Combustion analysis methodology question

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.

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, $$\begin{array}{crCl} & \pu{0.22955} & \pu{g\; \ce{CO2} theoretical} \\ -& \pu{0.20199} & \pu{g\; \ce{CO2} actual} \\ \hline & 0.02756 & \pu{g\; \ce{CO2} difference} \\ \end{array}$$

$$\frac{\pu{0.02756 g\; \ce{CO2}}}{\pu{0.22955 g\; \ce{CO2}}} \times 100\% = \pu{12.006\%\; \ce{H2O}}$$

1. The numerical values of a known-to-be-impure substance are used to calculate theoretical yield
2. The theoretical yield calculated is of an unrelated substance
3. I don't think the mass difference of carbon dioxide is proportional to the excess $$\ce{H2O}$$

Does the professor's answer make sense?

• I agree you are right but having a hard time formulating a reason. – A.K. Oct 16 '18 at 22:36
• @A.K. I spoke with a friend who had the following to say about it: 1. The professor states that the missing CO2 as a percent of of the amount of CO2 we expect is equal to the percent of water in the sample, but that's not true, because it's not proportional. However, if you multiply by 7/8th, you get 10.5%, which isn't quite right, but closer to right. 2. When you compare theoretical with actual, actual has an unknown quantity of H2O that's screwing up both sides. You have to find things off the actual CO2 and its share of the result H2O. – Liam White McShane Oct 16 '18 at 23:48
• Curious - What did the professor say about all this? Kudos to you for working through this. – MaxW Oct 19 '18 at 16:21
• @MaxW She said: "I figured out what I did wrong-In my excel macro I made a mistake with the grams of H2O that should have formed - for your exam 0.109186 grams of H2O, not 0.10674 grams. Your exam grade has been adjusted." I got a 105%. – Liam White McShane Oct 19 '18 at 18:43

I did this a little differently. If there were $$\pu{4.5897 \times 10^{-3}}$$ moles of $$\ce{CO2}$$ produced, that must mean that there were $$1/7$$ as many moles of $$\ce{C7H16O4}$$ present in the original sample, or $$\frac{4.5897\times 10^{-3}}{7}\times 164.1995=\pu{0.10766 g}$$ in the original sample. The rest of the original sample ($$\pu{0.01469 g}$$) must have been water. So the percentage of water must have been $$\frac{0.01469}{0.12235}\times 100\%= 12.006\%.$$ This agrees with your professor's answer.

• +1 Very straight forward solution! You could use better mw of 164.1995 – MaxW Oct 17 '18 at 1:01
• Thanks @MaxW I was just too lazy to work that out. – Chet Miller Oct 17 '18 at 1:07
• Good answer, but the professor's approach still bothers me. – A.K. Oct 17 '18 at 1:19
• Correct me if I'm wrong, but I think you are forgetting about the mass of oxygen that went into the reaction. You have 0.1075 g C7H16O4 plus about 0.1888 g O2. – Liam White McShane Oct 17 '18 at 1:19
• @LiamWhiteMcShane it's good, he used the number of moles of $\ce{CO2}$ extracted to get the number of moles of triethyleneglycol monomethylether in the sample – A.K. Oct 17 '18 at 1:22

For the sake of argument:

Let's assume that the professor is correct.

$$12.006 \% \text{ water} = 0.12235\text{ g} \times .12006 = 0.01469\text{ g}$$

which means that the sample has 0.12235 - 0.01469 grams of $$\ce{C7H16O4}$$, or 0.10766 grams of $$\ce{C7H16O4}$$.

