Atomic Structure
Principles of Chemistry
- (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:
- 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:
- This is why chlorine appears on the periodic table with Ar = 35.5, not a whole number
Common exam question
Calculating relative atomic mass from isotope abundances
Question: Calculate the relative atomic mass of a sample from the mass numbers and percentage abundances of its isotopes, rounded as instructed (2–4 marks).
Asked in 9 of the 23 papers. Multiply each mass number by its percentage and add the products, divide the total by 100, then round exactly as instructed. A correct rounded answer alone earns full marks in every scheme that asks for a value; the unrounded value alone drops the last mark when a rounding is specified, so show every line. Decimal fractions (0.752 rather than 75.2, with no division by 100) are accepted.
Two traps. Copying the periodic-table value scores zero: the question is about this sample's abundances. When protons and neutrons are given, add them for each mass number first: one scheme gives a separate mark for that, and neutron counts used as masses lose it. One paper asks you to show that the value is a given number, which is the same working.
Worked example
Relative atomic mass from a table of isotopes
The table gives the make-up of a sample of silicon. Use the information in the table to work out the relative atomic mass of the sample, to one decimal place.
| Isotope | Protons in each atom | Neutrons in each atom | Abundance in the sample (%) |
|---|---|---|---|
| 1 | 14 | 14 | 92.2 |
| 2 | 14 | 15 | 4.7 |
| 3 | 14 | 16 | 3.1 |
Solution:
- Mass numbers first: 14 + 14 = 28, 14 + 15 = 29, 14 + 16 = 30
- Multiply each by its abundance and add: (28 × 92.2) + (29 × 4.7) + (30 × 3.1) = 2581.6 + 136.3 + 93.0 = 2810.9
- Divide by 100: 2810.9 ÷ 100 = 28.109
- Round as asked: Ar = 28.1
The answer sits close to 28 because the lightest isotope dominates; an answer of about 14 means the neutron counts were used as the masses.