 mol O₂ is available, so H₂ is the limiting reactant. The equation's 2:2 H₂:H₂O ratio means 3 mol H₂ can form 3 mol H₂O.
- Higher
Scope: Higher · §4.3.4–4.3.5 Higher-only molar concentration and gas-volume calculations
Open revision guidance
What to know
- For Higher molar concentration, c (mol/dm³) = amount n (mol) ÷ solution volume V (dm³), so n = cV. With Mr, mass m (g) = nMr = cV × Mr. Thus concentration in mol/dm³ relates both mass and final solution volume. For two solutions that react completely, use the balanced equation's mole ratio to determine an unknown concentration from the measured volumes and known concentration.
- At room temperature and pressure (20°C and 1 atmosphere), one mole of any gas occupies 24 dm³; equal mole amounts of gases at the same temperature and pressure occupy equal volumes. The balanced equation gives gas-volume ratios under the same conditions.
What to do in an assessment
- For titration/solution calculations, convert cm³ to dm³ by dividing by 1000; calculate known moles cV; use the balanced equation to find unknown moles; divide by unknown volume to get c; multiply by Mr if mass of solute is asked. Rearrange the amount/concentration relationships when a different quantity is unknown, show the algebra, and never assume a 1:1 ratio unless the balanced equation shows it.
- For RTP gas volume, calculate moles from mass and Mr or use the coefficient ratio, then multiply by 24 dm³ mol⁻¹. Conversely, divide a gas volume in dm³ by 24 dm³ mol⁻¹ to obtain moles. Show conversions to/from cm³ and rearrange the equation to make the required quantity the subject.
In plain language
Concentration says how much solute is packed into a volume. The balanced equation matters because a reaction may use two particles of one substance for every one of another. For gas calculations, the 24 dm³ shortcut only applies at the stated room-temperature-and-pressure conditions.
Self-check
What volume at RTP is occupied by 0.25 mol of gas?
Show the answer
0.25 × 24 = 6.0 dm³, which is 6000 cm³.
- Higher
Scope: Higher · §4.3.2.5 Higher-only concentration relationship
Open revision guidance
What to know
- For the common mass-concentration form, c_m (g/dm³) = m (g) ÷ V (dm³), so m = c_mV. For molar concentration, c (mol/dm³) = n/V and n = cV; because m = nMr, c = m/(MrV) and m = cVMr. V is the final solution volume, not the volume of solvent before dissolving.
- Concentration describes how much solute is distributed through a known solution volume; the molar and mass forms have different units and must not be confused.
What to do in an assessment
- Select the correct relationship for the units requested, convert cm³ to dm³, substitute with units, and check dimensions: g/dm³ × dm³ = g; mol/dm³ × dm³ = mol. For a molar concentration, convert mass to moles using Mr or use m = cVMr.
- Explain how doubling solute mass at fixed volume doubles concentration, while doubling final volume at fixed solute mass halves it.
In plain language
Concentration is how much dissolved substance is packed into a particular solution volume. Changing mass or volume changes that packing, which is why the same amount of salt can make a weak or strong solution.
Self-check
A fixed 6 g of solute is dissolved to make 0.50 dm³ solution. What is the concentration, and what would it become if final volume doubled?
Show the answer
Initially 6 ÷ 0.50 = 12 g/dm³. At 1.00 dm³ with the same solute mass, it becomes 6 g/dm³.
- Higher
Scope: Higher · §4.3.3.1–4.3.3.2 Higher-only theoretical yield and pathway evaluation
Open revision guidance
What to know
- Theoretical yield is the maximum product mass predicted by the balanced equation for the amount of limiting reactant supplied; actual yield may be lower because of incomplete reaction, separation losses or competing reactions. The atom-economy equation measures the fraction of coefficient-weighted reactant Mr that becomes the desired product.
- A reaction route is not selected by yield alone: supplied evidence may include rate, equilibrium position, atom economy, energy/cost and usefulness of by-products. Higher-tier pathway questions may require a theoretical mass from a given reactant mass and balanced equation.
What to do in an assessment
- Convert the limiting reactant's mass to moles, use the coefficient ratio to calculate desired-product moles, then multiply by product Mr to obtain the theoretical mass; compare actual to theoretical mass for percentage yield. In route evaluation, compare the supplied values and state which priorities the recommendation assumes.
In plain language
The theoretical mass is the ideal amount if the equation's recipe runs exactly as expected. A real process has losses or side reactions. A factory may still choose a route with a lower yield if it is faster, cheaper or wastes fewer atoms.
Self-check
Route A has 92% yield but 35% atom economy; Route B has 78% yield and 82% atom economy. Which route is automatically best?
Show the answer
Neither is automatically best. Route A recovers a larger proportion of its theoretical product, but Route B uses a much larger fraction of reactant atoms in the desired product. The decision needs the question's priorities, costs, rate, energy and by-product data.
Topic group
Chemical changes
- Core
Scope: Core · §4.4.1.1–4.4.1.3 oxidation, reactivity and extraction
Open revision guidance
What to know
- Metals react with oxygen to form oxides; oxidation is gain of oxygen and reduction is loss. Unreactive metals such as gold can occur naturally as elements, while most metals occur as compounds. Metals less reactive than carbon can be extracted from their oxides by reduction with carbon; detailed industrial extraction processes are not required unless data are supplied.
- A more reactive metal forms positive ions more readily and can displace a less reactive metal from its compound. The required metal order is K > Na > Li > Ca > Mg > Zn > Fe > Cu. Hydrogen lies between iron and copper; carbon lies between magnesium and zinc in this comparison. Carbon and hydrogen are not metals but help predict extraction and acid reactions. With room-temperature water, potassium, sodium and lithium fizz vigorously to form hydrogen and soluble alkaline hydroxides; potassium may ignite. Calcium fizzes more slowly and forms calcium hydroxide, which can make the water cloudy. Magnesium reacts very slowly with cold water to give hydrogen and a white magnesium hydroxide suspension. Zinc, iron and copper show no reaction with cold water. With dilute acid, metals above hydrogen form a salt and hydrogen: potassium, sodium and lithium react extremely vigorously; calcium and magnesium react readily, zinc fizzes more slowly, iron more slowly still, and copper does not react. These outcomes relate observed vigour to the tendency to form positive ions; reactions with steam are excluded.
What to do in an assessment
- For K, Na, Li, Ca, Mg, Zn, Fe and Cu, state the room-temperature water observation/product and dilute-acid outcome, using evidence such as effervescence, cloudiness, hydrogen or no visible reaction to order reactivity. A positive test for evolved hydrogen is a squeaky pop. Compare safe, teacher-controlled results; do not treat reactive metal–acid combinations as a student practical.
- Write word equations from observed reactants and products, then use supplied formulas to balance representative equations. For extraction questions, identify which species gains or loses oxygen and evaluate any supplied method.
In plain language
A more reactive metal gives up electrons more readily to become a positive ion. Water or dilute acid may show that tendency through how readily hydrogen appears; a metal that reacts fastest is usually higher in the order. Carbon can remove oxygen from oxides of metals below it.
Self-check
At room temperature, comparable samples of zinc and iron both react with the same dilute acid. Zinc fizzes faster than iron, and hydrogen is identified. Which metal is more reactive, and what evidence supports that conclusion?
Show the answer
Zinc is more reactive than iron: under comparable conditions it produces hydrogen faster. Iron also reacts, but more slowly. A squeaky-pop test identifies hydrogen; the difference in reaction rate supports the comparison.
- Core
Scope: Core · §4.4.2.1–4.4.2.5 acids, salts, pH and shared titration preparation
Open revision guidance
What to know
- Acids produce hydrogen ions in aqueous solution; aqueous alkalis contain hydroxide ions. Acids are neutralised by soluble alkalis such as metal hydroxides, insoluble bases such as metal hydroxides/oxides, and metal carbonates. An acid + base/alkali forms salt + water; acid + carbonate forms salt + water + carbon dioxide; acids also react with some metals to form a salt and hydrogen. Hydrochloric acid produces chlorides, nitric acid nitrates and sulfuric acid sulfates; the positive ion in the salt comes from the metal base, alkali or carbonate.
- Soluble salts can be prepared by reacting an acid with a suitable solid metal, oxide, hydroxide or carbonate until no more reacts; filter off excess solid to obtain salt solution, then crystallise. With an insoluble solid, adding excess ensures all acid reacts. To predict a salt formula, combine common ions to give zero net charge: Na⁺, K⁺, NH₄⁺, Mg²⁺, Ca²⁺, Zn²⁺, Cu²⁺; Cl⁻, NO₃⁻, OH⁻, CO₃²⁻ and SO₄²⁻. Examples are Mg(NO₃)₂, CaCl₂ and Na₂SO₄.
- The pH scale runs from 0 to 14: aqueous pH below 7 is acidic, 7 is neutral and above 7 is alkaline. Universal or other wide-range indicator gives an approximate pH from its colour; a pH probe can also measure pH. In neutralisation, hydrogen ions and hydroxide ions form water: H⁺ + OH⁻ → H₂O.
- Titration preparation is shared Chemistry content for both tiers: use a pipette to measure alkali into a flask, add a suitable indicator, and deliver a strong acid from a burette while swirling. The strong acids in this assessment are hydrochloric, nitric and sulfuric acids. Near the endpoint add acid dropwise; the first permanent indicator colour change identifies the neutralisation volume. Repeat to obtain agreeing titres, then repeat the measured volumes without indicator if preparing pure salt crystals. Higher-only concentration calculations are taught separately.
What to do in an assessment
- Predict salt, water and (where relevant) carbon dioxide or hydrogen from the reactant types; name the salt from acid and cation; balance a word/symbol equation. For an insoluble base, warm dilute acid, add the solid until excess remains, filter, concentrate the filtrate, cool to crystallise, then collect and dry crystals. For a soluble alkali, titrate measured volumes to neutralise exactly because excess cannot be removed by filtration, then repeat without indicator to crystallise a pure salt.
