[StefR]

Hi

Last week we got a bunch of questions about the stoichiometry of reactions. I managed to solve everything except the following.

A group of students burns  0.300 mole of a gas mixture which consists of methane and butane to find its composition. (excess oxygen)
The reaction forms water vapor and carbon dioxide. After the reaction they capture all the carbon dioxide and measure it (27°C , 0.987atm), the result is 26.4g.

Determine the mass fraction of the methane in the gas mixture.

What i have tried:
Add an unknown, the mass fraction of methane.
0.300 mole * ω_CH4 * M_mixture/M_CH4 is the amount of methane in the mixture (in mole).

0.300 mole * ω_C4H10 * M_mixture/M_C4H10 is the amount of butane in the mixture.

26.4g / M_CO2 = 0.6 mole is the amount of formed CO₂

CH₄ + 2O₂  CO₂ + 2H₂O

2C₄H₁₀ + 13O₂  8CO₂ + 10H₂O

0.6 mole = 0.300 mole * ω_CH4 * M_mixture/M_CH4 + 8/2*(0.300 mole * ω_C4H10 * M_mixture/M_C4H10)


This equation gives the wrong answer, the right answer should be 35.6%

Can someone look at this and point me in the right direction?

Thanks in advance

[thetada]

First of all, thanks for this question. I'm going to be setting this for my students at the earliest opportunity 

Anyway, you've got what you need to make a simultaneous equation. Let's say that X = the number of moles (n) of methane, then Y = n (Butane). You know their sum. Next, express the number of moles of CO2 produced from the methane in terms of X, and  the number of moles of CO2 produced from butane in terms of Y. Their sum will equal the total number of moles of CO2. Hope that helps.

[StefR]

This solved the problem , thank you!
Can't see the fault in my reasoning though.

[StefR]

Okay , the molar mass of the mixture is unknown too, stupid me!

[thetada]

I wasn't too sure what your expression meant in your original message, but I'm glad you're feeling on top of it all now.