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: $\pu{0.04 mol})$


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