OxfordAQA IGCSE Chemistry: 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 available free from your board.

Related: Quantitative Chemistry revision notes

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Section A

1. Calculate the relative formula mass of Ca(OH)₂ and of CuSO₄·5H₂O. (A_r: H 1, O 16, S 32, Ca 40, Cu 64) [2]

2. Calculate the number of moles in 8.8 g of carbon dioxide. [2]

Section B

3. 4.8 g of magnesium is burned in excess oxygen: 2Mg + O₂ → 2MgO.

(a) Calculate the moles of magnesium. [2] (b) Calculate the maximum mass of magnesium oxide that could form. [3] (c) In practice, only 7.2 g of magnesium oxide is obtained. Give two reasons why less than the calculated maximum mass is obtained, even though no atoms are gained or lost. [2]

4. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass.

(a) Determine its empirical formula. [3] (b) Its relative molecular mass is 180. Determine the molecular formula. [2]

5. 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is neutralised by 20.0 cm³ of hydrochloric acid.

(a) Calculate the moles of sodium hydroxide. [2] (b) Calculate the concentration of the acid. [2]

6. Even though no atoms are gained or lost in a chemical reaction, explain why it is not always possible to obtain the calculated amount of a product. [3]

7. 2.50 g of calcium carbonate reacts completely with excess dilute hydrochloric acid, releasing carbon dioxide gas: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ (A_r: Ca = 40, C = 12, O = 16)

(a) Calculate the moles of calcium carbonate. [2]

(b) Calculate the volume of carbon dioxide gas produced at room temperature and pressure. [2]

8. 25.0 cm³ of 0.0500 mol dm⁻³ sulfuric acid is exactly neutralised by 20.0 cm³ of sodium hydroxide solution in a titration: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

(a) Calculate the moles of sulfuric acid used. [2]

(b) Using the mole ratio in the equation, calculate the concentration of the sodium hydroxide solution. [3]

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Answers

1. Ca(OH)₂ = 40 + 2(16 + 1) = 74 [1]. CuSO₄·5H₂O = 64 + 32 + 64 + 5(18) = 250 [1].

2. M_r(CO₂) = 44 [1]; moles = 8.8 ÷ 44 = 0.20 mol [1].

3. (a) 4.8 ÷ 24 [1] = 0.20 mol [1]. (b) Ratio Mg : MgO is 1 : 1, so 0.20 mol MgO [1]; M_r(MgO) = 40 [1]; mass = 0.20 × 40 = 8.0 g [1]. (c) The reaction may not go to completion because it is reversible [1]; some of the product may be lost when it is separated from the reaction mixture, or some of the reactants may react in unexpected side reactions [1].

4. (a) Divide by A_r: C 40.0 ÷ 12 = 3.33; H 6.7 ÷ 1 = 6.7; O 53.3 ÷ 16 = 3.33 [1]. Divide by the smallest (3.33): C 1, H 2, O 1 [1]. Empirical formula = CH₂O [1]. (b) Empirical mass = 30 [1]; 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆ [1].

5. (a) moles = c × V ÷ 1000 = 0.100 × 25.0 ÷ 1000 [1] = 2.50 × 10⁻³ mol [1]. (b) The ratio NaOH : HCl is 1 : 1, so moles of HCl = 2.50 × 10⁻³ [1]; concentration = 2.50 × 10⁻³ × 1000 ÷ 20.0 = 0.125 mol dm⁻³ [1].

6. The reaction may not go to completion because it is reversible [1]; some of the product may be lost when it is separated from the reaction mixture during purification [1]; some of the reactants may react in ways different from the expected reaction [1].

7. (a) M_r(CaCO₃) = 40 + 12 + 3(16) = 100 [1]; moles = 2.50 ÷ 100 = 0.0250 mol [1]. (b) The ratio CaCO₃ : CO₂ is 1 : 1, so n(CO₂) = 0.0250 mol [1]; V = 0.0250 × 24 = 0.600 dm³ [1].

8. (a) moles = c × V ÷ 1000 = 0.0500 × 25.0 ÷ 1000 [1] = 1.25 × 10⁻³ mol [1]. (b) The ratio H₂SO₄ : NaOH is 1 : 2, so n(NaOH) = 2 × 1.25 × 10⁻³ = 2.50 × 10⁻³ mol [1]; concentration = (2.50 × 10⁻³ × 1000) ÷ 20.0 [1] = 0.125 mol dm⁻³ [1]. Halving this ratio by mistake would give a concentration only half the correct value — exactly why the mole ratio must always come from the balanced equation, never assumed.

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Where marks are usually lost

• Forgetting the water of crystallisation in a hydrated salt’s formula mass.
• Using the mole ratio backwards.
• Rounding empirical formula ratios too early.
• Forgetting to convert cm³ to dm³ in concentration calculations.
• Using 24 dm³ mol⁻¹ for a substance that isn’t a gas, or forgetting that it applies specifically at room temperature and pressure.
• Assuming every acid-alkali titration has a 1 : 1 mole ratio — sulfuric acid reacting with sodium hydroxide is 1 : 2, and getting this wrong doubles or halves every answer that follows.
• Reading only one titre from a titration and using it directly, rather than averaging the concordant results (those within 0.10 cm³ of each other) and excluding any rough titre.

Questions 7 and 8 draw on the gas-volume and titration-ratio sections of the Quantitative Chemistry revision notes, material the earlier questions on this page don’t reach, since question 5 only tests a 1 : 1 titration.