Edexcel A-Level Chemistry Paper 3, June 2022: Question 1

4 marks · Easy difficulty · Short Open Response

Define relative atomic mass and calculate the relative atomic mass of a sample of neon given its isotopic percentage abundances.

Practise this question

Question

Question 1 asks to define relative atomic mass (2 marks) and calculate the relative atomic mass of neon to three significant figures using a provided table of isotopes (20Ne, 21Ne, 22Ne) and their percentage abundances (84.80, 2.26, and 12.94 respectively) (2 marks).
Question text

1 Relative atomic mass is an important concept in chemistry.

(a) Define the term relative atomic mass.

(2)

(b) A sample of neon consisted of three isotopes.

Isotope Percentage abundance

20Ne 84.80

21Ne 2.26

22Ne 12.94

Calculate the relative atomic mass of neon in this sample.

Give your answer to three significant figures.

(2)

(Total for Question 1 = 4 marks)

Mark scheme

Show the mark scheme The mark scheme for question 1(a) awards 1 mark for stating the weighted mean mass of atoms/isotopes and 1 mark for being relative to 1/12th of the mass of a carbon-12 atom. Question 1(b) awards 1 mark for the correct expression for the weighted mean and 1 mark for the calculated value of 20.3 rounded to 3 significant figures.

Question

Answer Additional guidance Mark

Number

1(a) An answer that makes reference to the following points: mass and atom only need to be mentioned once each (2)

in M1 or M2

• (the relative atomic mass of an element is) the weighted mean Accept the weighted mean mass of an atom

of the masses of its atoms / isotopes (1) Allow ‘average’ for ‘mean’ but not for ‘weighted’

Allow the mean mass of all atoms

Do not award just ‘element’ for atom / isotope

Do not award the weighted mean mass of an isotope

Ignore mention of mole.

• relative to 1/12 of the mass of carbon-12 / relative to carbon-12 Allow compared instead of relative

which has a mass of exactly 12 units (1) Do not award M2 if mass number mentioned

Note: the equation:

𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡𝑤𝑤𝑡𝑡 𝑚𝑚𝑤𝑤𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚 𝑚𝑚𝑡𝑡𝑜𝑜𝑚𝑚

𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑚𝑚 𝑐𝑐𝑚𝑚𝑐𝑐𝑐𝑐𝑜𝑜𝑚𝑚 12 𝑚𝑚𝑡𝑡𝑜𝑜𝑚𝑚

scores both marks

Question

Answer Additional guidance Mark

Number

1(b) Example of calculation (2)

(84.80 x 20) + (2.26 x 21) + (12.94 x 22)

• expression for weighted mean (1) 100

• calculation of relative atomic mass and correct rounding to 3 SF = (20.2814) = 20.3

(1) Correct answer with no / some working scores (2)

Allow TE only on a transcription error from data

Ignore units

(Total for Question 1 = 4 marks)

How to answer it

Relative Atomic Mass & Isotopic Abundance Study Guide

What this question tests

This core physical chemistry question assesses your precision with fundamental definitions and your ability to process mass spectrometry / isotopic abundance data mathematically. You must accurately state standard definitions referencing carbon-12 and execute multi-step weighted mean calculations while adhering strictly to significant figure instructions.

Question 1(a) — Definitions (2 Marks)

Define the term relative atomic mass.

✅ Correct Answer

The weighted mean mass of an atom of an element compared to 1/12th of the mass of an atom of carbon-12.

💡 Key Knowledge

  • Mark 1: Must mention weighted mean mass of an atom (or isotope). Saying "average" is allowed, but "weighted" is preferred.
  • Mark 2: Must reference 1/12th of the mass of a carbon-12 atom (or that carbon-12 has a mass of exactly 12 units).

❌ Common Errors

  • Using the word element instead of atom/isotope in the first marking point.
  • Stating "weighted mean mass of an isotope" (an isotope has a fixed mass, it cannot have a weighted mean).
  • Forgetting to include the fraction 1/12th when comparing to carbon-12.
Mark Scheme Breakdown: 1 mark for weighted mean mass of an atom/isotope + 1 mark for relative to 1/12 mass of carbon-12. (Equation using standard definitions also acceptable for both marks).
Question 1(b) — Calculations (2 Marks)

Calculate the relative atomic mass of neon from percentage abundance data.

📐 Step-by-Step Calculation

  1. Identify values: Isotope masses are 20, 21, 22 with respective percentages 84.80, 2.26, 12.94.
  2. Set up the expression:
    (84.80 × 20) + (2.26 × 21) + (12.94 × 22) / 100
  3. Calculate numerator: 1696 + 47.46 + 284.68 = 2028.14
  4. Divide by total (100): 2028.14 / 100 = 20.2814
  5. Apply significant figures: Round to 3 significant figures to get 20.3

🧠 Exam Technique & Examiner Tips

  • Significant Figures: The question explicitly asked for 3 SF. Failing to round (writing 20.2814) will lose you the final accuracy mark.
  • Working Out: Always write down your full unrounded calculation line before stating your final answered value. Even if rounding fails, showing the correct expression secures M1.
  • Units: Relative atomic mass (Ar) has no units. Do not write g or g mol⁻¹.
Mark Scheme Breakdown: 1 mark for correct expression for the weighted mean + 1 mark for the correct final value rounded to 3 significant figures (20.3).

Topics

Physical Chemistry · Topic 1: Atomic Structure and the Periodic Table

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.