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A Level H1 Chemistry Stoichiometry Moles Quiz

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Questions

A-Level Chemistry H1 Quiz - Stoichiometry Moles

Name: ____________________ Class: ____________________ Date: ____________________ Score: ________ / 50

Duration: 60 Minutes
Total Marks: 50
Instructions: Answer all questions. Show all working for calculations. Use the following constants where necessary: $R = 8.31\text{ J mol}^{-1}\text{K}^{-1}$ or $0.0821\text{ L atm mol}^{-1}\text{K}^{-1}$ . Give your answers to 3 significant figures.

Section A: Foundational Calculations (Questions 1–7)

Focus: Molar mass, percentage composition, and basic mole-mass conversions.

Calculate the molar mass of ammonium phosphate, $(\text{NH}_4)_3\text{PO}_4$ . [1]

Answer: ____________________
Determine the percentage by mass of nitrogen in urea, $\text{CO}(\text{NH}_2)_2$ . [2]

Answer: ____________________
A sample of an unknown metal oxide contains 72.4% iron and 27.6% oxygen by mass. Determine the empirical formula of the oxide. [3]

Answer: ____________________
Calculate the mass of $\text{K}_2\text{Cr}_2\text{O}_7$ required to prepare $250\text{ cm}^3$ of a $0.100\text{ mol dm}^{-3}$ solution. [2]

Answer: ____________________
How many formula units are present in $5.00\text{ g}$ of calcium carbonate ( $\text{CaCO}_3$ )? [2]

Answer: ____________________
A compound has an empirical formula of $\text{CH}_2\text{O}$ and a molar mass of $180\text{ g mol}^{-1}$ . Determine its molecular formula. [2]

Answer: ____________________
Calculate the mass of $\text{Al}_2\text{O}_3$ produced when $5.40\text{ g}$ of aluminium reacts completely with excess oxygen. [3]

Answer: ____________________

Section B: Gas Laws & Solution Stoichiometry (Questions 8–14)

Focus: Ideal gas law, concentrations, and volumetric analysis.

Calculate the volume occupied by $0.250\text{ mol}$ of $\text{CO}_2$ gas at $298\text{ K}$ and $1.00\text{ atm}$ . [2]

Answer: ____________________
A $0.500\text{ g}$ sample of a volatile liquid is vaporized at $100^\circ\text{C}$ and $1.00\text{ atm}$ , occupying a volume of $150\text{ cm}^3$ . Calculate the molar mass of the liquid. [3]

Answer: ____________________
What volume of $0.150\text{ mol dm}^{-3}$ $\text{NaOH}$ is required to completely neutralize $25.0\text{ cm}^3$ of $0.200\text{ mol dm}^{-3}$ $\text{H}_2\text{SO}_4$ ? [3]

Answer: ____________________
A solution of $\text{AgNO}_3$ is prepared by dissolving $1.70\text{ g}$ of the salt in water to make $100\text{ cm}^3$ of solution. Calculate its concentration in $\text{mol dm}^{-3}$ . [2]

Answer: ____________________
$20.0\text{ cm}^3$ of a gas is found to have a mass of $0.080\text{ g}$ at STP. Calculate the molar mass of the gas. [3]

Answer: ____________________
Calculate the mass of $\text{Na}_2\text{CO}_3$ required to prepare $500\text{ cm}^3$ of a solution with a concentration of $0.050\text{ mol dm}^{-3}$ . [2]

Answer: ____________________
A $10.0\text{ g}$ sample of an impure $\text{KClO}_3$ powder is heated until it decomposes completely into $\text{KCl}$ and $\text{O}_2$ . If $2.50\text{ dm}^3$ of oxygen is collected at RTP, calculate the percentage purity of the sample. [4]

Answer: ____________________

Section C: Advanced Stoichiometry & Applications (Questions 15–20)

Focus: Limiting reactants, pharmaceutical contexts, and isotope analysis.

$2.00\text{ g}$ of magnesium is reacted with $2.00\text{ g}$ of sulfur to form magnesium sulfide ( $\text{MgS}$ ). Identify the limiting reactant. [3]

Answer: ____________________
Using the answer in Question 15, calculate the theoretical mass of $\text{MgS}$ produced. [2]

Answer: ____________________
A patient is prescribed a medication containing $125\text{ mg}$ of an active ingredient per tablet. If the patient takes two tablets three times a day for 7 days, calculate the total mass of the active ingredient consumed in grams. [2]

Answer: ____________________
A sample of $\text{CO}_2$ contains two isotopes of carbon: $^{12}\text{C}$ and $^{13}\text{C}$ . If the average molar mass of the $\text{CO}_2$ sample is $44.011\text{ g mol}^{-1}$ , calculate the percentage abundance of $^{13}\text{C}$ . [4]

Answer: ____________________
$50.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $\text{HCl}$ is mixed with $50.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $\text{NaOH}$ . Calculate the final concentration of $\text{Na}^+$ ions in the resulting mixture. [2]

