# A 22.35*g mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained 7.13*g of potassium, 6.47*g of chlorine. What is its empirical formula?

Sep 19, 2016

$K C l {O}_{3}$

#### Explanation:

As always, we divide each constitiuent mass thru by the ATOMIC mass of each constituent atom.

$\text{Moles of potassium}$ $=$ $\frac{7.13 \cdot g}{39.10 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.182 \cdot m o l$.

$\text{Moles of chlorine}$ $=$ $\frac{6.47 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.182 \cdot m o l$.

$\text{Moles of oxygen}$ $=$ $\frac{\left(22.35 - 7.13 - 6.47\right) \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.546 \cdot m o l$.

And now we divide thru by the LOWEST molar quantity, that of potassium/chlorine, to give an empirical formula of $K C l {O}_{3}$.

How did I know that that there were $\left(22.35 - 7.13 - 6.47\right) \cdot g$ of $O$? The mass of oxygen was not given in the problem.