A Level H1 Chemistry Stoichiometry Moles Quiz

A-Level Chemistry H1 Quiz - Stoichiometry Moles: Answer Key

Total Marks: 40

---

Section A: Short Answer and Basic Calculations (10 marks)

1. Define the term mole.

• Answer: A mole is the amount of substance that contains the same number of particles (atoms, molecules, ions, or other entities) as there are atoms in exactly 12.0 g of carbon-12. [1]
• Marking note: Accept "Avogadro's number of particles (6.02 × 10²³)" or equivalent definition. Must reference number of particles or Avogadro's constant.

2. Calculate the relative molecular mass, Mr, of ammonium sulfate, (NH₄)₂SO₄.

• Answer: Mr = 2[14.0 + 4(1.0)] + 32.1 + 4(16.0) = 2(18.0) + 32.1 + 64.0 = 36.0 + 32.1 + 64.0 = 132.1 [1]
• Marking note: Award mark for correct answer. Allow 132 if rounded.

3. Calculate the number of moles of sodium hydroxide, NaOH, present in 8.00 g of the solid.

• Answer: Mr(NaOH) = 23.0 + 16.0 + 1.0 = 40.0; n = m/Mr = 8.00/40.0 = 0.200 mol [1]
• Marking note: Award mark for correct answer with or without working shown.

4. Calculate the mass of carbon dioxide, CO₂, produced when 0.500 mol of propane, C₃H₈, undergoes complete combustion. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

• Answer: Mole ratio C₃H₈ : CO₂ = 1 : 3 [1]; n(CO₂) = 3 × 0.500 = 1.50 mol; Mr(CO₂) = 44.0; mass = 1.50 × 44.0 = 66.0 g [1]
• Marking note: 1 mark for correct mole ratio/calculation of moles of CO₂; 1 mark for correct mass.

5. A sample of hydrated copper(II) sulfate, CuSO₄·xH₂O, has a mass of 4.99 g. After heating to constant mass, the anhydrous CuSO₄ has a mass of 3.19 g. Calculate the value of x.

• Answer: Mass of water lost = 4.99 – 3.19 = 1.80 g [1]; Mr(CuSO₄) = 63.5 + 32.1 + 64.0 = 159.6; n(CuSO₄) = 3.19/159.6 = 0.0200 mol [1]; n(H₂O) = 1.80/18.0 = 0.100 mol; Ratio H₂O : CuSO₄ = 0.100/0.0200 = 5; x = 5 [1]
• Marking note: 1 mark for mass of water; 1 mark for moles of CuSO₄ or H₂O; 1 mark for correct x value.

---

Section B: Stoichiometric Calculations (10 marks)

6. Calculate the volume occupied by 0.250 mol of nitrogen gas, N₂, at room temperature and pressure (RTP).

• Answer: Volume = n × 24.0 = 0.250 × 24.0 = 6.00 dm³ [1]
• Marking note: Accept 6.0 dm³. Unit must be stated or implied.

7. Calculate the concentration, in mol dm⁻³, of a solution prepared by dissolving 5.85 g of sodium chloride, NaCl, in water and making up to 250 cm³.

• Answer: Mr(NaCl) = 23.0 + 35.5 = 58.5; n = 5.85/58.5 = 0.100 mol; c = n/V = 0.100/(250/1000) = 0.400 mol dm⁻³ [1]
• Marking note: Award mark for correct final answer.

8. 2.70 g of aluminium reacts completely with excess dilute hydrochloric acid. 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

(a) Calculate the amount, in moles, of aluminium used.

• Answer: n(Al) = 2.70/27.0 = 0.100 mol [1]

(b) Calculate the amount, in moles, of hydrogen gas produced.

• Answer: Mole ratio Al : H₂ = 2 : 3; n(H₂) = 0.100 × (3/2) = 0.150 mol [1]

(c) Calculate the volume of hydrogen gas produced at RTP.

• Answer: Volume = 0.150 × 24.0 = 3.60 dm³ [1]

9. 25.0 cm³ of sulfuric acid required 20.0 cm³ of 0.100 mol dm⁻³ NaOH for neutralisation. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

(a) Calculate the amount, in moles, of NaOH used.

• Answer: n(NaOH) = 0.100 × (20.0/1000) = 0.00200 mol [1]

(b) Calculate the amount, in moles, of H₂SO₄ in 25.0 cm³ of the acid solution.

• Answer: Mole ratio NaOH : H₂SO₄ = 2 : 1; n(H₂SO₄) = 0.00200/2 = 0.00100 mol [1]

(c) Calculate the concentration of the sulfuric acid in mol dm⁻³.

• Answer: c = n/V = 0.00100/(25.0/1000) = 0.0400 mol dm⁻³ [1]

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

(a) Calculate the empirical formula of the compound.