- Build salt formulas by balancing ionic charges with the smallest whole-number ratio, using brackets for repeated polyatomic ions; for example Mg²⁺ + 2NO₃⁻ → Mg(NO₃)₂. Use the colour chart for universal/wide-range indicator or indicator paper to estimate pH; use a pH probe to measure it.
In plain language
The acid supplies one part of the salt's name and formula, while the metal or other positive ion supplies the other. Neutralisation removes the acid's hydrogen ions by joining them with hydroxide ions to make water.
Self-check
What products form when hydrochloric acid reacts with magnesium carbonate, and what is the salt called?
Show the answer
Magnesium chloride, water and carbon dioxide form. Hydrochloric acid supplies chloride ions and magnesium carbonate supplies magnesium ions. The carbonate ion is CO₃²⁻, not Mg²⁺; it reacts with hydrogen ions to form carbon dioxide and water.
- Higher
Scope: Higher · §4.4.1.4 Higher-only electron transfer
Open revision guidance
What to know
- In electron terms, oxidation is loss of electrons and reduction is gain. A species that loses electrons is oxidised; the species that gains those electrons is reduced.
- A metal displacement can be written as an ionic equation by removing spectator ions. The electron transfer must balance: electrons lost by one species equal electrons gained by the other.
What to do in an assessment
- For a metal-displacement reaction, start with the balanced symbol equation, remove unchanged spectator ions to write the ionic equation, then identify the oxidation and reduction half-changes. Add electrons so those lost equal those gained; check atom and total-charge balance in the ionic equation.
- Use OIL RIG only as a memory cue; justify the direction using the actual electron change in the equation. Identify the oxidised and reduced species in a supplied reaction, symbol equation or half-equation.
In plain language
Electrons move from the substance being oxidised to the one being reduced. The two labels happen together because the electrons lost by one reactant must be gained by another.
Self-check
In Zn + Cu²⁺ → Zn²⁺ + Cu, which species is oxidised?
Show the answer
Zinc is oxidised because each Zn atom loses two electrons to form Zn²⁺. Copper ions gain those electrons and are reduced.
- Higher
Scope: Higher · §4.4.2.1 and §4.4.2.5–4.4.2.6 / §4.7.2.4 Higher-only redox, titration calculations and weak-acid explanations
Open revision guidance
What to know
- For magnesium, zinc or iron reacting with hydrochloric or sulfuric acid, identify oxidation/reduction by electron loss/gain: the metal is oxidised as it forms positive ions and hydrogen ions are reduced to hydrogen. For example, Mg → Mg²⁺ + 2e⁻ and 2H⁺ + 2e⁻ → H₂. Hydrochloric, nitric and sulfuric acids are strong: they are completely ionised in aqueous solution. Ethanoic, citric and carbonic acids are weak: they are only partially ionised. Strong/weak describe degree of ionisation; concentrated/dilute describe amount of acid per volume. At the same solution concentration, the stronger acid has lower pH. A one-unit decrease in pH means hydrogen-ion concentration is multiplied by ten; pH 7 is neutral, with acidic and alkaline solutions below and above 7 respectively.
- Titration measures reacting volumes of strong acids and strong alkalis. For a calculation, known concentration × volume gives moles; use the balanced equation's mole ratio, then divide unknown moles by the final unknown-solution volume to find mol/dm³ concentration or convert with Mr to g/dm³.
What to do in an assessment
- For a titration calculation, convert measured cm³ to dm³; find known moles from concentration × volume; use the balanced equation to find unknown moles; then divide by the unknown solution's volume. Convert mass concentration using Mr when asked, and keep the endpoint volume distinct from an arbitrary excess.
- When comparing acids, keep strength (degree of ionisation) separate from concentration (amount per volume). A pH change is logarithmic: a two-unit decrease means 100 times the hydrogen-ion concentration; use partial ionisation to explain why a weak acid has fewer hydrogen ions than a strong acid of equal concentration.
In plain language
A dilute strong acid can contain fewer acid particles per litre than a concentrated weak acid. Strength describes how completely acid particles ionise; concentration describes how much acid is present. A titration finds the exact volume ratio that reacts.
Self-check
Which contains more hydrogen ions at equal concentration: a strong acid at pH 2 or a weak acid at pH 4? By what factor?
Show the answer
The pH 2 solution has 100 times the hydrogen-ion concentration of the pH 4 solution because it is two pH units lower.
- Core
Scope: Core · §4.4.3.1–4.4.3.4 electrolysis
Open revision guidance
What to know
- Electrolysis passes current through an electrolyte: a molten ionic compound or ionic solution whose mobile ions conduct electricity. Positive ions (cations) move to the negative cathode and negative ions (anions) to the positive anode; discharge at electrodes produces elements. In a molten binary ionic compound with inert electrodes, the metal forms at the cathode and the non-metal at the anode.
- For an aqueous solution electrolysed with inert electrodes, water molecules produce competing H⁺ and OH⁻ ions. At the cathode, hydrogen forms if the metal is more reactive than hydrogen; otherwise the metal can form. At the anode, oxygen forms unless the solution contains halide ions (chloride, bromide or iodide), when the halogen forms. Thus aqueous sodium chloride forms hydrogen at the cathode and chlorine at the anode.
- Very reactive metals such as aluminium are extracted electrolytically when carbon reduction is unsuitable. Aluminium oxide is mixed with cryolite to lower the operating temperature and energy needed to melt the electrolyte; the carbon anode is continually replaced because oxygen produced there reacts with it. Electrolysis therefore uses substantial energy to melt compounds and provide current.
What to do in an assessment
- List all ions from both the ionic compound and water, then apply the cathode/anode selection rules; include electrode signs and product state. For aqueous solution, explicitly state that H⁺ and OH⁻ compete with solute ions for discharge. Use gas tests when observations are requested.
- In an extraction comparison, explain the energy cost of melting and electrolysis and use supplied data; do not state that cryolite is the source of aluminium.
In plain language
The power supply pulls positive ions to the negative electrode and negative ions to the positive electrode. Once there, ions gain or lose electrons and become neutral atoms or molecules. Water adds competing ions in a solution, so aqueous rules differ from molten salts.
Self-check
Aqueous sodium chloride is electrolysed using inert electrodes. What gases are expected under AQA's general aqueous rules?
Show the answer
Hydrogen at the cathode because sodium is more reactive than hydrogen, and chlorine at the anode because chloride is a halide ion.
- Higher
Scope: Higher · §4.4.3.1 and §4.4.3.5 Higher-only electrode half-equations
Open revision guidance
What to know
- At a cathode (negative electrode), positive ions gain electrons and are reduced. At an anode (positive electrode), negative ions lose electrons and are oxidised. Electrons must appear on the side that accounts for the charge change.
- A half-equation represents one electrode process, not the whole cell reaction. The number of electrons must match the change in ionic charge and atoms must be conserved. Typical examples are 2H⁺ + 2e⁻ → H₂ at a cathode and 4OH⁻ → O₂ + 2H₂O + 4e⁻ at an anode.
What to do in an assessment
- For a metal ion Mⁿ⁺ forming a metal atom, write Mⁿ⁺ + n e⁻ → M; for hydrogen ions, write 2H⁺ + 2e⁻ → H₂. For a halide ion, show two halide ions forming a diatomic molecule and releasing two electrons (for example 2Cl⁻ → Cl₂ + 2e⁻). For hydroxide ions forming oxygen, balance atoms and charge as 4OH⁻ → O₂ + 2H₂O + 4e⁻.
- Complete or balance a supplied electrode half-equation, check atom count and total charge on each side, and label the electrode and oxidation/reduction.
In plain language
The equation balances the electrons that the power supply moves. A positive ion needs electrons to become neutral; a negative ion releases electrons when it becomes a neutral atom or molecule.
Self-check
Write the half-equation for Cu²⁺ becoming copper metal at the cathode.
Show the answer
Cu²⁺ + 2e⁻ → Cu. Two electrons balance the +2 charge; one copper atom is present on each side.
Topic group
Energy changes
- Core
Scope: Core · §4.5.1.1–4.5.1.2 energy transfer and profiles
Open revision guidance
What to know
- Energy is conserved: the total energy of the universe is unchanged by a reaction. If energy is transferred to the surroundings, the products have less energy than the reactants by the amount transferred. Exothermic reactions transfer energy to surroundings so their temperature rises; examples include combustion, many oxidation reactions and neutralisation. Self-heating cans and hand warmers use exothermic reactions. Endothermic reactions take in energy from surroundings, so their temperature falls; examples include thermal decomposition and citric acid reacting with sodium hydrogencarbonate. Some sports-injury cold packs use endothermic reactions. Evaluate an application using the information given about its benefits and disadvantages.
- Reacting particles must collide with each other with sufficient energy to react. The minimum energy required is the activation energy. A reaction profile has energy on the vertical axis and progress of reaction on the horizontal axis. Its curved path rises from reactants to a peak; activation energy is the vertical difference from reactant level to the peak. The product level is below reactants for an exothermic reaction and above for an endothermic reaction; the vertical separation is the overall energy change.
What to do in an assessment
- Draw a complete curved reaction profile with labelled energy and reaction-progress axes, reactant and product energy levels, a curve with a peak, an activation-energy arrow from reactants to the peak and an arrow showing the overall energy change between reactants and products. Use the relative levels to classify the reaction and explain why the activation energy is the minimum energy needed for a reaction to occur.
- For a practical temperature question, compare measured initial and maximum/minimum temperatures and connect the change to energy transfer to/from surroundings; evaluate whether a heat/cold application is effective using supplied information. AQA GCSE does not require a ΔH calculation here.
In plain language
A temperature rise means the reacting system released energy into its surroundings. A temperature fall means it took energy from them. On an energy profile, the hump is the minimum energy barrier that reacting particles must get over.
Self-check
A reaction mixture cools from 21°C to 15°C. Is the reaction exothermic or endothermic?
