# Question 5ce05

May 29, 2017

${\text{C"_8"H"_{16}"O}}_{4}$.

#### Explanation:

First figure out how much carbon there is per mole of the substance X:

({54.5% "C"}/{100%})×{176" g"}/{"mol X"}={96" g C"}/{"mol X"}

Then one mole of carbon is $12$ grams (atomic mass), so:

$\left\{96 \text{ g C"}/{"mol X"}×{1" mol C"}/{12"g C"}={8" mol C"}/{"mol X}\right\}$

So we have ${\text{C}}_{8}$ in the formula.

Do the same with hydrogen:

{9.09% "H"}/{100%}×{176" g"}/{"mol X"}={16" g H"}/{"mol X"}

$\left\{16 \text{ g H"}/{"mol X"}×{1" mol H"}/{1"g H"}={16" mol H"}/{"mol X}\right\}$

Thus ${\text{C"_8 "H}}_{16}$.

Now for the tricky part. When chemists report an elemental analysis and it adds up to less than 100%, the difference is generally attributed to oxygen. Our analytical methods, constrained by working in an oxygenated and water-laden world, do not in general detect or measure oxygen directly. Instead we infer it by difference after accounting for the other elements.

Here we found that $176 \text{ g X}$ contains $96 \text{ g C}$ and $16 \text{ g H}$. Subtracting then gives:

$176 \text{ g X"- 96" g C" - 16" g H} =$64" g O"

So then:

$\left\{64 \text{ g O"}/{"mol X"}×{1" mol O"}/{16"g O"}={4" mol O"}/{"mol X}\right\}$

So one mole of X contains four moles of oxygen to go with eight moles of carbon and 16 miles of hydrogen. Thus $C {\text{_8"H"_{16}"O}}_{4}$.