Total mass of organic compound = 0.468 g [Given]
Mass of barium sulphate formed = 0.668 g [Given]
1 mol of BaSO4 = 233 g of BaSO4 = 32 g of sulphur
Thus, 0.668 g of BaSO4 contains (32X0.668/233g) of sulphur = 0.0917 g of sulphur
Therefore, percentage of sulphur =(0.0197/0.468X100) = 19.59 %
Hence, the percentage of sulphur in the given compound is 19.59 %.