Weighted Average
Add a row per isotope; abundances should total 100%
Abundances entered: 0%
Average atomic mass = Σ (fraction × isotope mass). Methodology & sources →
Atomic structure practice? The General Chemistry Workbook's first chapter covers isotopes, average atomic mass and electron configuration.As an Amazon Associate we earn from qualifying purchases.
View on Amazon →Why It's a Weighted Average
Elements occur as a mixture of isotopes — atoms with the same number of protons but different numbers of neutrons. The periodic-table value is the average, weighted by how common each isotope is. Chlorine is roughly 75.77% ³⁵Cl (34.969 amu) and 24.23% ³⁷Cl (36.966 amu), giving 35.45 amu. The atomic-number and isotope notation side of this is covered by the Atomic Notation Calculator.
Worked Example — Weighted Average from Isotope Abundances
Question: Boron has two stable isotopes: boron-10 (mass 10.013 amu, 19.9% abundance) and boron-11 (mass 11.009 amu, 80.1% abundance). Calculate the average atomic mass and compare it with the periodic table value.
Step 1 — convert percentages to decimal fractions: 19.9% → 0.199, and 80.1% → 0.801. These must sum to 1 (0.199 + 0.801 = 1.000 ✓).
Step 2 — multiply each isotope's mass by its fractional abundance:
0.199 × 10.013 = 1.993
0.801 × 11.009 = 8.818
Step 3 — sum the contributions: 1.993 + 8.818 = 10.81 amu
Answer: 10.81 amu — matching boron's standard periodic table value, because the more abundant heavier isotope (¹¹B) pulls the weighted average closer to its own mass than a simple average of 10.013 and 11.009 would.
Common Mistakes
- Using a simple average instead of a weighted one. (10.013 + 11.009) ÷ 2 = 10.51, which is not boron's atomic mass. Each isotope must be weighted by how common it actually is.
- Forgetting to convert percent to a fraction. Multiplying by 19.9 instead of 0.199 gives an answer about 100× too large.
- Abundances not summing to 100%. If you're given abundances for only some of an element's isotopes, the remainder belongs to other isotopes — check that all listed abundances sum to (close to) 100% before computing.
Frequently Asked Questions
Multiply each isotope's mass by its fractional abundance (percent ÷ 100) and add the results.
Because it averages all the naturally occurring isotopes, weighted by abundance.
It's a weighted average of the masses of all naturally occurring isotopes of an element, each contributing in proportion to how common it is. Since isotope masses themselves are close to, but not exactly, whole numbers (due to nuclear binding energy), and the weighting mixes several such values, the result is essentially never a round number.
For an element's naturally occurring stable isotopes, yes — the abundances given in a problem should sum to 100% (or 1.000 as a fraction), since together they account for all atoms of that element found in nature. If a problem gives abundances that don't sum to 100%, either an isotope is missing from the data or there's a rounding discrepancy worth checking.