OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2021: Question 4

1 mark · Medium difficulty · Multiple Choice

Determine which of the four given samples of organic compounds contains the greatest number of moles of molecules.

Practise this question

Question

Multiple choice question 4 asks which sample contains the greatest number of molecules, with four options: A, 140.0 g of C2H2; B, 180.0 g of C2H6; C, 240.0 g of C4H10; and D, 400.0 g of C6H6. Below the options is an 'Your answer' box and a mark allocation of [1].
Question text

4 Which sample contains the greatest number of molecules?

A 140.0 g C2H2

B 180.0 g C2H6

C 240.0 g C4H10

D 400.0 g C6H6

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme table shows question number 4 with the correct answer as B, worth 1 mark, designated under assessment objective AO2.2.

4 B 1 AO2.2

How to answer it

Calculating the Greatest Number of Molecules

What this question tests

This question assesses your understanding of the mole concept, molar mass calculations, and the Avogadro constant relationship. Specifically, it tests competency in AO2.2: applying chemical knowledge and mathematical calculations to compare quantities of substance.

Question Analysis & Answer

Multiple Choice Question (Total: 1 mark)

✅ Correct Answer

B ( 180.0 g C₂H₆ )

Mark Awarded: 1 / 1 mark (AO2.2)

💡 Key Knowledge

  • Number of moles = Mass (g) ÷ Molar Mass (g mol⁻¹)
  • Number of molecules = Moles × Avogadro constant (Nₐ)
  • Since the Avogadro constant is a fixed number for every sample, finding the sample with the greatest number of molecules is identical to finding the sample with the greatest number of moles.

Step-by-Step Calculations

How to evaluate each option:

📐 Method breakdown

  • Step 1: Calculate the molar mass ($M_r$) for each compound using relative atomic masses ($C = 12.0$, $H = 1.0$).
  • Step 2: Divide the given mass by the molar mass to find the number of moles ($n = m / M$).
  • Step 3: Compare the mole values to identify the largest amount.

Option A: C₂H₂

  • Molar Mass = (2 × 12.0) + (2 × 1.0) = 26.0 g mol⁻¹
  • Moles = 140.0 g ÷ 26.0 g mol⁻¹ = 5.38 mol

Option B: C₂H₆ (Correct)

  • Molar Mass = (2 × 12.0) + (6 × 1.0) = 30.0 g mol⁻¹
  • Moles = 180.0 g ÷ 30.0 g mol⁻¹ = 6.00 mol

Option C: C₄H₁₀

  • Molar Mass = (4 × 12.0) + (10 × 1.0) = 58.0 g mol⁻¹
  • Moles = 240.0 g ÷ 58.0 g mol⁻¹ = 4.14 mol

Option D: C₆H₆

  • Molar Mass = (6 × 12.0) + (6 × 1.0) = 78.0 g mol⁻¹
  • Moles = 400.0 g ÷ 78.0 g mol⁻¹ = 5.13 mol

Exam Technique & Common Pitfalls

🧠 Exam Technique

  • Save time: Do not multiply every mole value by the Avogadro constant ($6.02 × 10²³$). Recognise that proportionality remains identical when comparing moles directly.
  • Show working in margins: Even in multiple-choice questions, jotting down quick molar mass and division steps prevents silly arithmetic slips under timed conditions.

❌ Common Errors

  • Mass confusion: Simply looking at the largest mass ( 400.0 g in option D) and assuming it has the most molecules without dividing by the molar mass.
  • Arithmetical slips: Incorrectly adding up atomic masses (e.g., miscounting hydrogen atoms in hydrocarbons like C₂H₆ or C₄H₁₀).

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.