OCR A-Level Chemistry AS Breadth in chemistry (01), November 2021: Question 4

1 mark · Medium difficulty · Multiple Choice

Identify which of the given alkane options has a carbon composition of 82.8% by mass

Practise this question

Question

Multiple choice question 4 asks which alkane is 82.8% carbon by mass. Four options are provided: A, CH4; B, C2H6; C, C4H10; D, C8H18. Below the options is an answer box and the mark allocation of [1].
Question text

4 Which alkane is 82.8% carbon by mass?

A CH4

B C2H6

C C4H10

D C8H18

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme table indicates question number 4 has the correct answer C, with 1 mark allocated and assessment objective AO1.2.

4 C 1 AO1.2

How to answer it

Percentage Composition of Alkanes

What this question tests

This question assesses your ability to calculate the percentage mass of an element within a given molecular formula (specifically carbon in alkanes) and apply quantitative chemistry skills under multiple-choice exam conditions (AO1.2).

Question 4 — Multiple Choice

Determining the Correct Alkane Formula

✅ Correct Answer: C (C₄H₁₀)

Option C is the correct answer. Butane has a total relative molecular mass of 58.1 , containing 48.0 units of carbon, yielding exactly 82.8% carbon by mass.

💡 Key Knowledge

  • Alkane General Formula: CₙH₂ₙ₊₂
  • Percentage Mass Formula:
    (Total mass of element / Mr of compound) × 100
  • Relative atomic masses: C = 12.0, H = 1.0

🧠 Exam Technique

Don't waste time calculating the percentage for all four options from scratch! Test options strategically, or use empirical thinking: larger alkanes have a higher carbon mass percentage, while methane ( CH₄ ) is only 74.9% carbon.

❌ Common Errors

  • Forgetting to multiply the carbon relative atomic mass by the number of carbon atoms in the formula (e.g., using 12 instead of 48 for C₄H₁₀ ).
  • Using incorrect molar masses by rounding inconsistently.

📐 Step-by-Step Calculation for Option C (C₄H₁₀)

  1. Find the total Mr of C₄H₁₀:
    (4 × 12.0) + (10 × 1.0) = 48.0 + 10.0 = 58.0 (or 58.1 with more precise Ar values).
  2. Find the total mass of carbon in the molecule:
    4 × 12.0 = 48.0 .
  3. Calculate the percentage:
    (48.0 / 58.0) × 100 = 82.758% , which rounds to 82.8% .
Mark Scheme Allocation: 1 mark awarded for selecting C. (AO1.2)

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), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.