AQA A-Level Chemistry Paper 3, June 2025: Question 12

1 mark · Medium difficulty · Multiple Choice

Calculate the percentage abundance of 235UF6 in a sample given the relative molecular mass of the sample and the isotopes present.

Practise this question

Question

Question 12 asks: 'A sample of UF6 consists of molecules containing 19F and either 235U or 238U atoms. The relative molecular mass of the sample is 351.43. What is the percentage abundance of 235UF6 in the sample?'. Four multiple-choice options are provided: A 19, B 33, C 50, and D 81.
Question text

12 A sample of UF consists of molecules containing 19F and either 235U or 238U atoms.

The relative molecular mass of the sample is 351.43

What is the percentage abundance of 235UF in the sample?

[1 mark]

A 19

B 33

C 50

D 81

Mark scheme

Show the mark scheme Mark scheme table showing question number 12, correct answer A, 1 mark for AO2, and the value 19.

12 A 1 (AO2) 19

How to answer it

Calculating Isotopic Abundances from Relative Molecular Mass

What this question tests

This question evaluates your ability to apply the concept of relative atomic/molecular mass (Ar and Mr) to a molecular system containing isotopes. You must deduce the relative atomic mass contribution of a multi-isotopic element (uranium) within a compound of known relative molecular mass and solve an algebraic simultaneous equation to find percentage abundance.

Question 12 • Multiple Choice • 1 Mark

Percentage Abundance of ²³⁵UF₆

Syllabus Reference: 3.1.1.1 (Fundamental Particles & Mass Spectrometry)

✅ Correct Answer

A — 19%

Award 1 mark for selecting option A (AO2)

💡 Key Knowledge

  • Relative Molecular Mass (Mr): The weighted average mass of a molecule compared to 1/12th of the mass of an atom of carbon-12.
  • Fluorine exists here solely as ¹⁹F, so 6 fluorine atoms contribute exactly:
    6 × 19 = 114 .
  • Because all molecules contain 6 fluorine atoms, the percentage abundance of ²³⁵UF₆ is identical to the percentage abundance of ²³⁵U atoms in the sample.

📐 Step-by-Step Calculation

Method 1: Isolating the Uranium contribution (Fastest)

  1. Calculate the total mass contribution of fluorine:
    Mass of 6 × ¹⁹F = 6 × 19 = 114
  2. Find the average relative atomic mass of Uranium, Ar(U):
    Ar(U) = Mr(UF₆) − mass of 6 × ¹⁹F
    Ar(U) = 351.43 − 114 = 237.43
  3. Set up an isotopic abundance equation:
    Let the percentage abundance of ²³⁵U be x .
    Therefore, the abundance of ²³⁸U is (100 − x) .
    Ar(U) = [(235 × x) + 238 × (100 − x)] / 100
  4. Solve for x:
    237.43 × 100 = 235x + 23800 − 238x
    23743 = 23800 − 3x
    3x = 23800 − 23743 = 57
    x = 57 / 3 = 19%

Method 2: Working with molecular masses directly

  • Mr(²³⁵UF₆) = 235 + 114 = 349
  • Mr(²³⁸UF₆) = 238 + 114 = 352
  • 351.43 = [349x + 352(100 − x)] / 100 → 35143 = 35200 − 3x → 3x = 57 → x = 19%

🧠 Exam Technique & Quick Sanity Checks

  • Estimation Trick: Notice that the sample's average mass (351.43) is much closer to the ²³⁸U molecule (352) than the ²³⁵U molecule (349).
    Distance from 349 to 352 is 3 units.
    Distance from 351.43 to 352 is only 0.57.
    This means ²³⁸U must be the dominant isotope (~81%), so ²³⁵U must be in the minority (~19%).
  • This instantly eliminates options C (50%) and D (81%) without doing full algebra!

❌ Common Traps & Mistakes

  • Finding the wrong isotope (selecting D): 81% is the abundance of ²³⁸UF₆, not ²³⁵UF₆. Always re-read which isotope the question asks for.
  • Forgetting to multiply Fluorine: Forgetting that there are 6 fluorine atoms and only subtracting 19 once from 351.43.
  • Mixing up atomic vs. mass numbers: Fluorine's mass number is 19 (given in the question as ¹⁹F). Do not confuse it with fluorine's atomic number 9.

Topics

Physical Chemistry · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.