Chapter 3, Problem 132AP

A mixture of methane $\left({\text{CH}}_{\text{4}}\right)$ and ethane $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right)$ of mass 13.43 g is completely burned in oxygen. If the total mass of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ produced is 64.84 g, calculate the fraction of ${\text{CH}}_{\text{4}}$ in the mixture.

Interpretation Introduction

Interpretation:

The fraction of ${\text{CH}}_4$ in the mixtureis to be calculated.

Concept introduction:

Thequantitative relationships between the various reactants and products in terms of moles, masses, molecules and volumes is called Stoichiometry. The coefficients of the balanced chemical equation are called stoichiometric coefficients.

Answer to Problem 132AP

Solution: $0.387$

Explanation of Solution

Given information:

$\text{Mass of the mixture}=13.43\text{ g}$

$\text{Mass of the products}=\text{64}\text{.84 g}$

The balanced chemical equation is -

$\begin{array}{l}{\text{CH}}_4+2{\text{O}}_2\to {\text{CO}}_2{\text{+2H}}_2\text{O}\\ 2{\text{C}}_2{\text{H}}_6{\text{+7O}}_2\to 4{\text{CO}}_2{\text{+6H}}_2\text{O}\end{array}$

Consider $x$ to be the mass of ${\text{CH}}_4$.

Then, the mass of ${\text{C}}_2{\text{H}}_6$ is $\left(13.43-x\right)\text{ g}$.

Calculate the mass of ${\text{CO}}_2$ from ${\text{CH}}_4$ as follows:

$\begin{array}{c}{\text{mass of CO}}_{\text{2}}=x\text{g}{\text{CH}}_{\text{4}}\text{×}\frac{\text{1}\text{mol}{\text{CH}}_{\text{4}}}{\text{16}\text{.04}\text{g}{\text{CH}}_{\text{4}}}\text{×}\frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}{\text{CH}}_{\text{4}}}\text{×}\frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}{\text{CO}}_{\text{2}}}\\ =2.744x\text{g}{\text{CO}}_{\text{2}}\end{array}$

Calculate the mass of ${\text{H}}_2\text{O}$ from ${\text{CH}}_4$ as follows:

$\begin{array}{c}{\text{mass of H}}_{\text{2}}\text{O}=x\text{g}{\text{CH}}_{\text{4}}\text{×}\frac{\text{1}\text{mol}{\text{CH}}_{\text{4}}}{\text{16}\text{.04}\text{g}{\text{CH}}_{\text{4}}}\text{×}\frac{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol}{\text{CH}}_{\text{4}}}\text{×}\frac{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ =2.247x\text{g}{\text{H}}_{\text{2}}\text{O}\\ \end{array}$

Calculate the mass of ${\text{CO}}_2$ from ${\text{C}}_2{\text{H}}_6$ as follows:

$\begin{array}{c}{\text{mass of CO}}_{\text{2}}=(13.43-x)\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{30}\text{.07}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{×}\frac{\text{4}\text{mol}{\text{CO}}_{\text{2}}}{\text{2}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{×}\frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}{\text{CO}}_{\text{2}}}\\ =2.927\left(13.43-x\text{g}\right){\text{CO}}_{\text{2}}\end{array}$

Calculate the mass of ${\text{H}}_2\text{O}$ from ${\text{C}}_2{\text{H}}_6$ as follows:

$\begin{array}{c}{\text{mass of H}}_2\text{O}=(13.43-x)\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{30}\text{.07}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{×}\frac{\text{6}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{2}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{×}\frac{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ =1.798\left(13.43-x\text{g}\right){\text{H}}_2\text{O}\end{array}$

The addition of all the masses of ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ gives $64.84\text{ g}$.

It is calculated as follows:

$\begin{array}{c}2.744x\text{ g}+2.247x\text{ g}+2.927(13.43-x)\text{ g}+1.798(13.43-x)\text{ g} = 64.84\text{ g}\\ 0.266x = 1.383\\ x = 5.20\text{ g}\end{array}$

The fraction of ${\text{CH}}_4$ is calculated by dividing the mass of ${\text{CH}}_4$ by the mass of the total mixture as follows:

$\begin{array}{c}{\text{fraction of CH}}_{\text{4}}=\frac{5.20\text{g}}{13.43\text{g}}\\ =0.387\end{array}$

Conclusion

The fraction of ${\text{CH}}_4$ in the mixture is $0.387$.