Question

Question #5ce05

Answer 1

`"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:

`{96" g C"}/{"mol X"}×{1" mol C"}/{12"g C"}={8" mol C"}/{"mol X"}`

So we have `"C"_8` in the formula.

Do the same with hydrogen:

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

`{16" g H"}/{"mol X"}×{1" mol H"}/{1"g H"}={16" mol H"}/{"mol X"}`

Thus `"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" g X"` contains `96" g C"` and `16" g H"`. Subtracting then gives:

`176" g X"- 96" g C" - 16" g H" =`64" g O"#

So then:

`{64" g O"}/{"mol X"}×{1" mol O"}/{16"g O"}={4" mol O"}/{"mol X"}`

So one mole of X contains four moles of oxygen to go with eight moles of carbon and 16 miles of hydrogen. Thus `C"_8"H"_{16}"O"_4`.