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

Short Answer

Expert verified
The fraction of methane (CH4) in the mixture is approximately 0.384 or 38.4%.

Step by step solution

01

Determine the molar mass

First, we calculate the molar mass of each compound. The molar mass of CH4 (methane) is \(16.04 \, g/mol\), of C2H6 (ethane) is \(30.07 \, g/mol\), of CO2 (carbon dioxide) is \(44.01 \, g/mol\), and H2O (water) is \(18.02 \, g/mol\).
02

Determine the fraction of CH4 and C2H6

Let the fraction of CH4 (methane) in the mixture be x and the fraction of C2H6 (ethane) be (1-x). The mass of the original mixture is \(13.43 \, g\). Therefore, the mass of methane is \(13.43x \, g\), and the mass of ethane is \(13.43(1-x) \, g\).
03

Calculate the mass of CO2 and H2O produced

On complete combustion, 1 mol of methane will produce 1 mol of carbon dioxide and 2 mol of water. Similarly, 1 mol of ethane will produce 2 mol of carbon dioxide and 3 mol of water. Hence by using these relations and the given molar masses we can calculate the total mass of CO2 and H2O produced by burning the methane and ethane.
04

Equate the calculated mass to the given total mass

We now have two equations, one for CO2, and one for H2O, which gives the calculated total mass when methane and ethane are burned. Add these equations together along with the mass equation from Step 2 and equate this result to the total given mass of \(64.84 \, g\).
05

Solve for x

Upon solving the equation for x, we find the value of x to be approximately 0.384, to three decimal places. Thus, the fraction of methane (CH4) in the mixture is approximately 0.384.

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