Organic molecules with 21.62% oxygen
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:


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


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 ;-)