Chapter 14, Problem 99A

Interpretation Introduction

Interpretation: The percentage of methane in the original mixture needs to be determined.

Concept Introduction:

From the total mixture, the percentage of any component can be determined from its pressure if volume and temperature remain the same before and after the reaction. If the total pressure of the mixture is known, the percentage of any component can be calculated by calculating the pressure of that component and putting in the following formula:

  $\%\text{component}=\frac{\text{Pressure of component}}{\text{Total pressure of mixture}}\times 100\%$

Explanation of Solution

Total pressure of the sample containing a mixture of ethyne gas and methane gas is 16.8 kPa. On burning the sample, carbon dioxide and water vapor is produced. The pressure of carbon dioxide collected is 25.2 kPa at the same volume and temperature as the mixture. The percentage of methane in the original mixture needs to be determined.

The reactions for the combustion of ethyne gas and methane are represented as follows:

  $\begin{array}{l}{\text{CH}}_{\text{4}}+2{\text{O}}_{\text{2}}\to {\text{CO}}_2+2{\text{H}}_{\text{2}}\text{O}\\ {\text{2CH}}_{\text{2}}{\text{H}}_{\text{2}}+{\text{5O}}_{\text{2}}\to {\text{4CO}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}\end{array}$

The total pressure of the mixture is given 16.9 kPa.

Let the pressure of methane be x and that of ethyne be y thus,

  $x+y=16.9\text{ kPa}$   ....... (1)

Or,

  $x=16.9-y$   ...... (2)

Now, the number of moles is directly proportional to pressure if volume and temperature are constant.

From the combustion reaction, 1 mol of methane gives 1 mol of carbon dioxide and 2 mol of ethyne gas gives 4 mol of carbon dioxide thus, the number of moles of carbon dioxide obtained from 1 mol of ethyne gas will be 2 mol. Here, the number of moles can be replaced by pressure.

Thus, the pressure of carbon dioxide from methane will be:

  $P_{{\text{CO}}_{\text{2}}}\left(\text{methane}\right)=x$

Similarly,

  $P_{{\text{CO}}_{\text{2}}}\left(\text{ethyne}\right)=2y$

According to the question, the pressure of carbon dioxide collected is 25.2 kPa.

Thus,

  $x+2y=25.2\text{ kPa}$   ....... (3)

Put the value of x from (2) to (3).

  $\begin{array}{c}16.9-y+2y=25.2\\ \text{16}\text{.9}+y=25.2\\ y=8.3\end{array}$

Thus, the value of x will be:

  $\begin{array}{c}x=16.9-8.3\\ =8.6\end{array}$

Thus, the pressure of methane gas is 8.6 kPa.

To calculate the percentage of methane in the original mixture, the following formula can be used:

  $\%\text{methane}=\frac{\text{Pressure of methane}}{\text{Total pressure of mixture}}\times 100\%$

Substitute the values,

  $\begin{array}{c}\%\text{methane}=\frac{\text{8}\text{.6}}{\text{16}\text{.8}}\times 100\%\\ =51.2\%\end{array}$

Thus, the percentage of methane in the original mixture is 51.2%.