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A Level H1 Chemistry Stoichiometry Moles Quiz

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A-Level Chemistry H1 Quiz - Stoichiometry Moles

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

Duration: 50 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.

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

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

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

3. What is the relative molecular mass, $M_r$ , of carbon dioxide, $CO_2$ ? (C = 12.0, O = 16.0) [1]

4. A sample contains $3.01 \times 10^{23}$ atoms of neon. Calculate the number of moles of neon. (Avogadro constant $N_A = 6.02 \times 10^{23}$ mol⁻¹) [2]

5. Calculate the mass of 0.250 mol of sodium chloride, NaCl. (Na = 23.0, Cl = 35.5) [2]

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

6. A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula. (C = 12.0, H = 1.0, O = 16.0) [3]

7. The empirical formula of a compound is CH₂O and its relative molecular mass is 180. Find its molecular formula. [2]

8. Write a balanced equation for the complete combustion of ethanol, $C_2H_5OH$ . [2]

9. Balance the following redox equation in acidic medium using half-equations: $MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}$ Show the half-equations and the final balanced equation. [3]

10. State the oxidation number of manganese in $MnO_4^-$ and of iron in $Fe^{2+}$ . [2]

Section C: Stoichiometric Calculations (Questions 11–15)

11. Calculate the volume of 0.100 mol dm⁻³ hydrochloric acid required to exactly neutralise 25.0 cm³ of 0.0800 mol dm⁻³ sodium hydroxide. [3]

12. In the reaction: $2Na + Cl_2 \rightarrow 2NaCl$ Calculate the mass of sodium chloride formed when 4.60 g of sodium reacts completely with excess chlorine. (Na = 23.0, Cl = 35.5) [3]

13. At room temperature and pressure (r.t.p.), 1 mol of gas occupies 24.0 dm³. What volume is occupied by 0.500 mol of oxygen gas, $O_2$ ? [2]

14. A solution is prepared by dissolving 5.00 g of anhydrous copper(II) sulfate, $CuSO_4$ , in water to make 250 cm³ of solution. Calculate the concentration in mol dm⁻³. (Cu = 63.5, S = 32.1, O = 16.0) [3]

15. Magnesium reacts with dilute sulfuric acid: $Mg + H_2SO_4 \rightarrow MgSO_4 + H_2$ If 0.120 g of magnesium is used, calculate the volume of hydrogen gas produced at r.t.p. (Mg = 24.3; molar gas volume = 24.0 dm³ mol⁻¹) [3]

Section D: Applied and Data-Based Stoichiometry (Questions 16–20)

16. A sample of a hydrocarbon burns completely in oxygen to give 4.40 g of $CO_2$ and 1.80 g of $H_2O$ . Determine the empirical formula of the hydrocarbon. (C = 12.0, H = 1.0, O = 16.0) [4]

17. The following data were obtained for the reaction: $A + 2B \rightarrow C$ Initial amounts: 0.040 mol A, 0.10 mol B. Identify the limiting reactant and calculate the amount of C formed. [3]

18. A tablet contains calcium carbonate, $CaCO_3$ . It is dissolved in excess hydrochloric acid: $CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$ The $CO_2$ evolved occupies 48.0 cm³ at r.t.p. Calculate the mass of $CaCO_3$ in the tablet. (Ca = 40.1, C = 12.0, O = 16.0; molar gas volume = 24.0 dm³ mol⁻¹) [3]

19. A student determines the purity of a sample of sodium carbonate, $Na_2CO_3$ . 1.06 g of impure sample requires 20.0 cm³ of 0.500 mol dm⁻³ HCl for complete reaction: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + CO_2 + H_2O$ Calculate the percentage purity of the sample. (Na = 23.0, C = 12.0, O = 16.0) [4]

20. A fertiliser labelled "15% nitrogen by mass" is analysed. A 2.00 g sample yields 0.300 g of $NH_3$ upon Kjeldahl digestion and distillation. Given that all nitrogen in the sample became $NH_3$ , calculate the experimental %N and comment on the label claim. (N = 14.0, H = 1.0) [4]

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Answers

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

Total Marks: 40
Topic: The Mole Concept and Stoichiometry (Syllabus 8873 Core Idea 3)

Section A: Foundations of the Mole Concept

Q1. [1 mark]
The mole is the amount of substance that contains as many entities (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12; numerically equal to the Avogadro constant, $6.02 \times 10^{23}$ mol⁻¹.
Teaching note: A mole links macroscopic mass to microscopic particle count via $N_A$ .

Q2. [2 marks]
$n = \frac{m}{M_r} = \frac{4.00}{4.00} = 1.00$ mol.
Marks: 1 for formula, 1 for correct answer.
Common mistake: Using wrong $M_r$ or forgetting units.

Q3. [1 mark]
$M_r(CO_2) = 12.0 + 2(16.0) = 44.0$ .

