Question

Microanalysis of an organic compound containing carbon, hydrogen, and oxygen, gives `62.6%*C`, and `4.21%*H`. What is the empirical formula of the compound?

Answer 1

`C_5H_4O_2` is the `"empirical formula....."`

Explanation:

We assume a `100*g` mass of compound, and thus....

`"Moles of carbon"=(62.6*g)/(12.011*g*mol^-1)=5.21*mol*C`

`"Moles of hydrogen"=(4.21*g)/(1.00794*g*mol^-1)=4.16*mol*H`

`"Moles of oxygen"=(33.3*g)/(15.999*g*mol^-1)=2.08*mol*H`

In each case, we have divided the mass of element by the ATOMIC mass of each element to give an elemental ratio.... How did I know that the compound was `33.3%` by mass with respect to oxygen?

And now we divide thru by the LOWEST molar quantity, that of oxygen to get a trial empirical formula of `C_(2.50)H_2O`. But by definition, the empirical formula is the lowest WHOLE number ratio defining constituent atoms in a species...so we gots `C_5H_4O_2`.

We cannot access the molecular formula inasmuch as we need a measurement of molecular mass. See here.

`"Molecular formula"="empirical formula"xxn`....