OCR A-Level Chemistry AS Breadth in chemistry (01), November 2020: Question 10

1 mark · Medium difficulty · Multiple Choice

Identify which chemical sample contains the greatest number of molecules by calculating moles for each given mass and formula.

Practise this question

Question

Multiple choice question 10 asking which sample contains the greatest number of molecules. Option A is 1 g of methanol, CH3OH. Option B is 2 g of nitrogen dioxide, NO2. Option C is 3 g of phosphorus, P4. Option D is 4 g of iodine, I2. An answer box is provided at the bottom.
Question text

10 Which sample contains the greatest number of molecules?

A 1 g of methanol, CH3OH

B 2 g of nitrogen dioxide, NO2

C 3 g of phosphorus, P4

D 4 g of iodine, I2

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme table indicates that the correct answer for question 10 is B, worth 1 mark.

10 B 1 2.6

How to answer it

Calculating and Comparing Moles of Molecules

📌 What this question tests

This multiple-choice question assesses your ability to interconvert between mass and amount of substance (moles) using molar mass ( Moles = Mass / Molar Mass ), and links the number of moles directly to the total number of particles (Avogadro constant) without requiring full numerical calculation of every particle.

Question 10: Full Solution & Breakdown

Exam Part: Multiple Choice Selection

✅ Correct Answer

B (2 g of nitrogen dioxide, NO₂)

Awarded 1 mark for selecting B.

💡 Key Knowledge

  • Number of molecules is directly proportional to the number of moles ( Molecules = Moles × 6.02 × 10²³ ).
  • To find the greatest number of molecules, calculate the number of moles for each option by dividing the given mass by the relative molecular mass (Mr).
  • Always check molecular formulas carefully (e.g., P₄ vs I₂).

🧠 Exam Technique

  • In multiple-choice questions involving "greatest number of...", work out the values systematically for each option until you find the highest.
  • You do not need to multiply by the Avogadro constant for each option; the highest number of moles automatically equates to the highest number of molecules!

❌ Common Errors

  • Multiplying mass by Mr instead of dividing ( Moles = Mass / Mr ).
  • Using atomic masses instead of molecular masses (e.g., using Mr of N = 14 instead of NO₂ = 46, or forgetting that phosphorus is P₄ and iodine is I₂).

📐 Step-by-Step Calculation Breakdown

Calculate the moles for each option to see which is largest:

  • A: Mass = 1 g, Mr of CH₃OH = (12.0 + (4 × 1.0) + 16.0) = 32.0 g mol⁻¹.
    Moles = 1 / 32.0 = 0.03125 mol
  • B: Mass = 2 g, Mr of NO₂ = (14.0 + (2 × 16.0)) = 46.0 g mol⁻¹.
    Moles = 2 / 46.0 = 0.04348 mol (Highest)
  • C: Mass = 3 g, Mr of P₄ = (4 × 31.0) = 124.0 g mol⁻¹.
    Moles = 3 / 124.0 = 0.02419 mol
  • D: Mass = 4 g, Mr of I₂ = (2 × 126.9) = 253.8 g mol⁻¹.
    Moles = 4 / 253.8 = 0.01576 mol

Conclusion: Option B yields the highest number of moles (0.0435 mol), and therefore contains the greatest number of molecules.

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 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.