Theory:

Atomic masses of some elements:

 Atomic number Name Symbol Atomic mass $$1$$ Hydrogen $$H$$ $$1.008$$ $$2$$ Helium $$He$$ $$4.003$$ $$3$$ Lithium $$Li$$ $$6.941$$ $$4$$ Beryllium $$Be$$ $$9.012$$ $$5$$ Boron $$B$$ $$10.811$$

How to calculate the average atomic mass?

Let us understand this with the help of the following examples.

Example: 1

Oxygen is the most abundant element in both the Earth’s crust and the human body. In nature, it exists as a combination of three stable isotopes, as shown in the table:

Isotopes of oxygen:

 Isotope Mass % abundance $$_{8}\textrm{O}^{16}$$ $$15.9949$$ $$99.757$$ $$_{8}\textrm{O}^{17}$$ $$16.9991$$ $$0.038$$ $$_{8}\textrm{O}^{18}$$ $$17.9992$$ $$0.205$$

$\begin{array}{l}\mathit{Atomic}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{oxygen}\\ =\left(15.9949×0.99757\right)+\left(16.9991×0.00038\right)+\left(17.9992×0.00205\right)\\ =15.999\phantom{\rule{0.147em}{0ex}}\mathit{amu}\end{array}$

Hence, the atomic mass of oxygen is $$15.999$$ $$amu$$.

Example: 2

Boron naturally found as a mixture of boron-$$10$$ ($$5$$ protons + $$5$$ neutrons) and boron-$$11$$ ($$5$$ protons + $$6$$ neutrons) isotopes. The percentage abundance of boron-$$10$$ is $$20$$, and that of boron-$$11$$ is $$80$$. The atomic mass of boron is then determined as follows:

$\begin{array}{l}\mathit{Atomic}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{boron}\\ =\left(10×\frac{20}{100}\right)+\left(11×\frac{80}{100}\right)\\ =\left(10×0.2\right)+\left(11×0.8\right)\\ =2+8.8\\ =10.8\phantom{\rule{0.147em}{0ex}}\mathit{amu}\end{array}$

Hence, the atomic mass of boron is $$10.8$$ $$amu$$.