Relative Atomic Mass — Atomic Structure | IGCSE Chemistry

Exam Frequency Analysis

Past paper frequency (2018 to 2024)

This topic accounts for approximately 9% of your exam marks.

stable

Medium

Stable9%

Subatomic particles and electronic configuration appear in every exam series.

(A_r): a weighted-average mass for an element, calculated from the masses and percentage abundances of all of its naturally occurring isotopes, expressed relative to 1/12 the mass of a carbon-12 atom

Calculation from isotope abundances

Multiply each isotope's percentage abundance by its mass number, sum the products, then divide by 100:

$A_r = \frac{(\%_A \times \text{mass}_A) + (\%_B \times \text{mass}_B) + \ldots}{100}$

Add one term to the top of the fraction for every isotope present
Example. Natural chlorine consists of 75% ³⁵Cl (mass 35) and 25% ³⁷Cl (mass 37); applying the formula:

$A_r = \frac{(75 \times 35) + (25 \times 37)}{100} = \frac{2625 + 925}{100} = \mathbf{35.5}$

This is why chlorine appears on the periodic table with A_r = 35.5, not a whole number

Exam tip

Calculating relative atomic mass from isotope abundances

What comes up: a 3-mark question provides the mass numbers and percentage abundances of two or three isotopes and asks you to calculate Ar, often requiring the answer to 1 decimal place.

Write (three marks): M1 multiply each mass number by its percentage and add the products (e.g. (79 × 52.8) + (81 × 47.2)); M2 divide that total by 100; M3 round the answer to the precision specified. Using decimal fractions directly (e.g. (79 × 0.528) + (81 × 0.472)) is also accepted, which skips the ÷ 100 step and still earns all three marks.

Watch out: M3 is dependent on using the correct mass numbers (e.g. 79 and 81 for bromine, not 78 and 80). If you use wrong mass numbers throughout, M3 is lost even if the arithmetic is correct. Also, the answer without working can score full marks only if the final value is correct and rounded appropriately.