A Level H1 Chemistry Stoichiometry Moles Quiz

A-Level Chemistry H1 Quiz - Stoichiometry Moles: ANSWER KEY

Total Marks: 40

Section A: Short Answer (10 marks)

1. Define the term mole. [1]

• Answer: The amount of substance that contains the same number of particles (atoms, molecules, ions, or electrons) as there are atoms in exactly 12 g of carbon-12. [1]
• Accept: The amount of substance that contains 6.02 × 10²³ specified particles.

2. State the value of the Avogadro constant and give its units. [1]

• Answer: 6.02 × 10²³ mol⁻¹ [1]
• Both value and units required for the mark.

3. Calculate the molar mass of ammonium sulfate, (NH₄)₂SO₄. [2]

• Working:
  • N: 2 × 14.0 = 28.0
  • H: 8 × 1.0 = 8.0
  • S: 1 × 32.1 = 32.1
  • O: 4 × 16.0 = 64.0
  • Total = 28.0 + 8.0 + 32.1 + 64.0 = 132.1 [1]
• Answer: 132.1 g mol⁻¹ [1]
• Award [1] for correct working, [1] for correct answer with units. Accept 132 g mol⁻¹.

4. A sample of magnesium contains 3.01 × 10²³ atoms. Calculate the amount, in moles, of magnesium present. [2]

• Working:
  • n = number of particles / Avogadro constant [1]
  • n = 3.01 × 10²³ / 6.02 × 10²³ = 0.500 mol [1]
• Answer: 0.500 mol

5. Calculate the number of oxygen atoms present in 0.500 mol of carbon dioxide, CO₂. [2]

• Working:
  • Each CO₂ molecule contains 2 oxygen atoms. [1]
  • Number of O atoms = 0.500 × 6.02 × 10²³ × 2 = 6.02 × 10²³ [1]
• Answer: 6.02 × 10²³

Section B: Structured Questions (10 marks)

6. A compound has the empirical formula CH₂O and a relative molecular mass of 180. Determine its molecular formula. [2]

• Working:
  • Mr of empirical formula CH₂O = 12.0 + 2(1.0) + 16.0 = 30.0 [1]
  • n = Mr(compound) / Mr(empirical formula) = 180 / 30.0 = 6
  • Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ [1]
• Answer: C₆H₁₂O₆

7. An organic compound contains 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen by mass.

(a) Calculate the empirical formula of the compound. [3]

• Working:
  • Assume 100 g sample:
    • C: 40.0 g / 12.0 = 3.33 mol
    • H: 6.67 g / 1.0 = 6.67 mol
    • O: 53.3 g / 16.0 = 3.33 mol [1]
  • Divide by smallest (3.33):
    • C: 3.33/3.33 = 1
    • H: 6.67/3.33 = 2
    • O: 3.33/3.33 = 1 [1]
  • Empirical formula: CH₂O [1]
• Answer: CH₂O

(b) The relative molecular mass of the compound is 60.0. Determine its molecular formula. [1]

• Working:
  • Mr(CH₂O) = 12.0 + 2.0 + 16.0 = 30.0
  • 60.0 / 30.0 = 2
  • Molecular formula = C₂H₄O₂ [1]
• Answer: C₂H₄O₂

8. Sodium hydrogencarbonate, NaHCO₃, decomposes on heating: 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g) A student heats 8.40 g of sodium hydrogencarbonate until decomposition is complete.

(a) Calculate the amount, in moles, of NaHCO₃ used. [2]

• Working:
  • Mr(NaHCO₃) = 23.0 + 1.0 + 12.0 + 3(16.0) = 84.0 [1]
  • n = m / Mr = 8.40 / 84.0 = 0.100 mol [1]
• Answer: 0.100 mol

(b) Calculate the mass of sodium carbonate, Na₂CO₃, produced. [2]

• Working:
  • Mole ratio NaHCO₃ : Na₂CO₃ = 2 : 1
  • n(Na₂CO₃) = 0.100 / 2 = 0.0500 mol [1]
  • Mr(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 106.0
  • m = n × Mr = 0.0500 × 106.0 = 5.30 g [1]
• Answer: 5.30 g

Section C: Structured Questions (10 marks)

9. A solution of hydrochloric acid, HCl, has a concentration of 0.200 mol dm⁻³.

(a) Calculate the mass of HCl present in 250 cm³ of this solution. [2]

