Secondary 3 Chemistry Stoichiometry Moles Quiz

Secondary 3 Chemistry Quiz - Stoichiometry Moles: Answer Key

Total Marks: 40

Section A: Multiple Choice (Questions 1–5)

1. Answer: B (2 marks)

• Explanation: The mole is a unit of amount of substance. One mole contains exactly $6.02 \times 10^{23}$ particles (atoms, molecules, ions, or formula units). This number is the Avogadro constant. Option A is wrong because the mass in grams equals the molar mass, which differs for each substance. Option C describes molar volume, which is only true for gases at room temperature and pressure. Option D is incorrect because the relative atomic mass is dimensionless, not expressed in kilograms.

2. Answer: B (2 marks)

• Explanation: Relative atomic mass is defined relative to carbon-12. A relative atomic mass of 24 means one mole of magnesium atoms has a mass of 24 g. Option A confuses atomic mass with molar mass. Option C confuses the mole concept. Option D is incorrect; the relative atomic mass compares to 1/12 of a carbon-12 atom, not the whole atom.

3. Answer: D (2 marks)

• Explanation: Molar mass of $\text{CaCO}_3$ = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100 g/mol

4. Answer: A (2 marks)

• Explanation:
  • Molar mass of $\text{NaOH}$ = 23 + 16 + 1 = 40 g/mol
  • Moles = mass ÷ molar mass = 8.0 ÷ 40 = 0.20 mol

5. Answer: A (2 marks)

• Explanation: Calculate moles in each sample:

  • A: $\text{H}_2$: 1.0 ÷ 2 = 0.50 mol
  • B: He: 4.0 ÷ 4 = 1.0 mol
  • C: $\text{O}_2$: 8.0 ÷ 32 = 0.25 mol
  • D: $\text{CH}_4$: 16.0 ÷ 16 = 1.0 mol

  Wait — rechecking: Option A gives 0.50 mol, not the greatest. Correct calculation shows B and D both give 1.0 mol. However, the question asks for greatest number of molecules. By Avogadro's law, equal moles contain equal molecules. But since B and D are equal at 1.0 mol each, and the question has only one answer, re-evaluation: The original intended answer is A if considering the question may have intended 2.0 g H₂, or this is a designed "trick" where students must recalculate carefully. Correct answer based on exact values: B or D (tie). Given standard exam design with one correct answer, B (helium, monatomic, simplest) is the most common intended answer if values differ slightly, or the question contains a slight error. Teaching note: Always recalculate carefully; if two options tie, re-read the question.

  Revised clarification: If the question stands as written, B and D are equal. In practice, exam papers avoid this. Assuming the intended question has 2.0 g H₂, answer is A. With values as given, B = D = 1.0 mol; select B as the first correct option, or flag for review.

  Standardized answer: B (accept D with working shown, or acknowledge tie if student identifies it).

Section B: Structured Questions (Questions 6–15)

6. (a) $6.02 \times 10^{23}$ mol⁻¹ (units: per mole or mol⁻¹) (2 marks: 1 for value, 1 for units)

• Explanation: The Avogadro constant is the number of particles in one mole of any substance. The units are "per mole" (mol⁻¹) because it is a count per unit amount.

(b) Number of atoms = 2.0 × $6.02 \times 10^{23}$ = $1.204 \times 10^{24}$ atoms (2 marks)

• Working: Number of particles = moles × Avogadro constant
• Teaching note: Iron is monatomic in these calculations, so atoms = particles.

7. Mass = 0.25 × 342 = 85.5 g (3 marks)

• Working:

  • Molar mass of $\text{Al}_2(\text{SO}_4)_3$ = (2 × 27) + 3 × [32 + (4 × 16)]
  • = 54 + 3 × [32 + 64]
  • = 54 + 3 × 96
  • = 54 + 288 = 342 g/mol
  • Mass = moles × molar mass = 0.25 × 342 = 85.5 g

• Marking: 1 mark for correct molar mass, 1 mark for correct method, 1 mark for final answer with unit.

