### Theory:

We have learnt the number of entities present in a given substance so far. However, the percentage of a specific element present in a compound is frequently required.
The percentage composition of a compound represents the mass of each element present in $$100$$ $$g$$ of the compound.
The formula for calculating the mass percent of an element is given below:

$\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{percent}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{an}\phantom{\rule{0.147em}{0ex}}\mathit{element}=\frac{\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{element}\phantom{\rule{0.147em}{0ex}}\mathit{in}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{compound}}{\mathit{Molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{compound}}×100$

Let us look at few examples for calculating percentage composition.

Example: $$1$$

Calculate the mass percent of water ($$H_2O$$).

Solution:

$\mathit{Molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{H}_{2}O=2\left(1\right)+16=18\phantom{\rule{0.147em}{0ex}}g$

$\begin{array}{l}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{percent}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{hydrogen}=\frac{2}{18}×100=11.11%\\ \mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{percent}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{oxygen}=\frac{16}{18}×100=88.89%\end{array}$

Example: $$2$$

Calculate the mass percent of methane ($$CH_4$$).

Solution:

$\mathit{Molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CH}}_{4}=12+4=16\phantom{\rule{0.147em}{0ex}}g$

$\begin{array}{l}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{percent}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{carbon}=\frac{12}{16}×100=75%\\ \mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{percent}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{hydrogen}=\frac{4}{16}×100=25%\end{array}$

This percentage composition is used for determining the empirical formula and molecular formula.