What is the percent abundance?
Type any three of the four values, such as the two isotope masses and the element’s average atomic mass. The percent abundance calculator gives the share of each isotope, or the average or a missing isotope mass, from the weighted-average formula.
- Isotope 1 abundance
- 75.7567%
With isotope masses 34.96885 u and 36.9659 u and an average of 35.453 u, isotope 1 is 75.7567% and isotope 2 is 24.2433%.
- Isotope 2 abundance
- 24.2433%
Isotope 1 abundance: 75.7567%. With isotope masses 34.96885 u and 36.9659 u and an average of 35.453 u, isotope 1 is 75.7567% and isotope 2 is 24.2433%.
How to calculate
Finds the percent abundance of each of an element’s two isotopes from their masses and the average atomic mass, or the average or an isotope’s mass from the others.
Example with the default inputs (Average atomic mass (u) 35.453, Isotope 1 mass (u) 34.96885, Isotope 2 mass (u) 36.9659): With isotope masses 34.96885 u and 36.9659 u and an average of 35.453 u, isotope 1 is 75.7567% and isotope 2 is 24.2433%.
Formula: M = m₁ × p₁ ÷ 100 + m₂ × p₂ ÷ 100 with p₂ = 100 − p₁, so p₁ = (M − m₂) ÷ (m₁ − m₂) × 100; m₁ = m₂ + (M − m₂) × 100 ÷ p₁; m₂ = m₁ + (M − m₁) × 100 ÷ p₂.
- The element has exactly two isotopes, and the abundances are percents of atoms (amount fractions).
- Masses are in unified atomic mass units (u, also called amu or Da); the molar mass in g/mol has the same number.
- An abundance is from 0% to 100%, so the average must lie between the two isotope masses.
Worked examples
Each example is checked against the calculator on every build.
- Average atomic mass (u) 35.453, Isotope 1 mass (u) 34.96885, Isotope 2 mass (u) 36.9659 gives Isotope 1 abundance 75.756741%, Isotope 2 abundance 24.243259%.Source: OpenStax, Chemistry 2e, §2.3 Atomic Structure and Symbolism (average mass = Σ(fractional abundance × isotopic mass); Example 2.5: 35Cl 34.96885 amu, 37Cl 36.96590 amu, average 35.453 amu give 75.76% and 24.24%; Check Your Learning: 63Cu 62.9296, 65Cu 64.9278, average 63.546 give 69.15% and 30.85%), https://openstax.org/books/chemistry-2e/pages/2-3-atomic-structure-and-symbolism (retrieved 2026-10-18)
- Average atomic mass (u) 63.546, Isotope 1 mass (u) 62.9296, Isotope 2 mass (u) 64.9278 gives Isotope 1 abundance 69.152237%, Isotope 2 abundance 30.847763%.Source: OpenStax, Chemistry 2e, §2.3 Atomic Structure and Symbolism (average mass = Σ(fractional abundance × isotopic mass); Example 2.5: 35Cl 34.96885 amu, 37Cl 36.96590 amu, average 35.453 amu give 75.76% and 24.24%; Check Your Learning: 63Cu 62.9296, 65Cu 64.9278, average 63.546 give 69.15% and 30.85%), https://openstax.org/books/chemistry-2e/pages/2-3-atomic-structure-and-symbolism (retrieved 2026-10-18)
- Isotope 1 mass (u) 10.0129, Isotope 2 mass (u) 11.0093, Isotope 1 abundance 19.9% gives Average atomic mass (u) 10.811016, Isotope 2 abundance 80.1%.Source: OpenStax, Chemistry 2e, §2.3 Atomic Structure and Symbolism (average mass = Σ(fractional abundance × isotopic mass); Example 2.5: 35Cl 34.96885 amu, 37Cl 36.96590 amu, average 35.453 amu give 75.76% and 24.24%; Check Your Learning: 63Cu 62.9296, 65Cu 64.9278, average 63.546 give 69.15% and 30.85%), https://openstax.org/books/chemistry-2e/pages/2-3-atomic-structure-and-symbolism (retrieved 2026-10-18) (boron, about 10.81 amu)
- Average atomic mass (u) 63.546, Isotope 1 mass (u) 62.9296, Isotope 1 abundance 69.15% gives Isotope 2 mass (u) 64.927655, Isotope 2 abundance 30.85%.Source: OpenStax, Chemistry 2e, §2.3 Atomic Structure and Symbolism (average mass = Σ(fractional abundance × isotopic mass); Example 2.5: 35Cl 34.96885 amu, 37Cl 36.96590 amu, average 35.453 amu give 75.76% and 24.24%; Check Your Learning: 63Cu 62.9296, 65Cu 64.9278, average 63.546 give 69.15% and 30.85%), https://openstax.org/books/chemistry-2e/pages/2-3-atomic-structure-and-symbolism (retrieved 2026-10-18)
How it works
An element with two isotopes, of masses m₁ and m₂, and abundances p₁ and p₂ in percent (p₁ + p₂ = 100), has the average atomic mass
M = m₁ × p₁ ÷ 100 + m₂ × p₂ ÷ 100
Type any three of M, m₁, m₂ and p₁, and the page solves for the fourth. Isotope 2’s abundance is always p₂ = 100 − p₁, from the exact p₁ (as typed, or as worked out from M, m₁ and m₂).
