Isotope abundances and relative atomic mass
Calculate a weighted mean and explain why isotopes have the same chemical properties.
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Identify isotopes from proton and neutron counts.
- Calculate percentages and averages.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What fraction represents an abundance of 25%?
What you will learn
- Calculate relative atomic mass from isotope masses and abundances.
- Connect isotope chemical similarity to the same electron arrangement.
Use a weighted mean
Relative atomic mass Aᵣ compares the mean mass of an element's atoms with one twelfth of the mass of a carbon-12 atom. It is a relative value with no unit. The mean depends on isotope abundance: a more common isotope makes a larger contribution. Multiply each isotope's relative mass by its fractional abundance, then add.
If abundances are percentages, multiply by each percentage and divide the total by 100. If data are counts or ratios, divide by the total count or total ratio parts instead. Check that the result lies between the smallest and largest isotope masses and nearer the more abundant isotope.
Same chemistry does not mean identical mass
Neutral atoms of an element's isotopes have the same proton count and therefore the same electron count and electron arrangement. Their chemical reactions depend mainly on these electrons, so the syllabus treats the isotopes as having the same chemical properties. The neutron difference affects mass rather than the electronic arrangement.
An average such as 35.5 does not mean a chlorine atom has half a neutron. Individual isotope nucleon counts are whole numbers; an average over many atoms can be fractional. Use isotope masses given in the question rather than silently replacing supplied measured masses with mass numbers.
Pause and explain
Why do neutral isotopes of one element have the same chemical properties in this model?
Worked example
An element has isotopes of relative masses 10 and 11 with abundances 20% and 80%. Find Aᵣ.
Show the worked solution
- Use the abundance-weighted contributions: 10×20 and 11×80.
- Add 200+880=1080 and divide by 100.
- The result 10.8 lies between 10 and 11, nearer the more common isotope.
Answer Aᵣ = 10.8 (no unit).
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Isotope masses 24, 25 and 26 have abundances 79%, 10% and 11%. Find Aᵣ.
Give me a hint
Form three weighted contributions before dividing by 100.
Compare my reasoning
- Contributions are 24×79=1896, 25×10=250 and 26×11=286.
- Their total is 2432, with total abundance 100%.
- 2432/100=24.32, close to the dominant mass-24 isotope.
24.32
Look for these in your work
- All isotope contributions are weighted.
- The mean is checked against the range and dominant isotope.
Isotopes of relative masses 63 and 65 occur in the ratio 7:3. Calculate Aᵣ and explain the denominator.
Give me a hint
The ratio has ten parts in total.
Compare my reasoning
- Weighted mass total is 63×7+65×3=636.
- Total abundance is 7+3=10 ratio parts.
- The mean is 636/10=63.6, rather than dividing by 100.
63.6; divide by 10 because the supplied ratio totals ten parts.
Look for these in your work
- Ratio parts are used consistently.
- The result is nearer the more abundant mass-63 isotope.
An element with isotope masses 10 and 11 has Aᵣ=10.2. Find the percentage of the mass-10 isotope and explain the element's chemical similarity across its isotopes.
Give me a hint
Let the fraction of mass-10 atoms be x and the other fraction be 1−x.
Compare my reasoning
- Set 10x+11(1−x)=10.2.
- This gives 11−x=10.2, so x=0.8 or 80%.
- Neutral isotopes share the same proton number and electron arrangement, so have the same chemical properties in this syllabus model.
80% mass-10; matching electron arrangements explain chemical similarity.
Look for these in your work
- Fractions sum to one.
- The mass calculation and electron-based explanation are distinct.
Common mistakes
- Taking an unweighted mean when abundances differ.
- Giving relative atomic mass units of grams or claiming a decimal neutron count.
What can you explain now?
Relative isotope masses 35 and 37 occur in the ratio 3:1. Find Aᵣ.
Compare with the explanation
35.5
The weighted mean is (35×3+37×1)/4 =142/4 =35.5.
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Make it stick
Make a two-isotope example where the simple average would be wrong. Calculate the weighted answer, then explain why a decimal mean is possible.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.