Relative Atomic Mass from Isotope Data

The relative atomic mass (Ar) of an element is not the mass of any single atom. It is the weighted average of the masses of all its naturally occurring isotopes. When the question gives you isotope data, you can calculate Ar using a straightforward formula.

The method

1. Multiply each isotope’s mass number by its percentage abundance.
2. Add all the results together.
3. Divide by 100 (the total percentage).

The formula

Ar = (mass1 x abundance1 + mass2 x abundance2 + …) / 100

Where:

• mass is the mass number of each isotope
• abundance is the percentage abundance of each isotope

If abundances are given as fractions or ratios rather than percentages, divide by the total of the abundances instead of 100.

Worked examples

Example 1: Two isotopes (chlorine)

Question: Chlorine has two isotopes: Cl-35 (75%) and Cl-37 (25%). Calculate the relative atomic mass of chlorine.

Working:

• Ar = (35 x 75 + 37 x 25) / 100
• Ar = (2625 + 925) / 100
• Ar = 3550 / 100 = 35.5

Answer: Ar = 35.5

Examiner note: This explains why the periodic table gives chlorine as 35.5 rather than a whole number. It is the weighted average.

Example 2: Two isotopes (copper)

Question: Copper has two isotopes: Cu-63 (69%) and Cu-65 (31%). Calculate the relative atomic mass of copper.

Working:

• Ar = (63 x 69 + 65 x 31) / 100
• Ar = (4347 + 2015) / 100
• Ar = 6362 / 100 = 63.62

Answer: Ar = 63.62 (or 63.6 to 1 d.p.)

Examiner note: The periodic table rounds this to 63.5 or 64 depending on the edition. Accept the calculated value.

Example 3: Three isotopes (silicon)

Question: Silicon has three isotopes: Si-28 (92.2%), Si-29 (4.7%) and Si-30 (3.1%). Calculate the relative atomic mass of silicon.

Working:

• Ar = (28 x 92.2 + 29 x 4.7 + 30 x 3.1) / 100
• Ar = (2581.6 + 136.3 + 93.0) / 100
• Ar = 2810.9 / 100 = 28.1

Answer: Ar = 28.1

Examiner note: With three isotopes, the calculation is longer but the method is identical. Be careful with the arithmetic.

Example 4: Using ratio data instead of percentages

Question: Boron has two isotopes: B-10 and B-11 in a ratio of 1:4. Calculate the relative atomic mass of boron.

Working:

• Total ratio = 1 + 4 = 5
• B-10 abundance = (1/5) x 100 = 20%, B-11 abundance = (4/5) x 100 = 80%
• Ar = (10 x 20 + 11 x 80) / 100
• Ar = (200 + 880) / 100 = 10.8

Answer: Ar = 10.8

Examiner note: Convert the ratio to percentages first, then apply the standard formula.

Example 5: Reverse calculation - finding abundance

Question: Bromine has two isotopes: Br-79 and Br-81. The relative atomic mass of bromine is 79.9. Calculate the percentage abundance of each isotope.

Working:

• Let the abundance of Br-79 = x%. Then abundance of Br-81 = (100 - x)%.
• 79.9 = (79x + 81(100 - x)) / 100
• 7990 = 79x + 8100 - 81x
• 7990 = 8100 - 2x
• 2x = 110
• x = 55
• Br-79 = 55%, Br-81 = 45%

Answer: Br-79: 55%, Br-81: 45%

Examiner note: Set up the equation with the unknown as x. The algebra is straightforward once the equation is written correctly.

Common mistakes

1. Dividing by the number of isotopes instead of 100. The formula requires dividing by the total percentage (100), not by the number of isotopes. This is a weighted average, not a simple mean.
2. Forgetting to convert ratios to percentages. If abundances are given as a ratio, convert to percentages first.
3. Using the wrong mass values. Use the mass number (whole number) of each isotope, not the Ar from the periodic table.
4. Arithmetic errors. These calculations involve large multiplications. Show your working clearly so partial credit is available even if the final arithmetic is wrong.

When this appears

This calculation appears on Extended tier papers and is tested on both Paper 2 (multiple choice) and Paper 4 (structured). It connects to atomic structure and the concept of isotopes. See the stoichiometry topic for the broader context.