## Chemistry: A Molecular Approach (3rd Edition)

$NiI_{2}$
Convert the mass of each constituent element into moles by dividing by its molar mass: $1.245g\ Ni\times\frac{1mol}{58.693g}=0.02121mol\ Ni$ $5.381g\ I\times\frac{1mol}{126.90g}=0.04240mol\ I$ Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close. Then multiply by whatever factor makes them all a whole number or very close to a whole number. $\frac{0.02121mol\ Ni}{0.02121}=1$ $\frac{0.04240mol\ I}{0.02121}=2$ So the empirical formula for this compound is: $NiI_{2}$