# What is the empirical formula for a compound that is 31.9% potassium, 28.9% chlorine, and 39.2% oxygen?

Dec 28, 2015

The empirical formula is $K C l {O}_{3}$, potassium chlorate.

#### Explanation:

In 100 g of stuff, there are 31.9 g potassium, 28.9 g of chlorine, and 39.2 g oxygen.

We divide these gram quantities by the respective atomic masses of each of the atoms to get the atomic ratio of constituents.

$K$: $\frac{31.9 \cdot g}{39.1 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.816 \cdot m o l$.

$C l$: $\frac{28.9 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.816 \cdot m o l$.

$O$: $\frac{39.2 \cdot g}{15.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.45 \cdot m o l$.

If we divide thru by the least molar quantity (that of potassium or oxygen), we get an empirical formula of $K C l {O}_{3}$, potassium chlorate.

Note that because this is an ionic, and not a molecular species, we can quote the empirical as the actual formula.