OCR A Level Chemistry: Amount of Substance — 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: Amount of Substance revision notes

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OCR’s real H432 Papers 1 and 2 open with a whole-paper Section A of 15 multiple-choice questions (15 marks, four-option A-D, single answer) before a longer Section B of short-answer and extended-response questions. Each of these papers is worth 100 marks in total, not 90 — this file’s mark totals are not, and are not intended to be, a reconstruction of that full 100-mark paper, since it covers only this one topic (Amount of Substance) rather than the whole specification. The set below now matches the real 15-mark weighting of Section A, but every question is drawn from this topic alone, rather than sampling across the whole specification as a genuine paper’s Section A would — so it is still not a reproduction of a full paper’s Section A.

Section A — Multiple choice (topic sample)

1. A sample of pure oxygen gas, O₂, has a mass of 8.00 g (M_r = 32, Avogadro constant = 6.02 × 10²³ mol⁻¹). How many molecules of O₂ does the sample contain? A) 1.51 × 10²³ B) 3.01 × 10²³ C) 2.41 × 10²⁴ D) 0.25 [1]

2. In the equation pV = nRT used at A Level, with R = 8.31 J K⁻¹ mol⁻¹, which set of units is correct for p, V and T respectively? A) atm, m³, °C B) kPa, cm³, K C) Pa, m³, K D) Pa, dm³, °C [1]

3. 25.0 cm³ of a solution of sodium carbonate, Na₂CO₃, is exactly neutralised by 20.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. What is the concentration, in mol dm⁻³, of the sodium carbonate solution? A) 0.0800 B) 0.0400 C) 0.0500 D) 0.100 [1]

4. A reaction has a percentage yield of 100% but a low atom economy. Which statement correctly explains how this is possible? A) All of the reactant mass converts into the desired product, but the reaction itself is inefficient in terms of energy B) The reaction produced more product than theoretically expected, but wasted reagents in a side reaction C) The percentage yield calculation does not account for the moles of reactant used, unlike atom economy D) All of the limiting reagent that reacted was converted into the desired product with no losses in the practical procedure, but the balanced equation itself produces significant mass in one or more by-products [1]

5. A student dissolves 4.00 g of NaOH (M_r = 40) in water and makes the solution up to 250 cm³ in a volumetric flask. What is the concentration of the resulting solution, in mol dm⁻³? A) 4.00 × 10⁻⁴ B) 16.0 C) 4.00 D) 0.400 [1]

6. What is the molar gas volume at RTP (room temperature and pressure), as given on the OCR Chemistry A Data Sheet? A) 22.4 dm³ mol⁻¹ B) 24.0 dm³ mol⁻¹ C) 24.0 cm³ mol⁻¹ D) 6.02 × 10²³ dm³ mol⁻¹ [1]

7. What is the concentration, in mol dm⁻³, of a solution formed by dissolving 0.500 mol of solute in 2.00 dm³ of solution? A) 4.00 B) 1.00 C) 0.250 D) 0.500 [1]

8. A compound has empirical formula CH₂O and a relative molecular mass of 180. What is its molecular formula? A) CH₂O B) C₂H₄O₂ C) C₆H₁₂O₆ D) C₃H₆O₃ [1]

9. In a reaction A + 2B → C, a chemist mixes 2.0 mol of A with 3.0 mol of B. Which reagent is the limiting reagent? A) A B) B C) Neither; both are used up exactly D) It cannot be determined without the molar masses [1]

10. What is the value of the Avogadro constant, and what does it represent? A) 6.02 × 10²³ mol⁻¹; the number of particles in one mole of a substance B) 6.02 × 10²³ g; the mass of one mole of a substance C) 1.38 × 10⁻²³ mol⁻¹; the number of particles in one gram of a substance D) 6.02 × 10²³ dm³; the volume of one mole of gas [1]

11. A standard solution is prepared using a volumetric flask rather than a beaker. What is the key reason for this? A) A volumetric flask is cheaper than a beaker B) A volumetric flask allows a solution of precisely known concentration to be made up to an exact, known volume C) A volumetric flask heats the solution more evenly D) A volumetric flask is required for filtration [1]

12. In a titration, what is the main purpose of using a burette rather than a pipette to add the acid or base being tested? A) A burette delivers a fixed, known volume every time B) A burette allows a variable, precisely measured volume to be added gradually and read off a graduated scale C) A burette is only ever used for the reagent in the conical flask, never for the titrant D) A burette does not need to be rinsed before use [1]

13. A compound has empirical formula CH₂ and a relative molecular mass of 42. What is its molecular formula? A) CH₂ B) C₂H₄ C) C₃H₆ D) C₄H₈ [1]

