Question

What is the empirical formula of a compound that is composed of 1.67 g of cerium and 4.54 g of iodine?

Answer 1

`CeI_3`

Explanation:

`"Moles of cerium"` `=` `(1.67*g)/(140.12*g*mol^-1)` `=` `0.0119*mol`

`"Moles of iodine"` `=` `(4.54*g)/(126.90*g*mol^-1)` `=` `0.0358*mol`

We divide thru by the lowest molar quantity, `0.0119*mol`, that of the metal, to get an empirical formula of `CeI_3`