Show the answer
Endothermic: the reaction takes in energy from the surroundings, so the mixture and nearby surroundings cool.
- Higher
Scope: Higher · §4.5.1.3 Higher-only bond-energy calculations
Open revision guidance
What to know
- Breaking bonds requires energy; forming bonds releases energy. The overall change is energy needed to break reactant bonds minus energy released when product bonds form.
- If more energy is released on bond formation than required for bond breaking, the reaction is exothermic. If breaking bonds requires more energy, it is endothermic.
What to do in an assessment
- Use the displayed formulae to count every bond broken and formed, multiply each count by the supplied bond energy, then calculate broken minus formed. Include the energy unit from the data and interpret the sign in words.
In plain language
A reaction first pays the energy cost of separating atoms, then gains energy when new bonds form. Compare the two totals: the larger side determines whether energy leaves the reaction or must be supplied.
Self-check
Breaking bonds needs 900 kJ and making product bonds releases 1100 kJ. What is the overall change and type?
Show the answer
900 − 1100 = −200 kJ, so 200 kJ is released overall and the reaction is exothermic.
- Core
Scope: Core · §4.5.2.1–4.5.2.2 cells and fuel cells
Open revision guidance
What to know
- Cells contain chemicals that react to produce electricity. A simple cell can be made by connecting two different metals in contact with an electrolyte; its voltage depends on factors including the electrode and electrolyte. Batteries connect two or more cells in series to provide a greater voltage.
- A non-rechargeable cell stops when a reactant is used up; an alkaline battery is non-rechargeable. A rechargeable cell can reverse its chemical changes when an external electric current is supplied. A hydrogen fuel cell is supplied continuously by an external source of hydrogen and oxygen or air; hydrogen is oxidised electrochemically and the overall reaction forms water and produces a potential difference.
What to do in an assessment
- Interpret supplied data about relative metal reactivity and compare cell voltages; use the data to evaluate cell use against relevant criteria. For cells, batteries and hydrogen fuel cells, compare voltage, lifetime, reactant supply, rechargeability, emissions at point of use, storage and infrastructure, and state the criterion behind your recommendation.
- For a hydrogen cell, identify oxidation of hydrogen in the overall reaction and distinguish the continuous external fuel supply from a battery's stored reactants.
In plain language
Both technologies turn chemical changes into electrical energy. A battery stores the chemicals inside itself; a fuel cell needs fuel delivered while it operates. Neither comparison is complete without considering how the fuel or electricity is produced and supplied.
Self-check
Why does an alkaline battery stop producing a useful voltage, while a rechargeable battery can work again after charging?
Show the answer
In a non-rechargeable alkaline battery a reactant is used up and the reaction cannot be restored in normal use. Charging a rechargeable cell supplies an external current that reverses its chemical changes.
- Higher
Scope: Higher · §4.5.2.2 Higher-only fuel-cell half-equations
Open revision guidance
What to know
- The overall hydrogen fuel-cell change oxidises hydrogen and forms water: 2H₂ + O₂ → 2H₂O. At the hydrogen electrode, hydrogen loses electrons; at the oxygen electrode, oxygen gains those electrons to form water. The half-equations use the ions appropriate to the cell's electrolyte.
- For an acidic/proton-conducting cell, the oxidation half-equation is 2H₂ → 4H⁺ + 4e⁻ and the reduction half-equation is O₂ + 4H⁺ + 4e⁻ → 2H₂O. For an alkaline cell, the corresponding equations are 2H₂ + 4OH⁻ → 4H₂O + 4e⁻ and O₂ + 2H₂O + 4e⁻ → 4OH⁻. In either case electrons cancel when the half-equations are added to give the overall reaction.
What to do in an assessment
- Use the electrolyte/conditions stated in a question to decide whether H⁺ or OH⁻ appears. Write oxidation at the hydrogen electrode and reduction at the oxygen electrode; balance atoms and charge in each half-equation, then confirm that electrons lost equal electrons gained when the halves are added.
- Do not mix acidic and alkaline half-equations in one cell reaction; use one internally consistent pair and check that their sum gives hydrogen and oxygen forming water.
In plain language
A half-equation accounts for what happens at one side of the cell. It must keep both atoms and charge balanced; the question's cell conditions determine which species carry charge through the electrolyte.
Self-check
What must be true about the electron counts when the fuel-cell oxidation and reduction half-equations are added?
Show the answer
The total electrons released in oxidation must equal those accepted in reduction, so they cancel from the overall equation.
Topic group
Rate and extent of chemical change
- Core
Scope: Core · §4.6.1.1–4.6.1.4 rates and catalysts
Open revision guidance
What to know
- Mean rate = quantity of reactant used or product formed ÷ time; quantity may be mass in g or gas volume in cm³, so rate units may be g/s or cm³/s. Draw and interpret graphs of quantity of product formed or reactant used up against time. The curve's gradient represents rate: a steeper curve means a faster rate; a tangent at a point represents the rate at that instant. A catalyst changes rate but is not used up; it provides a different pathway with lower activation energy. A catalyst can be identified by its effect on rate and because it is not included in the balanced equation. Different reactions need different catalysts, and enzymes are catalysts in biological systems.
- Increasing concentration of reactants in solution, pressure of reacting gases or the surface-area-to-volume ratio of a solid reactant increases collision frequency and rate. Increasing temperature makes collisions more frequent and more energetic, so a greater proportion have at least the activation energy. Use simple proportional comparisons carefully: changing the number of particles in a given volume changes collision frequency, but an exact rate multiple should only be claimed when supported by the question's information. In a closed system at dynamic equilibrium, forward and reverse reactions continue at equal rates and concentrations stay constant, not necessarily equal.
What to do in an assessment
- Draw labelled axes (quantity of product formed or reactant used against time), plot data with appropriate scales and draw a smooth curve. Find a mean rate by dividing overall quantity change by elapsed time. Draw a tangent to show the instantaneous rate at a point; use its slope as the rate, while numerical tangent-gradient calculations are a separate Higher-tier skill.
- Explain how changing concentration, pressure, temperature or solid-piece size affects rate using collision theory and, for solid pieces, surface-area-to-volume ratio. Explain catalyst action by its alternative lower-activation-energy pathway and identify a catalyst from its rate effect and absence from the balanced equation. Do not assume a catalyst name unless the specification names it. For a rate experiment, identify the changed concentration and measured gas volume or colour/turbidity; control other variables and use repeats.
In plain language
A reaction happens only when particles collide in a successful way. More collisions or a greater chance that a collision has enough energy gives a faster reaction. A catalyst offers a lower step over the energy barrier, not a different overall set of reactants and products.
Self-check
Why does powdered calcium carbonate react faster than the same mass as large chips, if the acid concentration and temperature are unchanged?
Show the answer
Powder has more exposed surface area. More acid particles can collide with the solid per second, increasing successful collision frequency and rate.
- Higher
Scope: Higher · §4.6.1.1 Higher-only tangent-gradient calculations
Open revision guidance
What to know
- A tangent touches a curve at the chosen instant. Its gradient estimates the instantaneous rate of change at that time; it differs from the average gradient between two distant points. Calculating the tangent gradient at a specific time is Higher Tier.
What to do in an assessment
- Draw a straight tangent at the stated time, choose two widely separated points on that tangent, calculate rise ÷ run and carry graph units through. If a graph shows reactant remaining, a negative slope is a positive rate of disappearance.
In plain language
A curved line changes steepness. To find its rate at one moment, draw the straight line that just matches the curve at that point and calculate that line's slope.
Self-check
A tangent to a product-volume graph rises 24 cm³ over 12 s. What instantaneous rate does it represent?
Show the answer
24 ÷ 12 = 2.0 cm³/s at the point where the tangent touches the curve.
- Higher
Scope: Higher · §4.6.1.1 Higher-tier amount-of-substance rates
Open revision guidance
What to know
- For Higher Tier, reaction quantities are also expressed in moles and the rate unit is mol/s. Molar rate = moles of reactant used (or product formed) ÷ elapsed time in seconds.
- If the amount is given as a mass, first calculate n = mass ÷ Mr. Use the mole change over the stated interval and include mol/s; this is a mean rate unless a specific instant/tangent is requested.
What to do in an assessment
- Convert mass to moles with n = m/Mr when necessary, then divide the reacting amount in moles by time in seconds. Example: 0.20 mol is used in 40 s, so mean rate = 0.20 ÷ 40 = 0.0050 mol/s. Do not confuse mol/s with g/s or cm³/s.
In plain language
Count how many moles react during a time interval, then divide by the seconds in that interval. The unit is moles per second.
Self-check
A reaction uses 0.30 mol of reactant in 60 s. What is its mean rate in mol/s?
Show the answer
0.30 mol ÷ 60 s = 0.0050 mol/s.
- Core
Scope: Core · §4.6.2.1–4.6.2.3 reversible reactions and equilibrium
Open revision guidance
What to know
- A reversible reaction can proceed in both directions; changing conditions can reverse which direction occurs. The specified example is NH₄Cl(s) ⇌ NH₃(g) + HCl(g): heating drives decomposition to ammonia and hydrogen chloride; cooling recombines them to form ammonium chloride. If one direction is exothermic, the reverse is endothermic by the same energy amount. For example, blue hydrated copper sulfate ⇌ white anhydrous copper sulfate + water; dehydration is endothermic and hydration is exothermic.
- Dynamic equilibrium is reached in a closed system when forward and reverse reactions occur at the same rate. Concentrations are then constant, although both reactions continue and the concentrations need not be equal.
What to do in an assessment
- Check whether the system is closed and whether both reaction directions continue at equal rate before labelling a situation equilibrium. Constant concentrations do not mean equal concentrations. Use the ammonium-chloride and hydrated-copper-sulfate examples to show that condition changes can reverse a reaction.
- Use the energy direction of one reaction to identify the opposite energy change for its reverse; connect heat supplied/removed to the observed reversible product.
