O Level Chemistry Stoichiometry Moles Quiz

O-Level Chemistry Quiz - Stoichiometry Moles — ANSWER KEY

Total Marks: 40

Section A: Short Answer and Structured Questions (20 marks)

1. Define the term relative atomic mass (Aᵣ).
[1 mark]

Answer: The relative atomic mass of an element is the average mass of one atom of the element compared to 1/12 the mass of one atom of carbon-12.
[Accept: ratio of the average mass of one atom to 1/12 the mass of a carbon-12 atom.]

2. Calculate the relative molecular mass (Mᵣ) of aluminium sulfate, Al₂(SO₄)₃.
[2 marks]

Working:

• Al: 2 × 27 = 54
• S: 3 × 32 = 96
• O: 12 × 16 = 192
• Mᵣ = 54 + 96 + 192 = 342

Answer: 342
[1 mark for correct atomic mass multiplication; 1 mark for correct total.]

3. A sample of iron(III) oxide, Fe₂O₃, has a mass of 8.00 g. Calculate the number of moles of Fe₂O₃ in this sample.
[2 marks]

Working:

• Mᵣ of Fe₂O₃ = 2(56) + 3(16) = 112 + 48 = 160
• n = m / Mᵣ = 8.00 / 160 = 0.0500 mol

Answer: 0.0500 mol
[1 mark for correct Mᵣ; 1 mark for correct calculation with units.]

4. Calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃.
[2 marks]

Working:

• Mᵣ of NH₄NO₃ = 14 + 4(1) + 14 + 3(16) = 14 + 4 + 14 + 48 = 80
• Mass of N in one formula unit = 2 × 14 = 28
• % N = (28 / 80) × 100 = 35.0%

Answer: 35.0%
[1 mark for correct Mᵣ; 1 mark for correct percentage.]

5. A compound has the empirical formula CH₂O and a relative molecular mass of 180. Determine the molecular formula of this compound.
[2 marks]

Working:

• Mᵣ of empirical formula CH₂O = 12 + 2(1) + 16 = 30
• n = Mᵣ(compound) / Mᵣ(empirical) = 180 / 30 = 6
• Molecular formula = (CH₂O)₆ = C₆H₁₂O₆

Answer: C₆H₁₂O₆
[1 mark for calculating ratio; 1 mark for correct molecular formula.]

Section B: Further Structured Questions (20 marks)

6. Magnesium reacts with hydrochloric acid according to the equation:
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Calculate the mass of magnesium needed to produce 4.80 dm³ of hydrogen gas at r.t.p.
[3 marks]

Working:

• Moles of H₂ = volume / molar volume = 4.80 / 24 = 0.200 mol
• From equation: 1 mol Mg produces 1 mol H₂
• Moles of Mg needed = 0.200 mol
• Mass of Mg = n × Aᵣ = 0.200 × 24 = 4.80 g

Answer: 4.80 g
[1 mark for moles of H₂; 1 mark for mole ratio; 1 mark for final mass with unit.]

7. A student dissolves 5.30 g of sodium carbonate, Na₂CO₃, in water to make 250 cm³ of solution. Calculate the concentration of this solution in mol/dm³.
[3 marks]

Working:

• Mᵣ of Na₂CO₃ = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106
• Moles of Na₂CO₃ = 5.30 / 106 = 0.0500 mol
• Volume in dm³ = 250 / 1000 = 0.250 dm³
• Concentration = n / V = 0.0500 / 0.250 = 0.200 mol/dm³

Answer: 0.200 mol/dm³
[1 mark for moles; 1 mark for volume conversion; 1 mark for correct concentration.]

8. 25.0 cm³ of sodium hydroxide solution required 20.0 cm³ of 0.100 mol/dm³ sulfuric acid for complete neutralisation. The equation is:
2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Calculate the concentration of the sodium hydroxide solution in mol/dm³.
[3 marks]

Working:

• Moles of H₂SO₄ = c × V = 0.100 × (20.0 / 1000) = 0.00200 mol
• From equation: 2 mol NaOH react with 1 mol H₂SO₄
• Moles of NaOH = 2 × 0.00200 = 0.00400 mol
• Concentration of NaOH = n / V = 0.00400 / (25.0 / 1000) = 0.00400 / 0.0250 = 0.160 mol/dm³

Answer: 0.160 mol/dm³
[1 mark for moles of H₂SO₄; 1 mark for mole ratio; 1 mark for final concentration.]

9. Explain what is meant by the term limiting reactant.
[2 marks]

Answer: The limiting reactant is the reactant that is completely used up in a chemical reaction. It determines/limits the amount of product(s) that can be formed. The other reactants are present in excess.
[1 mark for "completely used up"; 1 mark for "determines amount of product formed".]

