# Assume that 7.35 g of Cr reacts with oxygen to form 10.74 g of metal oxide. What's the empirical formula for the compound?

Apr 10, 2016

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

#### Explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of $\text{Cr}$ to $\text{O}$.

$\text{Mass of Cr = 7.35 g}$

$\text{Mass of chromium oxide = mass of Cr + mass of O}$

$\text{10.74 g = 7.35 g + mass of O}$

$\text{Mass of O = (10.74 – 7.35) g = 3.39 g}$

$\text{Moles of Cr" = 7.35 color(red)(cancel(color(black)("g Cr"))) × "1 mol Cr"/(52.00color(red)(cancel(color(black)( "g Mg")))) = "0.1413 mol Mg}$

$\text{Moles of O "= 3.39 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.2119 mol O}$

To get this into an integer ratio, we divide both numerator and denominator by the smaller value.

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

$\text{Element"color(white)(Mg) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mm)"×2"color(white)(mll)"Integers}$
stackrel(—————————————————-——)(color(white)(m)"Cr" color(white)(XXXm)7.35 color(white)(Xm)0.1413 color(white)(Xll)1color(white)(Xmmm)2color(white)(mmmml)2)
$\textcolor{w h i t e}{m} \text{O" color(white)(XXXXl)3.39 color(white)(mm)} 0.2119 \textcolor{w h i t e}{X l l} 1.499 \textcolor{w h i t e}{X X} 2.998 \textcolor{w h i t e}{m m l} 3$

There are 2 mol of $\text{Cr}$ for 3 mol of $\text{O}$.

The empirical formula of chromium oxide is ${\text{Cr"_2"O}}_{3}$.

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