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O Level Chemistry Stoichiometry Moles Quiz
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O-Level Chemistry Quiz - Stoichiometry Moles — Answer Key and Marking Scheme
Total Marks: 40
Section A: Short Answer and Basic Calculations (10 marks)
1. Define the term "mole."
(1 mark)
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 g of carbon-12. [Accept: the amount of substance containing 6.0 × 10²³ particles / Avogadro's number of particles.]
Marking: 1 mark for correct definition referencing Avogadro's number or the carbon-12 standard.
2. Calculate the relative molecular mass (Mr) of ammonium sulfate, (NH₄)₂SO₄.
(1 mark)
Answer:
N: 2 × 14 = 28
H: 8 × 1 = 8
S: 1 × 32 = 32
O: 4 × 16 = 64
Mr = 28 + 8 + 32 + 64 = 132
Marking: 1 mark for correct answer 132. Accept working showing correct summation.
3. How many atoms are present in 0.50 mol of carbon dioxide, CO₂?
(1 mark)
Answer:
1 molecule of CO₂ contains 3 atoms (1 C + 2 O).
Number of molecules = 0.50 × 6.0 × 10²³ = 3.0 × 10²³ molecules
Number of atoms = 3.0 × 10²³ × 3 = 9.0 × 10²³ atoms
Marking: 1 mark for correct answer 9.0 × 10²³. Accept 9 × 10²³.
4. A sample of magnesium contains 0.30 mol of Mg atoms. Calculate the mass of this sample.
(1 mark)
Answer:
Mass = moles × Ar = 0.30 × 24 = 7.2 g
Marking: 1 mark for correct answer 7.2 g.
5. Calculate the number of moles of water in 90.0 g of H₂O.
(1 mark)
Answer:
Mr of H₂O = (2 × 1) + 16 = 18
Moles = mass / Mr = 90.0 / 18 = 5.00 mol
Marking: 1 mark for correct answer 5.00 mol.
Section B: Stoichiometric Calculations (10 marks)
6. State the volume occupied by 2.0 mol of nitrogen gas at r.t.p.
(1 mark)
Answer:
Volume = moles × 24 = 2.0 × 24 = 48 dm³
Marking: 1 mark for correct answer 48 dm³.
7. Calculate the percentage by mass of oxygen in calcium carbonate, CaCO₃.
(2 marks)
Answer:
Mr of CaCO₃ = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100
Mass of oxygen in one formula unit = 3 × 16 = 48
Percentage of oxygen = (48 / 100) × 100 = 48.0%
Marking: 1 mark for correct Mr (100); 1 mark for correct percentage (48.0%). Accept 48%.
8. A compound has the empirical formula CH₂O and a relative molecular mass of 180. Determine its molecular formula.
(2 marks)
Answer:
Mr of empirical formula CH₂O = 12 + (2 × 1) + 16 = 30
n = Mr of compound / Mr of empirical formula = 180 / 30 = 6
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆
Marking: 1 mark for calculating n = 6; 1 mark for correct molecular formula C₆H₁₂O₆.
9. Magnesium reacts with oxygen to form magnesium oxide: 2Mg(s) + O₂(g) → 2MgO(s). Calculate the mass of magnesium oxide formed when 4.80 g of magnesium is burned completely in oxygen.
(3 marks)
Answer:
Moles of Mg = 4.80 / 24 = 0.200 mol
From equation, 2 mol Mg produces 2 mol MgO, so mole ratio Mg : MgO = 1 : 1
Moles of MgO = 0.200 mol
Mr of MgO = 24 + 16 = 40
Mass of MgO = 0.200 × 40 = 8.00 g
Marking: 1 mark for moles of Mg (0.200 mol); 1 mark for correct mole ratio and moles of MgO; 1 mark for correct mass (8.00 g). Award full marks for correct answer with working.
10. Zinc reacts with hydrochloric acid: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
(a) Calculate the number of moles of zinc in 3.25 g of zinc. (1 mark)
Answer: Moles of Zn = 3.25 / 65 = 0.0500 mol
Marking: 1 mark for correct answer 0.0500 mol.
