# A chromium chloride salt contains 32.8% metal content by mass. What is its empirical formula?

Mar 30, 2017

We assume a $100 \cdot g$ mass of compound, and get an empirical formula of $C r C {l}_{3}$.

#### Explanation:

In all problems of this sort, it is convenient to assume a $100 \cdot g$ mass of compound, and then work out the individual molar quantities of each element:

$\text{Moles of chromium} = \frac{32.8 \cdot g}{52.0 \cdot g \cdot m o {l}^{-} 1} = 0.631 \cdot m o l$

$\text{Moles of chlorine} = \frac{67.2 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1} = 1.90 \cdot m o l$

We divide thru by the SMALLEST molar quantity, that of the metal, to get an empirical formula of $C r C {l}_{3}$.