• Answer: Assume 100 g sample: C = 40.0 g, H = 6.7 g, O = 53.3 g [1]; Moles: C = 40.0/12.0 = 3.33, H = 6.7/1.0 = 6.7, O = 53.3/16.0 = 3.33 [1]; Divide by smallest (3.33): C = 1, H = 2, O = 1; Empirical formula = CH₂O [1]
• Marking note: 1 mark for correct mole calculations; 1 mark for correct ratio; 1 mark for correct formula.

(b) The relative molecular mass of the compound is 180. Determine its molecular formula.

• Answer: Mr(CH₂O) = 12.0 + 2.0 + 16.0 = 30.0; n = 180/30.0 = 6; Molecular formula = C₆H₁₂O₆ [1]

---

Section C: Limiting Reagents and Applied Stoichiometry (10 marks)

11. 1.20 g of magnesium reacts with 50.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid. Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

(a) Determine which reactant is the limiting reagent. Show your working.

• Answer: n(Mg) = 1.20/24.3 = 0.0494 mol [1]; n(HCl) = 2.00 × (50.0/1000) = 0.100 mol [1]; From equation, 1 mol Mg requires 2 mol HCl. 0.0494 mol Mg requires 0.0988 mol HCl. Available HCl (0.100 mol) > required (0.0988 mol), so Mg is the limiting reagent. [1]
• Marking note: 1 mark each for moles of Mg and HCl; 1 mark for correct identification with reasoning. Accept alternative method comparing mole ratios.

(b) Calculate the mass of magnesium chloride, MgCl₂, that can be formed.

• Answer: n(MgCl₂) = n(Mg) = 0.0494 mol [1]; Mr(MgCl₂) = 24.3 + 2(35.5) = 95.3; mass = 0.0494 × 95.3 = 4.71 g [1]
• Marking note: 1 mark for correct moles of MgCl₂; 1 mark for correct mass.

(c) Calculate the volume of hydrogen gas produced at RTP.

• Answer: n(H₂) = n(Mg) = 0.0494 mol; Volume = 0.0494 × 24.0 = 1.19 dm³ [1]

12. Titration of ethanoic acid with 0.100 mol dm⁻³ NaOH.

(a) Which titrations should be used to calculate the average volume of NaOH? Explain your answer.

• Answer: Titrations 1 and 2 should be used. [1] Titration 3 (23.60 cm³) is not concordant with titrations 1 and 2 (24.50 and 24.40 cm³). Concordant results should agree within ±0.10 cm³. Titration 3 differs by more than 0.10 cm³ from the others and is likely an outlier/rough titration. [1]
• Marking note: 1 mark for identifying titrations 1 and 2; 1 mark for valid explanation referencing concordancy.

(b) Calculate the average volume of NaOH used.

• Answer: Average = (24.50 + 24.40)/2 = 24.45 cm³ [1]

(c) Calculate the concentration of the ethanoic acid solution in mol dm⁻³.

• Answer: n(NaOH) = 0.100 × (24.45/1000) = 0.002445 mol [1]; n(CH₃COOH) = n(NaOH) = 0.002445 mol (1:1 ratio); c = 0.002445/(25.0/1000) = 0.0978 mol dm⁻³ [1]
• Marking note: 1 mark for correct moles of NaOH; 1 mark for correct concentration. Accept 0.098 mol dm⁻³.

13. Decomposition of impure NaHCO₃.

(a) Calculate the mass of NaHCO₃ that decomposed.

• Answer: Mass loss = mass of H₂O + CO₂ produced = 0.500 – 0.318 = 0.182 g [1]; Mr(H₂O + CO₂) = 18.0 + 44.0 = 62.0; n(H₂O + CO₂) = 0.182/62.0 = 0.002935 mol [1]; From equation, 2 mol NaHCO₃ produces 1 mol H₂O + 1 mol CO₂ (2 mol gases total). n(NaHCO₃) = n(gases) = 0.002935 mol; Mr(NaHCO₃) = 84.0; mass NaHCO₃ = 0.002935 × 84.0 = 0.247 g [1]
• Marking note: 1 mark for mass loss; 1 mark for moles of gases; 1 mark for mass of NaHCO₃. Accept alternative method using mass ratio.

(b) Calculate the percentage purity of NaHCO₃ in the original sample.

• Answer: Percentage purity = (0.247/0.500) × 100 = 49.4% [1] (Allow 49–50% depending on rounding)
• Marking note: 1 mark for correct percentage; 1 mark for using answer from (a). Error carried forward (ECF) applies.

14. Fertiliser containing NH₄NO₃.

(a) Calculate the percentage by mass of nitrogen in pure NH₄NO₃.

