AQA GCSE Chemistry 8462: Quantitative chemistry – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 4.3 Quantitative chemistry (4.3.1.1 to 4.3.5) of the AQA GCSE Chemistry (8462) specification, for teaching from September 2016 and exams from June 2018 onwards. The topic is assessed on Paper 1, set at Foundation and Higher Tier. Questions 6, 8 and 9, and the parts labelled below, test content the specification marks (HT only) and are labelled Higher tier only. Everything else is for both tiers.

Learn the content first in the Quantitative chemistry study guide and the revision notes. The course hub is AQA GCSE Chemistry, and the printable checklist lists every point. Titration technique is covered in Chemical changes.

Use these relative atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, K = 39, Fe = 56, Cu = 63.5, Zn = 65. One mole of any gas occupies 24 dm³ at room temperature and pressure. Avogadro constant = 6.02 × 10²³ per mole.

Questions

1. State the law of conservation of mass and explain why it means chemical equations must be balanced. [2]

2. This question is about balancing and formulae.

(a) Balance the equation: Al + O₂ → Al₂O₃ [1] (b) How many oxygen atoms are shown in 2Fe(NO₃)₃? [1]

3. Ammonium sulfate, (NH₄)₂SO₄, is used as a fertiliser.

(a) Calculate the relative formula mass (Mr) of ammonium sulfate. [2] (b) Calculate the percentage by mass of nitrogen in ammonium sulfate. Give your answer to 3 significant figures. [2]

4. A student heats 5.00 g of copper carbonate in an open crucible. The equation is CuCO₃ → CuO + CO₂. After strong heating the mass of solid left is 3.22 g.

(a) Explain, using the equation and the particle model, why the mass of solid decreases. [2] (b) Another student heats magnesium ribbon in an open crucible. The mass of solid increases. Explain why. [2] (c) Calculate the mass of gas that escaped from the copper carbonate. [1]

5. A student measures the volume of gas made in a reaction four times: 46, 49, 44 and 47 cm³.

(a) Calculate the mean volume. [1] (b) Use the range to estimate the uncertainty in the mean. Give the result in the form mean ± uncertainty. [2]

6. (Higher tier only) A sample contains 4.9 g of sulfuric acid, H₂SO₄.

(a) Calculate the number of moles of H₂SO₄ in the sample. [2] (b) Calculate the number of H₂SO₄ molecules in the sample. [1] (c) Calculate the number of hydrogen atoms in the sample. [1]

7. Iron is made in a reaction between iron(III) oxide and carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂

(a) (Higher tier only) Calculate the maximum mass of iron that can be made from 40.0 g of Fe₂O₃. [3] (b) In an experiment, 25.2 g of iron is obtained from 40.0 g of Fe₂O₃. Use your answer to (a) to calculate the percentage yield. [2] (c) Give two reasons why the yield is less than 100%. [2]

8. (Higher tier only) 2.30 g of sodium reacts with 0.80 g of oxygen, O₂, to make 3.10 g of sodium oxide, Na₂O. Use these masses to work out the balanced equation for the reaction. [4]

9. (Higher tier only) A student heats 11.2 g of iron with 8.0 g of sulfur. The equation is Fe + S → FeS.

(a) Show that iron is the limiting reactant. [3] (b) Calculate the mass of iron sulfide formed. [2] (c) Calculate the mass of sulfur left unreacted. [1]

10. A technician dissolves 5.04 g of potassium hydroxide, KOH, in water and makes the solution up to 500 cm³.

(a) Calculate the concentration of the solution in g/dm³. [2] (b) (Higher tier only) The technician makes a second solution with 5.04 g of KOH in 250 cm³. Explain how its concentration compares with the first. [1] (c) (Higher tier only) A 25.0 cm³ portion of the first solution is titrated. It is exactly neutralised by 18.75 cm³ of 0.120 mol/dm³ sulfuric acid. H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O Calculate the concentration of the KOH solution in mol/dm³. [4] (d) (Higher tier only) Show that the titration result agrees with your answer to (a). [1]

11. Hydrogen can be made by two routes.

Route A: Zn + H₂SO₄ → ZnSO₄ + H₂ Route B: CH₄ + H₂O → CO + 3H₂

	Route A	Route B
Yield	95%	70% (reversible reaction)
Rate	Slow	Fast at high temperature
By-product	Zinc sulfate, little use	Carbon monoxide, burned as a fuel on site

(a) Calculate the percentage atom economy for making hydrogen by Route A. Give your answer to 3 significant figures. [2] (b) Calculate the percentage atom economy for making hydrogen by Route B. Give your answer to 3 significant figures. [2] (c) (Higher tier only) Calculate the volume of hydrogen, at room temperature and pressure, that Route B could make from 8.0 g of methane. [3] (d) (Higher tier only) Using the data, explain which route a manufacturer would choose. [3]

Answers

1. No atoms are lost or made during a chemical reaction, so the mass of the products equals the mass of the reactants [1]. So each side of the equation must have the same number of atoms of each element [1]. Examiner insight: “Mass is conserved” alone restates the name of the law; the first mark needs the idea that atoms are neither lost nor made.

2. (a) 4Al + 3O₂ → 2Al₂O₃ [1] (b) 2 × 3 × 3 = 18 [1] Examiner insight: Only whole-number multipliers earn the balancing mark; fractions or changed subscripts such as AlO₃ score zero.

3. (a) 2 × (14 + 4 × 1) + 32 + 4 × 16 [1] = 132 [1] (b) (2 × 14) ÷ 132 × 100 [1] = 21.2% [1] Examiner insight: Using one nitrogen (14) instead of two loses the method mark, but a wrong Mr from (a) is usually carried forward if the method in (b) is right.

