A Level H1 Chemistry Stoichiometry Moles Quiz

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A-Level Chemistry H1 Quiz - Stoichiometry Moles

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 40

Duration: 60 minutes
Total Marks: 40

Instructions:

• Answer all 20 questions.
• Show all working clearly for calculation questions.
• Use the Data Booklet if needed.
• Write units where required.

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Section A: Foundations of the Mole Concept (Questions 1–5)

1. Define the mole in terms of the Avogadro constant. [1]

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2. Calculate the number of moles in 4.00 g of helium (He). (Relative atomic mass: He = 4.00) [2]

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3. What is the relative molecular mass, $M_r$, of $CuSO_4 \cdot 5H_2O$?
(Given: Cu = 63.5, S = 32.1, O = 16.0, H = 1.0) [2]

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4. A sample contains $3.01 \times 10^{23}$ atoms of carbon. Calculate the mass of this sample. (C = 12.0; $N_A = 6.02 \times 10^{23}$ mol⁻¹) [2]

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5. State the meaning of the term "empirical formula". [1]

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Section B: Formulae and Equations (Questions 6–10)

6. A compound has the following composition by mass: Na = 36.5%, S = 25.4%, O = 38.1%. Determine its empirical formula. (Na = 23.0, S = 32.1, O = 16.0) [3]

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7. A hydrocarbon has molecular formula $C_6H_{12}$. Given its empirical formula is $CH_2$, calculate its molecular mass. (C = 12.0, H = 1.0) [2]

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8. Balance the following equation:
$\_$ $Al$ + $\_$ $O_2$ → $\_$ $Al_2O_3$ [2]

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9. Write the balanced ionic half-equation for the oxidation of $Fe^{2+}$ to $Fe^{3+}$. [1]

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10. Construct the full redox equation from these half-equations:
$Zn \rightarrow Zn^{2+} + 2e^-$
$Cu^{2+} + 2e^- \rightarrow Cu$ [2]

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Section C: Stoichiometric Calculations (Questions 11–15)

11. Calculate the mass of $CO_2$ produced when 5.00 g of $CaCO_3$ is completely decomposed by heat.
$CaCO_3 \rightarrow CaO + CO_2$
(Ca = 40.1, C = 12.0, O = 16.0) [3]

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12. 25.0 cm³ of 0.100 mol dm⁻³ $HCl$ reacts exactly with 20.0 cm³ of $Ba(OH)_2$.
$2HCl + Ba(OH)_2 \rightarrow BaCl_2 + 2H_2O$
Calculate the concentration of $Ba(OH)_2$ in mol dm⁻³. [3]

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13. At room temperature and pressure (r.t.p.), 1 mole of gas occupies 24.0 dm³. What volume is occupied by 0.250 mol of $O_2$ gas? [2]

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14. In an experiment, 0.200 mol of $H_2$ reacts with excess $N_2$:
$N_2 + 3H_2 \rightarrow 2NH_3$
Calculate the amount, in moles, of $NH_3$ formed. [2]

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15. A solution contains 7.10 g of $Na_2SO_4$ in 250 cm³ of solution. Calculate its concentration in mol dm⁻³. (Na = 23.0, S = 32.1, O = 16.0) [3]

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Section D: Data Interpretation and Redox (Questions 16–20)

16. The following data were obtained for the combustion of a sample:

Image pending generation: table for Q16.

Using the table, calculate the mass of carbon and the mass of hydrogen in the 2.40 g sample. (C = 12.0, H = 1.0, O = 16.0) [3]

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17. From the data in Q16, determine the empirical formula of compound X, assuming it contains only C, H and O. [3]

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18. A redox reaction occurs:
$Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$
State the species reduced and the change in oxidation number of chlorine. [2]

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19. Calculate the volume of 0.500 mol dm⁻³ $AgNO_3$ solution needed to react completely with 0.0100 mol of $Cl^-$ ions:
$Ag^+ + Cl^- \rightarrow AgCl$ [2]

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20. 10.0 cm³ of a 0.200 mol dm⁻³ $KMnO_4$ solution contains excess acid. It oxidises $Fe^{2+}$ to $Fe^{3+}$. The half-equations are:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
$Fe^{2+} \rightarrow Fe^{3+} + e^-$
Calculate the amount, in moles, of $Fe^{2+}$ that can be oxidised. [3]

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

Total Marks: 40
Topic: Stoichiometry & Moles (Syllabus 8873, Core Idea 3)

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Section A: Foundations of the Mole Concept

Q1. [1 mark]
A mole is the amount of substance that contains the same number of specified entities as there are atoms in exactly 12 g of carbon-12; this number is the Avogadro constant, $N_A = 6.02 \times 10^{23}$ mol⁻¹.
Teaching note: The mole links microscopic particles to measurable mass. Do not confuse "mass of carbon-12" with just any carbon.

Q2. [2 marks]
Step 1: $n = \frac{m}{M_r} = \frac{4.00}{4.00} = 1.00$ mol.
Mark breakdown: 1 mark for correct formula, 1 mark for correct answer with unit implied.
Common mistake: Using wrong $M_r$ or forgetting it is an element (use Ar not Mr).