In plain language
Dynamic means the reactions have not stopped. Particles still change from reactants to products and back, but both changes happen at the same rate, so the amounts stay steady.
Self-check
At equilibrium, are there necessarily equal amounts of reactants and products?
Show the answer
No. Their concentrations are constant because the forward and reverse rates are equal; the actual amounts may be different.
- Higher
Scope: Higher · §4.6.2.4–4.6.2.7 Higher-only equilibrium changes
Open revision guidance
What to know
- Le Chatelier's Principle states that when a condition changes, an equilibrium system responds to counteract that change. The relative amounts of reactants and products at equilibrium depend on conditions. Increasing reactant concentration favours product formation; removing product favours the forward reaction until equilibrium is re-established.
- For temperature, treat heat as a reactant for an endothermic direction and a product for an exothermic direction. Increasing temperature increases the relative amount of products for an endothermic forward reaction and decreases it for an exothermic forward reaction; lowering temperature has the opposite effect.
- For gaseous equilibria, higher pressure favours the side with fewer gas molecules in the balanced equation; lower pressure favours the side with more. Count gas molecules from coefficients, not solid or liquid species. A catalyst changes how fast equilibrium is reached, not its position.
What to do in an assessment
- Mark the forward reaction as exothermic or endothermic, count gaseous molecules on each side (not coefficients for solids or liquids), then explain how the shift opposes the stated change.
- Use the supplied equilibrium data to support a qualitative prediction; do not claim pressure changes the position when both sides have equal gas-molecule counts.
In plain language
Imagine the system responding to a disturbance. Add a reactant and it uses more of it; remove product and it replaces it. Heat and pressure follow the same idea, but only after identifying whether heat is taken in or released and counting gas molecules.
Self-check
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat, what happens to the equilibrium amount of ammonia if pressure increases?
Show the answer
The equilibrium shifts right, toward the side with fewer gas molecules (2 rather than 4), so more ammonia is present at the new equilibrium.
Topic group
Organic chemistry
- Core
Scope: Core · §4.7.1.1–4.7.1.4 crude oil, fuels and cracking
Open revision guidance
What to know
- Crude oil is a finite resource found in rocks and formed from ancient biomass, mainly plankton buried in mud. It is a mixture of many compounds, most of which are hydrocarbons made only from carbon and hydrogen. Alkanes are saturated hydrocarbons with general formula CₙH₂ₙ₊₂; the first four are methane, ethane, propane and butane, for example ethane C₂H₆. Recognise an alkane from its formula or displayed structure.
- Fractional distillation separates fractions containing hydrocarbons with similar numbers of carbon atoms: crude oil is heated to vaporise, then vapours cool and condense at different heights because fractions have different boiling points. Fractions provide petrol, diesel, kerosene, heavy fuel oil and liquefied petroleum gas (LPG), as well as petrochemical feedstock for solvents, lubricants, polymers and detergents.
- As hydrocarbon molecules get larger, boiling point and viscosity rise while flammability falls. Combustion releases energy and oxidises the fuel's carbon and hydrogen; complete combustion produces carbon dioxide and water. Cracking breaks long hydrocarbons into smaller, more useful molecules because demand for small-molecule fuels is high. Catalytic cracking heats hydrocarbon vapour over a hot catalyst; steam cracking mixes hydrocarbon vapour with steam and heats strongly. Products include shorter alkanes and alkenes. Alkenes are more reactive than alkanes; bromine water changes from orange/brown to colourless. Short alkanes are useful fuels; alkenes are feedstocks for polymers and other chemicals used in modern life.
What to do in an assessment
- Recognise alkanes from molecular formulae or structural/displayed forms using CₙH₂ₙ₊₂; recall only methane, ethane, propane and butane by name. Recall the direction of the three specified hydrocarbon trends as molecular size increases.
- Balance complete-combustion equations from a formula by balancing carbon, then hydrogen, then oxygen. For cracking, conserve every atom in the supplied equation. Explain its usefulness by linking high demand for small fuel molecules with alkene feedstocks for polymers and other chemical manufacture.
- Compare the two cracking routes: catalytic cracking heats hydrocarbon vapour over a catalyst, while steam cracking heats hydrocarbon vapour mixed with steam. For the alkene test, state the bromine-water colour change and connect it to the C=C bond. Explain fractional distillation through evaporation and condensation, not chemical reaction.
In plain language
Crude oil is a mixture, not one compound. Distillation sorts it by boiling behaviour; cracking then changes some large hydrocarbon molecules into smaller useful ones. This makes a finite resource more flexible, but it does not make it renewable.
Self-check
Why can fractional distillation separate crude oil without chemically changing every hydrocarbon?
Show the answer
It is a physical separation: hydrocarbons evaporate and condense at different temperatures according to their boiling points, so the molecules remain the same substances.
- Core
Scope: Core · §4.7.2.1–4.7.2.4 alkenes, alcohols, carboxylic acids and esters
Open revision guidance
What to know
- Alkenes are unsaturated hydrocarbons with a C=C functional group and general formula CₙH₂ₙ; they have two fewer hydrogen atoms than the corresponding alkane. The first four are ethene, propene, butene and pentene. Hydrogen, water or a halogen adds across the double bond, making it single; alkenes are more reactive than alkanes and burn in air with smoky flames. The first four alcohols are methanol, ethanol, propanol and butanol, each with the –OH group. They react with sodium to form a sodium alkoxide and hydrogen, burn in air, dissolve in water and react with an oxidising agent to form a carboxylic acid. Main uses include fuels, solvents and chemical manufacture. Yeast ferments sugar solution to aqueous ethanol in warm, oxygen-free conditions.
- The first four carboxylic acids are methanoic, ethanoic, propanoic and butanoic acid; their functional group is –COOH. They dissolve in water, react with carbonates to form a salt, water and carbon dioxide, and react with alcohols to form esters and water. Ethyl ethanoate is the named ester example; esterification is reversible.
What to do in an assessment
- Recognise alkenes from the C=C group, names or formulae; draw fully displayed structures for ethene, propene, but-1-ene and pent-1-ene. In an addition reaction change C=C to C–C and add the reagent atoms across that bond: H₂ adds one H to each carbon; H₂O adds H and OH (ethene forms ethanol); Cl₂, Br₂ or I₂ adds one halogen to each carbon (ethene + Br₂ forms CH₂BrCH₂Br). Hydrogenation uses a nickel catalyst; hydration uses steam and a phosphoric-acid catalyst. Use the same rule to draw products for the first four alkenes.
- Recognise the first four alcohols by name or formula and identify –OH. Describe their reaction with sodium as sodium alkoxide plus hydrogen; combustion gives carbon dioxide and water, while oxidation with an oxidising agent forms a carboxylic acid. Describe dissolving in water and the warm, oxygen-free yeast fermentation of sugar to ethanol; give uses such as fuels, solvents and chemical manufacture. Balance combustion equations, but do not write balanced equations for the other alcohol reactions.
- Recognise carboxylic acids by their names/formulae and –COOH; describe their reactions with carbonates, water and alcohols. Identify alcohol + carboxylic acid → ester + water; ethyl ethanoate is the only ester name required. No balanced equations for carboxylic-acid reactions are required.
In plain language
A functional group is the reactive part of an organic molecule. An alkene's double bond can open so new atoms attach. Alcohol and acid groups make different reaction patterns; recognising the group is the first step to predicting the product.
Self-check
What happens to bromine water when it is shaken with an alkene, and what structural feature explains the test?
Show the answer
It changes from orange/brown to colourless because the alkene's C=C double bond undergoes addition with bromine.
- Core
Scope: Core · §4.7.3.1 and §4.7.3.4 addition polymers and natural polymers
Open revision guidance
What to know
- In addition polymerisation, many alkene monomers join when their C=C bonds open. The polymer repeating unit contains the same atoms as the monomer because no other molecule is formed; for example, ethene forms poly(ethene) with repeat unit [–CH₂–CH₂–]ₙ and propene forms poly(propene) with [–CH₂–CH(CH₃)–]ₙ.
- DNA is a large molecule essential for life and encodes genetic instructions for the development and functioning of living organisms and viruses. Most DNA has two polymer chains, made from four different nucleotide monomers, arranged as a double helix. Other naturally occurring polymers include proteins, starch and cellulose: amino acids form proteins, glucose monomers form starch and cellulose, and nucleotides form DNA.
What to do in an assessment
- Recognise an alkene monomer from its C=C group and an addition-polymer repeating unit from its chain and continuation bonds. To form the repeat unit, open the monomer double bond, connect the carbon chain and show continuation bonds at both ends; check that every monomer atom appears in the repeat unit. Practise with ethene and propene.
- Explain DNA's role as genetic instructions and identify the four nucleotide types as its monomer set (individual nucleotide/base names are not required). Identify amino acids as protein monomers and glucose as the monomer for starch and cellulose.
- When comparing polymers, identify monomer type and whether the reaction is addition or condensation; do not confuse a repeat-unit diagram with a full chain of fixed length.
In plain language
A polymer is a long molecule built from repeating smaller units. In addition polymerisation, the alkene double bond becomes links in the chain. Biological polymers also use repeated building blocks, but their monomers are not all the same chemical type.
Self-check
Which monomer forms cellulose, and how does an addition polymer differ in what happens to atoms during joining?
Show the answer
Cellulose is made from glucose monomers. In addition polymerisation the monomer atoms are all retained in the repeating unit; no small molecule is lost.
- Higher
Scope: Higher · §4.7.3.2–4.7.3.3 Higher-only condensation polymers and amino acids
Open revision guidance
What to know
- Condensation polymerisation joins monomers that each have two functional groups and usually eliminates a small molecule such as water. The simplest polyester is made from two different monomers, each with two of the same functional groups: ethanediol, HO–CH₂–CH₂–OH, and hexanedioic acid, HOOC–(CH₂)₄–COOH. Their –OH and –COOH groups form ester links in a repeating polyester chain; the repeat is –O–CH₂–CH₂–O–CO–(CH₂)₄–CO–.
