# What is the molecular formula of a compound that is 62.58% carbon, 9.63% hydrogen, 27.79% oxygen?

Mar 27, 2016

${C}_{6} {H}_{11} {O}_{2}$ is the simplest whole number ration that defines constituent atoms in a species, and is, therefore, the empirical formula.

#### Explanation:

As is the standard with these problems, we assume $100$ $g$ of compound, and break the percentages up into atoms.

In this mass, there are $\frac{62.58 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1}$ $C$ $=$ $5.21$ $m o l$ $C$.

And, $\frac{9.63 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $H$ $=$ $9.55$ $m o l$ $H$.

And, $\frac{27.79 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $O$ $=$ $1.740$ $m o l$ $O$.

We divide thru, by the smallest molar quantity ($O$), and get (almost) the empirical formula: ${C}_{3} {H}_{5.5} O$.

Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.

${C}_{6} {H}_{11} {O}_{2}$