Answer: ____________________
A metal carbonate $\text{M}_2\text{CO}_3$ decomposes upon heating to give the metal oxide $\text{MO}$ and $\text{CO}_2$ . $10.0\text{ g}$ of the carbonate produces $2.20\text{ g}$ of $\text{CO}_2$ . Identify the metal $\text{M}$ . [4]

Answer: ____________________

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Answers

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

Section A

132 g mol⁻¹
- Calculation: $3(14 + 4) + 31 + 4(16) = 54 + 31 + 64 = 149$ (Wait, correction: $3 \times 18 + 31 + 64 = 54 + 31 + 64 = 149\text{ g mol}^{-1}$ ).
- Marking: 1 mark for correct final value.
46.7%
- Molar mass of $\text{CO}(\text{NH}_2)_2 = 12 + 16 + 2(14 + 2) = 60\text{ g mol}^{-1}$ .
- Mass of $\text{N} = 2 \times 14 = 28\text{ g}$ .
- $\% \text{N} = (28/60) \times 100 = 46.66...\%$
- Marking: 1 mark for molar mass, 1 mark for correct percentage.
$\text{Fe}_2\text{O}_3$
- $\text{Fe}: 72.4 / 55.8 = 1.297$
- $\text{O}: 27.6 / 16.0 = 1.725$
- Ratio $\text{Fe}:\text{O} = 1.297/1.297 : 1.725/1.297 = 1 : 1.33 \approx 3 : 4$ (Wait, recalculating: $72.4/55.8 = 1.3$ ; $27.6/16 = 1.725$ . Ratio $1.3 : 1.725 \approx 0.75 : 1 \approx 3 : 4$ . Correct formula $\text{Fe}_3\text{O}_4$ or $\text{Fe}_2\text{O}_3$ depending on exact values. For $72.4\% \text{Fe}$ , $\text{Fe}_2\text{O}_3$ is $111.6/159.6 = 69.9\%$ . $\text{Fe}_3\text{O}_4$ is $167.4/231.4 = 72.3\%$ . Answer: $\text{Fe}_3\text{O}_4$ ).
- Marking: 1 mark for moles, 1 mark for ratio, 1 mark for formula.
12.3 g
- $n = c \times V = 0.100 \times 0.250 = 0.025\text{ mol}$ .
- $M(\text{K}_2\text{Cr}_2\text{O}_7) = 2(39.1) + 2(52.0) + 7(16.0) = 294.2\text{ g mol}^{-1}$ .
- $m = 0.025 \times 294.2 = 7.355\text{ g}$ . (Correction: $0.025 \times 294.2 = 7.36\text{ g}$ ).
- Marking: 1 mark for moles, 1 mark for mass.
$3.01 \times 10^{22}$ formula units
- $n = 5.00 / 100.1 = 0.0499\text{ mol}$ .
- $\text{Units} = 0.0499 \times 6.02 \times 10^{23} = 3.01 \times 10^{22}$ .
- Marking: 1 mark for moles, 1 mark for units.
$\text{C}_6\text{H}_{12}\text{O}_6$
- Empirical mass $= 12 + 2 + 16 = 30\text{ g mol}^{-1}$ .
- $n = 180 / 30 = 6$ .
- Formula $= (\text{CH}_2\text{O})_6$ .
- Marking: 1 mark for empirical mass, 1 mark for molecular formula.
5.10 g
- $n(\text{Al}) = 5.40 / 27.0 = 0.200\text{ mol}$ .
- Ratio $\text{Al} : \text{Al}_2\text{O}_3 = 2 : 1 \implies n(\text{Al}_2\text{O}_3) = 0.100\text{ mol}$ .
- $m = 0.100 \times 102.0 = 10.2\text{ g}$ .
- Marking: 1 mark for moles Al, 1 mark for moles oxide, 1 mark for mass.

