# Isotopes and Relative Atomic Mass

> CIE IGCSE Chemistry · 0620 (2026-2028)
> Source: https://www.owlsprep.com/study/cie-0620-u2-isotopes-and-relative-atomic-mass/

This guide covers isotope definitions, their core properties, and relative atomic mass calculations for CIE IGCSE Chemistry 0620 Core and Extended tiers, aligned to the 2026-2028 syllabus.

**Prerequisites:** [Knowledge of atomic structure (protons, neutrons, electrons)](https://www.owlsprep.com/study/cie-0620-u2-atomic-structure/); [Understanding of proton (atomic) and nucleon (mass) number](https://www.owlsprep.com/study/cie-0620-u2-subatomic-particles/)

## Learning objectives

- Define isotopes as atoms of the same element with equal proton number and different neutron number
- Explain why isotopes have identical chemical properties and differing physical properties
- Calculate relative atomic mass (Aᵣ) from percentage isotope abundances (Core)
- Calculate relative atomic mass (Aᵣ) from relative isotope abundances given as ratios (Extended)

## What Are Isotopes? (Core)

**Isotopes** — Atoms of the same element that have the same number of protons (equal proton number Z) but different numbers of neutrons (different nucleon number A)

*Example:* Carbon has two stable isotopes: carbon-12 (6 protons, 6 neutrons) and carbon-13 (6 protons, 7 neutrons)

Isotopes share identical chemical properties because they have the same number of outer shell electrons, so they take part in exactly the same reactions. Their physical properties (mass, density, boiling point) differ because they have different total masses.

**Worked example:** Three atoms have the following subatomic counts: Atom W: 17 protons, 18 neutrons, 17 electrons; Atom X: 17 protons, 20 neutrons, 17 electrons; Atom Y: 18 protons, 22 neutrons, 18 electrons. Identify which pair are isotopes, and justify your answer.

1. Recall that isotopes are atoms of the same element with the same proton number and different neutron number
2. Compare proton counts: Atom W has 17 protons, Atom X has 17 protons, Atom Y has 18 protons
3. Atom W and X are isotopes of chlorine: they have the same proton number (17) and different neutron counts (18 vs 20)

> **Exam tip:** When identifying isotopes in exam questions, always reference both proton and neutron numbers to earn full marks, do not only mention mass number.

## Relative Atomic Mass: Core Calculations

**Relative Atomic Mass** — The weighted average mass of one atom of an element, relative to 1/12 the mass of one atom of the carbon-12 isotope

*Notation:* A_r

Most elements exist as a mixture of isotopes in nature. The Aᵣ value listed on the periodic table accounts for both the mass and percentage abundance of each naturally occurring isotope of the element.

**Worked example:** Magnesium occurs naturally as three isotopes: 79% magnesium-24, 10% magnesium-25, 11% magnesium-26. Calculate the relative atomic mass of magnesium.

1. Use the weighted average formula for Aᵣ with percentage abundances:
2. $$A_r = \frac{(mass_1 \times \%_1) + (mass_2 \times \%_2) + (mass_3 \times \%_3)}{100}$$
3. $$A_r = \frac{(24 \times 79) + (25 \times 10) + (26 \times 11)}{100}$$
4. $$A_r = \frac{1896 + 250 + 286}{100} = \frac{2432}{100} = 24.32 \text{ (rounded to 24.3 for exam use)}$$

> **Exam tip:** Always show full working for Aᵣ calculations, even if you can compute the sum mentally, to access all available marks.

## Extended: Aᵣ from Relative Abundances

For Extended tier, you may be given the abundances of an element's isotopes as a simple ratio (for example 3:1) rather than as percentages. When the abundances are given as a ratio, you divide by the total of the ratio values instead of by 100.

**Worked example:** Rubidium occurs as two isotopes in the relative abundance ratio 3:1 — rubidium-85 (relative abundance 3) and rubidium-87 (relative abundance 1). Calculate the relative atomic mass of rubidium.

1. When abundances are given as a ratio instead of percentages, use the total of the ratio values as the denominator:
2. $$A_r = \frac{(mass_1 \times abundance_1) + (mass_2 \times abundance_2)}{abundance_1 + abundance_2}$$
3. $$A_r = \frac{(85 \times 3) + (87 \times 1)}{3 + 1} = \frac{255 + 87}{4} = \frac{342}{4} = 85.5$$

> **Exam tip:** If abundance values are given as a ratio (or arbitrary units) that does not add up to 100, always sum the abundance values to use as your denominator, do not divide by 100.

## Common pitfalls

- **Wrong:** Defining isotopes as atoms with the same mass number but different proton number
  - Why it fails: This confuses isotopes with atoms of different elements. Isotopes have the same proton number, different neutron/mass number.
  - Correct: Always state that isotopes are atoms of the same element, with equal proton number and different neutron number.
- **Wrong:** Calculating Aᵣ as a simple average of isotope masses instead of a weighted average
  - Why it fails: This ignores differences in isotope abundance, leading to an incorrect final value.
  - Correct: Multiply each isotope mass by its abundance, sum the products, then divide by total abundance (100 for percentages, sum of ratios for relative abundances).
- **Wrong:** Stating that isotopes have different chemical properties
  - Why it fails: Chemical properties depend only on outer shell electron arrangement, which is identical for all isotopes of the same element.
  - Correct: State that isotopes have identical chemical properties, and differ only in physical properties such as mass or density.
- **Wrong:** Using 100 as the denominator for Extended Aᵣ calculations when given relative abundance ratios
  - Why it fails: This incorrectly assumes abundances are percentages, leading to an over or under estimated Aᵣ value.
  - Correct: Sum all the relative abundance values given in the question, and use this total as your denominator.

## Cheatsheet

| Term | Core Requirement | Extended Requirement |
| --- | --- | --- |
| Isotope definition | Same element: same proton number, different neutron number; interpret atom and ion symbols | — |
| Isotope properties | Identical chemical properties; differing physical properties (mass, density) | Explain identical chemical properties: same number of electrons / same electronic configuration |
| Aᵣ calculation | Weighted average from percentage abundances | Calculate from relative abundances given as ratios (sum the ratios as the denominator) |

## What's next

Now that you have mastered isotopes and relative atomic mass, you are ready to progress to electron shell arrangement, the next core concept in atomic structure for CIE IGCSE Chemistry 0620. Understanding Aᵣ is also foundational for mole calculations, a key topic across both Core and Extended tiers that accounts for a large share of the exam marks. Extended learners should practise Aᵣ calculations from both percentage abundances and relative abundance ratios, as these appear frequently in structured exam papers.

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