AQA A-Level Chemistry AS Paper 1, June 2025: Question 10

📌 What this question tests

This question evaluates foundational quantitative chemistry skills required in physical and inorganic chemistry:

Deducing the correct chemical formula from systematic nomenclature with oxidation states (sulfate(VI)).
Determining the relative formula mass (Mᵣ) of an ionic compound using standard atomic masses (Aᵣ).
Calculating the percentage by mass of an element within a compound.

Question 10 • Multiple Choice [1 Mark]

The mass percentage of sodium in sodium sulfate(VI) is 32.4%.

Award 1 mark (AO2) for selecting option C.

Determine the chemical formula:
Sulfate(VI) has sulfur in oxidation state +6: SO₄²⁻.
Sodium forms Na⁺ ions.
Formula = Na₂SO₄
Look up Aᵣ values (from Data Sheet):
Aᵣ(Na) = 23.0, Aᵣ(S) = 32.1, Aᵣ(O) = 16.0
Calculate Mᵣ of Na₂SO₄:
Mᵣ = (2 × 23.0) + 32.1 + (4 × 16.0)
Mᵣ = 46.0 + 32.1 + 64.0 = 142.1
Calculate percentage mass of Na:
% mass = (Total mass of Na / Mᵣ) × 100
% mass = (46.0 / 142.1) × 100 = 32.37% ≈ 32.4%

Sulfate(VI) vs Sulfate(IV): The Roman numeral refers to the oxidation state of the central sulfur atom:
• Sulfate(VI) = SO₄²⁻
• Sulfate(IV) / Sulfite = SO₃²⁻
Balancing Charges: Two +1 sodium ions are required to balance the 2− charge of a single sulfate ion.
Total Element Mass: Always multiply the atomic mass by the subscript in the formula (here, 2 × Na).

Selecting A (19.3%): Occurs if the formula is written incorrectly as NaSO₄ without balancing ionic charges:
23.0 / 119.1 × 100 = 19.3%
Selecting B (22.3%): Calculating the mass percentage of sulfur in a related sulfur species, or using incorrect molar masses.
Selecting D (36.5%): Confusing sulfate(VI) with sodium sulfate(IV) / sodium sulfite (Na₂SO₃):
46.0 / 126.1 × 100 = 36.5%

Do not guess the formula: Take 5 seconds to write out the constituent ions (Na⁺ and SO₄²⁻) before touching your calculator.
Use standard AQA periodic table values: Always use 1 decimal place precision from the periodic table provided (e.g., S = 32.1, not 32).
Speed Check: Na contributes 46 out of ~140, which is approximately 1/3 (~33%). This immediately rules out A (19.3%) and B (22.3%).

Question