Answer
Empirical formula: $V_2S_3$
Work Step by Step
Atomic weights (g/mol):
$V: 50.9415, \ S: 32.06$
Number of moles consumed:
V: $2.04\ g\div50.9415=4.00\times10^{-2}\ mol\ of\ V$
S: $1.93\ g\div32.06=6.02\times10^{-2}\ mol\ of\ S$
Dividing by the smallest number of moles:
S: $6.02\times10^{-2}\div4.00\times10^{-2}=1.50=3/2$
Empirical formula: $V_2S_3$
