Question

A $\mathbf{9}\mathbf{.780}\mathbf{-}\mathbf{\text{g}}$gaseous mixture contains ethane and propane. Complete combustion to form carbon dioxide and water requires $\mathbf{1}\mathbf{.120}$moles of oxygen gas. Calculate the mass percent of ethane in the original mixture

Short answer

The mass percentage of ethane in the original mixture is$56.44\%$

Step-by-step solution

Definition of mass percentage

The mass of the element or solute divided by the mass of the compound or solute is defined as mass percentage

Calculate the mass of ethane

Moles of ethane $=\frac{\text{Mass}}{\text{Molar\hspace{0.17em}mass}}$

Mass of propane$=(9.780-x)\text{g}$

Moles of propane$=\frac{(9.780-x)}{44.1g/mol}$

Mass of ethane$=5.52\text{g}$

Mass percent of ethane in mixture

$\begin{array}{l}=\left(\frac{\text{Mass\hspace{0.17em}of\hspace{0.17em}ethane}}{\text{Mass\hspace{0.17em}of\hspace{0.17em}mixture}}\right)\times 100\\ =\left(\frac{5.52g}{9.78g}\right)\times 100\end{array}$

Mass percent of ethane is $56.44\%$