Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are `69.9` and `30.1`, respectively.

Explanation

Calculation of empirical formula `=Fe_(2)O_(3)`
Empirical formula mass of `Fe_(2)O_(3)`
`=2 xx 55.85 + 3 xx 16.00`
`=111.7 + 48.00 = 159.7 g "mol"^(-1)`
Molecular formula mass = 159.8 g `"mol.^(-1)`
`n=("Molecular formula mass")/("Empirical formula mass")`
`=159.8/159.7=1`
`therefore` Molecular formula `=(Fe_(2)O_(3)) = Fe_(2)O_(3)`