| Isotope | % Natural abundance | Molar mass |
| \[^{1}H\] | 99.985 | 1 |
| \[^{2}H\] | 0.015 | 2 |
Answer:
Average atomic mass
\[\text{= }\!\!\Sigma\!\!\text{
}\frac{\text{ }\!\!%\!\!\text{ Abundance}}{\text{100}}\text{
}\!\!\times\!\!\text{ Atomic}\,\text{mass}\]
\[=\frac{99.985}{100}\times
1+\frac{0.015}{100}\times 2\]
\[=0.99985+0.0003\]
\[=1.00015\,amu\]
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