(b) Calculate the volume of hydrogen gas produced at r.t.p. when 3.25 g of zinc reacts completely with excess hydrochloric acid. (2 marks)
Answer:
From equation, 1 mol Zn produces 1 mol H₂
Moles of H₂ = 0.0500 mol
Volume of H₂ = 0.0500 × 24 = 1.20 dm³
Marking: 1 mark for correct moles of H₂; 1 mark for correct volume (1.20 dm³ or 1200 cm³).
Section C: Limiting Reactants and Titration (10 marks)
11. 5.60 g of iron reacts with 3.65 g of hydrochloric acid: Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
(a) Determine which reactant is the limiting reactant. (3 marks)
Answer:
Moles of Fe = 5.60 / 56 = 0.100 mol
Moles of HCl = 3.65 / 36.5 = 0.100 mol
From equation, 1 mol Fe reacts with 2 mol HCl.
0.100 mol Fe would require 0.200 mol HCl, but only 0.100 mol HCl is available.
Therefore, HCl is the limiting reactant.
Marking: 1 mark for moles of Fe (0.100 mol); 1 mark for moles of HCl (0.100 mol); 1 mark for correct identification of HCl as limiting reactant with reasoning.
(b) Calculate the mass of iron(II) chloride, FeCl₂, formed. (2 marks)
Answer:
From equation, 2 mol HCl produces 1 mol FeCl₂.
Moles of FeCl₂ = 0.100 / 2 = 0.0500 mol
Mr of FeCl₂ = 56 + (2 × 35.5) = 127
Mass of FeCl₂ = 0.0500 × 127 = 6.35 g
Marking: 1 mark for correct moles of FeCl₂; 1 mark for correct mass (6.35 g).
12. Titration calculation: 25.0 cm³ NaOH requires 22.5 cm³ of 0.200 mol/dm³ HCl. NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l). Calculate the concentration of NaOH in mol/dm³.
(3 marks)
Answer:
Moles of HCl = (22.5 / 1000) × 0.200 = 0.00450 mol
From equation, 1 mol HCl reacts with 1 mol NaOH.
Moles of NaOH = 0.00450 mol
Concentration of NaOH = 0.00450 / (25.0 / 1000) = 0.00450 / 0.0250 = 0.180 mol/dm³
Marking: 1 mark for correct moles of HCl; 1 mark for correct moles of NaOH (1:1 ratio); 1 mark for correct concentration (0.180 mol/dm³).
13. A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. Determine its empirical formula.
(2 marks)
Answer:
C: 52.2 / 12 = 4.35 mol
H: 13.0 / 1 = 13.0 mol
O: 34.8 / 16 = 2.175 mol
Divide by smallest (2.175):
C: 4.35 / 2.175 = 2
H: 13.0 / 2.175 = 5.98 ≈ 6
O: 2.175 / 2.175 = 1
Empirical formula = C₂H₆O
Marking: 1 mark for correct mole calculations; 1 mark for correct empirical formula C₂H₆O.
Section D: Data Analysis and Application (10 marks)
14. Hydrated copper(II) sulfate: 5.00 g of CuSO₄·xH₂O heated to give 3.20 g anhydrous CuSO₄.
(a) Calculate the mass of water removed. (1 mark)
Answer: Mass of water = 5.00 – 3.20 = 1.80 g
Marking: 1 mark for correct answer 1.80 g.
(b) Calculate the number of moles of anhydrous CuSO₄ and the number of moles of water removed. (2 marks)
Answer:
Mr of CuSO₄ = 63.5 + 32 + (4 × 16) = 159.5
Moles of CuSO₄ = 3.20 / 159.5 = 0.0201 mol [Accept 0.020 mol]
Moles of H₂O = 1.80 / 18 = 0.100 mol
Marking: 1 mark for moles of CuSO₄; 1 mark for moles of H₂O.
