Calculating Relative Atomic Mass from Isotope Abundances

Why is the relative atomic mass of chlorine 35.45 and not a whole number? Because natural chlorine is a mixture of two isotopes, chlorine-35 and chlorine-37, and the periodic table value is their weighted average. This article shows exactly how that average is calculated, with worked examples from simple two-isotope elements to three-isotope elements and mass spectrometer data.

What relative atomic mass means

Relative atomic mass (Aᵣ) is the weighted average mass of the atoms of an element, taking into account the natural abundance of each isotope, compared with one-twelfth of the mass of a carbon-12 atom.

Because it’s a ratio of masses, it has no units. (The molar mass has the same numerical value, in g/mol. See what is molar mass? and atomic mass vs mass number vs molar mass.)

The formula

Aᵣ = Σ (isotope mass × abundance) ÷ Σ abundances

If abundances are percentages adding up to 100:

Aᵣ = (m₁ × %₁ + m₂ × %₂ + …) ÷ 100

If abundances are fractions adding up to 1, simply multiply and add.

Worked example 1: chlorine

Chlorine is 75.8% chlorine-35 and 24.2% chlorine-37. Calculate its relative atomic mass.

Aᵣ = (35 × 75.8 + 37 × 24.2) ÷ 100 = (2653 + 895.4) ÷ 100 = 3548.4 ÷ 100 = 35.5

This matches the periodic table value (35.45) to three significant figures. The answer is closer to 35 than 37 because chlorine-35 is more common.

Sense check: the average must lie between the lightest and heaviest isotope masses, and closer to the more abundant one.

Worked example 2: copper

Copper consists of 69.2% copper-63 and 30.8% copper-65.

Aᵣ = (63 × 69.2 + 65 × 30.8) ÷ 100 = (4359.6 + 2002.0) ÷ 100 = 63.6

The accepted value is 63.55. See the copper element page.

Worked example 3: boron, with fractions

Boron is 0.199 boron-10 and 0.801 boron-11 (as fractions).

Aᵣ = 10 × 0.199 + 11 × 0.801 = 1.99 + 8.811 = 10.8

Accepted value: 10.81.

Worked example 4: three isotopes (magnesium)

Magnesium: 78.99% magnesium-24, 10.00% magnesium-25, 11.01% magnesium-26.

Aᵣ = (24 × 78.99 + 25 × 10.00 + 26 × 11.01) ÷ 100 = (1895.76 + 250.00 + 286.26) ÷ 100 = 2432.02 ÷ 100 = 24.32

Accepted value: 24.305. The small difference is because we used mass numbers rather than exact isotopic masses (see below).

Worked example 5: from a mass spectrum

A mass spectrometer separates isotopes and shows each as a peak. Peak heights (or areas) are proportional to abundance, but they aren’t always given as percentages.

A mass spectrum of an element shows peaks at m/z 20 (height 90.5), 21 (height 0.3) and 22 (height 9.2).

Total = 90.5 + 0.3 + 9.2 = 100.0

Aᵣ = (20 × 90.5 + 21 × 0.3 + 22 × 9.2) ÷ 100.0 = (1810 + 6.3 + 202.4) ÷ 100.0 = 20.2

The element is neon (Aᵣ = 20.18).

Worked example 6: peak heights that don’t add to 100

A spectrum for gallium shows peaks at m/z 69 (height 12.0 cm) and 71 (height 7.9 cm).

Aᵣ = (69 × 12.0 + 71 × 7.9) ÷ (12.0 + 7.9) = (828 + 560.9) ÷ 19.9 = 1388.9 ÷ 19.9 = 69.8

Always divide by the total of the heights, not by 100. Accepted value: 69.72.

Using exact isotopic masses

Mass numbers are whole numbers, but the actual masses of isotopes aren’t quite whole (except carbon-12, which is exactly 12 by definition). This is because of nuclear binding energy: when protons and neutrons combine, a little mass is converted into energy (the “mass defect”). Precise calculations use exact isotopic masses.

Chlorine: ³⁵Cl mass 34.969, abundance 75.76%; ³⁷Cl mass 36.966, abundance 24.24%.

Aᵣ = (34.969 × 75.76 + 36.966 × 24.24) ÷ 100 = (2649.25 + 896.06) ÷ 100 = 35.45

This matches the periodic table value exactly. Most school questions use mass numbers; university and precise work use exact masses.

