Find the empirical formula of chromium oxide containing 68.4% of chromium. (Cr=52, O=16)

Explanation

Let the weight of chromium oxide be 100g.
`therefore` Weight of chromium= 68.4g and weight of oxygen =31.6g
`{:(therefore "moles of chromium"=(68.4)/(52)=1.32),("and moles of oxygen"=(31.6)/(16)=1.98):}}`
Relative number of moles of Cr and O atoms
`= ("moles of Cr")/("moles of O")= (1.32)/(1.98)= (132)/(198) = (2 xx 66)/(3 xx 66)= (2)/(3)` (by inspection only).
The empirical formula is `Cr_(2)O_(3)`.