Complete combustion of mixtures

I am working on my scholarship exam practice (around high school and first year university level). I could actually solve the problem below but my solution is pretty time consuming. It would be appreciated if I could find other aspects in solving this.



> 
> Exactly $\pu{4.32 g}$ of oxygen gas was required to completely burn a $\pu{2.16 g}$ sample of a mixture of methanol and ethanol. How many moles of ethanol are contained in the sample? (Answer: $\pu{0.04 mol})$
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What I did is setting $x$ and $y$ as the mass of methanol and ethanol, respectively. Then I formed 2 equations. First one was $(x + y = \pu{2.16 g})$ and another was the equation $(3x/64 + 3y/46 = \pu{0.135 mol})$ from the number of moles of oxygen gas, derived from two complete combustion equations. And then I solved simultaneous equations. I am wondering if this was the only way to solve this problem. If you think differently, please let me know.




Good news is you are in right track and set up two equations correctly. However, I think the calculations would get easier if you have chosen your units of variables in $\pu{mol}$. Also, you may not needed to do more calculations since the answer is in $\pu{mol}$. Thus, your two equations should be:
$$32x + 46y = 2.16 \;\text{.... (1), and}$$
$$\frac{3}{2}x + 3y = 0.135 \;\text{.... (2)}$$
Simplify equation (2) as:
$$x + 2y = 0.090 \;\text{ and }\therefore x=0.090-2y $$


Substitute $x$ in equation (1):
$$32(0.090-2y) + 46y = 2.16 $$
$$(64-46)y = 32 \times 0.090 - 2.16 = 0.72$$
$$\therefore y = \frac{0.72}{18} = 0.04$$


 3

I can think of another solution, which is somewhat more tedious, but as long as you asked to *think differently*, here it is: you can utilize the fact that both components are alcohols and share the same empirical formula $\ce{C\_yH\_{2y+2}O}$, $y\in (1,2)$.
This allows to find the unknown amount from the average molar mass $\bar{M}$ of the mix:


$$\ce{C\_yH\_{2y+2}O + 3/2$y$ O2 -> y CO2 + $(y + 1)$ H2O}$$ 


Let's denote methanol with 1, ethanol with 2.
Then amount of ethanol $n\_2$ in the mix of amount $n$ can be found via its mole fraction $x\_2$:


$$n\_2 = x\_2n \label{eqn:1}\tag{1}$$


Mole fraction of a component in the binary mix can be expressed in terms of molar masses:


$$x\_2 = \frac{\bar{M} - M\_1}{M\_2 - M\_1} \label{eqn:2}\tag{2}$$


Total amount of the mix can also be found from the total mass $m$ as well as reaction stoichiometry, expressing $n$ via the amount of oxygen:


$$n = \frac{m}{\bar{M}} = \frac{2}{3\cdot y}n(\ce{O2}) = \frac{2\cdot m(\ce{O2})}{3\cdot y\cdot M(\ce{O2})} \label{eqn:3}\tag{3}$$


Now, $y$ can be found by solving an equation with one unknown from the combination of average molar mass' definition and \eqref{eqn:3}:


$$
\begin{cases}
\bar{M} = y\cdot M(\ce{C}) + (2\cdot y + 2)M(\ce{H}) + M(\ce{O}) \\
\displaystyle\bar{M} = \frac{3\cdot y\cdot m\cdot M(\ce{O2})}{2\cdot m(\ce{O2})}
\end{cases}
$$


or


$$y\cdot M(C) + (2\cdot y + 2)M(H) + M(\ce{O}) = \frac{3\cdot y\cdot m\cdot M(\ce{O2})}{2\cdot m(\ce{O2})} \label{eqn:4}\tag{4}$$


Plugging in the numbers and solving \eqref{eqn:4}, one finds $y = 1.81$ and $\bar{M} = \pu{43.41 g mol-1}$.
From \eqref{eqn:2} $x\_2 = 0.81$; from \eqref{eqn:3} $n = \pu{4.98e-2 mol}$.
Finally,


$$n\_2 = 0.81\cdot \pu{4.98e-2 mol} = \pu{4.03e-2 mol}$$


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