# What is the empirical formula for fructose given its percent composition: 40.00% C, 6.72% H, and 53.29% O?

May 26, 2016

$C {H}_{2} O$

#### Explanation:

The empirical formula is the simplest whole number ratio defining constituent atoms in a species.

In all of these problems it is useful to assume $100$ $g$ of compund, and calculate the ATOMIC composition.

So, in $100 \cdot g$ fructose, there are:

$\frac{40.00 \cdot g}{12.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.33 \cdot m o l \cdot C$,

$\frac{6.72 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $=$ $6.66 \cdot m o l \cdot H$,

$\frac{53.29 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.33 \cdot m o l \cdot O$,

If we divide thru by the smallest molar quantity, we get a formula of $C {H}_{2} O$.

The molecuar mass of fructose is $180.16 \cdot g \cdot m o {l}^{-} 1$. How do we use these data to provide a molecular formula?