Question

An organic sample has composition by mass, `C:53.31%;H:11.18%;`, and the balance was oxygen. What is its empirical formula?

Answer 1

`C_2H_5O`.......Are you sure you quoted the elemental compositon properly?

Explanation:

We assume a `100*g` mass of compound....

And thus there are `(53.31*g)/(12.011*g*mol^-1)=4.44*mol` with respect to `C`....

And `(11.18*g)/(1.00794*g*mol^-1)=11.09*mol` with respect to `H`....

And `(35.51*g)/(16.00*g*mol^-1)=2.22*mol` with respect to `O`....

We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of `C_2H_5O`.

This organic formula is a bit whack, in that I think it contains a peroxo linkage (or it's a diether or diol); it must do if its MOLECULAR formula is `C_4H_10O_2`. The question has not been drawn from actual data, and the values have been interpolated.