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

Aug 20, 2016

$C e {I}_{3}$
$\text{Moles of cerium}$ $=$ $\frac{1.67 \cdot g}{140.12 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0119 \cdot m o l$
$\text{Moles of iodine}$ $=$ $\frac{4.54 \cdot g}{126.90 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0358 \cdot m o l$
We divide thru by the lowest molar quantity, $0.0119 \cdot m o l$, that of the metal, to get an empirical formula of $C e {I}_{3}$