OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2021: Question 3

1 mark · Medium difficulty · Multiple Choice

Determine which sample contains the greatest number of carbon atoms among four given organic substance options with specified masses.

Practise this question

Question

Multiple choice question 3 asks which sample contains the greatest number of carbon atoms, with four options: A, 20.0 g of C6H5OH; B, 30.0 g of C2H5COOH; C, 40.0 g of CH3CHO; D, 50.0 g of CH3OH. Below the options is a box for the answer and a mark allocation of [1].
Question text

3 Which sample contains the greatest number of carbon atoms?

A 20.0 g C6H5OH

B 30.0 g C2H5COOH

C 40.0 g CH3CHO

D 50.0 g CH3OH

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 3 is option C, worth 1 mark under assessment objective AO2.2.

3 C 1 AO2.2

How to answer it

Calculating Quantities: Number of Carbon Atoms

What this question tests

This question assesses your ability to apply the mole concept, calculate molar masses from chemical formulas, convert between mass and moles, and scale up from moles of molecules to the number of specific constituent atoms using the Avogadro constant.

Question Multiple Choice Analysis

Question 3

Determining the sample with the greatest number of carbon atoms

✅ Correct Answer: C

Option C yields the highest number of carbon moles and therefore the greatest total number of carbon atoms.

💡 Key Knowledge

  • Moles formula: Moles = Mass (g) / Molar Mass (g mol⁻¹)
  • Atomic scaling: Total carbon atoms = Moles of substance × Number of carbon atoms per molecule × Avogadro constant (Nₐ).
  • Since Nₐ is constant for all options, you only need to compare Moles of substance × Carbons per molecule .

🧠 Exam Technique

Don't waste time multiplying by 6.022 × 10²³ for every single option. Calculate the relative moles of carbon atoms (Moles × Number of C atoms) for each choice and find the maximum value.

❌ Common Errors

  • Stopping after calculating the moles of the molecule instead of multiplying by the number of carbon atoms in the formula.
  • Counting the total number of atoms (including H and O) instead of strictly focusing on carbon atoms.
  • Miscounting carbon atoms in condensed structural formulas like C₂H₅COOH (contains 3 carbons total).

📐 Step-by-Step Calculation Breakdown

  1. Option A: 20.0 g of C₆H₅OH (Molar Mass = 94.1 g mol⁻¹)
    Moles of molecules = 20.0 / 94.1 = 0.2126 mol
    Moles of carbon atoms = 0.2126 × 6 = 1.276 mol
  2. Option B: 30.0 g of C₂H₅COOH (Molar Mass = 74.0 g mol⁻¹)
    Moles of molecules = 30.0 / 74.0 = 0.4054 mol
    Moles of carbon atoms = 0.4054 × 3 = 1.216 mol
  3. Option C: 40.0 g of CH₃CHO (Molar Mass = 44.0 g mol⁻¹)
    Moles of molecules = 40.0 / 44.0 = 0.9091 mol
    Moles of carbon atoms = 0.9091 × 2 = 1.818 mol (Highest value)
  4. Option D: 50.0 g of CH₃OH (Molar Mass = 32.0 g mol⁻¹)
    Moles of molecules = 50.0 / 32.0 = 1.5625 mol
    Moles of carbon atoms = 1.5625 × 1 = 1.563 mol
Mark Scheme Allocation: 1 mark awarded for selecting C (AO2.2 - Application of quantitative chemistry concepts).

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