Study Guide

Isotopes and Relative Atomic Mass

CIE IGCSE ChemistryΒ· 2.3Β· 12 min read

1. What Are Isotopes? (Core)β˜…β˜…β˜†β˜†β˜†β± 3 min

πŸ“˜ Definition

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.

πŸ“ 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. 1

    Recall that isotopes are atoms of the same element with the same proton number and different neutron number

  2. 2

    Compare proton counts: Atom W has 17 protons, Atom X has 17 protons, Atom Y has 18 protons

  3. 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.

2. Relative Atomic Mass: Core Calculationsβ˜…β˜…β˜…β˜†β˜†β± 4 min

πŸ“˜ Definition

Relative Atomic Mass

ArA_r

The weighted average mass of one atom of an element, relative to 1/12 the mass of one atom of the carbon-12 isotope

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

πŸ“ 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. 1

    Use the weighted average formula for Aα΅£ with percentage abundances:

  2. 2
    Ar=(mass1Γ—%1)+(mass2Γ—%2)+(mass3Γ—%3)100A_r = \frac{(mass_1 \times \%_1) + (mass_2 \times \%_2) + (mass_3 \times \%_3)}{100}
  3. 3
    Ar=(24Γ—79)+(25Γ—10)+(26Γ—11)100A_r = \frac{(24 \times 79) + (25 \times 10) + (26 \times 11)}{100}
  4. 4
    Ar=1896+250+286100=2432100=24.32 (rounded to 24.3 for exam use)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 Aα΅£ calculations, even if you can compute the sum mentally, to access all available marks.

3. Extended: Aα΅£ from Relative Abundancesβ˜…β˜…β˜…β˜…β˜†Extended only⏱ 3 min

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.

πŸ“ Worked Example

Rubidium occurs as two isotopes in the relative abundance ratio 3:1 β€” rubidium-85 (relative abundance 3) and rubidium-87 (relative abundance 1). Calculate the relative atomic mass of rubidium.

  1. 1

    When abundances are given as a ratio instead of percentages, use the total of the ratio values as the denominator:

  2. 2
    Ar=(mass1Γ—abundance1)+(mass2Γ—abundance2)abundance1+abundance2A_r = \frac{(mass_1 \times abundance_1) + (mass_2 \times abundance_2)}{abundance_1 + abundance_2}
  3. 3
    Ar=(85Γ—3)+(87Γ—1)3+1=255+874=3424=85.5A_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.

4. Common Pitfalls

Wrong move:

Defining isotopes as atoms with the same mass number but different proton number

Why:

This confuses isotopes with atoms of different elements. Isotopes have the same proton number, different neutron/mass number.

Correct move:

Always state that isotopes are atoms of the same element, with equal proton number and different neutron number.

Wrong move:

Calculating Aα΅£ as a simple average of isotope masses instead of a weighted average

Why:

This ignores differences in isotope abundance, leading to an incorrect final value.

Correct move:

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 move:

Stating that isotopes have different chemical properties

Why:

Chemical properties depend only on outer shell electron arrangement, which is identical for all isotopes of the same element.

Correct move:

State that isotopes have identical chemical properties, and differ only in physical properties such as mass or density.

Wrong move:

Using 100 as the denominator for Extended Aα΅£ calculations when given relative abundance ratios

Why:

This incorrectly assumes abundances are percentages, leading to an over or under estimated Aα΅£ value.

Correct move:

Sum all the relative abundance values given in the question, and use this total as your denominator.

5. Quick Reference Cheatsheet

Term

Core Requirement

Extended Requirement

Isotope definition

Same element: same proton number, different neutron number; interpret atom and ion symbols

β€”

Isotope properties

Identical chemical properties; differing physical properties (mass, density)

Explain identical chemical properties: same number of electrons / same electronic configuration

Aα΅£ calculation

Weighted average from percentage abundances

Calculate from relative abundances given as ratios (sum the ratios as the denominator)

6. Frequently Asked

Why do isotopes have the same chemical properties?

Isotopes of the same element have exactly the same number of outer shell electrons, which control all chemical reactions. Neutron count does not affect reactivity.

Why is the Aα΅£ of chlorine 35.5 instead of a whole number?

Aα΅£ is a weighted average of all natural isotopes of chlorine. 75% of natural chlorine is chlorine-35, 25% is chlorine-37, leading to an average of 35.5.

Going deeper

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 Aα΅£ 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 Aα΅£ calculations from both percentage abundances and relative abundance ratios, as these appear frequently in structured exam papers.