91Ó°ÊÓ

By dividing the determined mass of the substance by the total mass of the sample multiplied by 100, the mass percentage of the compound can be obtained.

The formula may be used to compute the mass percent, and the formula can also be used to provide a solution.

$\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{\left(}\mathit{M}\mathit{\right)}\mathit{=}\frac{\mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{}\mathit{o}\mathit{f}\mathit{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathit{\left(}\mathit{i}\mathit{n}\mathit{m}\mathit{o}\mathit{l}\mathit{\right)}}{\mathit{V}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{ }\mathit{o}\mathit{f}\mathit{ }\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{(}\mathit{i}\mathit{n}\mathit{ }\mathit{L}\mathit{)}}$

The proportion of carbon and hydrogen in a given sample in terms of weight.

The amount of moles of carbon equals the number of moles of CO2.

The sample's carbon mass is computed as

$=(2.755×{10}^{-4}\mathrm{mol})(12.0106\mathrm{g}/\mathrm{molC})=3.308\mathrm{mg}$

The weight percentof carbon is computed based on the sample weight.

$$\begin{gathered} =\frac{3.308mgC}{6.234mgsample}×100\% \\ =53.06\% \end{gathered}$$

The number of moles of hydrogen equals two moles of water.

$=2\left(\frac{(2.529×{10}^{3}g{H}_{2}O)}{(18.015g/mol{H}_{2}O)}\right)=2.808×{10}^{-4}mol$

The sample's hydrogen mass is computed as

$=(2.808×{10}^{-4}mol)(1.008g/molH)=0.2830mg$

The weight percent of hydrogen is computed based on the sample weight.

$$\begin{gathered} =\frac{0.2830\mathrm{mgH}}{6.234\mathrm{mg}\mathrm{samplo}}×100\% \\ =4.54\% \end{gathered}$$

The carbon and hydrogen weight percentages in a specific sample are 53.06% and 4.53% respectively.