# A certain manganese oxide contains 72% manganese. What is the empirical formula?

May 1, 2017

Since % litterally means "per hundred" let's take a 100g sample.

#### Explanation:

We convert the mass ratio into mol ratio:

Then 72g of this is Manganese.
The molar mass of $M n = 55$, so $72 g \leftrightarrow \frac{72}{55} = 1.31 m o l$ of $M n$

The 28g of Oxygen (molar mass $O = 16$) breaks down to:
$28 g \leftrightarrow \frac{28}{16} = 1.75 m o l$ of $O$

The mol ratio $M n \div O = 1.31 \div 1.75 = 0.75 \div 1 = 3 \div 4$

Empirical formula: $M {n}_{3} {O}_{4}$

Note:
This works if you take any other sample size, but then there are more calculations to do.