Answer is: empirical formula is CrCl₄.
If we use 100 grams of compound:
ω(Cr) = 26.8% ÷ 100% = 0.268.
ω(Cl) = 73.2% ÷ 100% = 0.732.
m(Cr) = m(compound) · ω(Cr).
m(Cr) = 100 g · 0.268 = 26.8 g.
n(Cr) = m(Cr) ÷ M(Cr).
n(Cr) = 26.8 g ÷ 52 g/mol.
n(Cr) = 0.515 mol.
n(Cl) = 73.2 g ÷ 35.45 g/mol.
n(Cl) = 2.065 mol.
n(Cr) : n(Cl) = 0.515 mol : 2.065 mol /0.515.
n(Cr) : n(Cl) = 1 : 4.