# What is the empirical compound having the following composition: A 36 g sample of iron and oxygen that is 78% iron?

Nov 22, 2017

$\text{Fe"_2"O}$

#### Explanation:

• Mass of iron in given sample= 78/100 × "36 g" = "28.08 g"

Moles of iron$= \text{28.08 g"/"55.8 g/mol" ≈ "0.50 mol}$

• Mass of oxygen in sample$= \text{32 g" - "28.08 g" = "3.92 g}$

Moles of oxygen$= \text{3.92 g"/"16 g/mol" ≈ "0.25 mol}$

Now, ratio of their moles is

${n}_{\text{Iron"/n_"Oxygen" = "0.50 mol"/"0.25 mol}} = 2.0$

Empirical formula$= \text{Fe"_2"O}$