# Finding average Mr of a gaseous mixture

A 2 g mixture of cyclohexanol, acetone and pentanal is completely burnt in oxygen. When the gaseous product is passed through $\ce{P2O5}$, the mass of $\ce{P2O5}$ increased by 1.998 g. Find the average molecular mass of the mixture.

$\ce{P2O5}$ absorbs water, so the amount of water absorbed is 0.111 mol. I attempted constructing an equation for the combustion reaction but realized that the proportions and thus the mole ratios of each gas are unknown.

• Let's see your balanced reaction equations and your calculated molecular masses. – Chet Miller Oct 19 '17 at 1:05
• Are you asking about the average relative molecular mass $M_\mathrm r$ (as in the title), the average molecular mass (as in the text), or the average molar mass $M$ (not mentioned but reasonable)? These are different quantitites with different dimensions. – Faded Giant Oct 26 '17 at 17:28

I suspect one might utilize the fact that all three compounds – cyclohexanol $\ce{C6H12O}$, acetone $\ce{C3H6O}$ and pentanal $\ce{C5H10O}$ – share the same empirical formula $\ce{(CH2)_xO}$, which is also applicable to the mixture of reactants, and so instead of writing a set reaction equations

\begin{align} \ce{C6H12O + 8.5 O2 & -> 6 H2O + 6 CO2} \tag{R1}\\ \ce{C3H6O + 4 O2 & -> 3 H2O + 3 CO2} \tag{R2}\\ \ce{C5H10O + 7 O2 & -> 5 H2O + 5 CO2} \tag{R3} \end{align}

one can avoid finding volume fraction $\phi_i$ of each individual component and use a single reaction scheme such as

$$\ce{(CH2)_xO + \frac{3x-1}{2} O2 -> x H2O + x CO2} \tag{R4}$$

and find relation between average molecular weight $\bar{M}$ and $x$ solely from (R4). Lets denote the mass and the total amount of gas mixture components as $m$ and $n$, respectively. Then according to (R4) total amount of gas mixture is $n = \frac{n(\ce{H2O})}{x}$, and the average molecular weight is established as

$$\bar{M} = \frac{m}{n} = \frac{x \cdot m}{n(\ce{H2O})} = \frac{x \cdot m \cdot M(\ce{H2O})}{m(\ce{H2O})} = \frac{x \cdot \pu{2 g} \cdot \pu{18.015 g mol-1}}{\pu{1.998 g}} = (18.033 x) \,\pu{g mol-1} \tag{1}$$

At the same time using molecular weights of the elements:

$$\bar{M} = x \cdot (M(\ce{C}) + 2M(\ce{H})) + M(\ce{O}) = (14.027 x + 15.999)\,\pu{g mol-1} \tag{2}$$

Equating (1) and (2):

$$18.033 x = 14.027x + 15.999 \tag{3}$$

and solving the resulting equation (3) for $x$, one finds that $x = 3.994$. Solving either (1) or (2) with obtained $x$ yields in average molecular weight $\bar{M} = \pu{72.023 g mol-1}$.