Question

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

Answer 1

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 `Mn=55`, so `72gharr72/55=1.31mol` of `Mn`

The 28g of Oxygen (molar mass `O= 16`) breaks down to:
`28gharr28/16=1.75mol` of `O`

The mol ratio `MndivO=1.31div1.75=0.75div1=3div4`

Empirical formula: `Mn_3O_4`

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