- An amino acid has an amine and a carboxylic-acid functional group. For example, glycine is H₂N–CH₂–COOH. Amino acids join by condensation, eliminating water and forming peptide links in a polypeptide; different amino acids can occur in one chain to produce proteins.
What to do in an assessment
- For ethanediol + hexanedioic acid, match each alcohol –OH with a carboxylic-acid –COOH, form an ester link and eliminate water; identify the polyester repeat as –O–CH₂–CH₂–O–CO–(CH₂)₄–CO–. Check the repeat accounts for the atoms left after condensation and show continuation bonds to neighbouring units.
- Use glycine's H₂N–CH₂–COOH structure to identify its two functional groups. Explain how the amine group of one amino acid and carboxyl group of another join, with water lost, to make a peptide link and polypeptide. Recognise that different amino acids can be combined in the same protein chain.
In plain language
In condensation, two functional groups connect and a small molecule is removed as the link forms. A diol plus a diacid can build a polyester; amino acids can build a chain that is a polypeptide.
Self-check
What distinguishes a condensation polymer from an addition polymer in the monomer-to-repeat-unit atom count?
Show the answer
Condensation removes a small molecule, often water, when links form, so the repeat unit does not retain all atoms from the monomers. Addition retains the monomer atoms.
Topic group
Chemical analysis
- Core
Scope: Core · §4.8.1.1–4.8.1.3 purity, formulations and chromatography
Open revision guidance
What to know
- In chemistry, a pure substance is one element or compound, not a mixture. Pure elements and compounds melt and boil at specific temperatures; melting/boiling-point data can distinguish them from impure mixtures, which usually change state over a range. In everyday language, “pure” can mean unadulterated and natural (for example, milk), even though such a product is a mixture in chemistry. A formulation is a useful product made by mixing components in carefully measured quantities to give required properties. Examples include fuels, cleaning agents, paints, medicines, alloys, fertilisers and foods; the proprietary product's ingredient names are not required.
- Paper chromatography has a stationary phase (paper) and a mobile phase (solvent). Substances distribute differently between the phases, producing separated spots. Rf = distance moved by the centre of the spot from the origin ÷ distance moved by the solvent front. Different compounds may have different Rf values in different solvents; a pure compound produces one spot in all solvents, while a mixture may separate into different spots depending on the solvent.
What to do in an assessment
- Keep the origin above the solvent, mark the solvent front before it evaporates, measure from the baseline and use the centre of each spot. Calculate Rf with the distance moved by the substance divided by the distance moved by the solvent; it has no unit. Report an appropriate number of significant figures and compare Rf values only under the same solvent and conditions. To assess purity, repeat with different solvents: a pure compound produces a single spot in every solvent; a mixture may split into several spots.
- Identify a formulation from information about its carefully measured components, intended function and product properties; examples include a paint or medicine, not its proprietary recipe. Use melting/boiling-point data and chromatograms as evidence of purity, while avoiding overclaim from only one test.
In plain language
A solvent carries mixture components up paper. A substance that is more attracted to the moving solvent travels farther; one held more strongly by the paper travels less. Rf compares a spot's journey with the solvent's journey.
Self-check
A dye spot travels 3.0 cm while the solvent front travels 7.5 cm. What is its Rf?
Show the answer
Rf = 3.0 ÷ 7.5 = 0.40. It has no unit.
- Core
Scope: Core · §4.8.2.1–4.8.2.4 common gas tests
Open revision guidance
What to know
- Hydrogen gives a squeaky pop with a lit splint at the mouth of a test tube. Oxygen relights a glowing splint. Carbon dioxide turns limewater milky/cloudy. Chlorine bleaches damp litmus paper white.
- A positive test is the observation; name the gas only when its characteristic result is observed. Handle gases using the specified safe apparatus and do not smell unknown gases.
What to do in an assessment
- State the test, the expected positive observation and the inference. If a result differs, report that the test does not support the gas rather than forcing an identification.
In plain language
Each test looks for a different chemical behaviour: hydrogen burns rapidly, oxygen supports glowing combustion, carbon dioxide reacts with limewater, and chlorine bleaches damp dye.
Self-check
A glowing splint inserted into a sample relights. Which gas does the test support?
Show the answer
Oxygen. The characteristic result is a glowing splint relighting.
- Core
Scope: Core · §4.8.3.1–4.8.3.7 ion and instrumental tests
Open revision guidance
What to know
- Flame colours: lithium crimson, sodium yellow, potassium lilac, calcium orange-red and copper green; in mixtures, some colours may be masked. Sodium hydroxide gives white precipitates for Al³⁺, Ca²⁺ and Mg²⁺, with only aluminium hydroxide dissolving in excess; Cu²⁺ is blue, Fe²⁺ green and Fe³⁺ brown.
- Write balanced equations for insoluble metal hydroxides, for example AlCl₃ + 3NaOH → Al(OH)₃(s) + 3NaCl; CaCl₂ + 2NaOH → Ca(OH)₂(s) + 2NaCl; MgSO₄ + 2NaOH → Mg(OH)₂(s) + Na₂SO₄; CuCl₂ + 2NaOH → Cu(OH)₂(s) + 2NaCl; FeSO₄ + 2NaOH → Fe(OH)₂(s) + Na₂SO₄; and FeCl₃ + 3NaOH → Fe(OH)₃(s) + 3NaCl. The corresponding net ionic equations show the hydroxide precipitate; AQA does not require an equation for sodium aluminate.
- Carbonates react with dilute acid to release CO₂, identified by turning limewater cloudy. Halide ions with silver nitrate solution in dilute nitric acid give silver chloride (white), silver bromide (cream) and silver iodide (yellow). Sulfate ions with barium chloride solution in dilute hydrochloric acid give a white precipitate. Instrumental methods are accurate, sensitive and rapid. In flame-emission spectroscopy a sample is put in a flame and the emitted light passes through a spectroscope to form a line spectrum. A reference spectrum identifies metal ions; supplied reference data can also be used to measure their concentrations.
What to do in an assessment
- Perform a suitable test, record colour/state, compare with specified outcomes and conclude. Test a carbonate using dilute acid and confirm CO₂ with limewater; acidify halide samples with dilute nitric acid before silver nitrate, and sulfate samples with dilute hydrochloric acid before barium chloride. When asked for hydroxide precipitation equations, write and balance the full symbol or net ionic equation as required and include (s) for the precipitate; do not write an equation for sodium aluminate.
- For flame-emission data, compare the unknown's line positions with the reference spectrum to identify ions, then interpret supplied chart/table reference data to determine concentration. Explain advantages over chemical tests using accuracy, sensitivity and speed; a single flame colour in a mixture may be masked.
In plain language
A precipitate is a solid made when ions in solution combine into an insoluble compound. Its colour is evidence about which ions were present. More than one test gives a safer identification than relying on one ambiguous colour.
Self-check
An unknown solution forms a green precipitate with sodium hydroxide. Which ion is supported, and what additional note distinguishes it from iron(III)?
Show the answer
Iron(II) is supported. Iron(III) gives a brown precipitate under this test; copper(II) gives blue.
Topic group
Chemistry of the atmosphere
- Core
Scope: Core · §4.9.1.1–4.9.2.4 atmospheric composition and climate
Open revision guidance
What to know
- For about 200 million years the atmosphere has been roughly four-fifths nitrogen and one-fifth oxygen, with small amounts of carbon dioxide, water vapour and noble gases. Evidence about the atmosphere's formation is limited by the 4.6-billion-year timescale. One theory is that intense volcanic activity during Earth's first billion years released gases that formed an early atmosphere, mainly carbon dioxide with little or no oxygen, like the present atmospheres of Mars and Venus. As Earth cooled, water vapour condensed to form oceans; carbon dioxide dissolved in ocean water and carbonates precipitated as sediments. Volcanoes also released nitrogen, which gradually accumulated, and there may have been small amounts of methane and ammonia.
- Algae produced oxygen by photosynthesis about 2.7 billion years ago; plants later evolved and oxygen gradually increased over the next billion years to a level that enabled animals to evolve. Photosynthesis by algae and plants can be represented as 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ and lowered atmospheric carbon dioxide. Carbon dioxide also became locked in sedimentary rocks and fossil fuels: carbonate shells/skeletons of marine organisms accumulated as limestone; buried vegetation formed coal; buried plankton formed crude oil and natural gas over geological time.
- Water vapour, carbon dioxide and methane are greenhouse gases. Incoming short-wavelength solar radiation passes through the atmosphere and warms Earth's surface. The surface emits long-wavelength infrared radiation; greenhouse gases absorb some outgoing long-wave radiation and re-emit it in all directions, including back towards the surface, warming the lower atmosphere and surface. Some energy still escapes to space; the natural greenhouse effect keeps Earth warm enough to support life.
- Human sources include burning fossil fuels and deforestation for carbon dioxide, and cattle, rice farming, landfill and fossil-fuel extraction for methane. Based on peer-reviewed evidence, many scientists believe human activities will warm Earth's surface, but climate is complex and models have uncertainties; media accounts may use simplified models or only part of the evidence and may be biased. Potential effects include altered rainfall/drought, more extreme weather, rising sea levels, and ecosystem/food/water disruption. A carbon footprint is the total carbon dioxide and other greenhouse gases emitted over a product, service or event's full life cycle. Reducing carbon dioxide and methane emissions can lower it, but cost, technology, demand and lifestyle may limit action.
What to do in an assessment
- For an early-atmosphere question, distinguish a proposed theory from evidence that supports it; connect volcanic gases, cooling/ocean formation, photosynthesis and carbon storage in rocks/fossil fuels to changing gas proportions. Explain how limestone, coal, crude oil and natural gas deposits formed.
