Exam Questions

Stoichiometry Prelim Questions & Answers

Learn stoichiometry and the mole concept by solving problems taken from O Level prelim papers. Answers and explanations are included.

Basic Stoichiometry Questions

These basic stoichiometry questions that usually appear in Paper 1 (MCQ) test you on two things. Firstly, how well you know the formulae to convert number of particles, mass, and gaseous volume to number of moles. Secondly, how careful you are when reading the questions. Ask yourself if the question is asking for the amount of atoms, ions, or molecules.

QUESTION 1
Which one of the following contains the greatest amount of atoms.

A. 1 mol of helium
B. 18 g of water
C. 48 dm3 of hydrogen gas at rtp
D. 1.2×1024 atoms of oxygen

Answer: C

This question is about the amount of atoms. You can compare the amount in terms of number of particles or moles. However, comparing in moles is more straightforward, as you can directly convert mass and gaseous volume to number of moles. Let’s consider the four options in turn.

  • As helium is monoatomic, 1 mol of helium has exactly 1 mol of atoms.
  • We can divide the given mass of water by its molar mass to get the number of moles: 18 ÷ 18 = 1 mol of water molecules. However, each molecule of water, H2O comprises 3 atoms. Therefore, 1 mol of water molecules have 1 × 3 = 3 mol of atoms.
  • We can divide the given gaseous volume of hydrogen by the molar volume (24 dm3 at rtp) to get the number of moles: 48 ÷ 24 = 2 mol of hydrogen gas. However, as hydrogen gas is diatomic, each molecule of hydrogen has 2 hydrogen atoms. Therefore, 2 mol of hydrogen molecules have 2 × 2 = 4 mol of atoms.
  • We can divide the given number of oxygen atoms by the Avogadro constant (6×1023) to get the number of moles: 1.2×1024 ÷ 6×1023 = 2 mol of atoms.
QUESTION 2
Which of the following quantities is equal to the Avogadro constant?

A. The number of atoms in 24 dm3 of nitrogen gas at r.t.p. conditions.
B. The number of molecules in 1 dm3 of oxygen at 298 K and 1 atm.
C. The number of zirconium atoms in 962 g of tranquillityite, Fe3Zr2Ti3Si3O24, of molar mass 961.8 g mol-1.
D. The number of ions in 29.3 g of sodium chloride.

Answer: D

The Avogadro constant is 6×1023, which is equal to 1 mol. Essentially, this question is asking you which option has exactly 1 mol of the stated particle.

  • Nitrogen gas is diatomic. Thus, 24 dm3 of nitrogen gas has 1 mol of N2 molecules but 2 mol of N atoms.
  • The gaseous volume of 1 dm3 is way less than the molar volume of 24 dm3 at rtp, and hence there are fewer than 1 mol of molecules.
  • 962 g of tranquillityite has 1 mol of Fe3Zr2Ti3Si3O24, which has 2 mol of zirconium.
  • 29.3 g has 0.5 mol of NaCl. As it is a binary compound, 0.5 mol of NaCl has 1 mol of ions.

Stoichiometry Questions on Percentage Composition

QUESTION 3
Copper is found in the mineral chalcopyrites, CuFeS2. What mass of chalcopyrites must be melted to produce 100 tonnes of copper.

If you know the percentage mass of copper in chalcopyrites, you can work backwards to find the mass of chalcopyrites containing 100 tonnes of copper.

  • STEP 1: Find the relative molecular mass, Mr of CuFeS2

    Mr = 64 + 56 + 32×2 = 184 g

  • STEP 2: Compute the percentage mass of copper in CuFeS2

    % mass of Cu = (64÷184) × 100 = 34.782%

  • STEP 3: Find the mass of chalcopyrites needed

    Mass of chalcopyrites = (100÷34.782) × 100 = 288 tonnes (3 s.f.)

    There is no need to convert to grams here as percentage mass is a ratio. You can either have a ratio of mass in grams, or a ratio of mass in tonnes. Both are fine, so long as the units are consistent.