Why some elements have near-whole-number masses

Elements with one dominant isotope have relative atomic masses close to a whole number:

• Fluorine: 100% fluorine-19 → 19.00
• Sodium: 100% sodium-23 → 22.99
• Carbon: about 98.9% carbon-12 → 12.01

Elements with several significant isotopes can be far from whole numbers: chlorine (35.45), copper (63.55), bromine (79.90, from roughly equal ⁷⁹Br and ⁸¹Br).

Why abundances can vary

Isotope abundances vary slightly between samples from different sources, which is why IUPAC now gives some elements’ standard atomic weights as intervals (for example, for hydrogen, carbon and chlorine). These small variations are used to trace the origins of food, water and minerals.

Practice questions

1. Bromine is 50.7% bromine-79 and 49.3% bromine-81. Calculate Aᵣ.
2. Silver is 51.8% silver-107 and 48.2% silver-109. Calculate Aᵣ.
3. Silicon is 92.2% Si-28, 4.7% Si-29 and 3.1% Si-30. Calculate Aᵣ.
4. A spectrum of lithium shows peaks at m/z 6 (height 3.0) and 7 (height 37.0). Calculate Aᵣ.

Answers:

1. (79 × 50.7 + 81 × 49.3) ÷ 100 = 79.99 (≈ 80.0)
2. (107 × 51.8 + 109 × 48.2) ÷ 100 = 108.0
3. (28 × 92.2 + 29 × 4.7 + 30 × 3.1) ÷ 100 = (2581.6 + 136.3 + 93.0) ÷ 100 = 28.1
4. (6 × 3.0 + 7 × 37.0) ÷ 40.0 = (18 + 259) ÷ 40.0 = 6.93

For the reverse problem, finding abundances from Aᵣ, see working backwards: finding isotope abundances.

Where the isotope data come from

Modern isotope abundances are measured with high-precision mass spectrometers, often using isotope ratio techniques that compare a sample directly with an international reference material. IUPAC’s Commission on Isotopic Abundances and Atomic Weights reviews these measurements and publishes the standard atomic weights used on periodic tables, including the values on this site’s element pages.

Relative atomic mass in calculations

The periodic-table value of Aᵣ is what you use in almost every chemistry calculation involving moles. For example, the molar mass of sodium chloride is 22.99 + 35.45 = 58.44 g/mol. Using 35 or 37 instead of 35.45 would introduce an error of more than 1% in every calculation. See how to convert grams to moles and the molar mass calculator.

Worked example 7: a four-isotope element (strontium)

Strontium: ⁸⁴Sr 0.56%, ⁸⁶Sr 9.86%, ⁸⁷Sr 7.00%, ⁸⁸Sr 82.58%.

Aᵣ = (84 × 0.56 + 86 × 9.86 + 87 × 7.00 + 88 × 82.58) ÷ 100 = (47.04 + 847.96 + 609.00 + 7267.04) ÷ 100 = 8771.04 ÷ 100 = 87.71

Accepted value: 87.62; the difference comes from using mass numbers rather than exact isotopic masses (each strontium isotope is about 0.09 u lighter than its mass number). The method is the same however many isotopes there are: multiply each mass by its abundance, add, and divide by the total abundance.

A quick estimation habit

Before calculating, estimate. If an element’s two isotopes are 10 and 11 and the heavier one is about four times as common, the answer must be about four-fifths of the way from 10 to 11, roughly 10.8. Estimating first makes it easy to spot a slip in the arithmetic, such as a misplaced decimal point.

Common mistakes

• Taking a simple (unweighted) average: (35 + 37) ÷ 2 = 36 is wrong for chlorine.
• Dividing by 100 when peak heights don’t add up to 100.
• Giving units (Aᵣ has none; molar mass is in g/mol).
• Rounding too early in multi-step calculations.

Key takeaways

• Relative atomic mass is the abundance-weighted average of an element’s isotope masses.
• Aᵣ = Σ(mass × abundance) ÷ Σ(abundance).
• The result must lie between the isotope masses, nearer the most abundant one.
• For mass spectra, divide by the total peak height.
• Exact isotopic masses (not whole numbers) give the precise periodic-table values.