- For climate claims, assess evidence quality, peer review, uncertainty and possible media bias. Describe four possible impacts, then discuss their scale, risk and environmental implications. Describe concrete carbon-footprint actions: replace fossil-fuel electricity with renewable generation, insulate buildings to reduce fuel burned for heating, use public transport or walk instead of making individual car journeys, and protect or plant trees so carbon is stored rather than released by deforestation. Methane emissions can be reduced by collecting landfill gas instead of releasing it and by reducing leaks from natural-gas equipment. For each action, explain the emission source it reduces, then discuss cost, technology, infrastructure or lifestyle limitations; do not just write 'reduce the carbon footprint'.
In plain language
Earth's atmosphere changed as the planet cooled and living things evolved. Some gases dissolved or became locked into rocks and fuels; photosynthesis raised oxygen. Today's climate evidence combines observations and models, so conclusions should reflect both the evidence and its uncertainty.
Self-check
Name two processes that lowered atmospheric carbon dioxide and one process that increased oxygen.
Show the answer
CO₂ fell through photosynthesis and formation of carbonate sedimentary rocks or fossil fuels. Photosynthesis by algae and plants increased oxygen.
- Core
Scope: Core · §4.9.3.1–4.9.3.2 atmospheric pollutants
Open revision guidance
What to know
- Complete combustion of carbon-containing fuel forms carbon dioxide; hydrogen forms water. Limited oxygen can form toxic carbon monoxide or soot, sulfur impurities form sulfur dioxide, and high-temperature reactions involving nitrogen and oxygen form nitrogen oxides.
- Carbon monoxide is colourless and odourless and can prevent oxygen transport in the body. Sulfur dioxide and nitrogen oxides cause respiratory problems and acid rain; particulates can harm health and cause global dimming. Unburned hydrocarbons can also form particulates.
What to do in an assessment
- Use fuel composition and combustion conditions to predict possible products, then connect each pollutant to an effect. Do not say every pollutant forms in every flame; use the question's oxygen supply and fuel information.
In plain language
A fuel's atoms and burning conditions determine what leaves the exhaust. With enough oxygen, carbon mainly becomes carbon dioxide; with too little, dangerous carbon monoxide or carbon particles can form. Sulfur and nitrogen oxides cause different environmental harms.
Self-check
Why might a fuel burner producing carbon monoxide be especially dangerous indoors?
Show the answer
Carbon monoxide is toxic, colourless and odourless, so people may not notice it. Inadequate ventilation can allow it to build up.
Topic group
Using resources
- Core
Scope: Core · §4.10.1.1–4.10.1.3 resources, potable water and wastewater
Open revision guidance
What to know
- People use resources for warmth, shelter, food and transport. Natural resources include water, timber, cotton, crops and crude oil; natural products can be supplemented or replaced by agricultural and synthetic products—for example, cultivated crops/timber can supplement wild harvest, and synthetic fibres or synthetic rubber can replace some natural fibres/rubber. Finite resources are limited; renewable resources are replaced naturally on a relevant timescale. Sustainable development meets current needs without compromising future generations' ability to meet theirs.
- Potable water is safe to drink, with sufficiently low dissolved salts and microbes; it is not chemically pure because it contains dissolved substances. Treatment depends on available supplies and local conditions. In the UK most potable water comes from fresh water collected from rainfall in the ground, lakes and rivers: choose an appropriate source, pass water through filter beds and sterilise it using chlorine, ozone or ultraviolet light. Salt water is harder to make potable because desalination by distillation or membrane processes such as reverse osmosis removes dissolved salts and uses large amounts of energy.
- Wastewater treatment screens solids and grit, allows sedimentation into sludge and effluent, digests sludge anaerobically and treats effluent aerobically. Sewage/agricultural waste water requires organic matter and harmful microbes removed; industrial waste water may also contain harmful chemicals. Relative ease is source-dependent, but ground/fresh water usually needs less treatment than waste water, and salty water needs energy-intensive desalination.
What to do in an assessment
- For a given source, compare how easy potable water is to obtain from waste water, ground/fresh water and salt water, citing contaminants and treatment: fresh/ground water is filtered and sterilised; waste water needs screening, sedimentation, sludge digestion and aerobic effluent treatment; salt water needs energy-intensive distillation or reverse osmosis.
- Distinguish finite from renewable resources using information given. Extract and interpret resource information from charts, graphs and tables; translate graph data to numbers and use orders of magnitude to judge the significance of the quantities. Choose treatment based on source and contaminants, and distinguish measured resource data from value judgements.
In plain language
Safe drinking water is not the same as pure water. The goal is to remove enough harmful microbes and dissolved material for the water to be safe. Salt water takes more processing than fresh water, while sewage needs biological treatment as well as separating solids.
Self-check
Why can potable water contain dissolved substances and still not be pure water?
Show the answer
Potable means safe to drink, not chemically pure. It may retain harmless dissolved minerals or salts while having sufficiently low levels of harmful microbes and contaminants.
- Higher
Scope: Higher · §4.10.1.4 Higher-only biological extraction of copper
Open revision guidance
What to know
- Phytomining uses plants to absorb metal compounds; harvested plants are burned and the ash contains metal compounds. Bioleaching uses bacteria to produce a leachate containing metal compounds.
- Copper can then be recovered from solutions by displacement with scrap iron or electrolysis. The approaches can use low-grade ores and reduce the amount of rock moved and discarded compared with traditional mining.
What to do in an assessment
- Compare the supplied land, time, energy, yield, cost and environmental data; state both advantages and limits. Explain the recovery step after the copper compound enters solution.
In plain language
Plants or bacteria help gather copper compounds from material that is not rich enough for ordinary mining to be attractive. The metal still has to be separated from the resulting ash or solution.
Self-check
Which method uses living plants, and what happens after harvesting?
Show the answer
Phytomining. The plants are harvested and burned; their ash contains metal compounds that can be processed to obtain copper.
- Core
Scope: Core · §4.10.2.1–4.10.3.3 life-cycle assessment and materials
Open revision guidance
What to know
- A life-cycle assessment considers extracting/processing raw materials, manufacture/packaging, use/operation and end-of-life disposal, including transport/distribution at each stage. It compares the same functional task across a product's whole life. Water, energy, resources and waste can be quantified; assigning values to pollutant effects involves value judgements. Selective/abbreviated studies can be misused to support a predetermined claim if they omit an inconvenient stage.
- For the required paper-versus-plastic shopping-bag comparison, relevant evidence may include raw-material source and extraction, manufacturing energy/water, bag mass and transport, number of reuses, and end-of-life reuse/recycling/disposal and waste. The conclusion depends on the supplied data, local disposal and how many times each bag is used; there is not one universal winner independent of assumptions.
- Reduce, reuse and recycle to conserve limited materials and energy. Metals, glass, building materials, clay ceramics and most plastics use limited raw materials; mining and quarrying affect the environment. Glass bottles can be reused, crushed and melted into different glass products; metals can be melted, recast or reformed, and scrap steel can be added to blast-furnace iron to reduce new iron extraction. The separation needed for recycling depends on the material and the properties required of the final product.
- Corrosion is the destruction of materials by chemical reactions with substances in the environment; rusting is corrosion of iron, and both air and water are needed. A dry iron nail kept in air without water and an iron nail in boiled water protected from air by oil should not rust; an iron nail exposed to both air and water does. Grease, paint and electroplating form barrier coatings; aluminium's oxide coat protects it. Galvanising iron with more-reactive zinc gives sacrificial protection because zinc corrodes preferentially.
- Most metals used in everyday life are alloys. Bronze is copper/tin and is used for medals or statues; brass is copper/zinc and is used for fittings or instruments. Gold jewellery is usually an alloy with silver, copper and zinc; 24-carat gold is pure and 18-carat is 75% gold. Steels are iron alloys with carbon and other metals: high-carbon steel is strong but brittle and suits cutting tools; low-carbon steel is softer/easier to shape and suits car bodies; chromium/nickel stainless steel is hard, corrosion-resistant and suits cutlery. Low-density aluminium alloys are used in aircraft.
- Soda-lime glass is made by heating sand, sodium carbonate and limestone; borosilicate glass uses sand and boron trioxide and melts at a higher temperature. Clay ceramics such as pottery and bricks are shaped from wet clay and heated in a furnace. Polymer properties depend on monomer and production conditions: high-pressure production makes more branched low-density poly(ethene), while a catalyst at lower pressure makes straighter, more densely packed high-density poly(ethene); both start from ethene. Thermosoftening polymers melt when heated; thermosetting polymers do not because covalent links hold chains together. A composite has a matrix/binder that surrounds and binds reinforcing fibres/fragments; examples include glass-fibre composite, carbon-fibre composite and reinforced concrete.
What to do in an assessment
- For the paper/plastic bag LCA, use a comparison table for the same number/capacity of shopping trips; compare source/extraction, manufacture, transport/distribution, reuse, and disposal/recycling using supplied energy, water, resource, waste and pollutant data. State missing stages, assumptions (especially reuse count/local disposal), uncertainty and value judgements before deciding which option has lower impact.
- For a rusting investigation, compare iron exposed to both air and water with controlled conditions that exclude water or air; use the observations to show both are necessary. For sacrificial protection, compare metal reactivity and explain why the more reactive coating corrodes first.
- For alloy/material choices, connect composition and structure to supplied quantitative properties, compare values using consistent units, explain the link to use, and select a suitable material from the evidence. Recall a use of each specified alloy and evaluate unfamiliar alloy compositions from supplied information.
- Explain the polymer contrast structurally: thermosoftening polymers have separate chains held to neighbouring chains by intermolecular forces, with no covalent cross-links between chains. Heating overcomes these attractions so chains can move past one another and the polymer melts; the covalent bonds along each chain remain. Thermosetting polymers have covalent cross-links joining chains into a network, preventing this movement when heated, so they do not melt like a thermosoftening polymer.
In plain language
A product's impact does not end at the factory. Raw materials, transport, energy during use and disposal all count. One product can look better if a study leaves out the stage where it has the largest impact.
