Exam Questions

2019 O Level Stoichiometry TYS Questions

Fresh from the furnace of O Level


The stoichiometry questions from the 2019 O Level paper are less routine. Familiar formulas are used in new ways. For example, instead of finding the percentage mass of an element in a compound, you are asked to find the percentage mass of a polyatomic ion in a compound.

The questions also test you more on the concepts rather than memorised procedures. In the MCQ about doubling the product, you do not need to convert everything to number of moles and then apply the mole ratio. You simply need to recognise that both the reactants must be doubled, lest there is a limiting reactant.

Try out the stoichiometry TYS questions to see it for yourself!

On Molar Mass

What is the Mr for ammonia, NH3, and hydrochloric acid, HCl, respectively.

Mr of NH3 = 14 + 3 × 1 = 17
Mr of HCl = 1 + 35.5 = 36.5

On Percentage Mass and Empirical Formula

4.00 g of metal oxide, XO2, is reduced to form 3.15 g of metal X. What is the identity of X?

Answer: Sn, tin

STEP 1: Since nothing is known about X, we will work from O first
Mass of oxygen = 4 – 3.15 = 0.85 g
Number of moles of oxygen atom in XO2 = 0.85 ÷ 16 = 0.05313

STEP 2: Use mole ratio to relate the no. of moles of X to that of O
Mole ratio of X to O in XO2 = 1:2
Number of moles of X atom in XO2 = 0.05313 × 1/2 = 0.02657

STEP 3: Find molar mass and hence the relative atomic mass of X to identify it
Molar mass of X = mass ÷ no. of moles = 3.15 ÷ 0.02657 = 119 g/mol
Relative atomic mass of X = 119
Chemical identity of X = Sn, tin

Chlorate(I), ClO, is an oxyanion. It contains a single covalent bond between the oxygen and chlorine atoms. The ion also has an overall negative charge.

Show by calculation that the percentage by mass of chlorate(I) ions in calcium chlorate(I) is greater than the percentage by mass of chlorate(I) ions in sodium chlorate(I), NaOCl.

Since chlorate(I) has a charge of 1-, the formula of calcium chlorate is Ca(OCl)2.

Mr of Ca(OCl)2 = 40 + 2×16 + 2×35.5 = 143
Relative mass of chlorate(I) ions in Ca(OCl)2 = 2×16 + 2×35.5 = 103
Percentage mass of chlorate in Ca(OCl)2 = 103/143 × 100 = 72.0%

Mr of NaOCl = 23 + 16 + 35.5 = 74.5
Relative mass of chlorate(I) ion in NaOCl = 16 + 35.5 = 51.5
Percentage mass of chlorate(I) in NaOCl = 51.5/74.5 × 100 = 69.1% < 72.0%

Marker’s comment: when calculating the percentage mass of chlorate in Ca(OCl)2, the mass of both chlorate ions per formula unit must be added to give the numerator of 103.

On Mole Ratio and Limiting Reactant

A sugar has the formula (CH2O)6.

One mole of this sugar is burned in excess oxygen and the gas produced is collected at room temperature and pressure. What is the volume collected?

Answer: 144 dm3

(CH2O)6 (s) + 6O2 (g) ⟶ 6CO2 (g) + 6H2O (l)

As excess oxygen was used, the sugar underwent complete combustion to give carbon dioxide and water. At the room temperature of 25 °C, only carbon dioxide but not water exists as a gas. This means that the volume of gas produced is solely contributed by carbon dioxide gas.

Mole ratio of (CH2O)6 to CO2 = 1:6

Using the mole ratio, 1 mol of sugar will give 6 mol of carbon dioxide. Multiplying this by the molar volume of 24 dm3, the volume of gas is 6 × 24 = 144 dm3.

Magnesium reacts with dilute sulfuric acid to form magnesium sulfate and hydrogen.

Mg + H2SO4 ⟶ MgSO4 + H2

2.4 g of magnesium react with exactly 100 cm3 of 1 mol/dm3 sulfuric acid to give 2.4 dm3 of hydrogen, at room temperature and pressure. Which combination gives 4.8 dm3 of hydrogen at room temperature and pressure?

A) 2.4 g of Mg with 100 cm3 of 2 mol/dm3 sulfuric acid
B) 2.4 g of Mg with 200 cm3 of 1 mol/dm3 sulfuric acid
C) 4.8 g of Mg with 100 cm3 of 1 mol/dm3 sulfuric acid
D) 4.8 g of Mg with 100 cm3 of 2 mol/dm3 sulfuric acid

Answer: D

The question wants us to produce 2 times the amount of product. To do this, we need to double the amount of both reactants.

For a solid like magnesium, we simply double the mass. For a solution, we can either double the concentration or the volume to double the amount. In option D, the volume of sulfuric acid remains constant as the concentration is doubled, allowing the amount to be doubled.

In options A and B, magnesium becomes the limiting reactant. Even though the amount of sulfuric acid is doubled by doubling either the concentration or volume, the same amount of magnesium used will prevent further reaction, causing the amount of product to still be 2.4 dm3 of hydrogen. On the other hand, sulfuric acid is the limiting reactant in option C.

On Percentage Yield

Sodium burns in air to form oxides.

(a) Given that 0.50 g of sodium was burnt, calculate the theoretical yield of sodium oxide, Na2O.
(b) When the product was weighed, the mass was only 0.32 g. Calculate the percentage yield.
(c) Suggest a reason why the percentage yield is less than 100%

4Na + O2 ⟶ 2Na2O

Molar mass of Na = 23
No. of moles of sodium = 0.50 ÷ 23 = 0.01667

Mole ratio of Na to Na2O = 4:2 = 2:1

No. of moles of Na2O formed theoretically = 0.01667 × 1/2 = 0.008335
Molar mass of Na2O formed theoretically = 23×2 + 16 = 62
Theoretical yield of Na2O = 0.008335 × 62 = 0.517 g

Percentage yield of Na2O = 0.32/0.517 × 100 = 61.9%

The yield was less than 100% as the sodium used might have impurities. Therefore, the actual mass and amount of sodium reacted was lower. Furthermore, incomplete reaction might further reduce the yield.

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