Finding average Mr of a gaseous mixture
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 &amp; -&gt; 6 H2O + 6 CO2} \tag{R1}\\
\ce{C3H6O + 4 O2 &amp; -&gt; 3 H2O + 3 CO2} \tag{R2}\\
\ce{C5H10O + 7 O2 &amp; -&gt; 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

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

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 $\bar{M} = \pu{72.023 g mol-1}$.