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

1 Answer
May 1, 2017

Answer:

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.