4. (a) Carbon dioxide gas is produced [1] and its particles escape from the open crucible into the air, so they are not weighed; only the copper oxide is left as a solid [1]. (b) Oxygen from the air reacts with the magnesium [1]; the oxygen atoms are added to the solid, so magnesium oxide weighs more than the magnesium did [1]. (c) 5.00 − 3.22 = 1.78 g [1] Examiner insight: Saying mass was “used up” or “destroyed” contradicts conservation of mass and earns nothing; name the gas and say where it went.

5. (a) (46 + 49 + 44 + 47) ÷ 4 = 46.5 cm³ [1] (b) Range = 49 − 44 = 5 cm³ [1]; uncertainty = ± 5 ÷ 2, so 46.5 ± 2.5 cm³ [1] Examiner insight: Quoting ± 5 (the full range) loses the second mark; the uncertainty about the mean is half the range.

6. (a) Mr = 2 + 32 + 64 = 98 [1]; moles = 4.9 ÷ 98 = 0.050 mol [1] (b) 0.050 × 6.02 × 10²³ = 3.01 × 10²² [1] (c) 2 × 3.01 × 10²² = 6.02 × 10²² [1] Examiner insight: Answers in standard form must keep the correct power of ten; 3.01 × 10²³ for (b) scores zero even though the digits match.

7. (a) Moles Fe₂O₃ = 40.0 ÷ 160 = 0.250 mol [1]; ratio 1 : 2, so moles Fe = 0.500 mol [1]; mass = 0.500 × 56 = 28.0 g [1] (b) 25.2 ÷ 28.0 × 100 [1] = 90% [1] (c) Any two from: some iron is lost when it is separated from the reaction mixture [1]; the reaction may not go to completion, or some reactants react in a different, unexpected way [1]. Examiner insight: Foundation candidates are usually given the theoretical mass; on either tier the yield is actual ÷ theoretical, and a value above 100% signals the fraction was inverted.

8. Moles Na = 2.30 ÷ 23 = 0.100 mol; moles O₂ = 0.80 ÷ 32 = 0.025 mol [1]; moles Na₂O = 3.10 ÷ 62 = 0.050 mol [1]. Divide by 0.025: ratio 4 : 1 : 2 [1]. 4Na + O₂ → 2Na₂O [1] Examiner insight: A correct equation written from memory with no mole working does not get full credit; the question asks you to use the masses, so show all three mole values.

9. (a) Moles Fe = 11.2 ÷ 56 = 0.200 mol [1]; moles S = 8.0 ÷ 32 = 0.250 mol [1]; ratio is 1 : 1, so 0.200 mol Fe needs only 0.200 mol S and iron runs out first [1]. (b) Moles FeS = 0.200 mol [1]; mass = 0.200 × 88 = 17.6 g [1] (c) (0.250 − 0.200) × 32 = 1.6 g [1] Examiner insight: “Iron has the larger mass” is not a reason; the third mark needs a comparison of moles against the 1 : 1 ratio.

10. (a) 500 cm³ = 0.500 dm³ [1]; 5.04 ÷ 0.500 = 10.08 g/dm³ (10.1 g/dm³) [1] (b) The same mass of solute is in half the volume, so the concentration is doubled (20.16 g/dm³) [1]. (c) Moles H₂SO₄ = 0.120 × 18.75 ÷ 1000 = 0.00225 mol [1]; moles KOH = 2 × 0.00225 = 0.00450 mol [1]; concentration = 0.00450 ÷ 0.0250 [1] = 0.180 mol/dm³ [1] (d) 0.180 × 56 = 10.08 g/dm³, the same as (a) [1]. Examiner insight: Forgetting the 1 : 2 ratio gives 0.090 mol/dm³; the ratio step earns its own mark, and later marks can still be awarded by error carried forward.

11. (a) Sum of reactants = 65 + 98 = 163 [1]; 2 ÷ 163 × 100 = 1.23% [1] (b) Sum of reactants = 16 + 18 = 34; desired product = 3 × 2 = 6 [1]; 6 ÷ 34 × 100 = 17.6% [1] (c) Moles CH₄ = 8.0 ÷ 16 = 0.50 mol [1]; moles H₂ = 3 × 0.50 = 1.5 mol [1]; volume = 1.5 × 24 = 36 dm³ [1] (d) Route B, because: it has the higher atom economy [1]; it is faster [1]; its by-product is useful as a fuel, which outweighs its lower yield from the reversible reaction [1]. Examiner insight: In (b), using 2 instead of 3 × 2 for the hydrogen loses the first mark; in (d), each mark needs a point taken from the data, and “Route B is better” with no reason scores zero.

Where marks are usually lost

• Balancing by changing subscripts or leaving fractions in the final equation.
• Counting one group inside brackets, e.g. one NH₄ in (NH₄)₂SO₄.
• Explaining an open-crucible mass change without saying a gas entered or left.
• Giving the full range instead of half the range as the uncertainty.
• (Higher tier only) Going from mass to mass without converting to moles and using the ratio.
• (Higher tier only) Picking the limiting reactant from masses rather than moles.
• Forgetting to convert cm³ to dm³ before a concentration calculation.
• Missing the ratio step in a titration calculation where the acid is H₂SO₄.
• Leaving a balancing number out of an atom economy calculation.
• Rounding too early, e.g. writing 1.2% in 11(a) when 3 significant figures are asked for.

Next steps

• Quantitative chemistry revision notes
• Quantitative chemistry study guide
• AQA GCSE Chemistry course hub
• Printable AQA GCSE Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Chemistry (8462) specification, Version 1.1 (October 2019), for teaching from September 2016 and exams from June 2018 onwards (AQA), section 4.3 Quantitative chemistry.