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 #"Cr"# to #"O"#.

#"Mass of Cr = 7.35 g"#

#"Mass of chromium oxide = mass of Cr + mass of O"#

#"10.74 g = 7.35 g + mass of O"#

#"Mass of O = (10.74 – 7.35) g = 3.39 g"#

#"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"#

#"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.

#"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)#

#color(white)(m)"O" color(white)(XXXXl)3.39 color(white)(mm)"0.2119 color(white)(Xll)1.499 color(white)(XX)2.998color(white)(mml)3#

There are 2 mol of #"Cr"# for 3 mol of #"O"#.

The empirical formula of chromium oxide is #"Cr"_2"O"_3#.

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