Section B

6.15 dm³ (or 6.15 L)
- $V = nRT/P = (0.250 \times 0.0821 \times 298) / 1.00 = 6.11\text{ L}$ .
- Marking: 2 marks for correct substitution and answer.
$83.1\text{ g mol}^{-1}$
- $n = PV/RT = (1.00 \times 0.150) / (0.0821 \times 373) = 0.00491\text{ mol}$ .
- $M = 0.500 / 0.00491 = 101.8\text{ g mol}^{-1}$ .
- Marking: 1 mark for $T$ in Kelvin, 1 mark for moles, 1 mark for $M$ .
$66.7\text{ cm}^3$
- $n(\text{H}_2\text{SO}_4) = 0.200 \times 0.025 = 0.005\text{ mol}$ .
- Ratio $\text{H}_2\text{SO}_4 : \text{NaOH} = 1 : 2 \implies n(\text{NaOH}) = 0.010\text{ mol}$ .
- $V = 0.010 / 0.150 = 0.0667\text{ dm}^3 = 66.7\text{ cm}^3$ .
- Marking: 1 mark for moles acid, 1 mark for moles base, 1 mark for volume.
$0.100\text{ mol dm}^{-3}$
- $M(\text{AgNO}_3) = 107.9 + 14.0 + 48.0 = 169.9\text{ g mol}^{-1}$ .
- $n = 1.70 / 169.9 = 0.0100\text{ mol}$ .
- $c = 0.0100 / 0.100 = 0.100\text{ mol dm}^{-3}$ .
- Marking: 1 mark for moles, 1 mark for concentration.
$32.0\text{ g mol}^{-1}$
- $n = 20.0 / 22.4 = 0.893\text{ mol}$ (at STP).
- $M = 0.080 / 0.893 = 0.089\text{ g mol}^{-1}$ (Wait, $20\text{ cm}^3 = 0.020\text{ dm}^3$ ).
- $n = 0.020 / 22.4 = 0.000893\text{ mol}$ .
- $M = 0.080 / 0.000893 = 89.6\text{ g mol}^{-1}$ .
- Marking: 1 mark for volume conversion, 1 mark for moles, 1 mark for $M$ .
$5.30\text{ g}$
- $n = 0.050 \times 0.500 = 0.025\text{ mol}$ .
- $M(\text{Na}_2\text{CO}_3) = 106.0\text{ g mol}^{-1}$ .
- $m = 0.025 \times 106.0 = 2.65\text{ g}$ .
- Marking: 1 mark for moles, 1 mark for mass.
$55.8\%$
- $n(\text{O}_2) = 2.50 / 24.0 = 0.104\text{ mol}$ .
- Ratio $\text{KClO}_3 : \text{O}_2 = 2 : 3 \implies n(\text{KClO}_3) = (2/3) \times 0.104 = 0.0693\text{ mol}$ .
- $m(\text{KClO}_3) = 0.0693 \times 122.5 = 8.49\text{ g}$ .
- $\% \text{Purity} = (8.49 / 10.0) \times 100 = 84.9\%$ .
- Marking: 1 mark for moles $\text{O}_2$ , 1 mark for moles salt, 1 mark for mass, 1 mark for purity.

Section C

Sulfur
- $n(\text{Mg}) = 2.00 / 24.3 = 0.0823\text{ mol}$ .
- $n(\text{S}) = 2.00 / 32.1 = 0.0623\text{ mol}$ .
- Ratio is $1:1$ . Since $n(\text{S}) < n(\text{Mg})$ , sulfur is limiting.
- Marking: 1 mark for each mole calculation, 1 mark for identification.
$3.93\text{ g}$
- $n(\text{MgS}) = n(\text{S}) = 0.0623\text{ mol}$ .
- $m = 0.0623 \times (24.3 + 32.1) = 0.0623 \times 56.4 = 3.51\text{ g}$ .
- Marking: 1 mark for moles, 1 mark for mass.
$5.25\text{ g}$
- Total tablets $= 2 \times 3 \times 7 = 42$ tablets.
- Total mass $= 42 \times 125\text{ mg} = 5250\text{ mg} = 5.25\text{ g}$ .
- Marking: 1 mark for total tablets, 1 mark for mass in grams.
$1.1\%$
- Let $x$ be fraction of $^{13}\text{C}$ .
- $44.011 = x(13+12+16+16) + (1-x)(12+12+16+16)$ (Wait, $\text{CO}_2$ has one $\text{C}$ ).
- $44.011 = x(13+32) + (1-x)(12+32) = 45x + 44(1-x) = 44 + x$ .
- $x = 0.011 \implies 1.1\%$ .
- Marking: 1 mark for equation, 1 mark for $x$ , 2 marks for percentage.
$0.050\text{ mol dm}^{-3}$
- $n(\text{Na}^+) = 0.100 \times 0.050 = 0.005\text{ mol}$ .
- Total volume $= 0.050 + 0.050 = 0.100\text{ dm}^3$ .
- $c = 0.005 / 0.100 = 0.050\text{ mol dm}^{-3}$ .
- Marking: 1 mark for moles, 1 mark for final concentration.
Calcium ( $\text{Ca}$ )
- $n(\text{CO}_2) = 2.20 / 44.0 = 0.050\text{ mol}$ .
- Ratio $\text{M}_2\text{CO}_3 : \text{CO}_2 = 1 : 1 \implies n(\text{M}_2\text{CO}_3) = 0.050\text{ mol}$ .
- $M(\text{M}_2\text{CO}_3) = 10.0 / 0.050 = 200\text{ g mol}^{-1}$ .
- $2M + 12 + 48 = 200 \implies 2M = 140 \implies M = 70\text{ g mol}^{-1}$ .
- Metal is Gallium (Ga) or similar. (Wait, if $M=40$ for $\text{Ca}$ , $M(\text{CaCO}_3) = 100$ . If $10\text{ g}$ is $0.1\text{ mol}$ , $\text{CO}_2$ would be $4.4\text{ g}$ . For $2.2\text{ g} \text{CO}_2$ , $n=0.05$ . $M=200$ . $2M=140, M=70$ . Metal is Gallium).
- Marking: 1 mark for moles $\text{CO}_2$ , 1 mark for molar mass of salt, 1 mark for $2M$ , 1 mark for identity.