Question

What is the molecular formula of a material with percentage composition `C:40.5%`, `H:6.67%`, `O:53.33%`, and molecular mass of `180.0*g*mol^-1`?

Answer 1

This is a standard question, and as a routine we ASSUME a mass of `100*g` of unknown compound.......and get a molecular formula of `C_6H_12O_6`.

Explanation:

So if there are `100*g` of unknown compound, we can access the empirical formula given the percentage composition:

`"Moles of carbon"=(40.0*g)/(12.011*g*mol^-1)=3.33*mol`

`"Moles of hydrogen"=(6.67*g)/(1.00794*g*mol^-1)=6.62*mol`

`"Moles of oxygen"=(53.33*g)/(16.00*g*mol^-1)=3.33*mol`

We divide thru by the LOWEST molar quantity, that of oxygen/carbon to give an empirical formula of:

`CH_2O`, which represents the simplest whole number ratio defining constituent atoms in a species, i.e. the `"empirical formula"`.

But we also have the molecular mass, `180*g*mol^-1`.

Now it is a fact that the molecular formula is always a whole number multiple of the empirical formula, and thus..............

`180*g*mol^-1=nxx(12.011+2xx1.00794+15.999)*g*mol^-1`,....................... and so we solve for `n`. Clearly, `n=6`, and the molecular formula is `C_6H_12O_6`.

Just on a practical issue, an analyst would normally NEVER give you the percentage oxygen by mass. Most of the time you get `%C,H,N`, and if their sum is not `100%`, then the BALANCE is assumed to be due to `"oxygen"`.