So we start off with

$$\begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & 0.10766 & \ldots & 0 & 0.01469\\ \hline C & & \ldots & & \\ \hline A & & \ldots& & \\ \hline \end{array}$$

However the grams of $$\ce{CO2}$$ that 0.10766 grams of $$\ce{C7H16O4}$$ should produce is:

$$\dfrac{0.10766}{164.1995}\times 7 \times 44.009 = 0.20199$$

The grams of $$\ce{H2O}$$ that 0.10766 grams of $$\ce{C7H16O4}$$ should produce is:

$$\dfrac{0.10766}{164.1995}\times 8 \times 18.015 = 0.09449$$

So now we have:

$$\begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & 0.10766 & \ldots & 0 & 0.01469\\ \hline C & & \ldots & 0.20199 & 0.09449 \\ \hline A & & \ldots& & \\ \hline \end{array}$$

Finishing the table we have

$$\begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & 0.10766 & \ldots & 0 & 0.01469\\ \hline C & & \ldots & 0.20199 & 0.09449 \\ \hline A & & \ldots& 0.20199 & 0.10918 \\ \hline \end{array}$$

Thus the professor's solution is inconsistent with the given amount of water which was 0.10674 g.

Now let's assume that Liam is correct.

$$10.007 \% \text{ water} = 0.12235\text{ g} \times .10007 = 0.01224\text{ g}$$

which means that the sample has 0.12235 - 0.01224 grams of $$\ce{C7H16O4}$$, or 0.11011 grams of $$\ce{C7H16O4}$$.

So we start off with

$$\begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & 0.11011 & \ldots & 0 & 0.01224\\ \hline C & & \ldots & & \\ \hline A & & \ldots& & \\ \hline \end{array}$$

However the grams of $$\ce{CO2}$$ that 0.11011 grams of $$\ce{C7H16O4}$$ should produce is:

$$\dfrac{0.11011}{164.1995}\times 7 \times 44.009 = 0.20658$$

The grams of $$\ce{H2O}$$ that 0.11011 grams of $$\ce{C7H16O4}$$ should produce is:

$$\dfrac{0.11011}{164.1995}\times 8 \times 18.015 = 0.09664$$

So now we have:

$$\begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & 0.11011 & \ldots & 0 & 0.01224\\ \hline C & & \ldots & 0.20658 & 0.09664 \\ \hline A & & \ldots& & \\ \hline \end{array}$$

Finishing the table we have

$$\begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & 0.11011 & \ldots & 0 & 0.01224\\ \hline C & & \ldots & 0.20658 & 0.09664 \\ \hline A & & \ldots& 0.20658 & 0.10888 \\ \hline \end{array}$$

Thus Liam's solution is inconsistent with the given amount of carbon dioxide which was 0.20199 g and the given amount of water which was 0.10674 g.

Methodology

Both methods of computing the percent water are completely accurate and logically sound, and will give the same result when the problem quantities are in the correct ratios.

Mass of oxygen based on mass balance

The sample mass is $$\pu{0.12235 g sample}. \tag{1}$$ The sum of the product masses is $$\pu{0.10674 g \ce{H2O}} + \pu{0.20199 g \ce{CO2}} = \pu{0.30873 g products}. \tag{2}$$ The mass of the final reactant, $$\ce{O2}$$, must therefore be $$\pu{0.30873 g products} - \pu{0.12235 g sample} = \pu{0.18638 g \ce{O2}}. \tag{3}$$

Mass of oxygen based on based on the reaction ratios

There are 9 moles of $$\ce{O2}$$ per 7 moles of $$\ce{CO2}$$. We start with no $$\ce{CO2}$$ and end with no $$\ce{O2}$$.

We ended with $$\pu{4.5897 \times 10^{-3} mol \ce{CO2}}. \tag{4}$$

Therefore, we must have started with $$\pu{4.5897 \times 10^{-3} mol \ce{CO2}} \times \frac{\pu{9 mol \ce{O2}}}{\pu{7 mol \ce{CO2}}} = \pu{5.9010 \times 10^{-3} mol \ce{O2}}. \tag{5}$$

