The sample contains
• (0.321 g C) (1/12.011 mol/g) ≈ 0.0267 mol C
• (0.044 g H) (1/1.008 mol/g) ≈ 0.044 mol H
• (0.285 g S) (1/32.06 mol/g) ≈ 0.00889 mol S
Given that the sample has mass 0.650 g and molecular weight 146 g/mol, this means we have
(0.650 g sample) (1/146 mol/g) ≈ 0.00445 mol sample
So, one mole of the sample would contain
• (1 mol sample) (0.0267 mol C) / (0.00445 mol sample) ≈ 6 mol C
• (1 mol sample) (0.044 mol H) / (0.00445 mol sample) ≈ 10 mol H
• (1 mol sample) (0.00889 mol S) / (0.00445 mol sample) ≈ 2 mol S
which means the compound's formula is C₆H₁₀S₂.
