# What is the empirical formula for the compound containing 13.5 g Ca, 10.8 g O, and 0.675 g H?

Dec 27, 2016

$C a {\left(O H\right)}_{2}$

#### Explanation:

We divide each given mass by the atomic mass of each individual element:

$\text{Calcium:}$ $\frac{13.5 \cdot g}{40.1 \cdot g \cdot m o {l}^{-} 1} = 0.337 \cdot m o l .$

$\text{Oxygen:}$ $\frac{10.8 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 0.675 \cdot m o l .$

$\text{Hydrogen:}$ $\frac{0.675 \cdot g}{1.008 \cdot g \cdot m o {l}^{-} 1} = 0.675 \cdot m o l .$

If we divide thru by the smallest atomic quantity, we get an empirical formula of, $C a {H}_{2} {O}_{2}$, i.e. $C a {\left(O H\right)}_{2}$.