Average Atomic Mass Calculator
Compute the weighted average atomic mass of an element from its isotopes, solve for a missing abundance (or both abundances, given only a two-isotope element and its known average), and identify an unknown element from its average mass. Every mode shows a visual, full step-by-step math, and a callout explaining what the result means.
Background
The atomic mass listed on the periodic table is a weighted average of an element's naturally occurring isotopes — each isotope's mass counts in proportion to how abundant it is in nature. Chlorine, for example, is about 75.78% chlorine-35 (34.969 amu) and 24.22% chlorine-37 (36.966 amu), which averages to 35.45 amu — much closer to 35 than to 36, because the lighter isotope is far more common.
How to use this calculator
- Choose Compute the Average to enter every isotope's mass and abundance and get the weighted average atomic mass.
- Choose Solve One Abundance when you know every isotope's mass, all-but-one abundance, and the target average — it solves the missing percentage.
- Choose Solve Both Abundances for a two-isotope element where you don't know either percentage yet — only the two isotope masses and the standard average. Since the two abundances must add to 100%, that's enough information to solve both.
- Choose Identify the Element to match an average atomic mass against real elements and find the closest fit.
- Click Calculate to see the diagram, the full step-by-step math, and a callout explaining what the result means.
How Average Atomic Mass Works
Average atomic mass is a weighted average: multiply each isotope's mass by its fractional abundance (percent ÷ 100), then add up every isotope's contribution.
Because it's weighted, the average always lands closer to whichever isotope is more abundant — not simply halfway between the isotope masses.
All the abundances for one element must add up to 100%, since every atom of that element in nature is one isotope or another.
If you know every abundance but one, and the target average, you can solve for the missing abundance directly: multiply out the knowns, subtract from 100×average, and divide by the missing isotope's mass.
For a two-isotope element, knowing just the two masses and the average is enough to solve both abundances at once, since p₁ + p₂ = 100 and m₁p₁ + m₂p₂ = 100×average are two equations in two unknowns.
A useful shortcut for two isotopes is the lever rule: the ratio of the two abundances equals the inverse ratio of their distances from the average on a number line — the isotope closer to the average is more abundant.
Every element's average atomic mass is unique enough that, given just that number, you can usually identify which element it belongs to by comparing it to the periodic table.
Formulas & Equations Used
Weighted average: Average = Σ mᵢ × (pᵢ / 100)
Solving one missing abundance: p_missing = (100 × Average − Σ knownᵢ mᵢpᵢ) / m_missing
Solving both abundances (2 isotopes): p₁ = 100 × (Average − m₂) / (m₁ − m₂), p₂ = 100 − p₁
Lever rule check: p₁ / p₂ = (m₂ − Average) / (Average − m₁)
Molar mass link: 1 amu of average atomic mass = 1 gram per mole of that element
Example Problems & Step-by-Step Solutions
Example 1 — Computing the average (chlorine)
Cl-35 = 34.969 amu (75.78%), Cl-37 = 36.966 amu (24.22%).
Step: (34.969 × 0.7578) + (36.966 × 0.2422) = 26.500 + 8.955.
Result: Average = 35.45 amu.
Example 2 — Solving one missing abundance (boron)
B-10 = 10.013 amu (unknown %), B-11 = 11.009 amu (80.1%), average = 10.81 amu.
Step: Known contribution = 11.009 × 0.801 = 8.818. p_missing = (100×10.81 − 881.8) / 10.013 ≈ 19.9%.
Result: B-10 abundance ≈ 19.9%.
Example 3 — Solving both abundances (bromine)
Br-79 = 78.9183 amu, Br-81 = 80.9163 amu, average = 79.904 amu.
Step: p₁ = 100 × (79.904 − 80.9163) / (78.9183 − 80.9163) ≈ 50.70%. p₂ = 100 − 50.70 = 49.30%.
Result: Br-79 ≈ 50.7%, Br-81 ≈ 49.3%.
Example 4 — Identifying an element
An unknown sample averages 63.55 amu.
Step: Compare against known standard atomic weights; copper is 63.546 amu, a difference of only 0.004.
Result: The sample is almost certainly copper (Cu).
Frequently Asked Questions
Why isn't the average just the midpoint between the two isotope masses?
Because it's a weighted average, not a simple one. Whichever isotope is more abundant pulls the average closer to itself — chlorine's average (35.45) sits much closer to 35 than to 37 because ³⁵Cl is about three times more common.
Do abundances have to add up to exactly 100%?
In theory yes, since every atom of an element is some isotope. In practice, measured values often round to something like 99.9% or 100.1% — this calculator allows a small ±0.5% tolerance before warning you.
Can I solve for an abundance if I don't know any of them?
Yes, but only for a two-isotope element. With two isotopes, the two abundances must add to 100%, which gives you exactly enough equations to solve for both from the masses and the known average alone.
What if an element has three or more isotopes and I'm missing two abundances?
That's underdetermined — two unknowns need two independent equations, and with three or more isotopes you'd only have one (they sum to 100%) unless you're given more data. This calculator only solves the two-unknown case for exactly two isotopes.
How does the lever rule help double-check an answer?
It reframes the same math visually: picture the two isotope masses on a number line with the average marked between them. The isotope closer to the average is more abundant, and the ratio of abundances equals the inverse ratio of those two distances.
How reliable is the "identify the element" match?
Very reliable for common elements, since standard atomic weights are usually unique to several decimal places. If your value doesn't land close to anything in the comparison list, the calculator will flag it as a weak match instead of guessing confidently.
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