# Atomic Mass From Atomic Abundance Chemistry Problem

## Worked Atomic Abundance Chemistry Problem The atomic weight of an element is a weighted ratio of atomic weights. For boron, this means the number of neutrons in an atom isn't always 5. ROGER HARRIS/SCIENCE PHOTO LIBRARY / Getty Images

You may have noticed the atomic mass of an element isn't the same as the sum of the protons and neutrons of a single atom. This is because elements exist as multiple isotopes. While each atom of an element has the same number of protons, it can have a variable number of neutrons. The atomic mass on the periodic table is a weighted average of the atomic masses of atoms observed in all samples of that element. You can use the atomic abundance to calculate the atomic mass of any element sample if you know the percentage of each isotope.

## Atomic Abundance Example Chemistry Problem

The element boron consists of two isotopes, 105B and 115B. Their masses, based on the carbon scale, are 10.01 and 11.01, respectively. The abundance of 105B is 20.0% and the abundance of 115B is 80.0%.

What is the atomic mass of boron?

Solution:

The percentages of multiple isotopes must add up to 100%. Apply the following equation to the problem:

atomic mass = (atomic mass X1) · (% of X1)/100 + (atomic mass X2) · (% of X2)/100 + ...
where X is an isotope of the element and % of X is the abundance of the isotope X.

Substitute the values for boron in this equation:

atomic mass of B = (atomic mass of 105B · % of 105B/100) + (atomic mass of 115B · % of 115B/100)
atomic mass of B = (10.01· 20.0/100) + (11.01· 80.0/100)
atomic mass of B = 2.00 + 8.81
atomic mass of B = 10.81