14. In the ideal gas equation pV = nRT, the pressure and number of moles of gas are kept constant while the temperature is increased. What happens to the volume? A) It increases B) It decreases C) It stays the same D) It cannot be determined from the information given [1]

15. A reaction has a theoretical yield of 12.0 g of product. The actual yield obtained is 9.0 g. What is the percentage yield? A) 133% B) 75.0% C) 25.0% D) 108% [1]

Section B — Short answer and extended questions

16. State the Avogadro constant and define the mole. [2]

17. Calculate the number of molecules in 3.2 g of methane. [3]

18. A gas occupies 250 cm³ at 100 kPa and 298 K.

(a) State the ideal gas equation, defining each term with its unit. [3] (b) Calculate the amount, in moles, of gas present. [3] (c) If the mass of the gas is 0.44 g, calculate its relative molecular mass and suggest its identity. [3]

19. In a titration, 25.0 cm³ of sodium hydroxide is neutralised by 22.40 cm³ of 0.0500 mol dm⁻³ sulfuric acid.

(a) Write the equation for the reaction. [1] (b) Calculate the concentration of the sodium hydroxide. [4] (c) Explain why a rough titration is carried out first and why concordant titres are used. [3]

20. 5.00 g of calcium carbonate is heated: CaCO₃ → CaO + CO₂. 2.24 g of calcium oxide is obtained.

(a) Calculate the theoretical yield of calcium oxide. [3] (b) Calculate the percentage yield. [2]

21. A compound is 85.7% carbon and 14.3% hydrogen by mass, and has a relative molecular mass of 56. Determine its molecular formula, showing your working. [4]

22. A student heats 5.00 g of hydrated copper(II) sulfate, CuSO₄·xH₂O, to constant mass, leaving 3.20 g of anhydrous copper(II) sulfate.

(a) Calculate the moles of anhydrous CuSO₄ and of water lost. [3] (b) Determine the value of x. [2]

23. For the reaction CaCO₃ → CaO + CO₂, with CaO as the desired product, calculate the atom economy of the reaction. Explain why this value is different from the percentage yield calculated in Question 20(b). [4]

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Answers

1. A) M_r(O₂) = 32; moles = 8.00 ÷ 32 = 0.25 mol; molecules = 0.25 × 6.02 × 10²³ = 1.51 × 10²³ [1].

2. C) In the SI form of the ideal gas equation used with R = 8.31 J K⁻¹ mol⁻¹, pressure must be in pascals (Pa), volume in cubic metres (m³) and temperature in kelvin (K) [1].

3. B) Moles HCl = 0.100 × 20.0 ÷ 1000 = 2.00 × 10⁻³ mol. The equation shows 1 mol Na₂CO₃ reacts with 2 mol HCl, so moles Na₂CO₃ = 1.00 × 10⁻³ mol. Concentration = 1.00 × 10⁻³ × 1000 ÷ 25.0 = 0.0400 mol dm⁻³ [1].

4. D) A 100% percentage yield means none of the limiting reagent that reacted was lost in the practical procedure (no losses to side reactions, incomplete reaction or purification); a low atom economy instead reflects the stoichiometry of the balanced equation itself, where a large proportion of the total reactant mass is built into unwanted by-products rather than the desired product [1].

5. D) Moles NaOH = 4.00 ÷ 40 = 0.100 mol; concentration = 0.100 ÷ (250 ÷ 1000) = 0.100 ÷ 0.250 = 0.400 mol dm⁻³ [1].

6. B) The Data Sheet gives the molar gas volume as 24.0 dm³ mol⁻¹ at room temperature and pressure, RTP [1]. 22.4 dm³ mol⁻¹ is the molar volume of an ideal gas at 273 K and 101.3 kPa (1 atm), a colder reference condition, so it is not the RTP value; C has the wrong unit, and D confuses molar volume with the Avogadro constant.

7. C) Concentration = moles ÷ volume (in dm³) = 0.500 ÷ 2.00 = 0.250 mol dm⁻³ [1].

8. C) M_r(CH₂O) = 12 + 2 + 16 = 30. Number of empirical units = 180 ÷ 30 = 6, so the molecular formula is (CH₂O)₆ = C₆H₁₂O₆ [1].

9. B) The equation requires 2 mol B for every 1 mol A. To react completely with 2.0 mol A would need 4.0 mol B, but only 3.0 mol B is available, so B runs out first and is the limiting reagent [1].

10. A) The Avogadro constant is 6.02 × 10²³ mol⁻¹, the number of particles (atoms, molecules or ions) in exactly one mole of a substance [1].

11. B) A volumetric flask is calibrated to hold one precise, known volume when filled to its graduation mark, which is essential for making up a standard solution of accurately known concentration; a beaker’s volume markings are far too imprecise for this [1].