QUESTION 4
A garden fertiliser is said to have a phosphorous content of 30% P2O5. What is the percentage by mass of phosphorous in the fertiliser?

We can first find the percentage mass of phosphorous in P2O5. Then using the percentage of P2O5 in the fertiliser, we finally get the percentage mass of phosphorous in the fertiliser.

  • STEP 1: Find the relative molecular mass, Mr of P2O5

    Mr = 31×2 + 16×5 = 142

  • STEP 2: Find the relative mass of all the phosphorous atoms in P2O5

    Relative mass of phosphorous in P2O5 = 31×2 = 62

  • STEP 3: Compute the percentage mass in P2O5

    % mass of P in P2O5 = (62÷142) × 100 = 43.662%

  • STEP 4: Compute the percentage mass in the fertiliser

    % mass of P in the fertiliser = 43.662% × 30% = 13.1%

Stoichiometry Questions on Empirical Formula

QUESTION 5
Compound X is an organic liquid which can be combined with other substances to be used as a cleaning agent. It contains 53.3% carbon, 11.1% hydrogen and 35.6% oxygen by mass. Deduce the empirical formula of X.

It is found that 75 cm3 of liquid X contains 1 mol of compound X. Given that the density of X is 1.20 g/cm3, find the molecular formula of X.
CarbonHydrogenOxygen
Mass in 100 g sample/ g53.311.135.6
Ar12116
No. of moles/ mol53.3 ÷ 12 = 4.4411.1 ÷ 1 = 11.135.6 ÷ 16 = 2.225
Divide by smallest value4.44 ÷ 2.225 = 211.1 ÷ 2.225 = 52.225 ÷ 2.225 = 1
Simplest ratio251

Therefore, the empirical formula is C2H5O. To find its molecular formula:

  • STEP 1: Find the relative mass of atoms in the empirical formula of C2H5O

    Relative mass of atoms = 2×12 + 5×1 + 16 = 45

  • STEP 2: Find the molar mass of compound X

    Molar mass = 75 cm3 × 1.20 g/cm3 = 90 g

  • STEP 3: Given that the molecular formula is C2nH5nOn, find the value of the multiplier, n

    n × 45 = 90
    n = 90 ÷ 45 = 2

  • STEP 4: Write the molecular formula

    C4H10O2

QUESTION 6
In an experiment, 15 g of sodium sulfate crystals on being heated left a residue of 7.95 g of the anhydrous salt. Calculate the percentage of water of crystallisation in the crystals. Hence, find the value of x in the formula Na2SO4.xH2O.

Answer: 47%

Na2SO4.xH2O (s) ⟶ Na2SO4 (s) + xH2O (g)

During the heating, water was lost as steam. This contributed to the decrease in mass from 15 g to 7.95 g. Therefore, we can find the mass of water of crystallisation by taking the difference between the initial mass and final mass.

Mass of water of crystallisation = 15 – 7.95 = 7.05 g
% mass of water in crystals = (7.05÷15) × 100 = 47%

In the second part, the unknown x can be found by finding the ratio between sodium sulfate and water in the crystals, much like how we find the ratio between elements in a compound.

Na2SO4H2O
Mass in 15 g sample/ g7.957.05
Mr23×2 + 32 + 16×4 = 1421×2 + 16 = 18
No. of moles/ mol7.95 ÷ 142 = 0.05607.05 ÷ 18 = 0.392
Divide by smallest value0.0560 ÷ 0.0560 = 10.392 ÷ 0.0560 = 7
Simplest ratio17

Therefore, x = 7 and the chemical formula for the hydrated crystal is Na2SO4.7H2O.

Stoichiometry Questions Involving Chemical Reactions

QUESTION 7
10 g sample of copper(II) oxide was heated with aluminium powder, to produce copper and aluminium oxide. Calculate the minimum mass of aluminium that was required for the reaction.

By asking for the “minimum mass“, the question wants you to use the mole ratio to find the exact amount of aluminium needed to react with 10 g of copper(II) oxide. This is like a titration, whereby both reactants are limiting and there is no excess.