8. (a) Moles = 4.8 ÷ 24 = 0.20 mol (2 marks)

• Working: For gases at r.t.p.: moles = volume (in dm³) ÷ 24

(b) Mass = 0.20 × 28 = 5.6 g (2 marks)

• Working:

  • Nitrogen gas is $\text{N}_2$, so M$_r$ = 2 × 14 = 28
  • Mass = moles × molar mass = 0.20 × 28 = 5.6 g

• Common error: Using atomic mass 14 instead of molecular mass 28 (lose 1 mark).

9. Mass of copper = 6.4 g (3 marks)

• Working:

  • Moles of CuO = 8.0 ÷ (64 + 16) = 8.0 ÷ 80 = 0.10 mol
  • Ratio CuO : Cu = 1 : 1, so moles of Cu = 0.10 mol
  • Mass of Cu = 0.10 × 64 = 6.4 g

• Marking: 1 mark for moles of CuO, 1 mark for using correct ratio, 1 mark for final answer.

10. (a) Mass of CaO = 11.2 g (2 marks)

• Working:
  • Moles of $\text{CaCO}_3$ = 20.0 ÷ (40 + 12 + 48) = 20.0 ÷ 100 = 0.20 mol
  • Ratio $\text{CaCO}_3$ : CaO = 1 : 1, so moles of CaO = 0.20 mol
  • Mass of CaO = 0.20 × (40 + 16) = 0.20 × 56 = 11.2 g

(b) Volume of CO₂ = 4.8 dm³ (2 marks)

• Working:
  • Moles of CO₂ = 0.20 mol (1:1 ratio)
  • Volume at r.t.p. = 0.20 × 24 = 4.8 dm³

11. Volume of H₂ = 2.4 dm³ (3 marks)

• Working:

  • Moles of Fe = 5.6 ÷ 56 = 0.10 mol
  • Ratio Fe : H₂ = 1 : 1, so moles of H₂ = 0.10 mol
  • Volume at r.t.p. = 0.10 × 24 = 2.4 dm³

• Marking: 1 mark each for moles of Fe, moles of H₂, final volume.

12. (a) The empirical formula is the simplest whole number ratio of atoms present in a compound. (1 mark)

(b) Empirical formula = CH₂O (3 marks)

• Working:

• Empirical formula = CH₂O

• Marking: 1 mark for correct moles, 1 mark for ratio, 1 mark for formula.

13. Mass of O₂ = 40.0 g (3 marks)

• Working:

  • Molar mass of $\text{C}_3\text{H}_8$ = (3 × 12) + (8 × 1) = 36 + 8 = 44 g/mol
  • Moles of $\text{C}_3\text{H}_8$ = 11.0 ÷ 44 = 0.25 mol
  • Ratio $\text{C}_3\text{H}_8$ : O₂ = 1 : 5, so moles of O₂ = 0.25 × 5 = 1.25 mol
  • Mass of O₂ = 1.25 × 32 = 40.0 g

• Marking: 1 mark for moles of propane, 1 mark for moles of oxygen using ratio, 1 mark for mass.

14. (a) Empirical formula = NO₂ (2 marks)

• Working:

(b) Molecular formula = N₂O₄ (2 marks)

• Working:
  • Mass of empirical formula unit = 14 + (2 × 16) = 46
  • $n$ = 92 ÷ 46 = 2
  • Molecular formula = 2 × (NO₂) = N₂O₄

15. (a) Moles of Zn = 6.5 ÷ 65 = 0.10 mol (1 mark)

(b) Volume of H₂ = 2.4 dm³ (2 marks)

• Working:
  • Ratio Zn : H₂ = 1 : 1, so moles of H₂ = 0.10 mol
  • Volume = 0.10 × 24 = 2.4 dm³

Section C: Application and Data Analysis (Questions 16–20)

16. (a) Mass of Mg = 27.8 – 25.4 = 2.4 g; Mass of MgO = 29.4 – 25.4 = 4.0 g (2 marks)

• Working: Mass of magnesium = (mass before) – (mass of empty container)
• Mass of MgO = (mass after heating) – (mass of empty container)

(b) Moles of Mg = 2.4 ÷ 24 = 0.10 mol (1 mark)

(c) Any valid reason: Some magnesium oxide may have escaped as smoke/sparks; not all magnesium reacted; magnesium nitride also formed (reaction with nitrogen in air); incomplete heating. Accept any reasonable physical source of error. (1 mark)

• Common errors: Students often suggest "heat loss" which is incorrect for mass increase reactions. This is a combustion where product mass should exceed reactant mass if all Mg converts to MgO.

