Question

An organic compound is `40.0%, C`, `6.70%, H`, `53.3%, O`; what is its empirical formula?

Answer 1

`CH_2O` is the empirical formula........

Explanation:

By definition, the empirical formula is the simplest whole number ratio defining constituent atoms in a species. And for simplicity we ASSUME that there are `100*g` of compound, and then we interrogate the molar quantities. And if there is `40.0%` carbon by mass, do you agree that there are `40*g` carbon in the assumed mass?

So.....

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

`"Moles of hydrogen"=(6.7*g)/(1.00794*g*mol^-1)=6.70*mol`

`"Moles of oxygen"=(53.3*g)/(16.00*g*mol^-1)=3.33*mol`

And now we divide thru by the SMALLEST molar quantity, that of carbon and oxygen to give an empirical formula of `CH_2O`. Do you agree?

Now we know that the `"molecular formula"` is ALWAYS a whole number multiple of the `"empirical formula"`. Do we know this?

`"molecular formula"=nxx"empirical formula"`

And so we can solve for `n`.

`nxx{12.011+2xx1.00794+16}*g*mol^-1=180.1*g*mol^-1`

And thus `n=6`, and the molecular formula is `C_6H_12O_6`.

You will be doing a lot of these problems over the years, so if there is something you cannot grasp, now is the time to address it. Good luck. Sometimes these problems DO NOT QUOTE the percentage of oxygen in that this element is difficult to measure by microanalysis, and its percentage is assessed by difference.