Also, because the chemical equation is not given, you need to write it down first to make sense of the reactants, products, and their mole ratio.

  • STEP 1: Write the chemical equation and highlight the substances you are interested in

    3CuO (s) + 2Al (s) ⟶ 3Cu (s) + Al2O3 (s)

  • STEP 2: Convert any given data to number of moles

    Molar mass of CuO = 64 + 16 = 80 g/mol
    No. of moles of CuO = 10 ÷ 80 = 0.125 mol

  • STEP 3: State the mole ratio of reactants

    Ratio of Al to CuO = 2:3

  • STEP 4: Use the mole ratio to find the number of moles of Al needed

    No. of moles of Al = 2/3 × 0.125 = 0.083333 mol

  • STEP 5: Convert number of moles to mass, as required

    Molar mass of Al = 27 g/mol
    Mass of Al = 0.083333 × 27 = 2.25 g

QUESTION 8
Nickel is a transition metal. During its extraction, 200 g of nickel(II) sulfide was reacted with oxygen in air to form nickel(II) oxide and sulfur dioxide.

Write the chemical equation and calculate the volume of sulfur dioxide formed, at room temperature and pressure.

Air has an unlimited supply of oxygen. So we shall assume that oxygen is in excess, while nickel(II) sulfide is limiting. Let’s get cracking!

  • STEP 1: Write the chemical equation and highlight the substances you are interested in

    2NiS + 3O2 ⟶ 2NiO + 2SO2

  • STEP 2: Convert any given data to number of moles

    Molar mass of NiS = 59 + 32 = 91
    No. of moles of NiS = 200 ÷ 91 = 2.1978 mol

  • STEP 3: State the mole ratio of reactants

    Ratio of SO2 to NiS = 2:2 = 1:1

  • STEP 4: Use the mole ratio to find the number of moles of SO2 produced

    No. of moles of Al = 1/1 × 2.1978 = 2.1978 mol

  • STEP 5: Convert number of moles to volume of gas, as required

    Molar volume at rtp = 24 dm3
    Volume of SO2 gas = 24 × 2.1978 = 52.7 dm3 (or 52700 cm3)

QUESTION 9
The exothermic thermite reaction is applied in the welding of railway tracks as represented by the chemical reaction below. In the process, the molten iron solidifies on cooling, thus welding the tracks together.

3Fe3O4 (s) + 8Al (s) ⟶ 4Al2O3 (s) + 9Fe (l)

In a given reaction, 60 kg of the iron oxide were reacted with 15 kg of aluminium. Calculate the mass of molten iron formed in the reaction.

Noticed that this question provides us with the mass of both reactants? That is a big hint to first identify the limiting reactant. It is the limiting reactant that will react fully to directly determine the amount of product formed.

  • STEP 1: Convert any given data to number of moles

    Molar mass of Fe3O4 = 56×3 + 16×4 = 232 g/mol
    No. of moles of Fe3O4 = 60(1000) ÷ 232 = 259 mol

    Molar mass of Al = 27 g/mol
    No. of moles of Al = 15(1000) ÷ 27 = 556 mol

  • STEP 2: State the mole ratio of reactants

    Ratio of Fe3O4 to Al = 3:8

  • STEP 3: Determine the limiting reactant

    Consider the case that Fe3O4 were completely reacted:

    No. of moles of Al used = 8/3 × 259 = 691 mol

    Only 556 mol of Al was provided. Therefore, there was insufficient Al to completely react with Fe3O4. Therefore, Al was the limiting reactant while Fe3O4 was in excess.

  • STEP 4: State the mole ratio of limiting reactant to product

    Ratio of Al to Fe = 8:9

  • STEP 5: Use the limiting reactant to calculate the amount of product formed

    No. of moles of Fe formed = 9/8 × 556 = 625.5 mol

  • STEP 6: Convert number of moles to mass, as required

    Mass of Fe formed = 625.5 × 56 = 35000 g (or 35 kg)

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