Q4. [2 marks]
$n = \frac{N}{N_A} = \frac{3.01 \times 10^{23}}{6.02 \times 10^{23}} = 0.500$ mol.
Marks: 1 for method, 1 for answer.

Q5. [2 marks]
$M_r(NaCl) = 23.0 + 35.5 = 58.5$
$m = n \times M_r = 0.250 \times 58.5 = 14.6$ g.
Marks: 1 for molar mass, 1 for mass.

Section B: Formulae and Equations

Q6. [3 marks]
Assume 100 g: C = 40.0 g, H = 6.7 g, O = 53.3 g.
Moles: C = 40.0/12.0 = 3.33; H = 6.7/1.0 = 6.7; O = 53.3/16.0 = 3.33.
Ratio = 1 : 2 : 1 → $CH_2O$ .
Marks: 1 for moles, 1 for ratio, 1 for formula.

Q7. [2 marks]
Empirical mass = 12+2+16 = 30.
$n = 180/30 = 6$ . Molecular formula = $(CH_2O)_6 = C_6H_{12}O_6$ .

Q8. [2 marks]
$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$
Marks: 1 for products, 1 for balancing.

Q9. [3 marks]
Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Multiply oxidation by 5 and add:
$MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$
Marks: 1 each for half-eq, 1 for final.

Q10. [2 marks]
Mn in $MnO_4^-$ : $x + 4(-2) = -1 \Rightarrow x = +7$ .
Fe in $Fe^{2+} = +2$ .
Marks: 1 each.

Section C: Stoichiometric Calculations

Q11. [3 marks]
$HCl + NaOH \rightarrow NaCl + H_2O$ (1:1)
$n(NaOH) = 0.0800 \times 25.0/1000 = 2.00 \times 10^{-3}$ mol
$V(HCl) = n/c = 2.00\times10^{-3}/0.100 = 0.0200$ dm³ = 20.0 cm³.
Marks: 1 mole NaOH, 1 volume, 1 unit.

Q12. [3 marks]
$n(Na) = 4.60/23.0 = 0.200$ mol
From eq: 2 mol Na → 2 mol NaCl, so 0.200 mol NaCl.
$M_r(NaCl)=58.5$ , $m = 0.200 \times 58.5 = 11.7$ g.

Q13. [2 marks]
$V = n \times 24.0 = 0.500 \times 24.0 = 12.0$ dm³.

Q14. [3 marks]
$M_r(CuSO_4) = 63.5+32.1+64.0 = 159.6$
$n = 5.00/159.6 = 0.0313$ mol
$c = 0.0313 / (250/1000) = 0.125$ mol dm⁻³.

Q15. [3 marks]
$n(Mg) = 0.120/24.3 = 4.94\times10^{-3}$ mol
1 mol Mg → 1 mol $H_2$ , so $n(H_2)=4.94\times10^{-3}$
$V = 4.94\times10^{-3} \times 24.0 = 0.118$ dm³ = 118 cm³.

Section D: Applied and Data-Based Stoichiometry

Q16. [4 marks]
From $CO_2$ : $n(C) = 4.40/44.0 = 0.100$ mol
From $H_2O$ : $n(H_2O)=1.80/18.0=0.100$ mol → $n(H)=0.200$ mol
Ratio C:H = 1:2 → $CH_2$ .
Marks: 1 C, 1 H, 1 ratio, 1 formula.

Q17. [3 marks]
For 0.040 mol A need 0.080 mol B. Available B = 0.10 mol, so A limiting.
From eq, 1 mol A → 1 mol C, so C = 0.040 mol.
Marks: 1 limiting, 1 calc, 1 amount.

Q18. [3 marks]
$n(CO_2) = 48.0/24000 = 2.00\times10^{-3}$ mol
1:1 with $CaCO_3$ → $n=2.00\times10^{-3}$
$M_r=100.1$ , $m = 2.00\times10^{-3}\times100.1 = 0.200$ g.

Q19. [4 marks]
$n(HCl)=0.500\times20.0/1000=0.0100$ mol
From eq, 2 HCl : 1 $Na_2CO_3$ → $n(Na_2CO_3)=0.00500$ mol
$M_r=106.0$ , pure mass = 0.530 g
% purity = 0.530/1.06 × 100 = 50.0%.
Marks: 1 HCl, 1 Na2CO3, 1 mass, 1 %.

Q20. [4 marks]
$M_r(NH_3)=17.0$ , $n(NH_3)=0.300/17.0=0.01765$ mol → $n(N)=0.01765$ mol
$m(N)=0.01765\times14.0=0.247$ g
%N = 0.247/2.00 × 100 = 12.4%
Label claims 15%; experimental lower, so label overstates N content.
Marks: 1 mole NH3, 1 mass N, 1 %, 1 comment.