OCR A-Level Chemistry AS Breadth in chemistry (01), June 2019: Question 3

1 mark · Easy difficulty · Multiple Choice

Calculate the percentage composition by mass of nitrogen in ammonium carbonate, (NH4)2CO3.

Practise this question

Question

Multiple-choice question asking for the percentage composition by mass of nitrogen in (NH4)2CO3, with four options A (14.58%), B (17.95%), C (29.17%), and D (37.50%).
Question text

3 What is the percentage composition by mass of nitrogen in (NH4)2CO3?

A 14.58%

B 17.95%

C 29.17%

D 37.50%

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is C for question 3, worth 1 mark.

3 C 1 AO1.2

How to answer it

Calculating Percentage Composition by Mass

What this question tests

This question assesses your ability to calculate the percentage mass of a specific element within a compound using its chemical formula and relative atomic masses (Ar). It tests foundational stoichiometry skills required for quantitative chemistry in AS Level OCR Chemistry.

Question 3 Multiple Choice

Solution & Breakdown

✅ Correct Answer: C (29.17%)

Option C is the only correct calculated percentage of nitrogen mass in ammonium carbonate, (NH₄)₂CO₃ .

💡 Key Knowledge

  • Formula for percentage mass:
    (Total mass of element / Mr of compound) × 100
  • Brackets in chemical formulas multiply everything inside them by the subscript outside.
  • Standard relative atomic masses (Ar): H = 1.0, C = 12.0, N = 14.0, O = 16.0.

🧠 Exam Technique

Always count individual atoms carefully before calculating the molar mass. Writing out the breakdown of each element prevents silly multiplication errors under timed exam conditions.

❌ Common Errors

  • Forgetting to multiply the nitrogen atoms by 2 due to the outside bracket (NH₄)₂ .
  • Dividing by the mass of just the nitrogen or just the ammonium ions instead of the total Mr of the whole compound.

📐 Step-by-Step Calculation

  1. Find the total mass of Nitrogen (N):
    There are 2 ammonium groups, each containing 1 nitrogen atom: 2 × 14.0 = 28.0 g mol⁻¹ .
  2. Calculate the relative molecular mass (Mr) of (NH₄)₂CO₃:
    - Nitrogen (N): 2 × 14.0 = 28.0
    - Hydrogen (H): 8 × 1.0 = 8.0
    - Carbon (C): 1 × 12.0 = 12.0
    - Oxygen (O): 3 × 16.0 = 48.0
    Total Mr = 28.0 + 8.0 + 12.0 + 48.0 = 96.0 g mol⁻¹ .
  3. Calculate the percentage composition by mass:
    (28.0 / 96.0) × 100 = 29.166...%
  4. Round appropriately:
    Rounding to 2 decimal places gives 29.17% .
🎯 Mark Awarded: 1 mark for selecting C (AO1.2 demonstration of quantitative chemical knowledge).

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.