Key Concepts: Balanced equations conserve mass. Amount of substance is measured in moles using $n = m/M_r$. Mole ratios from equations allow reacting mass calculations. Empirical and molecular formulae can be deduced from composition data.
Section A — Relative Atomic Mass and Mr
1. Define relative atomic mass (Ar). [2]
2. Calculate the relative formula mass (Mr) for each of the following. Show your working. [4]
a) H₂SO₄
b) Ca(OH)₂
c) (NH₄)₂SO₄
d) Al₂(SO₄)₃
Section B — Balancing Equations
3. Write balanced symbol equations (with state symbols) for the following reactions. [8]
a) Magnesium + hydrochloric acid → magnesium chloride + hydrogen
b) Calcium carbonate → calcium oxide + carbon dioxide
c) Iron + oxygen → iron(III) oxide
d) Sodium hydroxide + sulfuric acid → sodium sulfate + water
Section C — Mole Calculations
4. Calculate the number of moles in each of the following. Show working. [4]
a) 9 g of water (H₂O)
b) 4.0 g of sodium hydroxide (NaOH)
c) 27 g of aluminium (Al)
d) 22 g of carbon dioxide (CO₂)
5. Calculate the mass of each of the following. Show working. [3]
a) 2.0 mol of NaCl
b) 0.5 mol of CaCO₃
c) 0.25 mol of H₂SO₄
Section D — Reacting Masses and Percentage Yield
6. Consider: 2H₂ + O₂ → 2H₂O. If 8 g of oxygen reacts completely, calculate the mass of water produced. [3]
7. Consider: CaCO₃ → CaO + CO₂. Calculate the mass of calcium oxide produced from 50 g of calcium carbonate. (Ar: Ca = 40, C = 12, O = 16) [3]
8. Define percentage yield and explain why the actual yield is often less than the theoretical yield. [3]
9. The theoretical yield of a reaction is 10 g. The actual yield is 7.5 g. Calculate the percentage yield. [2]
Section E — Empirical and Molecular Formulae
10. Distinguish between empirical formula and molecular formula. [2]
11. A compound contains 24 g of carbon and 4 g of hydrogen. Determine the empirical formula. [3]
12. The empirical formula of a compound is CH₂. Its Mr is 56. Find the molecular formula. [3]
13. A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine the empirical formula. (Ar: C = 12, H = 1, O = 16) [3]
Total marks: 42
Mark Scheme
1. The weighted average mass of one atom of the element [1] relative to 1/12 the mass of a carbon-12 atom [1] [2]
2. H₂SO₄: (2×1) + 32 + (4×16) = 98 [1]; Ca(OH)₂: 40 + 2(16+1) = 74 [1]; (NH₄)₂SO₄: 2(14+4) + 32 + 64 = 132 [1]; Al₂(SO₄)₃: 2(27) + 3(32+64) = 342 [1] [4]
3. a) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) [2]; b) CaCO₃(s) → CaO(s) + CO₂(g) [2]; c) 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) [2]; d) 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l) [2] [8]
4. a) 9/18 = 0.5 mol; b) 4.0/40 = 0.1 mol; c) 27/27 = 1.0 mol; d) 22/44 = 0.5 mol [4]
5. a) 2.0 × 58.5 = 117 g; b) 0.5 × 100 = 50 g; c) 0.25 × 98 = 24.5 g [3]
6. Mr O₂ = 32; moles O₂ = 8/32 = 0.25 mol [1]; ratio O₂:H₂O = 1:2 → moles H₂O = 0.5 mol [1]; mass = 0.5 × 18 = 9 g [1] [3]
7. Mr CaCO₃ = 100; moles = 50/100 = 0.5 mol [1]; ratio 1:1 → moles CaO = 0.5 mol [1]; mass = 0.5 × 56 = 28 g [1] [3]
8. % yield = (actual yield / theoretical yield) × 100 [1]; actual yield is often less because: some reactants may not react; product lost during transfer/filtering/evaporation; side reactions occur [2] [3]
9. (7.5/10) × 100 = 75% [2]
10. Empirical formula: simplest whole-number ratio of atoms [1]; molecular formula: actual number of each atom in one molecule [1] [2]
11. C: 24/12 = 2; H: 4/1 = 4; ratio C:H = 2:4 = 1:2 → empirical formula CH₂ [3]
12. Empirical mass of CH₂ = 14; 56/14 = 4; molecular formula = C₄H₈ [3]
13. C: 40/12 = 3.33; H: 6.7/1 = 6.7; O: 53.3/16 = 3.33 [1]; ratio 3.33:6.7:3.33 = 1:2:1 [1]; empirical formula CH₂O [1] [3]