Complete combustion of mixtures

Question

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.

Answer 1

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

Answer 2

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,

$$n_2 = 0.81\cdot \pu{4.98e-2 mol} = \pu{4.03e-2 mol}$$