Answer
Empirical formula: $ICl_3$
Molecular formula: $I_2Cl_6$
Work Step by Step
Atomic weights (g/mol):
$Cl: 35.45, \ I: 126.904$
Number of moles consumed:
I: $0.678\ g\div126.904=5.34\times10^{-3}\ mol\ of\ I$
Cl: $(1.246-0.678)\ g\div35.45=1.602\times10^{-2}\ mol\ of\ Cl$
Dividing by the smallest number of moles:
Cl: $1.602\times10^{-2}\div5.34\times10^{-3}=3.00$
Empirical formula: $ICl_3$
Molar mass of the empirical formula:
$228.11\ g/mol$
Molar mass of the compound: $467\ g/mol$
Ratio of the two: $2.0$
Molecular formula: $I_2Cl_6$
