Average Atomic Mass Calculator
ChemistryCalculate the average atomic mass of an element from its isotope masses and natural abundances. Enter up to 3 isotopes with mass and percent abundance.
Average Atomic Mass
What is a Average Atomic Mass?
The Average Atomic Mass Calculator computes the weighted average atomic mass of an element from its isotope masses and natural abundances: Ā = Σ(mᵢ × xᵢ) where xᵢ = abundance/100. Enter up to three isotopes with their masses (in amu) and abundance percentages to get the average atomic mass.
Average atomic mass is the value shown on periodic tables — not the mass of any single isotope, but the weighted mean over the element's natural mixture of isotopes. Chlorine's 35.453 amu reflects its 3:1 mixture of ³⁵Cl and ³⁷Cl; bromine's 79.904 amu reflects its near-equal mixture of ⁷⁹Br and ⁸¹Br. The Atomic Mass Calculator provides these pre-computed tabulated values; this calculator shows how they are derived from isotope data.
Understanding average atomic mass connects atomic physics (isotopes and nuclear structure) to macroscopic chemistry (periodic table values and stoichiometry). The same calculation applies to calculating blend averages in mixtures and weighted averages in statistics — but in chemistry it has the specific meaning of the periodic table's atomic weight.
How to use this Average Atomic Mass calculator
- Enter Isotope 1 Mass (amu) — the mass of the first isotope (close to but not equal to its mass number). For ³⁵Cl: 34.969 amu.
- Enter Isotope 1 Abundance (%) — the natural abundance percentage. For ³⁵Cl: 75.77%.
- Repeat for Isotope 2. The calculator works with just 2 isotopes for elements like H, C, N, Cl, Br.
- For three-isotope elements (O, Mg, Si, S, Ca), enter Isotope 3 data.
- Check Total Abundance — must equal 100% for the average to be valid.
- Read Average Atomic Mass and compare to the periodic table value.
Formula & Methodology
Weighted average atomic mass:Ā = m₁ × (x₁/100) + m₂ × (x₂/100) + m₃ × (x₃/100) where Σxᵢ = 100%Inverse (two-isotope system):x₁ = (Ā − m₂) / (m₁ − m₂) × 100% x₂ = 100% − x₁Worked example — Boron (two isotopes): ¹⁰B: mass = 10.013 amu, abundance = 19.9% ¹¹B: mass = 11.009 amu, abundance = 80.1%Ā = 10.013 × 0.199 + 11.009 × 0.801 = 1.9926 + 8.8182 = 10.811 amuThis matches the IUPAC standard atomic mass of boron (10.811 g/mol). The predominantly ¹¹B abundance means boron's average mass is closer to 11 than 10. Boron-10 has a very high thermal neutron capture cross-section and is used in boron neutron capture therapy (BNCT) for brain tumours — an active research area at Bhabha Atomic Research Centre (BARC) and the Tata Memorial Centre.
Frequently Asked Questions