Oxygen is composed of three isotopes 16/8 O (15.995 u), 17/8 O (16.999 u) and 18/8 O (17.999 u). One of these isotopes, 17/8 O, comprises of 0.037% of oxygen. What is the percentage abundance of the other two isotopes, using the average atomic mass of 15.9994 u.

Md Ariful Islam

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The abundance of $""_8^16"O"$ is 99.762 %, and the abundance of $""_8^18"O"$ is 0.201 %.

Assume you have 100 000 atoms of O. Then you have 37 atoms of $""_8^17"O"$ and 99 963 atoms of the other .

Let x = the number of atoms of $""_8^16"O"$. Then the number of atoms of $""_8^18"O"$ = 99 963 - x

The total mass of 100 000 atoms is

x × 15.995 u + (99 963 – x) × 17.999 u + 37 × 16.999 u = 100 000 × 15.9994 u

15.995 x + 1 799 234.037 – 17.999 x + 628.963 = 1 599 940

2.004 x = 199 123

x = 199 123/2.004 = 99 762

So there are 99 762 atoms of $""_8^16"O"$ or 99.762 %.

The number of $""_8^18"O"$ atoms is 99 963 – 99 762 = 201 atoms or 0.201 %.

Hope this helps.

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