Self-check
A paper-bag LCA counts manufacture but ignores transport and disposal. Why is its conclusion incomplete?
Show the answer
It has omitted life-cycle stages. A fair comparison considers extraction, processing, manufacture, distribution, use and end-of-life for both products, and should disclose uncertain or qualitative impacts.
- Core
Scope: Core · §4.10.4.1–4.10.4.2 Haber process and NPK fertilisers
Open revision guidance
What to know
- The Haber process manufactures ammonia for nitrogen-based fertilisers: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Nitrogen is obtained from air and hydrogen commonly from natural gas. Purified gases pass over an iron catalyst at about 450°C and 200 atmospheres. The reaction is reversible, so some ammonia decomposes back into nitrogen and hydrogen. The reaction mixture is cooled so ammonia liquefies and is removed; unreacted nitrogen and hydrogen are recycled.
- NPK fertilisers are formulations of salts containing compounds of nitrogen, phosphorus and potassium in chosen percentages. Ammonia can make ammonium salts and nitric acid. Potassium chloride, potassium sulfate and phosphate rock are mined; phosphate rock cannot be used directly as a fertiliser. Treatment converts it into soluble fertiliser salts: nitric acid gives calcium nitrate; sulfuric acid gives calcium sulfate and calcium dihydrogen phosphate; phosphoric acid gives calcium dihydrogen phosphate.
What to do in an assessment
- Write and balance N₂ + 3H₂ ⇌ 2NH₃; identify feedstocks, an iron catalyst, about 450°C and 200 atmospheres, cooling/liquefaction and removal of ammonia, and recycling of unreacted gases.
- Prepare an ammonium salt by neutralising aqueous ammonia with the appropriate dilute acid: for ammonium sulfate, 2NH₃(aq) + H₂SO₄(aq) → (NH₄)₂SO₄(aq). Measure ammonia solution with a pipette, add indicator and titrate with acid from a burette to establish the neutralising volume. Repeat using the measured volumes without indicator so it does not contaminate the salt. Gently evaporate some water, leave the concentrated solution to crystallise, then filter and dry the crystals; do not strongly heat to dryness. Use eye protection and avoid inhaling ammonia. Unlike insoluble copper oxide, excess dissolved ammonia cannot be removed by filtration, which is why a measured neutralisation is appropriate.
- Name the salts from each phosphate-rock treatment, then compare industrial production with laboratory preparation of the same compound using supplied scale, purity, waste or energy information.
In plain language
A factory combines nitrogen and hydrogen and removes ammonia as it forms, then reuses the gases that did not react. Fertilisers combine plant nutrients in measured proportions; phosphate rock must be processed into soluble salts.
Self-check
What are the main Haber-process conditions and what happens to ammonia and the unreacted gases after the reaction mixture leaves the reactor?
Show the answer
The mixture passes over an iron catalyst at about 450°C and 200 atmospheres. Cooling liquefies ammonia so it can be removed; nitrogen and hydrogen that did not react are recycled.
- Higher
Scope: Higher · §4.10.4.1 Higher-only Haber equilibrium and commercial trade-offs
Open revision guidance
What to know
- For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the product side has fewer gas molecules (2 vs 4), so higher pressure increases rate and shifts equilibrium toward ammonia. The forward reaction is exothermic: lower temperature increases equilibrium ammonia yield, while higher temperature increases rate but reduces equilibrium yield. An iron catalyst increases both rates and reaches equilibrium sooner without changing its position.
- Commercial conditions are a compromise: about 450°C gives a useful rate while retaining a reasonable yield; about 200 atmospheres favours ammonia but avoids the much higher plant/equipment cost, energy demand and safety burden of extreme pressures. Nitrogen is available from air; hydrogen is commonly obtained from natural gas, whose cost and supply and the energy required for processing matter. Rate, equilibrium position, raw-material availability and energy/equipment costs all influence the chosen conditions.
What to do in an assessment
- Read a rate-versus-condition graph by identifying the axes, units, controlled variables and trend; explain how increasing temperature, pressure or adding an iron catalyst changes rate, without claiming that the catalyst shifts equilibrium. For equilibrium questions, count gaseous molecules and use whether the forward reaction is exothermic to predict which way the position moves.
- Explain the trade-off for commercial Haber conditions: higher pressure favours the side with fewer molecules; lower temperature favours the exothermic product direction; faster rate often favours higher temperature/pressure and catalyst. Relate the actual compromise (about 450°C, 200 atm, iron) to production rate, yield, raw-material supply, energy, equipment cost and safety.
In plain language
The factory does not simply choose the setting that makes the most ammonia at equilibrium. It needs ammonia quickly, and extreme pressure or a very low temperature can make production impractical or expensive.
Self-check
Why does an iron catalyst not increase the final equilibrium amount of ammonia even though it speeds production?
Show the answer
It speeds both forward and reverse reactions by the same principle, so equilibrium is reached sooner but its position and final yield do not change.
Topic group
Required practicals and skills
- Core
Scope: Core · §8.1 apparatus and techniques AT1–AT8
Open revision guidance
What to know
- All students taking GCSE Chemistry are expected to carry out the required practical activities. AT1: use apparatus to measure mass, time, temperature and liquid/gas volume accurately and record results. AT2: safely use heating devices/techniques, including a Bunsen burner and water bath or electric heater. AT3: choose apparatus/reagents and techniques to conduct and monitor reactions, including measuring pH in different situations. AT4: safely purify/separate mixtures by evaporation, filtration, crystallisation, chromatography and distillation.
- AT5: make and record observations, including temperature changes and reaction rates measured by gas production or colour change. AT6: safely handle gases, liquids and solids; mix reagents carefully under controlled conditions to investigate chemical changes/products. AT7: select apparatus/techniques to draw, set up and use electrochemical cells to separate or produce elements and compounds. AT8 (Chemistry only): use qualitative reagents/techniques to identify unknown samples/products by gas, flame and precipitation tests, and determine strong-acid/strong-alkali concentrations.
- Safety overrides other considerations: centres must carry out full risk assessments for practical work. Written examinations assess practical knowledge and skills; practical questions are not a separate practical examination. Use appropriate scientific diagrams to show apparatus/procedure where helpful. AT1–AT7 are shared with Combined Science, and AT8 is Chemistry-only.
What to do in an assessment
- Match apparatus to the measured quantity and required precision: balance for mass, thermometer for temperature, stopwatch for time, graduated cylinder/pipette/burette for liquid volume, gas syringe or displacement equipment for gas, and a suitable probe for pH. Set up heating, separation and electrochemical equipment appropriate to the task.
- Describe a clear sequence that another student could repeat; draw and label the apparatus, identify controlled and measured variables, select reagents and measurement techniques, record observations accurately, and include a relevant hazard control and risk-reduction step. Choose a separation method to suit properties such as particle size, solubility or boiling point.
In plain language
Practical questions test whether you understand why a piece of equipment or step is there. A method is strongest when another student could repeat it safely and collect the same kind of evidence.
Self-check
A student must measure a reacting volume accurately for a strong acid–alkali titration. Which apparatus and endpoint evidence are appropriate?
Show the answer
Use a burette for a measured, controlled delivery of one solution and a pipette for a fixed aliquot of the other; use a suitable indicator and record the endpoint titre. Repeat to obtain concordant readings.
- Core
Scope: Core · §8.2.1 Required Practical 1: soluble salt
Open revision guidance
What to know
- To prepare a pure, dry soluble salt, heat dilute acid with a Bunsen burner, then add an insoluble oxide or carbonate a little at a time, stirring and adding excess so all acid is used. Filter off the unreacted solid. Use a water bath or electric heater to evaporate some solvent, leave the concentrated solution to crystallise, then collect and dry the crystals. Example: warm sulfuric acid + excess copper oxide → copper sulfate solution + water; CuO + H₂SO₄ → CuSO₄ + H₂O.
- The method uses controlled heating, appropriate reagents, filtration and crystallisation; select and manipulate apparatus accurately, mix reagents carefully and follow risk controls. The salt name depends on the acid (hydrochloric → chloride, sulfuric → sulfate, nitric → nitrate) and the metal/base used.
What to do in an assessment
- Explain why insoluble solid is added in excess (ensure all acid reacts), why filtration follows (remove surplus solid), and why only some solvent is evaporated before cooling (crystals form as the solution cools/concentrates). Collect and dry the crystals to obtain the pure, dry product; avoid evaporating to complete dryness when crystallisation is intended.
- Choose apparatus and reagents for the specified salt, carry out each step in order, measure accurately, handle hot acid/solids safely and record the procedure. For the copper sulfate example, name the salt, write CuO + H₂SO₄ → CuSO₄ + H₂O and explain the separation.
In plain language
The insoluble base is a chemical sponge for the acid. Once extra solid remains, filter it away; the dissolved salt passes through. Removing some water lets salt crystals grow from the solution.
Self-check
Why is excess copper oxide filtered off before the salt solution is evaporated?
Show the answer
Copper oxide is insoluble, so excess solid can be removed by filtration. The dissolved copper salt remains in the filtrate and can crystallise as water is removed.
- Core
Scope: Core · §8.2.2 Required Practical 2: titration
Open revision guidance
What to know
- A titration determines the reacting volumes of strong acid and strong alkali. Rinse a pipette with one solution, transfer a fixed aliquot to a conical flask, rinse/fill a burette with the other solution and record its initial reading. Add a suitable indicator, deliver titrant while swirling, add it dropwise near the endpoint and record the final burette reading; titre = final − initial.
- Repeat until concordant titres are obtained and calculate their mean using only the concordant results. The practical develops accurate volume measurement, observations and safe apparatus manipulation. Higher-tier work can use reacting volumes and the known solution concentration to calculate an unknown mol/dm³ and g/dm³ concentration.
