The empirical formula is the simplest whole number molar ratio of the elements in the compound.
We must convert the masses of "Fe"Fe and "O"O to moles and then find their ratio.
"Mass of O" = "Mass of oxide - mass of Fe" = "1000 g - 700 g" = "300 g"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 Fe=700g Fe×1 mol Fe55.84g 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"Moles of O=300g O×1 mol O16.00g 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.