10. Empirical formula of magnesium oxide.

(a) Calculate the mass of magnesium used.
[1 mark]

Answer: Mass of Mg = 26.08 - 25.36 = 0.72 g

(b) Calculate the mass of oxygen that combined with the magnesium.
[1 mark]

Answer: Mass of oxygen = 26.56 - 26.08 = 0.48 g

(c) Determine the empirical formula of magnesium oxide.
[3 marks]

Working:

Empirical formula = MgO

Answer: MgO
[1 mark for correct moles; 1 mark for ratio calculation; 1 mark for correct formula.]

Section C: Data-Based and Diagram Interpretation Questions (20 marks)

11. Calcium carbonate and hydrochloric acid reaction.

(a) Write a balanced chemical equation, with state symbols.
[2 marks]

Answer: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
[1 mark for correct formulae; 1 mark for correct state symbols and balancing.]

(b) Using the graph, state the final volume of carbon dioxide produced.
[1 mark]

Answer: 480 cm³

(c) Calculate the number of moles of carbon dioxide produced.
[1 mark]

Working: n = V (dm³) / 24 = (480/1000) / 24 = 0.480 / 24 = 0.0200 mol

Answer: 0.0200 mol

(d) Hence, calculate the mass of calcium carbonate that reacted. Comment on the purity of the sample.
[3 marks]

Working:

• From equation: 1 mol CaCO₃ produces 1 mol CO₂
• Moles of CaCO₃ = 0.0200 mol
• Mᵣ of CaCO₃ = 40 + 12 + 3(16) = 100
• Mass of CaCO₃ that reacted = 0.0200 × 100 = 2.00 g
• The sample mass was 2.00 g and the mass that reacted was 2.00 g, so the sample is 100% pure.

Answer: 2.00 g; The sample is pure (100% calcium carbonate) because the calculated mass equals the sample mass.
[1 mark for moles of CaCO₃; 1 mark for mass calculation; 1 mark for purity comment.]

12. Water of crystallisation in copper(II) sulfate.

(a) Calculate the mass of water removed.
[1 mark]

Answer: Mass of water = 2.50 - 1.60 = 0.90 g

(b) Calculate the number of moles of anhydrous CuSO₄ and the number of moles of water removed.
[2 marks]

Working:

• Mᵣ of CuSO₄ = 63.5 + 32 + 4(16) = 159.5
• Moles of CuSO₄ = 1.60 / 159.5 = 0.0100 mol (to 3 s.f.)
• Moles of H₂O = 0.90 / 18 = 0.0500 mol

Answer: CuSO₄ = 0.0100 mol; H₂O = 0.0500 mol
[1 mark for each correct mole calculation.]

(c) Determine the value of x in CuSO₄·xH₂O.
[2 marks]

Working:

• Ratio H₂O : CuSO₄ = 0.0500 : 0.0100 = 5 : 1
• Therefore, x = 5

Answer: x = 5 (CuSO₄·5H₂O)
[1 mark for ratio; 1 mark for correct x value.]

13. Combustion of propane.

(a) Calculate the volume of oxygen required for the complete combustion of 12.0 dm³ of propane, measured at the same temperature and pressure.
[2 marks]

Working:

• From equation: 1 volume C₃H₈ reacts with 5 volumes O₂
• Volume of O₂ = 12.0 × 5 = 60.0 dm³

Answer: 60.0 dm³
[1 mark for identifying 1:5 ratio; 1 mark for correct calculation.]

(b) Calculate the mass of water produced when 22.0 g of propane is completely burned.
[3 marks]

Working:

• Mᵣ of C₃H₈ = 3(12) + 8(1) = 44
• Moles of C₃H₈ = 22.0 / 44 = 0.500 mol
• From equation: 1 mol C₃H₈ produces 4 mol H₂O
• Moles of H₂O = 0.500 × 4 = 2.00 mol
• Mᵣ of H₂O = 2(1) + 16 = 18
• Mass of H₂O = 2.00 × 18 = 36.0 g

Answer: 36.0 g
[1 mark for moles of propane; 1 mark for mole ratio; 1 mark for final mass.]

14. Define the term molar mass and state its units.
[2 marks]

Answer: Molar mass is the mass of one mole of a substance. Its units are grams per mole (g/mol).
[1 mark for definition; 1 mark for correct units.]

15. Calculate the number of molecules in 0.50 mol of carbon dioxide gas. (Avogadro constant = 6.02 × 10²³ mol⁻¹)
[1 mark]

Working: Number of molecules = moles × Avogadro constant = 0.50 × 6.02 × 10²³ = 3.01 × 10²³

Answer: 3.01 × 10²³
[1 mark for correct answer with unit.]

