OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2020: Question 11

1 mark · Medium difficulty · Multiple Choice

Determine the empirical formula of a hydrocarbon given the mass of the hydrocarbon and the mass of carbon dioxide produced upon complete combustion.

Practise this question

Question

Multiple-choice question 11 states: Complete combustion of 1.00g of a hydrocarbon gives 3.38g carbon dioxide. What is the empirical formula of the hydrocarbon? Four options are given: A: CH, B: CH2, C: C2H5, D: C3H8. Below the options is a box labelled 'Your answer' and the mark [1] on the right.
Question text

11 Complete combustion of 1.00 g of a hydrocarbon gives 3.38 g carbon dioxide.

What is the empirical formula of the hydrocarbon?

A CH

B CH2

C C2H5

D C3H8

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme for question 11 showing the correct answer is A, with 1 mark.

11 A 1 2.6

How to answer it

Determining Empirical Formula from Combustion Data

What this question tests

This question assesses your ability to calculate the empirical formula of a hydrocarbon using combustion mass data. You must demonstrate mastery of moles calculations (moles = mass ÷ molar mass), conservation of mass for elements (all carbon in CO₂ originates from the hydrocarbon), and determining simplest whole-number ratios.

Question 11: Full Worked Solution

Multiple Choice Question (1 Mark)

✅ Correct Answer: A (CH)

The correct option is A. Through mass and mole calculations, the ratio of carbon to hydrogen simplifies to a 1:1 ratio, giving the empirical formula CH .

💡 Key Knowledge

  • Combustion products: Hydrocarbons burn in oxygen to form CO₂ and H₂O.
  • Carbon source: All the carbon in the carbon dioxide produced came directly from the original hydrocarbon sample.
  • Hydrocarbon definition: A compound containing hydrogen and carbon only.

🧠 Exam Technique

Don't panic if an empirical formula looks unusually simple like CH (often associated with aromatic rings or ethyne derivatives like benzene, C₆H₆). Trust your calculated mole ratio rather than guessing familiar alkane formulas like C₃H₈!

📐 Step-by-Step Calculation

  1. Find moles of CO₂ produced:
    Molar mass of CO₂ = 12.0 + (16.0 × 2) = 44.0 g mol⁻¹
    Moles of CO₂ = 3.38 g ÷ 44.0 g mol⁻¹ = 0.07682 mol
  2. Find moles of Carbon atoms:
    Since each CO₂ molecule contains 1 carbon atom, moles of C = moles of CO₂ = 0.07682 mol
    Mass of Carbon = 0.07682 mol × 12.0 g mol⁻¹ = 0.9219 g
  3. Find mass and moles of Hydrogen atoms:
    Total mass of hydrocarbon = 1.00 g
    Mass of Hydrogen = Total mass - Mass of Carbon = 1.00 - 0.9219 = 0.0781 g
    Moles of H = 0.0781 g ÷ 1.01 g mol⁻¹ = 0.07733 mol
  4. Determine the simplest whole-number ratio:C : H
    C : H = 0.07682 : 0.07733
    Divide by the smallest value (~0.0768):
    C : H = 1 : 1.006 ≈ 1 : 1

❌ Common Errors & Examiner Pitfalls

  • Treating it as a molecular formula: Students sometimes get confused thinking empirical formulas must resemble standard homologous series (like alkanes).
  • Forgetting oxygen subtraction: If a question involves an oxygenated organic compound, students forget to subtract carbon and hydrogen masses from the total sample mass to find oxygen. (Though here it's just a hydrocarbon!).
  • Rounding too early: Rounding intermediate mole values to 1 sig fig will completely ruin your ratio. Always keep at least 3 significant figures until the final step.
Examiner Insight: Top-level responses quickly recognized that finding the mass of carbon directly via carbon dioxide is the most efficient route. Distractor D ( C₃H₈ ) catches out students who mistakenly try to work backwards using trial molar masses instead of executing the mole calculation step-by-step.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.