OCR A-Level Chemistry AS Breadth in chemistry (01), June 2023: Question 2

1 mark · Easy difficulty · Multiple Choice

Determine the empirical formula of a hydrocarbon containing 85.71% carbon by mass.

Practise this question

Question

Multiple-choice question 2 asks for the empirical formula of a hydrocarbon containing 85.71% carbon by mass, with four options provided: A CH, B CH2, C CH4, and D C2H4, alongside an answer box and [1] mark allocation.
Question text

2 A hydrocarbon contains 85.71% carbon by mass.

What is the empirical formula of the hydrocarbon?

A CH

B CH2

C CH4

D C2H4

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme for question 2 showing the correct answer as B, with 1 mark allocated under assessment objective AO1.2.

2 B 1 AO1.2

How to answer it

Determining Empirical Formula from Percentage Mass

What this question tests: This question assesses your ability to calculate the empirical formula of a compound using percentage by mass data. You must demonstrate core quantitative chemistry skills: finding the percentage of missing elements (hydrogen), converting mass percentages to moles using relative atomic masses, and finding the simplest whole-number ratio.
Question 2 (Multiple Choice)

Exam Breakdown

✅ Correct Answer

B: CH₂

Mark Awarded: 1 / 1 (AO1.2)

💡 Key Knowledge

  • A hydrocarbon is a compound containing only carbon and hydrogen.
  • If carbon is 85.71% by mass, hydrogen must make up the remainder: 100% - 85.71% = 14.29% .
  • Empirical formula represents the simplest whole-number ratio of atoms of each element in a compound.

🧠 Exam Technique

For multiple-choice calculations, set out your working quickly using the standard table method (Element, Percentage, Ar, Moles, Ratio) to avoid careless arithmetic errors under timed conditions.

❌ Common Errors

  • Forgetting to calculate hydrogen: Assuming the given percentage applies to carbon only without realizing the second element is hydrogen.
  • Inverting the ratio: Dividing Ar by mass instead of mass by Ar when calculating moles.
  • Selecting D (C₂H₄): Choosing a molecular formula instead of simplifying it down to the lowest whole-number empirical ratio ( CH₂ ).

📐 Step-by-Step Calculation

  1. Find the percentage of Hydrogen:
    Since it's a hydrocarbon, %H = 100 - 85.71 = 14.29% .
  2. Calculate moles of each element (assume 100g sample):
    Moles of C = 85.71 / 12.0 = 7.1425 mol
    Moles of H = 14.29 / 1.0 = 14.29 mol
  3. Find the simplest whole-number ratio:
    Divide both by the smallest number of moles (7.1425):
    C: 7.1425 / 7.1425 = 1
    H: 14.29 / 7.1425 = 2.001 ≈ 2
  4. Write the final empirical formula: CH₂ (Matches option B).

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 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.