Section D: Integrated Problem-Solving Questions (20 marks)

16. A sample of unknown mass of potassium chloride, KCl, is dissolved in water. Excess silver nitrate solution is added, and 2.87 g of silver chloride, AgCl, precipitate is formed. The equation for the reaction is:
KCl(aq) + AgNO₃(aq) → AgCl(s) + KNO₃(aq)

Calculate the mass of potassium chloride in the original sample.
[4 marks]

Working:

• Mᵣ of AgCl = 108 + 35.5 = 143.5
• Moles of AgCl = mass / Mᵣ = 2.87 / 143.5 = 0.0200 mol
• From equation: 1 mol KCl produces 1 mol AgCl
• Moles of KCl = 0.0200 mol
• Mᵣ of KCl = 39 + 35.5 = 74.5
• Mass of KCl = moles × Mᵣ = 0.0200 × 74.5 = 1.49 g

Answer: 1.49 g
[1 mark for Mᵣ of AgCl; 1 mark for moles of AgCl; 1 mark for mole ratio; 1 mark for final mass with unit.]

17. A student reacts 3.27 g of zinc with excess dilute sulfuric acid. The equation for the reaction is:
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Calculate the volume of hydrogen gas produced at r.t.p.
[3 marks]

Working:

• Aᵣ of Zn = 65
• Moles of Zn = mass / Aᵣ = 3.27 / 65 = 0.0503 mol (approx. 0.0500 mol if using 65.4, but 65 is given)
• Using Aᵣ = 65: 3.27 / 65 = 0.0503 mol (keep as 0.0503 for accuracy)
• From equation: 1 mol Zn produces 1 mol H₂
• Moles of H₂ = 0.0503 mol
• Volume of H₂ = moles × 24 = 0.0503 × 24 = 1.2072 dm³ ≈ 1.21 dm³ (3 s.f.)

Answer: 1.21 dm³
[1 mark for moles of Zn; 1 mark for mole ratio; 1 mark for correct volume with unit.]

18. 10.0 cm³ of a solution of nitric acid, HNO₃, is neutralised by 25.0 cm³ of 0.200 mol/dm³ potassium hydroxide, KOH, solution. The equation for the reaction is:
HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l)

Calculate the concentration of the nitric acid in mol/dm³.
[4 marks]

Working:

• Moles of KOH = c × V = 0.200 × (25.0 / 1000) = 0.00500 mol
• From equation: 1 mol HNO₃ reacts with 1 mol KOH
• Moles of HNO₃ = 0.00500 mol
• Volume of HNO₃ in dm³ = 10.0 / 1000 = 0.0100 dm³
• Concentration of HNO₃ = n / V = 0.00500 / 0.0100 = 0.500 mol/dm³

Answer: 0.500 mol/dm³
[1 mark for moles of KOH; 1 mark for mole ratio; 1 mark for volume conversion; 1 mark for final concentration.]

19. A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its relative molecular mass is 180. Determine both the empirical and molecular formula of the compound.
[5 marks]

Working:

• Assume 100 g of compound:
  • C: 40.0 g / 12 = 3.33 mol
  • H: 6.7 g / 1 = 6.7 mol
  • O: 53.3 g / 16 = 3.33 mol
• Divide by smallest (3.33):
  • C: 3.33 / 3.33 = 1
  • H: 6.7 / 3.33 ≈ 2
  • O: 3.33 / 3.33 = 1
• Empirical formula = CH₂O
• Mᵣ of empirical formula = 12 + 2(1) + 16 = 30
• n = Mᵣ(compound) / Mᵣ(empirical) = 180 / 30 = 6
• Molecular formula = (CH₂O)₆ = C₆H₁₂O₆

Answer: Empirical formula: CH₂O; Molecular formula: C₆H₁₂O₆
[1 mark for correct moles of each element; 1 mark for correct ratio; 1 mark for empirical formula; 1 mark for n value; 1 mark for molecular formula.]

20. Explain the difference between empirical formula and molecular formula. Give an example to illustrate your answer.
[4 marks]

Answer:

• The empirical formula is the simplest whole-number ratio of atoms of each element in a compound.
• The molecular formula gives the actual number of atoms of each element in one molecule of the compound.
• For example, glucose has the molecular formula C₆H₁₂O₆. The simplest ratio of C:H:O is 1:2:1, so its empirical formula is CH₂O.
• Another example: ethane (C₂H₆) has an empirical formula of CH₃.

[1 mark for definition of empirical formula; 1 mark for definition of molecular formula; 2 marks for a clear, correct example illustrating the difference.]

END OF ANSWER KEY