Atoms, elements, compounds & mixtures – GCSE Chemistry exam questions with mark scheme

1Grade 1–3Foundation5 marks

The diagram shows the particles in four boxes, A, B, C and D.

Each circle represents an atom. Circles with different shading represent atoms of different elements. Circles that touch are chemically bonded together.

(a) Which box contains only one element? [1]

(b) Which box contains only one compound? [1]

(d) Box C contains a mixture of two elements. Explain how the diagram shows this. [2]

(a) A(1)
(b) B(1)
(c) D(1)
(d) two different types of particle / two different substances are present allow molecules made of white atoms and separate black atoms; allow each particle contains only one type of atom(1)
(d) the two substances are not chemically combined / not bonded to each other allow the white and black atoms are not joined together; do not accept 'not mixed'(1)
Answer: (a) A (b) B (c) D (d) two different substances, each made of one type of atom, not chemically combined with each other
Examiner tip: In a mixture the different substances are not chemically bonded to each other: say this explicitly, because 'two different atoms' alone also describes a compound.

2Grade 1–3Foundation4 marks

Compounds can be named from their formulae.

(a) Complete the table. [3]

Formula	Name of compound
$\text{MgCl}_2$	…
$\text{K}_2\text{SO}_4$	…
$\text{Ca(NO}_3\text{)}_2$	…

(b) Give the total number of atoms in the formula $\text{Ca(NO}_3\text{)}_2$ [1]

magnesium chloride do not accept magnesium chlorine(1)
potassium sulfate allow potassium sulphate; do not accept potassium sulfide / potassium sulfite(1)
calcium nitrate do not accept calcium nitride / calcium nitrite(1)
(b) 9(1)
Answer: (a) magnesium chloride; potassium sulfate; calcium nitrate (b) 9
Examiner tip: The bracket multiplies everything inside it: $\text{(NO}_3\text{)}_2$ is 2 N and 6 O, so the total is 1 + 2 + 6 = 9, not 5.

3Grade 4–54 marks

Mixtures can be separated by physical processes.

Suggest the most suitable method to separate each mixture to obtain the substance shown in bold. [4]

Mixture	Substance wanted	Method
powdered chalk stirred into water	chalk (insoluble)	…
copper(II) chloride dissolved in water	copper(II) chloride crystals	…
black ink containing several dissolved dyes	the separate dyes	…
ethanol mixed with water (both liquids)	ethanol	…

filtration allow filtering; ignore decanting(1)
crystallisation allow evaporate (some of the water) and leave to crystallise; do not accept evaporate to dryness / distillation alone(1)
(paper) chromatography(1)
fractional distillation do not accept distillation / simple distillation(1)
Answer: filtration; crystallisation; chromatography; fractional distillation
Examiner tip: Two miscible liquids need fractional distillation: writing just 'distillation' does not get the mark.

4Grade 4–54 marks

Air contains the elements nitrogen and oxygen.

Carbon dioxide contains the elements carbon and oxygen.

Explain why air is a mixture but carbon dioxide is a compound. [4]

(air) the nitrogen and oxygen are not chemically combined / not bonded to each other(1)
(air) can be separated by a physical process allow named method e.g. fractional distillation (of liquid air); allow the proportions of gases (in air) can vary(1)
(carbon dioxide) carbon and oxygen are chemically combined / bonded together(1)
(carbon dioxide) in fixed proportions allow always 1 carbon atom to 2 oxygen atoms / formula $\text{CO}_2$ is fixed; allow can only be separated by a chemical reaction(1)
Answer: Air: N₂ and O₂ not chemically combined, separable by physical means (fractional distillation). CO₂: C and O chemically combined in fixed proportions (1 C : 2 O), separable only by a chemical reaction.
Examiner tip: Both substances contain oxygen, so the answer must be about whether the elements are chemically combined, not just which elements are present.

5Grade 4–54 marks

Calcium reacts with water. The products are calcium hydroxide, $\text{Ca(OH)}_2$, and hydrogen gas.

