# Another compound found on Mars contains iron and oxygen. The compound contains 70% by mass of iron and 30% by mass of oxygen. How do you calculate the empirical formula of this compound?

Jun 19, 2017

${\text{Fe"_2"O}}_{3}$

#### Explanation:

Your ultimate goal here is to figure out the smallest whole number ratio that exists between iron and oxygen in this unknown compound, i.e. the empirical formula of the compound.

Start by converting the percent composition to grams by using a $\text{100-g}$ sample of compound.

You will have

• $\text{70% Fe " stackrel(color(white)(acolor(blue)("100 g sample")aaa))(->) " 70 g Fe}$
• $\text{30% O " stackrel(color(white)(acolor(blue)("100 g sample")aaa))(->) " 30 g O}$

Next, use the molar masses of the two elements to convert the masses to moles

$\text{For Fe: " 70 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.2535 moles Fe}$

$\text{For O: " 30 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.8750 moles O}$

To find the mole ratio that exists between iron and oxygen, divide both values by the smallest one.

You will have

"For Fe: " (1.2535 color(red)(cancel(color(black)("moles"))))/(1.2535color(red)(cancel(color(black)("moles")))) = 1

"For O: " (1.8750color(red)(cancel(color(black)("moles"))))/(1.2535color(red)(cancel(color(black)("moles")))) = 1.496 ~~ 1.5

Now, you are looking for the smallest whole number ratio that exists between the two elements, so multiply both values by $2$ to get

${\text{Fe"_((2 * 1))"O}}_{\left(2 \cdot 1.5\right)}$

You can thus say that the empirical formula of the compound is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{Fe"_2"O}}_{3}}}} \to$ empirical formula