# What is the empirical formula for the compound consisting of 63% manganese and 37% oxygen by mass?

Aug 21, 2016

$M n {O}_{2}$

#### Explanation:

As with all empirical formula calculations, a starting mass of $100$ $g$ is assumed.

There are thus $\frac{63 \cdot g}{54.94 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.15 \cdot m o l$ $M n$, and $\frac{37 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.132 \cdot m o l$ with respect to $O$.

Now, we simply divide thru by the LOWEST molar quantity, which here is the metal to give an empirical formula of $M n {O}_{2} \text{, manganese(IV) oxide}$.

AS always, the empirical formula is the simplest, whole number ratio defining constituent atoms in a species.