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

Sep 14, 2017

${C}_{2} {H}_{5} O$.......Are you sure you quoted the elemental compositon properly?

#### Explanation:

We assume a $100 \cdot g$ mass of compound....

And thus there are $\frac{53.31 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 4.44 \cdot m o l$ with respect to $C$....

And $\frac{11.18 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 11.09 \cdot m o l$ with respect to $H$....

And $\frac{35.51 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 2.22 \cdot m o l$ with respect to $O$....

We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of ${C}_{2} {H}_{5} O$.

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 ${C}_{4} {H}_{10} {O}_{2}$. The question has not been drawn from actual data, and the values have been interpolated.