(a) Write a word equation for this reaction. [1]

(b) Write a balanced symbol equation for this reaction. [2]

$$\ldots\text{Al} + \ldots\text{O}_2 \rightarrow \ldots\text{Al}_2\text{O}_3$$

(a) calcium + water → calcium hydroxide + hydrogen ignore state symbols; do not accept formulae in a word equation(1)
(b) correct formulae: $\text{Ca} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2$(1)
(b) correctly balanced: $\text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2$ allow multiples; ignore state symbols(1)
(c) $4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$ allow multiples / fractions e.g. $2\text{Al} + 1\tfrac{1}{2}\text{O}_2 \rightarrow \text{Al}_2\text{O}_3$(1)
Answer: (a) calcium + water → calcium hydroxide + hydrogen (b) Ca + 2H₂O → Ca(OH)₂ + H₂ (c) 4, 3, 2
Examiner tip: Never change a formula to make an equation balance: only change the big numbers in front of the formulae.

6Grade 6–76 marks

A student heated different masses of copper in air until all the copper had reacted to form copper oxide.

The student measured the mass of copper oxide formed each time. The graph shows the results and the student's line of best fit.

(a) One result is anomalous. Give the mass of copper used for this result and suggest one reason for it. [2]

(b) Use the line of best fit to find the mass of copper oxide formed from 1.20 g of copper. [1]

Calculate the simplest whole-number ratio of mass of copper : mass of oxygen in copper oxide. [2]

(d) Explain how the results show that copper oxide is a compound and not a mixture of copper and oxygen. [1]

(a) 1.50 g (of copper)(1)
(a) not all the copper reacted / heated for too short a time / not enough oxygen reached the copper allow some copper oxide was lost (from the crucible); reason must explain a mass that is too low(1)
(b) 1.50 g allow 1.48–1.52 g(1)
(c) mass of oxygen = 2.50 − 2.00 = 0.50 g(1)
(c) 4 : 1 do not accept 1 : 4; an unsimplified 2.00 : 0.50 scores the first mark only(1)
(d) (mass of copper oxide is directly proportional to mass of copper / straight line through the origin, so) copper and oxygen always combine in a fixed proportion / fixed ratio(1)
Answer: (a) 1.50 g; e.g. not all the copper reacted (b) 1.50 g (c) 4 : 1 (d) the elements combine in a fixed ratio by mass
Examiner tip: A ratio asked for 'copper : oxygen' must be written in that order: 4 : 1, not 1 : 4.

7Grade 4–55 marks

A student used the apparatus shown to obtain pure water from a purple solution of potassium manganate(VII) in water.

(a) Name the piece of apparatus labelled Q. [1]

(b) Explain how this apparatus produces pure water from the purple solution. [3]

What does this show about liquid R? [1]

(a) (Liebig) condenser(1)
(b) (the solution is heated and) the water evaporates / boils allow water turns to (water) vapour / steam(1)
(b) the potassium manganate(VII) / purple solute / dye does not evaporate / stays in the flask(1)
(b) the water vapour is cooled (in the condenser) and condenses (to liquid water) allow steam condenses in Q / in the condenser(1)
(c) it is (pure) water allow it is a pure substance / it boils at the boiling point of water(1)
Answer: (a) condenser (b) water evaporates, the solute stays in the flask, the vapour is cooled in the condenser and condenses to pure water (c) R is pure water
Examiner tip: 'Evaporates' and 'condenses' are both needed: describing only the heating step leaves out how the liquid is collected.

8Grade 8–9Higher5 marks

When lead(II) nitrate solution is added to potassium iodide solution, a yellow precipitate of lead(II) iodide forms.

(a) Balance the equation. [1]

$$\text{Pb(NO}_3\text{)}_2\text{(aq)} + \ldots\text{KI(aq)} \rightarrow \text{PbI}_2\text{(s)} + \ldots\text{KNO}_3\text{(aq)}$$

(b) Write the ionic equation for this reaction. Include state symbols. [2]

(d) Iodide ions can also be oxidised to iodine. Complete the half equation. [1]

$$2\text{I}^- \rightarrow \text{I}_2 + \ldots\ldots$$

(a) 2 (KI) and 2 (KNO₃) – both needed(1)
(b) $\text{Pb}^{2+} + 2\text{I}^- \rightarrow \text{PbI}_2$ (correct species and balanced) allow multiples(1)
(b) state symbols: $\text{Pb}^{2+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow \text{PbI}_2\text{(s)}$ second mark only if equation is correct(1)
(c) they do not take part in the reaction / are unchanged / stay in solution (on both sides) allow they are spectator ions(1)
(d) $2\text{e}^-$ do not accept $\text{e}^-$ alone / $2\text{e}$ without the negative sign / $-2\text{e}^-$ on the right(1)
Answer: (a) 2, 2 (b) Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s) (c) they are spectator ions: unchanged in solution (d) 2e⁻
Examiner tip: An ionic equation must balance for charge as well as atoms: Pb²⁺ needs two I⁻ ions to give neutral PbI₂.

