Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Explanation

`|{:("Element","Symbol",%"by mass","Atomic mass","Moles of the element","Simplest molar ratio","Simplest whole number molar rate"),("Iron",Fe,69.9,55.85,69.9/55.85=1.25,1.25/1.25=1,2),("Oxygen",O,30.1,16.00,30.1/16.00=1.88,1.88/1.25=1.5,3):}|`
`:.` Empirical formula `=Fe_(2)O_(3)`