We can see this by filling in $$\ce{O2}$$ in the BCA table: $$\begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & \ldots & 5.9010 \times 10^{-3} & 0 & 6.7964 \times 10^{-4}\\ \hline C & \ldots & -5.9010 \times 10^{-3} & +4.5897 \times 10^{-3} & +5.2454 \times 10^{-3} \\ \hline A & 0 & 0 & 4.5897 \times 10^{-3} & 5.9251 \times 10^{-3}\\ \hline \end{array}$$

$$\pu{5.9010 \times 10^{-3} mol \ce{O2}}$$ is, in grams, $$\pu{5.9010 \times 10^{-3} mol \ce{O2}} \times \frac{\pu{31.998 g}}{\pu{1 mol \ce{O2}}} = \pu{0.18882 g \ce{O2}}. \tag{6}$$

Conclusion

$$(1)$$ is clearly not equal to $$(6)$$. Hence, this is a contradiction. $$\pu{0.18882 g \ce{O2}} \neq \pu{0.18638 g \ce{O2}}.$$ They are $$1.3\%$$ different.

Finding the missing mass

The difference in the oxygen mass is also exactly equal to the difference in the calculated water masses from the solutions presented here. $$\pu{0.18882 g \ce{O2} - 0.18638 g \ce{O2}} = \pu{0.00244 g \ce{O2}} \tag{1}$$

Let's expand out the BCA table a little further: $$\begin{array}{|c|c|c|c|c|} \hline & \ce{C7H16O4} & \ce{9O2} & \ce{7CO2} & \ce{8H2O} \\ \hline B & 6.5567 \times 10^{-4} & \ldots & 0 & 6.7964 \times 10^{-4}\\ \hline C & -6.5567 \times 10^{-4} & \ldots & +4.5897 \times 10^{-3} & +5.2454 \times 10^{-3} \\ \hline A & 0 & \ldots & 4.5897 \times 10^{-3} & 5.9251 \times 10^{-3}\\ \hline \end{array}$$

Now, using the values from the table: $$\pu{6.7964 \times 10^{-4} mol \ce{H2O}} \times \frac{\pu{18.015 g}}{\pu{1 mol \ce{H2O}}} = \pu{0.012244 g \ce{H2O}}\tag{2}$$ $$\pu{6.5567 \times 10^{-4} mol \ce{C7H16O4}} \times \frac{\pu{164.1995 g}}{\pu{1 mol \ce{C7H16O4}}} = \pu{0.10766 g \ce{C7H16O4}} \tag{3}$$ $$\pu{0.12235 g brake fluid} - \pu{0.10766 g \ce{C7H16O4}} = \pu{0.014689 g \ce{H2O}}\tag{4}$$

Finally, we can see $$(5)$$ is approximately equal to $$(1)$$: $$\pu{0.014689 g \ce{H2O}} - \pu{0.012244 g \ce{H2O}} = \pu{0.0024451 g \ce{H2O}} \tag{5}$$ $$\pu{0.0024451 g \ce{H2O}} \approx \pu{0.00244 g \ce{H2O}}.$$

To find the amount of water we should have had at the end, we will assume there were $$\pu{0.014689 g \ce{H2O}}$$ before the reaction, and that the amount of $$\ce{C7H16O4}$$ is still accurate. In the BCA table, we see the reaction created $$\pu{5.2454 \times 10^{-3} mol \ce{H2O}}$$, or $$\pu{5.2454 \times 10^{-3} mol \ce{H2O}} \times \frac{\pu{18.015 g}}{\pu{1 mol \ce{H2O}}} = \pu{0.094496 g \ce{H2O}}.$$

The amount of water that should have been recovered at the end is therefore $$\pu{0.014689 g} + \pu{0.094496 g} = \pu{0.10919 g \ce{H2O}}.$$

This gives a mass percentage of $$12\%$$, as was intended by the professor.

• You could go a bit further in the conclusion and point out that the difference in mass of the oxygen is exactly the difference between the different masses of water. – MaxW Oct 17 '18 at 17:41
• You should accept your own answer. You correctly have pointed out the contradiction with the problem. – MaxW Oct 17 '18 at 20:01
• @MaxW I have updated the post. – Liam White McShane Oct 17 '18 at 20:06