The empirical formula is the simplest whole number molar ratio of the elements in the compound.
We must convert the masses of #"Fe"# and #"O"# to moles and then find their ratio.
#"Mass of O" = "Mass of oxide - mass of Fe" = "1000 g - 700 g" = "300 g"#
#"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"#
#"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.
#bbul("Element"color(white)(m) "Mass/g"color(white)(mm) "Moles"color(white)(m) "Ratio"color(white)(m)×2color(white)(m)color(white)(m)"Integers")#
#color(white)(mm)"Fe" color(white)(XXXm)700 color(white)(Xmmm)12.5 color(white)(Xml)1color(white)(mmmll)2color(white)(mmmmml)2#
#color(white)(mm)"O" color(white)(XXXXl)300 color(white)(mmmm)18.8 color(white)(Xml)1.50 color(white)(mml)3.00color(white)(mmmll)3#
The molar ratio is #"Fe:O = 2:3"#.
The empirical formula is #"Fe"_2"O"_3#.
Here is a video that illustrates how to determine an empirical formula.
