# What is the empirical formula for a compound that is 36.1% Ca and 63.9% CI?

Nov 23, 2016

${\text{CaCl}}_{2}$

#### Explanation:

As with all these problems, we assume a $100 \cdot \text{g}$ of compound, and divide through by the ATOMIC masses of each component element:

$\text{Moles of calcium}$ =(36.1*"g")/(40.08*"g"*"mol"^-1)=0.901 ·"mol".

$\text{Moles of chlorine}$ =(63.9*"g")/(35.45*"g"*"mol"^-1)=1.802·"mol".

We divide each molar quantity through by the smallest molar quantity (that of calcium) to give an empirical formula of ${\text{CaCl}}_{2}$.

For ionic materials, we do not speak of the molecular formula, and the empirical formula is the formula we quote for reference.