AQA A-Level Chemistry AS Paper 1, June 2022: Question 14

1 mark · Easy difficulty · Multiple Choice

Determine the empirical formula of a molybdenum oxide given that 5.0 g of the oxide contains 4.0 g of molybdenum.

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Question

Question 14 asks: '5.0 g of an oxide contains 4.0 g of molybdenum. What is the empirical formula of this oxide? [1 mark]' with four multiple-choice options: A MoO2, B MoO5, C Mo2O3, and D Mo3O2.
Question text

14 5.0 g of an oxide contains 4.0 g of molybdenum.

What is the empirical formula of this oxide?

[1 mark]

A MoO2

B MoO5

C Mo2O3

D Mo3O2

Mark scheme

Show the mark scheme Mark scheme for Question 14 indicating that the correct answer is C (AO2), representing Mo2O3, worth 1 mark.

14 C (AO2) 1 Mo2O3

How to answer it

Empirical Formula of Molybdenum Oxide

📌 What this question tests

This question assesses fundamental quantitative chemistry skills from Topic 3.1.2 (Amount of Substance):

  • Deducing the mass of an unstated element in a binary compound using conservation of mass.
  • Looking up relative atomic masses (Ar) from the AQA Periodic Table (Mo = 95.9 or approx. 96).
  • Converting mass to moles ( n = m / Ar ).
  • Simplifying fractional mole ratios (e.g. 1 : 1.5) to whole-number empirical ratios.
Question 14 • Multiple Choice • 1 Mark

Question Breakdown & Walkthrough

"5.0 g of an oxide contains 4.0 g of molybdenum. What is the empirical formula of this oxide?"

📐 Step-by-Step Calculation

  1. Find the mass of Oxygen (O):
    Mass of compound = 5.0 g
    Mass of Mo = 4.0 g
    Mass of O = 5.0 g − 4.0 g = 1.0 g
  2. Find the moles of each element:
    From data sheet: Ar(Mo) = 95.9, Ar(O) = 16.0
    Moles of Mo = 4.0 / 95.9 = 0.0417 mol
    Moles of O = 1.0 / 16.0 = 0.0625 mol
  3. Determine the simplest molar ratio:
    Divide both by the smaller number (0.0417):
    Mo = 0.0417 / 0.0417 = 1.00
    O = 0.0625 / 0.0417 = 1.50
  4. Convert to whole numbers:
    Multiply both by 2 to remove the decimal:
    Mo : O = (1 × 2) : (1.5 × 2) = 2 : 3
    Empirical Formula = Mo₂O₃

✅ Correct Answer

C — Mo₂O₃

Mark Scheme Reference: 14 • Key: C [AO2, 1 mark]

The mole ratio is 2 moles of molybdenum atoms to every 3 moles of oxygen atoms.

💡 Key Knowledge

  • Empirical Formula: The simplest whole-number ratio of atoms of each element present in a compound.
  • Atomic vs Molecular mass: Always divide the mass of oxygen by its atomic mass (Ar = 16.0), NOT molecular mass (O₂ = 32.0), because the formula counts individual atoms.
  • Periodic Table Lookups: Molybdenum is a transition metal in Period 5, Group 6 (atomic number 42, Ar = 95.9).

🧠 Exam Technique & Speed Tips

  • Recognise 1.5 instantly: A ratio ending in .5 means you multiply everything by 2. If it ends in .33 or .67, multiply by 3; if .25 or .75, multiply by 4.
  • Eliminate impossible distractors: Options B (MoO₅) and D (Mo₃O₂) are non-standard transition metal oxidation states for this stoichiometry and are easily ruled out once you spot the rough 1 : 1.5 relationship.
  • Rough mental math check: 4/96 ≈ 1/24 ≈ 0.042; 1/16 ≈ 0.0625. Notice that 0.0625 / 0.042 = 1.5, pointing straight to a 2:3 ratio!

❌ Common Errors & Traps

  • Using 32 for Oxygen: Dividing by 32 instead of 16 yields 0.03125 mol O, giving a ratio of Mo : O ≈ 1.33 : 1 (or Mo₄O₃), causing confusion and wasted exam time.
  • Rounding too early: Rounding 1.5 to 1 or 2 leads to choosing A (MoO₂) or B. Never round halves to the nearest integer!
  • Inverting the ratio: Confusing the numerator/denominator can lead students to select D (Mo₃O₂). Keep element headings clearly labelled on your scrap paper.

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.