# What is the empirical formula of a compound if it has 80g Ca and 70g Cl?

Aug 12, 2017

$C a C l$....

#### Explanation:

We determine the number of moles of calcium and chlorine respectively....

$\text{Moles of calcium} = \frac{80 \cdot g}{40.1 \cdot g \cdot m o {l}^{-} 1} = 2.0 \cdot m o l$

$\text{Moles of chlorine} = \frac{70 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1} = 2.0 \cdot m o l$

We divide thru by the smallest molar quantity to get $C a C l$; a $C a \left(+ I\right)$ salt that I don't think is known.