I am questioning the answer key of an exam question my chemistry professor has given, because I think there is a flaw in the methodology of the exam answer. I have worked out my solution using my assumptions here, and would like to know if I have erred in my reasoning, or if my professor has.
The question asks:
Almost all brake fluids are hygroscopic which means they absorb water from the atmosphere. This is why you should change your brake fluid yearly. Assume for this exam brake fluid is comprised of only $\ce{C7H16O4}$ (triethylene glycol monomethyl ether).
You send a $\pu{0.12235 g}$ sample of your brake fluid, $\ce{C7H16O4}$, for combustion analysis and the results give $\pu{0.10674 g}$ of $\ce{H2O}$ and $\pu{0.20199 g}$ $\ce{CO2}$. What is the mass percentage of water in your brake fluid?
Because we don't know the purity of the sample of the brake fluid, I started by assuming that the generated $\ce{CO2}$ could be used to determine the amount of extra water, if water was the only impurity.
I started by converting grams to moles:
$$ \pu{0.20199 g \ce{CO2}} \times \frac{\pu{1 mol \ce{CO2}}}{\pu{44.009 g \ce{CO2}}} = \pu{4.5897 \times 10^-3 mol \ce{CO2}} $$ $$ \pu{0.10674 g \ce{H2O}} \times \frac{\pu{1 mol \ce{H2O}}}{\pu{18.015 g \ce{H2O}}} = \pu{5.9251 \times 10^-3 mol \ce{H2O}} $$
Then I balanced the equation: $$ \ce{C7H16O4 + 9O2 -> 7CO2 + 8H2O} $$
I inputted the known values into a BCA table: $$ \begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & & & &\\ \hline C & & & &\\ \hline A & & & 4.5897 \times 10^{-3} & 5.9251 \times 10^{-3}\\ \hline \end{array} $$
If we assume that $100\%$ of the $\ce{C7H16O4}$ combusted, and that there was extra $\ce{H2O}$ in the reaction mixture, then we can base the rest of this table off of the amount of $\ce{CO2}$ generated, because heating water does not produce $\ce{CO2}$.
So, filling out the necessary components, we get: $$ \begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & & & 0 & 6.7964 \times 10^{-4}\\ \hline C & & & +4.5897 \times 10^{-3} & +5.2454 \times 10^{-3} \\ \hline A & & & 4.5897 \times 10^{-3} & 5.9251 \times 10^{-3}\\ \hline \end{array} $$
Based on this, we can see that there were $\pu{6.7964 \times 10^{-4} mol \ce{H2O}}$ of extra water before the reaction occurred.
Converting moles to grams: $$ \pu{6.7964 \times 10^{-4} mol \ce{H2O}} \times \frac{\pu{18.015 g \ce{H2O}}}{\pu{1 mol \ce{H2O}}} = \pu{0.012244 g \ce{H2O}} $$
Finally, we can divide by the total mass of the contaminated sample: $$ \frac{\pu{0.012244 g \ce{H2O}}}{\pu{0.12235 g sample}} \times 100\% = \pu{10.007\%\; \ce{H2O}} $$
However, my professor's exam solution doesn't agree with this. I am skeptical as to whether the given exam solution is correct, because the reasoning of it seems off to me:
Theoretical yield: $$ \begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & 7.4513 \times 10^{-4} & \ldots & 0 & 0\\ \hline C & -7.4513 \times 10^{-4} & \ldots & +5.2159 \times 10^{-3} & +5.9610 \times 10^{-3}\\ \hline A & 0 & \ldots & 5.2159 \times 10^{-3} & 5.9610 \times 10^{-3}\\ \hline \end{array} $$ or $\pu{0.22955 g \ce{CO2}}$ and $\pu{0.10739 g \ce{H2O}}$.
Since carbon does not change due to the addition of water, $\Delta \ce{CO2}$ is proportional to the percent $\ce{H2O}$.
Therefore, $$ \begin{array}{crCl} & \pu{0.22955} & \pu{g\; \ce{CO2} theoretical} \\ -& \pu{0.20199} & \pu{g\; \ce{CO2} actual} \\ \hline & 0.02756 & \pu{g\; \ce{CO2} difference} \\ \end{array} $$
$$ \frac{\pu{0.02756 g\; \ce{CO2}}}{\pu{0.22955 g\; \ce{CO2}}} \times 100\% = \pu{12.006\%\; \ce{H2O}} $$
Three things are extremely off-putting about this explanation:
- The numerical values of a known-to-be-impure substance are used to calculate theoretical yield
- The theoretical yield calculated is of an unrelated substance
- I don't think the mass difference of carbon dioxide is proportional to the excess $\ce{H2O}$
Does the professor's answer make sense?