# What is the percent abundance?

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.

- Page: https://www.acalculator.org/chemistry/percent-abundance-calculator
- JSON spec: https://www.acalculator.org/chemistry/percent-abundance-calculator.json
- Version: b128a245df6e

## Default answer

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%.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| avg | Average atomic mass (u) | The element’s average atomic mass, in unified atomic mass units. |
| m1 | Isotope 1 mass (u) | The mass of the first isotope, in unified atomic mass units. |
| m2 | Isotope 2 mass (u) | The mass of the second isotope, in unified atomic mass units. |
| p1 | Isotope 1 abundance | The percent of the element’s atoms that are isotope 1. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| avg | Average atomic mass (u) | m₁ × p₁ ÷ 100 + m₂ × p₂ ÷ 100. |
| m1 | Isotope 1 mass (u) | m₂ + (M − m₂) × 100 ÷ p₁. |
| m2 | Isotope 2 mass (u) | m₁ + (M − m₁) × 100 ÷ (100 − p₁). |
| p1 | Isotope 1 abundance | (M − m₂) ÷ (m₁ − m₂) × 100: the percent of atoms that are isotope 1. |
| p2 | Isotope 2 abundance | 100% − isotope 1’s abundance. |

## Method

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.

## Assumptions

- 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

1. avg = 35.453, m1 = 34.96885, m2 = 36.9659 gives p1 = 75.756741%, p2 = 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).
2. avg = 63.546, m1 = 62.9296, m2 = 64.9278 gives p1 = 69.152237%, p2 = 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).
3. m1 = 10.0129, m2 = 11.0093, p1 = 19.9% gives avg = 10.811016, p2 = 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).
4. avg = 63.546, m1 = 62.9296, p1 = 69.15% gives m2 = 64.927655, p2 = 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).

## FAQ

### 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.

## Sources

- OpenStax, Chemistry 2e, §2.3 Atomic Structure and Symbolism: average mass = Σ(fractional abundance × isotopic mass); Example 2.5, Calculation of Percent Abundance (chlorine, 75.76% 35Cl and 24.24% 37Cl); Check Your Learning (copper, 69.15% Cu-63 and 30.85% Cu-65). https://openstax.org/books/chemistry-2e/pages/2-3-atomic-structure-and-symbolism (retrieved 2026-10-18)