17. (a) To remove all air/oxygen from the apparatus and prevent an explosive mixture with hydrogen. (1 mark)

• Teaching note: Hydrogen forms explosive mixtures with air. Purging ensures only hydrogen and copper oxide are present.

(b) (i) Mass of copper = 24.58 – 18.42 = 6.16 g (1 mark)

(ii) Mass of oxygen = 26.10 – 24.58 = 1.52 g (or total oxide mass 7.68 g – 6.16 g = 1.52 g) (1 mark)

(c) Empirical formula = CuO (2 marks)

• Working:

• Ratio ≈ 1:1, so CuO

• Marking: 1 mark for method (moles and ratio), 1 mark for correct formula.

18. (a) Mass of water lost = 4.99 – 3.19 = 1.80 g (1 mark)

(b) x = 5 (3 marks)

• Working:

  • Molar mass of anhydrous $\text{CuSO}_4$ = 64 + 32 + (4 × 16) = 160 g/mol
  • Moles of $\text{CuSO}_4$ = 3.19 ÷ 160 = 0.0199375 mol ≈ 0.0200 mol
  • Molar mass of water = 18 g/mol
  • Moles of water = 1.80 ÷ 18 = 0.10 mol
  • Ratio: $\frac{\text{moles H}_2\text{O}}{\text{moles CuSO}_4} = \frac{0.10}{0.0200} = 5$
  • Therefore x = 5, formula is $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$

• Marking: 1 mark for moles of CuSO₄, 1 mark for moles of water, 1 mark for ratio and value of x.

19. (a) Mass of carbon = 2.4 g (2 marks)

• Working:

  • Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol
  • Moles of CO₂ = 8.8 ÷ 44 = 0.20 mol
  • Each CO₂ contains one C atom, so moles of C = 0.20 mol
  • Mass of C = 0.20 × 12 = 2.4 g

• Alternative: Mass fraction of C in CO₂ = 12/44; mass of C = 8.8 × (12/44) = 2.4 g

(b) Mass of hydrogen = 0.60 g (2 marks)

• Working:

  • Molar mass of H₂O = (2 × 1) + 16 = 18 g/mol
  • Moles of H₂O = 5.4 ÷ 18 = 0.30 mol
  • Each H₂O contains two H atoms, so moles of H = 0.30 × 2 = 0.60 mol
  • Mass of H = 0.60 × 1 = 0.60 g

• Alternative: Mass fraction of H in H₂O = 2/18; mass of H = 5.4 × (2/18) = 0.60 g

(c) Empirical formula = C₂H₆O (2 marks)

• Working:
  • Mass of O in compound = 4.6 – 2.4 – 0.60 = 1.6 g
  • Moles of O = 1.6 ÷ 16 = 0.10 mol

• Empirical formula = C₂H₆O

• Marking: 1 mark for correct moles and method, 1 mark for correct formula (must include oxygen detection and calculation).

20. (a) Moles of Na₂S₂O₃ = 0.0050 mol or 5.0 × 10⁻³ mol (2 marks)

• Working:
  • Moles = concentration × volume (in dm³)
  • Volume = 50 cm³ = 50/1000 = 0.050 dm³
  • Moles = 0.10 × 0.050 = 0.0050 mol or 5.0 × 10⁻³ mol

(b) Time = 50 s (1 mark)

• Working: From graph: at 0.10 mol/dm³, 1/time = 0.020 s⁻¹, so time = 1 ÷ 0.020 = 50 s
• Teaching note: Read y-axis carefully; reciprocal relationship means higher concentration gives faster reaction (shorter time), and 1/time is directly proportional to rate.

(c) The rate of reaction is directly proportional to the concentration of sodium thiosulfate. OR "As concentration doubles, rate doubles." (1 mark)

• Key phrase needed: Directly proportional / straight line through origin / first order with respect to Na₂S₂O₃.

Marking Summary