4CH1

Atomic Structure

Principles of Chemistry · 1 question type

Practise

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.

  • (Ar): 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:

Ar=(%A×massA)+(%B×massB)+…100A_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:

Ar=(75×35)+(25×37)100=2625+925100=35.5A_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 Ar = 35.5, not a whole number
Worked example

Relative atomic mass from isotope abundances

Bromine has two isotopes: bromine-79 (abundance 52.8%) and bromine-81 (abundance 47.2%). Calculate its relative atomic mass to 1 decimal place.

Solution:

  • Multiply each mass number by its percentage abundance and add the products: (79 × 52.8) + (81 × 47.2) = 4171.2 + 3823.2 = 7994.4
  • Divide the total by 100: 7994.4 ÷ 100 = 79.944
  • Round to 1 decimal place: Ar = 79.9

Using decimal fractions directly — (79 × 0.528) + (81 × 0.472) — skips the ÷ 100 step and gives the same answer. Use the correct mass numbers (79 and 81 for bromine): a wrong mass number costs the final mark even if the arithmetic is right.