- Abundance from the masses and the average: p₁ = (M − m₂) ÷ (m₁ − m₂) × 100.
- Average from the masses and p₁: M = m₂ + (m₁ − m₂) × p₁ ÷ 100.
- An isotope mass from the average, the other mass and p₁: m₁ = m₂ + (M − m₂) × 100 ÷ p₁, or m₂ = m₁ + (M − m₁) × 100 ÷ (100 − p₁).
Rules. Masses are above 0 and at most 300 u; the abundance p₁ is from 0% to 100%. A solved abundance outside 0% to 100% (an average that is not between the two isotope masses), or a solved mass outside 0 to 300 u, has no answer. Equal isotope masses give no abundance, and an abundance of 0% gives no mass for that isotope. A result too large or too small to hold in a 64-bit float (beyond about 10³⁰⁸, or a nonzero value below about 10⁻³²³) has no answer rather than showing infinity or 0. Typed values are read as exact decimals, and each result is worked as an exact fraction before it is rounded for display.
Output format. Abundances show as percents to 4 decimal places, and masses in u to 6 decimal places, with trailing zeros dropped.
Worked examples by hand
Chlorine (OpenStax Example 2.5). m₁ = 34.96885 u (35Cl), m₂ = 36.96590 u (37Cl), M = 35.453 u. p₁ = (35.453 − 36.96590) ÷ (34.96885 − 36.96590) × 100 = −1.51290 ÷ −1.99705 × 100 = 75.7567%; p₂ = 24.2433%. OpenStax rounds them to 75.76% and 24.24%.
Copper (OpenStax Check Your Learning). m₁ = 62.9296 u, m₂ = 64.9278 u, M = 63.546 u. p₁ = −1.3818 ÷ −1.9982 × 100 = 69.1522%; p₂ = 30.8478% (the book: 69.15% and 30.85%).
Boron, the average. m₁ = 10.0129 u at 19.9%, m₂ = 11.0093 u. M = 11.0093 + (10.0129 − 11.0093) × 19.9 ÷ 100 = 11.0093 − 0.1982836 = 10.8110164 u, and p₂ = 80.1%.
Copper, a missing mass. M = 63.546 u, m₁ = 62.9296 u, p₁ = 69.15%, so p₂ = 30.85%. m₂ = 62.9296 + 0.6164 × 100 ÷ 30.85 = 62.9296 + 1.998055 = 64.927655 u.
Other questions people ask
How do I calculate percent abundance?
Let x be the fraction of isotope 1. The average atomic mass is x × m₁ + (1 − x) × m₂, so x = (average − m₂) ÷ (m₁ − m₂). Multiply by 100 for the percent; isotope 2 has the rest. For chlorine: (35.453 − 36.96590) ÷ (34.96885 − 36.96590) = 0.7576, so 75.76% 35Cl and 24.24% 37Cl.
Why is the average atomic mass closer to one isotope?
The average is weighted by abundance, so it lies nearer the more common isotope. Chlorine’s average of 35.453 u is much closer to 35Cl (34.969 u) because about three quarters of chlorine atoms are 35Cl.
How do I find the average atomic mass from the abundances?
Multiply each isotope’s mass by its abundance as a fraction and add the results. Boron with 19.9% 10B (10.0129 u) and 80.1% 11B (11.0093 u) averages 10.0129 × 0.199 + 11.0093 × 0.801 = 10.811 u.
Can I find the mass of one isotope?
Yes. Type the average, the other isotope’s mass and isotope 1’s abundance. With copper’s average 63.546 u and 63Cu at 62.9296 u and 69.15%, 65Cu is the other 30.85%, and its mass is 62.9296 + (63.546 − 62.9296) × 100 ÷ 30.85 = 64.928 u.
What if the element has three or more isotopes?
One average cannot give three unknown shares. Use the atomic mass calculator with all the masses and abundances, or this page when only two isotopes are unknown and the others are known.
Why does the page give no answer for my numbers?
The average must lie between the two isotope masses, because each share is between 0% and 100%. An average outside that range, or two equal isotope masses, has no answer.