9Grade 6–76 marks

A student has a beaker containing a mixture of sand, water and dissolved magnesium sulfate.

Sand is insoluble in water. Magnesium sulfate is soluble in water and forms crystals.

Describe how the student could obtain:

a dry sample of sand
pure crystals of magnesium sulfate
a sample of pure water.

In your answer, name the apparatus used and explain why each step works. [6]

filter the mixture (using filter paper and a funnel)(1)
sand is the residue (left in the filter paper) because it is insoluble / its particles cannot pass through the filter paper(1)
magnesium sulfate solution (and water) is the filtrate / passes through the filter paper(1)
wash the sand with (a little) distilled / pure water to remove dissolved magnesium sulfate(1)
dry the sand in a warm oven / leave to dry / pat dry between filter papers(1)
(to get pure water) distil some of the filtrate: heat the filtrate in a flask so water evaporates / boils; the magnesium sulfate stays behind(1)
water vapour passes into a (Liebig) condenser where it is cooled and condenses; collect the pure water in a beaker(1)
(crystals) heat / evaporate the remaining solution until it is saturated / until crystals start to form (on a glass rod) – do not evaporate to dryness(1)
leave the concentrated solution to cool (and crystallise) because magnesium sulfate is less soluble at lower temperature(1)
filter off / remove the crystals and dry them between filter papers / in a warm (not hot) place(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically ordered method that gives all three products, with the correct technique for each (filtration, distillation, crystallisation) and reasons linked to the properties of the substances (insolubility, boiling, condensing, crystals forming as the solution cools). Practical details such as washing/drying are included.
Level 2	3–4	Correct techniques for at least two of the products, in a mostly sensible order, with some explanation of why they work. The method may miss some steps or contain a minor error.
Level 1	1–2	Names one or two relevant techniques or simple steps (e.g. filter the mixture, evaporate the water) with little or no explanation. The order may be unclear.
No marks	0	No relevant content.

Answer: Filter: sand is the residue (wash with distilled water, dry). Distil part of the filtrate: water evaporates, condenses in a condenser, collected as pure water. Heat the remaining filtrate to the point of crystallisation, leave to cool, filter off and dry the crystals.
Examiner tip: Each of the three products needs its own method: filtration for the sand, distillation for the water, crystallisation for the salt. Answers that evaporate everything to dryness cannot get pure water and lose Level 3.

10Grade 6–75 marks

A student separated a mixture of propanone and water by fractional distillation.

Propanone boils at 56 °C. Water boils at 100 °C.

The graph shows the temperature recorded by the thermometer at the top of the fractionating column during the experiment.

(a) Give the time interval during which pure propanone was being collected. [1]

(b) The student changed the collecting beaker at 12 minutes. Use the graph to explain why. [2]

(a) 4 to 12 minutes allow between 4 and 12 minutes; do not accept a shorter interval inside this range or one starting before 4 min(1)
(b) after 12 minutes the temperature rises above 56 °C / rises towards 100 °C(1)
(b) so water (vapour) starts to reach the top / distil over, and changing beaker stops the propanone being contaminated with water / keeps the propanone pure(1)
(c) the column is hotter at the bottom and cooler at the top / there is a temperature gradient(1)
(c) water (higher boiling point) condenses (in the column) and runs back into the flask, but propanone (lower boiling point) stays as vapour and reaches the top / passes into the condenser(1)
Answer: (a) 4–12 min (b) the temperature rises above 56 °C after 12 min, so water begins to distil and would contaminate the propanone (c) the column is cooler at the top; water vapour condenses and drips back, propanone vapour reaches the top and passes to the condenser
Examiner tip: Link the flat part of the graph to the boiling point: a constant 56 °C means only propanone vapour is reaching the thermometer.

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