# A compound contains 63g Mn and 37g O, what is the empirical formula?

Sep 19, 2016

$M n {O}_{2}$.

#### Explanation:

We divide each constituent mass thru by the ATOMIC mass of each element:

$\text{Moles of manganese}$ $=$ $\frac{63 \cdot g}{54.94 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.15 \cdot m o l$.

$\text{Moles of oxygen}$ $=$ $\frac{37 \cdot g}{15.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.31 \cdot m o l$.

We divide thru by the smallest molar quantity, that of the metal, to give an empirical formula of $M n {O}_{2}$.