OCR A Level Chemistry: Amount of Substance — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — 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

---

Section A

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

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

Section B

3. 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]

4. 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]

5. 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]

6. 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]

7. 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]

8. 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 5(b). [4]

---

Answers

1. 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.)

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

3. (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].

4. (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].

5. (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].

6. 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].

7. (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].

8. 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].

---

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.