# Question #819a0

Aug 31, 2017

The empirical formula is ${\text{Fe"_2"O}}_{3}$.

#### Explanation:

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

We must convert the masses of $\text{Fe}$ and $\text{O}$ to moles and then find their ratio.

$\text{Mass of O" = "Mass of oxide - mass of Fe" = "1000 g - 700 g" = "300 g}$

$\text{Moles of Fe" = 700 color(red)(cancel(color(black)("g Fe"))) × "1 mol Fe"/(55.84 color(red)(cancel(color(black)("g Fe")))) = "12.5mol Fe}$

$\text{Moles of O" = 300 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "18.8mol O}$

From this point on, I like to summarize the calculations in a table.

$\boldsymbol{\underline{\text{Element"color(white)(m) "Mass/g"color(white)(mm) "Moles"color(white)(m) "Ratio"color(white)(m)×2color(white)(m)color(white)(m)"Integers}}}$
$\textcolor{w h i t e}{m m} \text{Fe} \textcolor{w h i t e}{X X X m} 700 \textcolor{w h i t e}{X m m m} 12.5 \textcolor{w h i t e}{X m l} 1 \textcolor{w h i t e}{m m m l l} 2 \textcolor{w h i t e}{m m m m m l} 2$
$\textcolor{w h i t e}{m m} \text{O} \textcolor{w h i t e}{X X X X l} 300 \textcolor{w h i t e}{m m m m} 18.8 \textcolor{w h i t e}{X m l} 1.50 \textcolor{w h i t e}{m m l} 3.00 \textcolor{w h i t e}{m m m l l} 3$

The molar ratio is $\text{Fe:O = 2:3}$.

The empirical formula is ${\text{Fe"_2"O}}_{3}$.

Here is a video that illustrates how to determine an empirical formula.