(c) Determine the value of x in CuSO₄·xH₂O. (1 mark)
Answer:
Ratio H₂O : CuSO₄ = 0.100 / 0.0201 = 4.98 ≈ 5
x = 5
Marking: 1 mark for x = 5.
15. Ammonia production: N₂(g) + 3H₂(g) → 2NH₃(g). 280 g N₂ and 60 g H₂ used.
(a) Calculate the number of moles of nitrogen and hydrogen used. (2 marks)
Answer:
Moles of N₂ = 280 / 28 = 10.0 mol
Moles of H₂ = 60 / 2 = 30.0 mol
Marking: 1 mark for each correct mole value.
(b) Identify the limiting reactant and explain your reasoning. (2 marks)
Answer:
From equation, 1 mol N₂ reacts with 3 mol H₂.
10.0 mol N₂ would require 30.0 mol H₂.
Available H₂ is exactly 30.0 mol, so neither reactant is in excess; both are completely consumed. [Accept: both are limiting / neither is in excess / the reactants are in the exact stoichiometric ratio.]
Marking: 1 mark for correct comparison of mole ratio; 1 mark for correct conclusion with reasoning.
(c) Calculate the maximum mass of ammonia that can be produced. (2 marks)
Answer:
From equation, 1 mol N₂ produces 2 mol NH₃.
Moles of NH₃ = 2 × 10.0 = 20.0 mol
Mr of NH₃ = 14 + (3 × 1) = 17
Mass of NH₃ = 20.0 × 17 = 340 g
Marking: 1 mark for correct moles of NH₃; 1 mark for correct mass (340 g).
16. A student dissolves 8.40 g of sodium hydrogen carbonate, NaHCO₃, in water to make 250 cm³ of solution.
(a) Calculate the concentration of the solution in mol/dm³. (2 marks)
Answer:
Mr of NaHCO₃ = 23 + 1 + 12 + (3 × 16) = 84
Moles of NaHCO₃ = 8.40 / 84 = 0.100 mol
Volume in dm³ = 250 / 1000 = 0.250 dm³
Concentration = 0.100 / 0.250 = 0.400 mol/dm³
Marking: 1 mark for correct moles of NaHCO₃; 1 mark for correct concentration (0.400 mol/dm³).
17. The actual mass of ammonia produced in question 15 is 255 g. Calculate the percentage yield. (2 marks)
Answer:
Theoretical yield = 340 g (from 15c)
Percentage yield = (255 / 340) × 100 = 75.0%
Marking: 1 mark for correct formula; 1 mark for correct answer (75.0%). Accept 75%.
18. Calculate the number of moles of sodium ions in 250 cm³ of 0.400 mol/dm³ sodium carbonate solution, Na₂CO₃. (2 marks)
Answer:
Moles of Na₂CO₃ = (250 / 1000) × 0.400 = 0.100 mol
Each Na₂CO₃ provides 2 Na⁺ ions.
Moles of Na⁺ = 0.100 × 2 = 0.200 mol
Marking: 1 mark for moles of Na₂CO₃; 1 mark for correct moles of Na⁺ (0.200 mol).
19. A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56. Determine its molecular formula. (2 marks)
Answer:
C: 85.7 / 12 = 7.14 mol
H: 14.3 / 1 = 14.3 mol
Divide by smallest (7.14):
C: 7.14 / 7.14 = 1
H: 14.3 / 7.14 = 2.00 ≈ 2
Empirical formula = CH₂
Mr of CH₂ = 12 + 2 = 14
n = 56 / 14 = 4
Molecular formula = C₄H₈
Marking: 1 mark for correct empirical formula CH₂; 1 mark for correct molecular formula C₄H₈.
20. Explain why the actual yield of a chemical reaction is often less than the theoretical yield. Give one specific reason. (1 mark)
Answer: Any one of the following:
- Some product may be lost during purification (e.g., filtration, recrystallisation).
- The reaction may be reversible and not go to completion.
- Side reactions may occur, producing unwanted by-products.
- Some reactants may not react completely.
Marking: 1 mark for any valid reason with a brief explanation.