# Organic molecules with 21.62% oxygen

I have been told to answer the following question:

Seven different compounds, all either alcohols or ethers, contained 21.62% oxygen. Find the structural formula for all of these compounds and name the alcohols.

So far I've tried using empirical formula calculations to work out a simple formula but this has yielded no results because I do not know the exact ratio of hydrogen to carbon in the formula nor do I know any masses which I could use to scale up the empirical formula into a molecular formula. It seems to me that there just isn't enough information to answer the question - after hours of searching I can't even find one, let alone seven. I'm sure the answer is stupidly easy but it's evaded me. Does anyone have any suggestions of things I might be able to try? This question is UK A-Level standard.

• I am not sure if the precision that is implied in the question is correct. For example, butanol actually contains 21.58 % oxygen, which is suspiciously close to the given 21.62 %. – Faded Giant Aug 4 '18 at 11:22
• With this many significant digits, check the atomic weights used in this 21.62% figure. Do they agree with your sources also to four significant digits? If not you may be chasing your tail. – Oscar Lanzi Aug 4 '18 at 11:44
• I also get butanol using empirical formulas. Note that UK A-Level uses one decimal place atomic masses. – obackhouse Aug 4 '18 at 11:45
• "Alcohols or ethers," assuming no double bonds, suggests a structural formula $\ce{C_nH_{2n+2}O}$. Find the percentage of oxygen in such a compound as a function of $n$, set it equal to 21.62%, then solve for $n$. If a non-integral solution emerges, assume one double bond and repeat, etc. Then form seven isomers from the identified structural formula. – a-cyclohexane-molecule Aug 4 '18 at 12:58

Disclaimer:

Here I take the molar masses of carbon, hydrogen and oxygen to be $\pu{12g, 1g, 16g}$ respectively, which is what would be expeccted of you at school.

For the sake of simplicity, let's assume you have $\pu{100g}$ of the said compound. Which would mean it contains $\pu{21.62g}$ of oxygen.

Employing a bit of basic math in the analysis of that fact, leads us to similar (but more useful) re-statement:

$\pu{74g}$ of the said compound will contain $\pu{16g}$ of oxygen (the number also being the same as the molar mass of oxygen).

Now our substance possess a general formula of $\ce{C_{x}H_{y}O_{z}}$.

So in $\pu{74g}$ of it (of which $\pu{16g}$ is oxygen), the molar masses of $\pu{C}$ ( times the number of moles of it, which is $x$ ) and $\pu{H}$ ( times the number of moles of it, which is $y$ ) must add up to $\pu{74-16=58}$. Which gives us the equation:

$$\ce{x(12) + y(1) = 58}$$

Now neither $x$ nor $y$ can carry the value $\pu{0}$ (else we won't be dealing with a hydrocarbon anymore). This gives us three mathematically (not necessarily physically) plausible options:

1. $\pu{x= 2; y= 34}$
2. $\pu{x= 3; y= 22}$
3. $\pu{x= 4; y= 10}$

Even if you've only just started studying organic chem. you should be able to tell that the first two options are absolutely ridiculous (that's way too many hydrogen atoms that can be accommodated around 2 or 3 carbon atoms). So we go with the third option: $\pu{x= 4; y= 10}$, which sounds promising.

Which gives us the formula $\ce{C_{4}H_{10}O}$

You may now work out what alcohols or ethers that could give you ;-)