At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 16.5 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

Question

abb0t · Answer

Keep in mind that combustion is defined by: C$\sf \color{orange }{_n}$H$\sf \color{orange }{_{2n+2}}$ + O$_2$ $\rightarrow$ H$_2$O + CO$_2$ 
where $\sf \color{red}{n}$= number of carbon(s)!
However, notice that for your problem, you have TWO organics. Methane and Propane!

Next, assume that the gases are ideal, so that you can use the formular $\sf \color{blue}{PV = nRT}$, where
P = pressure [atm]
V = volume [L]
n = mol
R = gas constant  [0.0821 $\frac{liter~ \times ~atm}{mol ~ \times~K}$]
T = temperature [K]

Now, from algebra, rearrange to solve for $\sf \color{purple}{n}$!! Which means you divide presure and volume by the ideal gas constant and temperature. (YOU'RE DOING THI FOR $\sf \color{red}{BOTH}$ propane $and$ methane

[NOTE #2: or you you could use the fact that an ideal gas at STP = 22.4 $\frac{L}{mol}$
Eiehter method, you get same amount of mol value.

Now, this will get tricky, you have: $\sf \color{green}a$CH$\sf \color{orange }{_{4}}$ + $\sf \color{purple}b$C$\sf \color{red }{_3}$H$\sf \color{red}{_{8}}$ + O$_2$ 
$$\left\{\begin{matrix}
a+b = n &  & \ 
4a + 8b =n&  & 
\end{matrix}\right.$$

the TOP = # of carbon on reactant side
BOTTOM = #  hydrogen on reactant side. 

and n = mol that you calculated above either using ideal gas law formula OR ideal volume @ STP. 

fROM ALGEBRA, you should know how to solve for two unknowns. You can use $\sf \color{red}{substitution}$ method or $\sf \color{green}{elimination}$ method.

Finally, to get the mole fraction IS mol of methane divide by total mol. And you're done.