91Ó°ÊÓ

The Weight Percentage is simply the ratio of a solute's mass to the mass of a solution multiplied by 100. The Weight Percentageis also referred to as the Mass Percentage.

$\mathrm{Percentbyweight}=\frac{\mathrm{gram}\mathrm{of}\mathrm{solute}}{100\mathrm{g}\mathrm{of}\mathrm{solution}}$

In this task we have a mixture which weighs 7.290mg and contains only cyclohexane and oxirane. When we analyze the mixture by combustion analysis, 21.999 mg of ${\mathrm{CO}}_2$are produced. Here we will find the weight percent of oxirane in the mixture.

Consider the following reaction of combustion:

$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_{12}+{\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}→{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{x}=\mathrm{m}\left({\mathrm{C}}_6{\mathrm{H}}_{12}\right)\mathrm{y}=\mathrm{m}\left({\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}\right) \\ \mathrm{x}+\mathrm{y}=7.290\mathrm{mg} \end{gathered}$$

Next write the equation we will use in order to determine the value of both and :

$$\begin{gathered} \mathrm{n}\left({\mathrm{CO}}_2\right)=6\mathrm{n}\left({\mathrm{C}}_6{\mathrm{H}}_{12}\right)+2\mathrm{n}\left({\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}\right) \\ \frac{21.999\mathrm{mg}}{44.010\mathrm{g}/\mathrm{mol}}=6\frac{\mathrm{x}}{84.159\mathrm{g}/\mathrm{mol}}+2\frac{\mathrm{y}}{44.053\mathrm{g}/\mathrm{mol}} \end{gathered}$$

Then make the substitution in order to get the value of :

$$\begin{gathered} \frac{21.999\mathrm{mg}}{44.010\mathrm{g}/\mathrm{mol}}=6\frac{7.290-\mathrm{y}}{84.159\mathrm{g}/\mathrm{mol}}+2\frac{\mathrm{y}}{44.053\mathrm{g}/\mathrm{mol}} \\ \mathrm{y}=0.767\mathrm{mg} \end{gathered}$$

Also make the substitution in order to get the value of :

$$\begin{gathered} \mathrm{x}+\mathrm{y}=7.290\mathrm{mg} \\ \mathrm{x}+\mathrm{y}0.767\mathrm{mg}=7.290\mathrm{mg} \\ \mathrm{x}=7.290\mathrm{mg}-0.767\mathrm{mg} \\ \mathrm{x}=6.523\mathrm{mg} \end{gathered}$$

The final step is to find the weight percent of oxirane in the mixture:

$\mathrm{wt}\%=\frac{\mathrm{y}}{\mathrm{m}\left(\mathrm{mixture}\right)}$

$$\begin{gathered} \mathrm{wt}\%=\frac{0.767\mathrm{mg}}{7.290\mathrm{mg}} \\ \mathrm{wt}\%=0.1052 \end{gathered}$$

Therefore the weight percent of oxirane in the mixture is 0.1052%.