12. B) A burette is graduated along its length so that a variable volume can be released gradually, drop by drop, and read off precisely — exactly what is needed to find an unknown, variable titre; a pipette, by contrast, delivers only one fixed volume [1].

13. C) M_r(CH₂) = 12 + 2 = 14. Number of empirical units = 42 ÷ 14 = 3, so the molecular formula is (CH₂)₃ = C₃H₆ [1].

14. A) Rearranging pV = nRT gives V = nRT ÷ p; with n and p constant, V is directly proportional to T, so increasing the temperature increases the volume [1].

15. B) Percentage yield = (actual yield ÷ theoretical yield) × 100 = (9.0 ÷ 12.0) × 100 = 75.0% [1].

16. 6.02 × 10²³ mol⁻¹ [1]. Since the 2018 SI redefinition, one mole is defined as the amount of substance containing exactly 6.02214076 × 10²³ specified elementary entities [1] — rounded to 6.02 × 10²³ for calculations at this level. (The older “as many particles as atoms in 12 g of carbon-12” definition gives the same value, but is no longer the exact defining statement.)

17. M_r(CH₄) = 16 [1]; moles = 3.2 ÷ 16 = 0.20 mol [1]; molecules = 0.20 × 6.02 × 10²³ = 1.2 × 10²³ [1].

18. (a) pV = nRT [1], where p is pressure in Pa, V is volume in m³, n is moles, R is 8.31 J K⁻¹ mol⁻¹ and T is temperature in K [1] [1]. (b) p = 100 000 Pa; V = 250 × 10⁻⁶ = 2.50 × 10⁻⁴ m³ [1]. n = pV ÷ RT = (100 000 × 2.50 × 10⁻⁴) ÷ (8.31 × 298) [1] = 1.01 × 10⁻² mol [1]. (c) M_r = mass ÷ moles = 0.44 ÷ (1.01 × 10⁻²) [1] = 43.6 ≈ 44 [1]; the gas is likely to be carbon dioxide [1].

19. (a) 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O [1]. (b) moles H₂SO₄ = 0.0500 × 22.40 ÷ 1000 = 1.12 × 10⁻³ [1]. Ratio NaOH : H₂SO₄ = 2 : 1, so moles NaOH = 2.24 × 10⁻³ [1] [1]. Concentration = 2.24 × 10⁻³ × 1000 ÷ 25.0 = 0.0896 mol dm⁻³ [1]. (c) The rough titration locates the approximate end point quickly so subsequent runs can be added dropwise near it [1]. Concordant titres are those within 0.10 cm³ of each other [1], showing the results are precise and repeatable, so the mean is reliable [1].

20. (a) M_r(CaCO₃) = 100; moles = 5.00 ÷ 100 = 0.0500 [1]. Ratio 1 : 1, so 0.0500 mol CaO [1]; M_r(CaO) = 56, so mass = 0.0500 × 56 = 2.80 g [1]. (b) (2.24 ÷ 2.80) × 100 [1] = 80.0% [1].

21. Assume 100 g: C = 85.7 ÷ 12 = 7.14 mol; H = 14.3 ÷ 1 = 14.3 mol [1]. Divide by the smallest (7.14): C = 1, H = 2.00 [1], giving empirical formula CH₂ (mass 14). Since 56 ÷ 14 = 4 [1], the molecular formula is C₄H₈ [1].

22. (a) Moles CuSO₄ = 3.20 ÷ 160 = 0.0200 mol [1]. Mass of water lost = 5.00 − 3.20 = 1.80 g [1], so moles H₂O = 1.80 ÷ 18 = 0.100 mol [1]. (b) Ratio H₂O : CuSO₄ = 0.100 ÷ 0.0200 = 5 [1], so x = 5 (CuSO₄·5H₂O) [1].

23. Atom economy = (M_r of CaO ÷ sum of M_r of reactants) × 100 = (56 ÷ 100) × 100 [1] = 56% [1]. This differs from percentage yield because atom economy is a fixed property of the reaction’s stoichiometry — how much of the total reactant mass ends up in the desired product, with the rest going to CO₂ as an unavoidable by-product [1] — whereas percentage yield depends on how the reaction was actually carried out, including losses to side reactions, incomplete reaction, or purification [1].

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Where marks are usually lost

• Not converting cm³ to m³ and kPa to Pa in the ideal gas equation.
• Ignoring the 2 : 1 ratio in the sulfuric acid titration.
• Including the rough titre in the mean.
• Confusing theoretical and actual yield in the percentage calculation.
• Not dividing by the smallest mole value when finding an empirical formula ratio, or stopping at the empirical formula when the molecular formula was asked for.
• Confusing atom economy (a property of the reaction’s stoichiometry) with percentage yield (a property of how the reaction was actually carried out) — a reaction can have 100% yield but poor atom economy, or vice versa.