• Answer: Mr(NH₄NO₃) = 14.0 + 4.0 + 14.0 + 48.0 = 80.0 [1]; Mass of N = 2 × 14.0 = 28.0; Percentage N = (28.0/80.0) × 100 = 35.0% [1]
• Marking note: 1 mark for correct Mr; 1 mark for correct percentage.

(b) A 5.00 g sample of the fertiliser was analysed and found to contain 1.40 g of nitrogen. Calculate the percentage purity of NH₄NO₃ in the fertiliser.

• Answer: Mass of NH₄NO₃ needed to provide 1.40 g N = 1.40 × (80.0/28.0) = 4.00 g [1]; Percentage purity = (4.00/5.00) × 100 = 80.0% [1]
• Marking note: 1 mark for mass of NH₄NO₃; 1 mark for correct percentage.

15. Calculate the mass of calcium carbonate, CaCO₃, required to produce 1.20 dm³ of carbon dioxide gas at RTP when reacted with excess hydrochloric acid. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

• Answer: n(CO₂) = 1.20/24.0 = 0.0500 mol [1]; Mole ratio CaCO₃ : CO₂ = 1 : 1; n(CaCO₃) = 0.0500 mol; Mr(CaCO₃) = 40.1 + 12.0 + 48.0 = 100.1; mass = 0.0500 × 100.1 = 5.01 g [1]
• Marking note: 1 mark for moles of CO₂; 1 mark for correct mass.

---

Section D: Data Interpretation and Advanced Stoichiometry (10 marks)

16. Iron ore analysis by redox titration.

(a) Calculate the amount, in moles, of MnO₄⁻ used in the titration.

• Answer: n(MnO₄⁻) = 0.0200 × (24.0/1000) = 0.000480 mol [1]

(b) Calculate the amount, in moles, of Fe²⁺ in the solution.

• Answer: Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5; n(Fe²⁺) = 5 × 0.000480 = 0.00240 mol [1]

(c) Calculate the mass of iron in the ore sample.

• Answer: m(Fe) = n × Ar = 0.00240 × 55.8 = 0.134 g [1]

(d) Calculate the percentage by mass of iron in the ore.

• Answer: Percentage Fe = (0.134/2.50) × 100 = 5.36% [1]

17. A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56.0.

(a) Determine the empirical formula of the hydrocarbon.

• Answer: Assume 100 g: C = 85.7 g, H = 14.3 g; Moles: C = 85.7/12.0 = 7.14, H = 14.3/1.0 = 14.3 [1]; Divide by smallest (7.14): C = 1, H = 2; Empirical formula = CH₂ [1]

(b) Determine the molecular formula of the hydrocarbon.

• Answer: Mr(CH₂) = 14.0; n = 56.0/14.0 = 4; Molecular formula = C₄H₈ [1]

18. Calculate the concentration of sodium carbonate solution. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

• Answer: n(HCl) = 0.500 × (15.0/1000) = 0.00750 mol [1]; Mole ratio Na₂CO₃ : HCl = 1 : 2; n(Na₂CO₃) = 0.00750/2 = 0.00375 mol; c = n/V = 0.00375/(10.0/1000) = 0.375 mol dm⁻³ [1]

19. Standard solution and titration.

(a) Calculate the concentration of the sodium carbonate standard solution in mol dm⁻³.

• Answer: Mr(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 106.0; n = 2.65/106.0 = 0.0250 mol [1]; c = 0.0250/(250/1000) = 0.100 mol dm⁻³ [1]

(b) Calculate the concentration of the hydrochloric acid in mol dm⁻³.

• Answer: n(Na₂CO₃) in 25.0 cm³ = 0.100 × (25.0/1000) = 0.00250 mol [1]; Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂; n(HCl) = 2 × 0.00250 = 0.00500 mol; c(HCl) = 0.00500/(20.0/1000) = 0.250 mol dm⁻³ [1]

20. Gravimetric analysis of chloride mixture.

(a) Calculate the amount, in moles, of silver chloride formed.

• Answer: Mr(AgCl) = 107.9 + 35.5 = 143.4; n(AgCl) = 1.44/143.4 = 0.0100 mol [1]

(b) Calculate the mass of sodium chloride in the original mixture.

• Answer: n(Cl⁻) = n(AgCl) = 0.0100 mol; n(NaCl) = 0.0100 mol [1]; Mr(NaCl) = 58.5; mass NaCl = 0.0100 × 58.5 = 0.585 g [1]

(c) Calculate the percentage by mass of sodium nitrate in the mixture.

• Answer: Mass of NaNO₃ = 1.00 – 0.585 = 0.415 g; Percentage NaNO₃ = (0.415/1.00) × 100 = 41.5% [1]

---

END OF ANSWER KEY