• Working:
  • n = c × V = 0.200 × (250/1000) = 0.0500 mol [1]
  • Mr(HCl) = 1.0 + 35.5 = 36.5
  • m = n × Mr = 0.0500 × 36.5 = 1.83 g [1]
• Answer: 1.83 g

(b) This solution is diluted by adding water to make 500 cm³ of solution. Calculate the concentration of the diluted acid. [2]

• Working:
  • n remains constant = 0.0500 mol [1]
  • c = n / V = 0.0500 / (500/1000) = 0.100 mol dm⁻³ [1]
• Answer: 0.100 mol dm⁻³

10. A student carries out a titration: 25.0 cm³ of NaOH solution required 22.50 cm³ of 0.100 mol dm⁻³ H₂SO₄ for complete neutralisation. 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

(a) Calculate the amount, in moles, of H₂SO₄ used. [1]

• Working:
  • n = c × V = 0.100 × (22.50/1000) = 0.00225 mol [1]
• Answer: 0.00225 mol (or 2.25 × 10⁻³ mol)

(b) Calculate the amount, in moles, of NaOH in 25.0 cm³ of the solution. [1]

• Working:
  • Mole ratio H₂SO₄ : NaOH = 1 : 2
  • n(NaOH) = 2 × 0.00225 = 0.00450 mol [1]
• Answer: 0.00450 mol (or 4.50 × 10⁻³ mol)

(c) Calculate the concentration, in mol dm⁻³, of the NaOH solution. [2]

• Working:
  • c = n / V = 0.00450 / (25.0/1000) [1]
  • c = 0.180 mol dm⁻³ [1]
• Answer: 0.180 mol dm⁻³

11. Calculate the volume of carbon dioxide gas produced when 8.40 g of sodium hydrogencarbonate is heated until decomposition is complete, measured at room temperature and pressure (r.t.p.). [Molar volume at r.t.p. = 24.0 dm³ mol⁻¹] [2]

• Working:
  • Mr(NaHCO₃) = 84.0; n(NaHCO₃) = 8.40 / 84.0 = 0.100 mol [1]
  • Mole ratio NaHCO₃ : CO₂ = 2 : 1
  • n(CO₂) = 0.100 / 2 = 0.0500 mol
  • Volume = n × 24.0 = 0.0500 × 24.0 = 1.20 dm³ [1]
• Answer: 1.20 dm³

Section D: Data Interpretation and Application (10 marks)

12. A sample of hydrated magnesium sulfate, MgSO₄·xH₂O, has a mass of 6.15 g. On heating to constant mass, 3.00 g of anhydrous magnesium sulfate, MgSO₄, remains.

(a) Calculate the mass of water lost on heating. [1]

• Answer: 6.15 - 3.00 = 3.15 g [1]

(b) Calculate the amount, in moles, of anhydrous MgSO₄. [1]

• Working:
  • Mr(MgSO₄) = 24.3 + 32.1 + 4(16.0) = 120.4
  • n = 3.00 / 120.4 = 0.0249 mol [1]
• Answer: 0.0249 mol

(c) Calculate the amount, in moles, of water lost. [1]

• Working:
  • Mr(H₂O) = 18.0
  • n = 3.15 / 18.0 = 0.175 mol [1]
• Answer: 0.175 mol

(d) Determine the value of x in MgSO₄·xH₂O. [2]

• Working:
  • Ratio n(H₂O) : n(MgSO₄) = 0.175 : 0.0249 [1]
  • 0.175 / 0.0249 = 7.03 ≈ 7
  • x = 7 [1]
• Answer: x = 7 (MgSO₄·7H₂O)

13. A student dissolved 5.30 g of anhydrous sodium carbonate, Na₂CO₃, in distilled water and made the solution up to 250 cm³.

(a) Calculate the concentration, in mol dm⁻³, of the sodium carbonate solution. [2]

• Working:
  • Mr(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 106.0
  • n = 5.30 / 106.0 = 0.0500 mol [1]
  • c = n / V = 0.0500 / (250/1000) = 0.200 mol dm⁻³ [1]
• Answer: 0.200 mol dm⁻³

(b) The student titrated 25.0 cm³ portions of this Na₂CO₃ solution against HCl of unknown concentration. Average titre = 24.60 cm³. Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g) Calculate the concentration, in mol dm⁻³, of the hydrochloric acid. [3]

• Working:
  • n(Na₂CO₃) in 25.0 cm³ = 0.200 × (25.0/1000) = 0.00500 mol [1]
  • Mole ratio Na₂CO₃ : HCl = 1 : 2
  • n(HCl) = 2 × 0.00500 = 0.0100 mol [1]
  • c(HCl) = n / V = 0.0100 / (24.60/1000) = 0.407 mol dm⁻³ [1]
• Answer: 0.407 mol dm⁻³ (or 0.4065 mol dm⁻³)

14. A compound containing only carbon, hydrogen, and oxygen undergoes complete combustion. When 0.440 g of the compound is burned, 0.880 g of CO₂ and 0.360 g of H₂O are produced.