Q3. [2 marks]
$M_r = 63.5 + 32.1 + 4(16.0) + 5(2(1.0)+16.0) = 63.5+32.1+64.0+90.0 = 249.6$
Mark breakdown: 1 mark for summing Cu+S+4O, 1 mark for adding 5H₂O correctly.
Note: Hydrated salt includes water of crystallisation.

Q4. [2 marks]
$n = \frac{3.01 \times 10^{23}}{6.02 \times 10^{23}} = 0.500$ mol
$m = n \times M_r = 0.500 \times 12.0 = 6.00$ g
Mark breakdown: 1 mark for moles, 1 mark for mass.

Q5. [1 mark]
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound.

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Section B: Formulae and Equations

Q6. [3 marks]
Divide % by Ar:
Na: $36.5 / 23.0 = 1.59$
S: $25.4 / 32.1 = 0.791$
O: $38.1 / 16.0 = 2.38$
Ratio ÷ smallest (0.791): Na = 2.01 ≈ 2, S = 1, O = 3.01 ≈ 3
Empirical formula = $Na_2SO_3$
Mark breakdown: 1 mark for mole ratios, 1 mark for simplification, 1 mark for formula.

Q7. [2 marks]
Empirical mass $CH_2 = 12.0 + 2(1.0) = 14.0$
Molecular mass = $6 \times 14.0 = 84.0$
Mark breakdown: 1 mark for empirical mass, 1 mark for total.

Q8. [2 marks]
$4Al + 3O_2 \rightarrow 2Al_2O_3$
Mark breakdown: 1 mark for Al coefficient, 1 mark for O₂ and product balanced.

Q9. [1 mark]
$Fe^{2+} \rightarrow Fe^{3+} + e^-$

Q10. [2 marks]
Add half-equations (electrons cancel):
$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$
Mark breakdown: 1 mark for correct species, 1 mark for balanced equation.

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Section C: Stoichiometric Calculations

Q11. [3 marks]
$M_r(CaCO_3) = 40.1+12.0+48.0 = 100.1$
$n(CaCO_3) = 5.00/100.1 = 0.0500$ mol
1:1 ratio → $n(CO_2)=0.0500$ mol
$m(CO_2) = 0.0500 \times 44.0 = 2.20$ g
Mark breakdown: 1m moles CaCO₃, 1m moles CO₂, 1m mass.

Q12. [3 marks]
$n(HCl) = 0.100 \times (25.0/1000) = 0.00250$ mol
From eq: 2 mol HCl : 1 mol Ba(OH)₂ → $n = 0.00125$ mol
$c = 0.00125 / (20.0/1000) = 0.0625$ mol dm⁻³
Mark breakdown: 1m moles HCl, 1m moles Ba(OH)₂, 1m concentration.

Q13. [2 marks]
$V = n \times 24.0 = 0.250 \times 24.0 = 6.00$ dm³
Mark breakdown: 1m method, 1m answer.

Q14. [2 marks]
3 mol H₂ → 2 mol NH₃
$0.200 \times (2/3) = 0.133$ mol NH₃
Mark breakdown: 1m ratio, 1m answer.

Q15. [3 marks]
$M_r(Na_2SO_4)=2(23.0)+32.1+64.0=142.1$
$n = 7.10/142.1 = 0.0500$ mol
$c = 0.0500 / (250/1000) = 0.200$ mol dm⁻³
Mark breakdown: 1m Mr, 1m moles, 1m concentration.

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Section D: Data Interpretation and Redox

Q16. [3 marks]
Mass C = $(12.0/44.0) \times 5.28 = 1.44$ g
Mass H = $(2.0/18.0) \times 2.16 = 0.240$ g
Mark breakdown: 1m for C calc, 1m for H calc, 1m both correct values.
Image note: Table must show 2.40 g X, 5.28 g CO₂, 2.16 g H₂O.

Q17. [3 marks]
Mass O = 2.40 – 1.44 – 0.240 = 0.720 g
Moles: C = 1.44/12 = 0.120; H = 0.240/1 = 0.240; O = 0.720/16 = 0.0450
Ratio ÷0.0450 → C=2.67≈8/3, H=5.33≈16/3, O=1 → ×3 → C₈H₁₆O₃
Mark breakdown: 1m O mass, 1m mole ratios, 1m empirical formula.

Q18. [2 marks]
Species reduced: $Cl_2$ (0 to –1)
Change: oxidation number decreases from 0 to –1.
Mark breakdown: 1m species, 1m change.

Q19. [2 marks]
1:1 ratio → $n(Ag^+) = 0.0100$ mol
$V = n/c = 0.0100 / 0.500 = 0.0200$ dm³ = 20.0 cm³
Mark breakdown: 1m moles, 1m volume.

Q20. [3 marks]
$n(MnO_4^-) = 0.200 \times (10.0/1000) = 0.00200$ mol
1 mol $MnO_4^-$ accepts 5 mol e⁻ → total e⁻ = 0.0100 mol
1 mol $Fe^{2+}$ gives 1 e⁻ → $n(Fe^{2+}) = 0.0100$ mol
Mark breakdown: 1m moles MnO₄⁻, 1m electron calc, 1m Fe²⁺ moles.