OCR A-Level Chemistry AS Breadth in chemistry (01), June 2024: Question 3

1 mark · Medium difficulty · Multiple Choice

Calculate the empirical formula of a nitrogen oxide containing 36.84% nitrogen by mass.

Practise this question

Question

Multiple-choice question asking for the empirical formula of a nitrogen oxide containing 36.84 percent nitrogen by mass, with options A (NO), B (NO2), C (N2O), and D (N2O3), followed by a box for the student's answer and a mark allocation of 1 mark.
Question text

3 A nitrogen oxide contains 36.84% of nitrogen by mass.

What is the empirical formula of the nitrogen oxide?

A NO

B NO2

C N2O

D N2O3

Your answer [1]

Mark scheme

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

3 D 1

How to answer it

Calculating the Empirical Formula of a Nitrogen Oxide

What this question tests

This question assesses your ability to determine the empirical formula of a compound from percentage mass data. Core skills include calculating moles from masses or percentages, finding the simplest whole-number ratio, and working with relative atomic masses (Ar).

Question 3 (Multiple Choice)

Exam Breakdown & Worked Solution

✅ Correct Answer

D: N₂O₃

Marks: 1 / 1

💡 Key Knowledge

  • Empirical formula: The simplest whole-number ratio of atoms of each element in a compound.
  • Atomic masses: N = 14.0, O = 16.0.
  • Always assume a 100 g sample when working with percentages.

🧠 Exam Technique

For multiple-choice calculations, you can either calculate the empirical formula from scratch or work backwards by finding the percentage composition of nitrogen for options A, B, C, and D to see which matches 36.84% .

❌ Common Errors

  • Dividing masses by the wrong relative atomic mass (e.g., using atomic number instead of mass number).
  • Rounding intermediate numbers too early, leading to incorrect final whole-number ratios.
  • Failing to find the percentage of oxygen first (assuming 36.84% is the only number needed).

📐 Step-by-Step Calculation

  1. Find the percentage of oxygen:
    100% − 36.84% = 63.16% oxygen.
  2. Assume a 100 g sample and find the mass of each element:
    Mass of N = 36.84 g
    Mass of O = 63.16 g
  3. Calculate moles of each element (Moles = Mass ÷ Ar):
    Moles of N = 36.84 ÷ 14.0 = 2.6314 mol
    Moles of O = 63.16 ÷ 16.0 = 3.9475 mol
  4. Find the simplest whole-number ratio (divide by the smallest number of moles):
    Ratio of N = 2.6314 ÷ 2.6314 = 1
    Ratio of O = 3.9475 ÷ 2.6314 = 1.5
  5. Scale up to get integers:
    Multiply both sides by 2 to clear the decimal: N : O = 2 : 3 .
    Therefore, the empirical formula is N₂O₃ .

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 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.