OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2017: Question 3

1 mark · Medium difficulty · Multiple Choice

Deduce the molecular formula of an organic compound given the volumes of carbon dioxide and water vapour produced upon its complete combustion.

Practise this question

Question

Multiple choice question 3 asks to identify the molecular formula of an organic compound that forms 40 cm3 of carbon dioxide and 40 cm3 of water vapour upon complete combustion under the same conditions. Four options are provided: A, C3H8; B, C2H2O; C, C2H4O; D, C2H3N. A blank box for the answer is shown at the bottom left with [1] mark indicated at the bottom right.
Question text

3 Complete combustion of an organic compound forms 40 cm3 of carbon dioxide and 40 cm3 of

water vapour, under the same conditions of temperature and pressure.

Which molecular formula could the organic compound have?

A C3H8

B C2H2O

C C2H4O

D C2H3N

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme table shows the correct answer for question 3 is option C, worth 1 mark.

3 C 1

How to answer it

Combustion Volumes and Molecular Formulae

What this question tests

This question assesses your understanding of gas stoichiometry, Avogadro's Law (that equal volumes of gases under the same conditions contain equal numbers of moles), and how to interpret reacting molar ratios from complete combustion data to determine an unknown molecular formula.

Question 3 Analysis

Determining Molecular Formula from Combustion Gas Volumes

✅ Correct Answer

C ( C₂H₄O )

Option C is the only compound that yields a 1:1 molar ratio of carbon dioxide to water vapour upon complete combustion.

💡 Key Knowledge

  • Avogadro's Law: Under the same temperature and pressure, volume is directly proportional to moles ( V ∝ n ).
  • Equal volumes mean equal moles. Therefore, 40 cm³ of CO₂ and 40 cm³ of H₂O(g) represent a 1:1 mole ratio of products.

🧠 Exam Technique

Instead of balancing full equations for all four options, look at the ratio of carbon atoms to hydrogen atoms implied by the product volumes: equal volumes of CO₂ and H₂O mean the compound must contain equal numbers of carbon and hydrogen atoms (or a 2:4 ratio simplifying to 1:2 in terms of C to H₂ units).

❌ Common Errors

  • Assuming mass conservation instead of volume/mole stoichiometry for gases.
  • Failing to account for the subscript in H₂O when relating moles of water molecules to moles of hydrogen atoms.

📐 Step-by-Step Working Out

  1. Compare product volumes:
    Volume of CO₂ = 40 cm³
    Volume of H₂O(g) = 40 cm³
  2. Convert volumes to a molar ratio:
    Since moles are proportional to volume, the ratio of CO₂ : H₂O produced is 40 : 40 , which simplifies to 1 : 1 .
  3. Test the options based on stoichiometry:
    • A ( C₃H₈ ): Forms 3 CO₂ and 4 H₂O (Ratio 3:4 — Incorrect)
    • B ( C₂H₂O ): Forms 2 CO₂ and 1 H₂O (Ratio 2:1 — Incorrect)
    • C ( C₂H₄O ): Forms 2 CO₂ and 2 H₂O (Ratio 2:2 simplifies to 1:1 — Correct)
    • D ( C₂H₃N ): Contains nitrogen, which would form NOₓ or N₂ gas, altering simple hydrocarbon-style combustion ratios.
Mark Scheme Allocation: 1 mark awarded for selecting option C.

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.