Why abundance changes the average

Isotopes of an element have the same number of protons but different numbers of neutrons, and therefore different masses.

The relative atomic mass, $A_{r}$, of an element is an average mass for its atoms. It is a weighted mean: isotopes that are more abundant affect the average more than isotopes that are rare.

Think of a sample where 90 out of every 100 atoms have relative mass 36 and only 10 out of every 100 atoms have relative mass 34. The average must be much closer to 36 than to 34, because most atoms in the sample have the larger relative mass.

The calculation method

To calculate relative atomic mass from isotope data, multiply each isotope relative mass by its abundance. Add these values, then divide by the total abundance.

For a complete isotope mixture expressed as percentages, the abundances add to 100. If the abundances are given as relative numbers, such as 3 and 2, the total abundance is their sum, so 5 in that example.

Use this layout to make the weighting clear:

1. Write each relative mass × abundance term.
2. Add the terms.
3. Divide by the total abundance.
4. Give the final $A_{r}$. Relative atomic mass has no unit.

Worked example with percentages

Suppose element X has two isotopes:

First multiply each relative mass by its percentage abundance:

$35 \times 75 = 2625$

$37 \times 25 = 925$

Then add the values and divide by 100:

$A_{r} = (2625 + 925) / 100$

$A_{r} = 3550 / 100$

$A_{r} = 35.5$

This answer makes sense because it lies between 35 and 37, and it is closer to 35 because the isotope with relative mass 35 is more abundant.

When abundances are not percentages

Sometimes abundance data is not written as a percentage. The method is the same, but the denominator is the total of the given abundances.

For example, an element has these isotope data:

The total abundance is $3 + 7 = 10$.

$A_{r} = ((10 \times 3) + (11 \times 7)) / 10$

$A_{r} = (30 + 77) / 10$

$A_{r} = 10.7$

The answer is closer to 11 because the isotope with relative mass 11 has the larger relative abundance.

Check for common calculation errors

A good answer can be checked before you move on:

• The calculated $A_{r}$ should be between the smallest and largest isotope relative masses.
• For a mixture of two isotopes, it should be closer to the mass of the more abundant isotope. With three or more, consider all the weighted contributions.
• If percentage abundances add to 100, divide by 100.
• If relative abundances do not add to 100, divide by their total.
• Do not add a unit to $A_{r}$.

One common error is to find the simple mean of the isotope masses. For relative masses 10 and 11, the simple mean is 10.5, but that is only correct if the isotopes are equally abundant. If the abundances are 3 and 7, the weighted mean is 10.7.

Another error is to round too early. Keep the full value in the calculation and round only at the end if the question asks for a particular number of significant figures.

Show your working when requested. Keeping the weighted terms visible also helps you find an error if your final value lies outside the isotope mass range.