What to do in an assessment
- Read the burette at eye level, record initial/final readings consistently in decimal cm³, calculate titre by subtraction and compare repeats for concordance; report appropriate significant figures. Wear eye protection and handle acid/alkali carefully. For Higher concentration work, use the balanced equation coefficients and convert cm³ to dm³.
In plain language
The indicator signals that enough acid and alkali have reacted. A burette lets you add the second solution slowly near that point, so the volume needed is measured accurately.
Self-check
A burette starts at 0.40 cm³ and ends at 24.65 cm³. What is the titre?
Show the answer
24.65 − 0.40 = 24.25 cm³.
- Core
Scope: Core · §8.2.3 Required Practical 3: electrolysis
Open revision guidance
What to know
- Investigate what happens when aqueous solutions are electrolysed using inert electrodes; the activity must involve developing a hypothesis. Connect the proposed products to the ions and discharge rules, assemble an electrochemical cell and record observations at both electrodes. Test gases where appropriate: hydrogen gives a pop with a lit splint, oxygen relights a glowing splint and chlorine bleaches damp litmus paper.
- The activity uses reaction-monitoring and electrochemical-cell equipment (AT3/AT7), and qualitative tests including gas tests (AT8). Select apparatus/reagents, plan a controlled procedure, manipulate equipment safely, and record observations and measurements.
What to do in an assessment
- Use the ions/electrolysis rules to formulate a testable prediction; choose the solution as the independent variable and hold electrode type, exposed area, voltage, duration and solution volume/concentration as appropriate. Use small quantities, eye protection and safe ventilation, particularly if chlorine may form.
- Record the positive and negative electrode observations separately (including electrode sign, deposits, bubbles and test results), distinguish observations from product inferences, and evaluate the method or suggest a further test.
In plain language
The current moves ions, and the electrode products show which particles were discharged. A hypothesis predicts an observation; the experiment checks that prediction under controlled conditions.
Self-check
A gas forms at the negative electrode during electrolysis of aqueous sodium chloride. Which gas is expected, and why?
Show the answer
Hydrogen is expected. Sodium is more reactive than hydrogen, so hydrogen ions/water are reduced in preference to sodium ions at the cathode.
- Core
Scope: Core · §8.2.4 Required Practical 4: temperature change
Open revision guidance
What to know
- Investigate variables that affect temperature change in reacting solutions, including acid + metal, acid + carbonate, neutralisation and metal displacement. Use suitable apparatus to measure mass, solution volume and temperature accurately; record the initial and highest/final temperature and calculate ΔT = final − initial. Repeat measurements so a mean can be found.
- Compare a chosen variable fairly while controlling other conditions such as reactant amounts/concentrations, solution volume, insulation and apparatus. Record data with suitable units/significant figures and be prepared to translate the results to a graph or compare two plotted variables.
What to do in an assessment
- Write a hypothesis, choose a fair comparison, measure reactant mass/volume and temperature accurately, mix safely, record the greatest temperature reached and calculate the change. Repeat each condition and find the mean; plot or interpret temperature change against the independent variable when data are provided.
- Evaluate heat loss to surroundings, thermometer resolution, mixing time and repeat spread; propose insulation or a temperature probe as improvements. State safety controls and identify a further investigation where relevant.
In plain language
The reaction's energy transfer changes the solution temperature, but some heat can escape to the room. A fair comparison measures the same way each time and reduces heat loss.
Self-check
A solution warms from 19°C to 27°C. What is the temperature change and what does it indicate?
Show the answer
The change is +8°C. The surroundings/solution warmed, indicating an exothermic reaction.
- Core
Scope: Core · §8.2.5 Required Practical 5: rate and concentration
Open revision guidance
What to know
- Investigate how concentration affects reaction rate using both a method that measures gas volume and a method involving a colour change or turbidity; develop a hypothesis from collision theory. Measure gas volume at known times with suitable apparatus, or record the time to the same defined colour/turbidity endpoint for every trial. Repeat each concentration to compare rates fairly.
- Control other variables such as total volume, temperature, reactant amounts and apparatus. Tabulate observations, use suitable significant figures and units, find means, plot/interpret data and use a tangent gradient when the curve and question call for an instantaneous rate. Use ratios/fractions/percentages in simple comparisons, estimate calculation results, and interpret a linear relationship y = mx + c when relevant.
What to do in an assessment
- Choose concentration as the independent variable and specify dilution steps; keep total volume, temperature and other reactants constant. For each concentration, collect gas-volume/time data or time the same endpoint, repeat trials and compare initial rate/gradient or time to the same amount formed.
- Identify uncontrolled temperature, gas leakage, subjective colour judgement or inconsistent mixing as limitations; suggest repeats, a probe, improved sealing or a fixed observation method. Label graph axes and explain the rate trend using collision theory.
- Apply y = mx + c to a straight-line region of gas volume y against time x: m = change in volume/change in time is the rate in cm³/s, and c is the extrapolated volume at time zero. For example, a line through (0 s, 0 cm³) and (10 s, 20 cm³) has m = 2 cm³/s and c = 0 cm³. A non-zero intercept may prompt examination of the start-time or volume-zero procedure, rather than silently forcing the line through the origin. If the graph curves, its rate changes: use a tangent at the required time rather than treating the whole curve as one y = mx + c line.
In plain language
A more concentrated solution has more reactant particles in the same volume. They collide more often, so the reaction should be faster. The two required ways of seeing that are gas made and a visible change in colour or cloudiness.
Self-check
In a gas-volume experiment, why should the apparatus be checked for leaks before measuring rates?
Show the answer
A leak lets product gas escape, so the recorded volume is too small and may not represent the reaction. It also makes comparisons between concentrations unreliable.
- Core
Scope: Core · §8.2.6 Required Practical 6: chromatography
Open revision guidance
What to know
- Apply small sample spots to a pencil baseline, place the paper in solvent below the baseline and let solvent rise. The mobile solvent carries components over the stationary paper; different attraction causes separation.
- Mark the solvent front immediately, dry/measure spots, and calculate Rf as distance from baseline to spot centre divided by distance from baseline to solvent front.
What to do in an assessment
- Keep the starting line above solvent, use pencil rather than soluble ink and measure consistently. Compare spots or Rf values under the same solvent conditions to tell coloured substances apart.
In plain language
The solvent should move up through the paper without dissolving your starting mark. Each dye travels a fraction of the solvent distance; that ratio lets you compare its movement.
Self-check
Why must the sample spot start above the solvent level?
Show the answer
If the spot is submerged, the sample dissolves directly into the solvent reservoir instead of being carried up the paper for separation.
- Core
Scope: Core · §8.2.7 Required Practical 7: ion tests
Open revision guidance
What to know
- Identify ions in an unknown single ionic compound using the flame, hydroxide, carbonate, halide and sulfate tests. Flame-test a clean sample to identify specified metal cations; add sodium hydroxide to a fresh solution portion to compare precipitate colour and whether it dissolves in excess. Add dilute acid to a carbonate and test the gas with limewater. For halides, acidify with dilute nitric acid before silver nitrate; for sulfate, acidify with dilute hydrochloric acid before barium chloride.
- Use separate small portions and clean apparatus to avoid contamination; record exact flame/precipitate colours, gas observations and whether a solid dissolves. Use a Bunsen burner safely and handle test reagents carefully; the unknown is one ionic compound, so combine cation and anion evidence.
What to do in an assessment
- Test the cation and anion systematically and compare results with the AQA test table: use a flame test or NaOH precipitate for the cation; use acid/limewater for carbonate, nitric acid/silver nitrate for halides, or hydrochloric acid/barium chloride for sulfate. Combine the observations to identify the compound.
- If a colour is ambiguous or masked, report uncertainty and use another suitable test; distinguish cation from anion evidence, keep test portions separate, and include relevant Bunsen/reagent safety controls.
In plain language
An unknown salt has both a positive ion and a negative ion. One test can reveal one side of the formula; a second test identifies the other. Clean samples matter because contaminating ions can create misleading colours or precipitates.
Self-check
Why use separate portions of an unknown for a flame test and a precipitation test?
Show the answer
The reagents and heating can change or contaminate the sample. Separate aliquots preserve a clean sample for each test and make observations easier to interpret.
- Core
Scope: Core · §8.2.8 Required Practical 8: water analysis and purification
Open revision guidance
What to know
- AQA requires analysis and purification of water samples from different sources, including pH, dissolved solids and distillation. Representative sampling, safe heating and separation apparatus matter.
- Distillation boils water and condenses the vapour, leaving many dissolved solids behind. Measuring the residue or mass of dissolved solids and comparing pH helps characterise source samples.
What to do in an assessment
- Take equal, representative volumes, use the same calibrated pH method and evaporate a measured portion carefully to compare residue. Set up distillation with a thermometer and condenser and collect the condensed liquid safely.
- Link a limitation to a specific improvement: use several representative samples and repeat measurements to reduce dependence on one sample; calibrate the pH probe with buffer solutions before measuring; weigh the dry evaporation dish and then the dish with residue to obtain dissolved-solid mass for a known sample volume. Check that the condenser has a steady cooling-water flow and that apparatus does not leak, without sealing the heated system. Suggest a further investigation, such as comparing dissolved-solid mass and pH at different sampling locations or after rainfall, using equal volumes and the same method. Volatile contaminants can pass into the distillate, so pH and residue results alone cannot establish drinking-water safety.
In plain language
Distillation separates water by changing it into vapour and cooling it back to liquid. Many dissolved solids do not evaporate with the water, so the collected liquid is less salty than the starting sample.
Self-check
Why can distillation lower dissolved-solids content but fail to remove every possible contaminant?
Show the answer
Non-volatile dissolved solids remain in the flask, but volatile contaminants can evaporate and travel with water vapour. Distillation does not guarantee removal of every substance.
Earlier topic checklist
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Chemistry topic list
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These ticks keep your earlier topic checklist. They do not count towards the new reviewed requirements above. Its numbering may be internal; use the linked references above for the exact reviewed source requirements.
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