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CIE iGCSE Co-Ordinated Science C3.2 Relative masses of atoms and molecules Exam Style Questions Paper 2

Question

Which description of mass is used to define relative atomic mass, \(A_r\)?

A. the mass of \(\frac{1}{12}\) of a mole of \(^{12}\text{C}\) atoms
B. the mass of one mole of \(^{12}\text{C}\) atoms
C. the mass of \(\frac{1}{12}\) the mass of one atom of \(^{12}\text{C}\)
D. the mass of one atom of \(^{12}\text{C}\)

▶️ Answer/Explanation
Relative atomic mass is defined as the average mass of an atom of an element compared to one-twelfth of the mass of an atom of carbon-12. The correct wording is “the mass of \(\frac{1}{12}\) the mass of one atom of \(^{12}\text{C}\)”. Option C is correct.
Answer: (C)

Question

Lead sulfate, PbSO₄, is prepared by reacting lead oxide, PbO, with excess dilute sulfuric acid.

The equation for the reaction is shown.

PbO + H₂SO₄ → PbSO₄ + H₂O

What is the mass of lead oxide required to produce 22.6 g of lead sulfate?

A. 15.4 g
B. 16.6 g
C. 17.6 g
D. 17.8 g

▶️ Answer/Explanation
From the balanced equation, 1 mole of PbO produces 1 mole of PbSO₄. The molar masses are: PbO = 207 + 16 = 223 g/mol, and PbSO₄ = 207 + 32 + (4 × 16) = 303 g/mol. To find the mass of PbO needed for 22.6 g of PbSO₄: (223/303) × 22.6 = 16.6 g. The stoichiometry is 1:1, so the same number of moles is involved.
Answer: (B)

Question

Nonane, \(C_{9}H_{20}\), burns in oxygen to form carbon dioxide and water.

The equation for this reaction is shown.

\(C_{9}H_{20} + 14O_{2} → 9CO_{2} + 10H_{2}O\)

What is the mass of oxygen required for the complete combustion of 64 g of nonane?

A. 32 g
B. 224 g
C. 396 g
D. 448 g

▶️ Answer/Explanation
The molar mass of \( C_9H_{20} \) is \( (9\times12)+(20\times1)=128 \text{ g/mol} \), so 64 g is \( \frac{64}{128}=0.5 \) mol of nonane.
From the equation, each mole of nonane needs 14 mol of \( O_2 \), so 0.5 mol needs \( 0.5\times14=7 \) mol of \( O_2 \).
Mass of oxygen \( = 7 \times 32 = 224 \text{ g} \) (using \( M_r(O_2)=32 \)).
Answer: (B)

Question

1 g of hydrogen contains \(6 \times 10^{23}\) atoms.

The relative atomic mass of helium is 4.

How many atoms does 1 g of helium contain?

A. \(1.5 \times 10^{23}\)
B. \(3 \times 10^{23}\)
C. \(6 \times 10^{23}\)
D. \(2.4 \times 10^{24}\)

▶️ Answer/Explanation
Hydrogen has a relative atomic mass of 1, so 1 g of hydrogen (= 1 mole) contains \(6 \times 10^{23}\) atoms, which is the Avogadro constant.
Helium has a relative atomic mass of 4, meaning each helium atom is 4 times heavier than a hydrogen atom.
So 1 g of helium contains 4 times fewer atoms than 1 g of hydrogen: \(\frac{6 \times 10^{23}}{4} = 1.5 \times 10^{23}\).
Answer: (A)

Question

What is the definition of the relative atomic mass, Ar, of an element?
(A)  the average mass of atoms of the element on a scale in which an atom of  $^{12}$C has a mass of exactly 12 units
(B)  the average mass of atoms of the element on a scale in which an atom of  $^{1}$H has a mass of exactly 1 unit
(C)  the average mass of atoms of the element on a scale in which an atom of  $^{12}$C has a mass of exactly 1 unit
(D)  the mass in grams of one mole of atoms of the element

▶️Answer/Explanation

Ans: A

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