(a) Calculate the mass of carbon present in the original compound. [1]

• Working:
  • Mr(CO₂) = 44.0; mass of C in CO₂ = (12.0/44.0) × 0.880 = 0.240 g [1]
• Answer: 0.240 g

(b) Calculate the mass of hydrogen present in the original compound. [1]

• Working:
  • Mr(H₂O) = 18.0; mass of H in H₂O = (2.0/18.0) × 0.360 = 0.0400 g [1]
• Answer: 0.0400 g

(c) Hence, determine the empirical formula of the compound. [3]

• Working:
  • Mass of O = 0.440 - (0.240 + 0.0400) = 0.160 g [1]
  • Moles:
    • C: 0.240 / 12.0 = 0.0200 mol
    • H: 0.0400 / 1.0 = 0.0400 mol
    • O: 0.160 / 16.0 = 0.0100 mol [1]
  • Divide by smallest (0.0100):
    • C: 0.0200/0.0100 = 2
    • H: 0.0400/0.0100 = 4
    • O: 0.0100/0.0100 = 1
  • Empirical formula: C₂H₄O [1]
• Answer: C₂H₄O

15. A student reacts 2.70 g of aluminium with excess hydrochloric acid according to the equation: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) Calculate the volume of hydrogen gas produced at r.t.p. [Molar volume at r.t.p. = 24.0 dm³ mol⁻¹] [3]

• Working:
  • Mr(Al) = 27.0; n(Al) = 2.70 / 27.0 = 0.100 mol [1]
  • Mole ratio Al : H₂ = 2 : 3
  • n(H₂) = (3/2) × 0.100 = 0.150 mol [1]
  • Volume = n × 24.0 = 0.150 × 24.0 = 3.60 dm³ [1]
• Answer: 3.60 dm³

16. Define the term molar mass. [1]

• Answer: The mass of one mole of a substance, expressed in grams per mole (g mol⁻¹). [1]

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

• Working:
  • Mr(NH₄NO₃) = 14.0 + 4(1.0) + 14.0 + 3(16.0) = 80.0 [1]
  • Mass of N = 2 × 14.0 = 28.0
  • % N = (28.0 / 80.0) × 100 = 35.0% [1]
• Answer: 35.0%

18. A solution contains 1.50 mol of glucose, C₆H₁₂O₆, dissolved in 500 cm³ of water. Calculate the concentration of the solution in mol dm⁻³. [2]

• Working:
  • V = 500 cm³ = 0.500 dm³ [1]
  • c = n / V = 1.50 / 0.500 = 3.00 mol dm⁻³ [1]
• Answer: 3.00 mol dm⁻³

19. In a titration, 20.0 cm³ of 0.150 mol dm⁻³ potassium hydroxide, KOH, is neutralised by 15.0 cm³ of nitric acid, HNO₃. Calculate the concentration of the nitric acid. [2]

• Working:
  • n(KOH) = 0.150 × (20.0/1000) = 0.00300 mol [1]
  • KOH + HNO₃ → KNO₃ + H₂O (1:1 ratio)
  • n(HNO₃) = 0.00300 mol
  • c(HNO₃) = 0.00300 / (15.0/1000) = 0.200 mol dm⁻³ [1]
• Answer: 0.200 mol dm⁻³

20. A compound has the composition 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56.0. Determine its molecular formula. [3]

• Working:
  • Assume 100 g: C: 85.7 g / 12.0 = 7.14 mol; H: 14.3 g / 1.0 = 14.3 mol [1]
  • Divide by smallest (7.14): C: 1; H: 2 → Empirical formula: CH₂ [1]
  • Mr(CH₂) = 12.0 + 2.0 = 14.0
  • n = 56.0 / 14.0 = 4
  • Molecular formula: C₄H₈ [1]
• Answer: C₄H₈

END OF ANSWER KEY