The active ingredient in disulfiriam, a drug for the treatment of chronic alcoho

Question

The active ingredient in disulfiriam, a drug for the treatment of chronic alcoholism, is tetraethylthiurram disulfide. This drug has a molecule weight of 296.55, and contains four (4) sulfur atoms per molecule. The sulfur in a 0.4329 g sample of disulfiriam preparation was oxidized to SO_2, which was absorbed in H_2 O_2 to give H_2 SO_4 .The acid was treated with 22.13 mL of 0.03736 M NaOH. Calculate the percentage tetraethylthiurram disulfide in the preparation. 91 sample containing (NH_4)_2 SO_4, NH_4 NO_3, and inert material was in a volumetric flask. A 50.00 mL aliquot was men made

Explanation / Answer

Moles of NaOH used = 0.02213*0.03736 = 0.0008267768 moles

Since 2 moles NaOH react with 1 mole H2SO4, so moles of acid present = 0.0008267768*0.5 = 0.0004133884 moles

So, moles of SO2 which were formed = 0.0004133884

Since 1 mole SO2 has 1 mole S, so

Moles of S present = 0.0004133884

Since 4 moles S are present in 1 mole tetraethylthiurram disulfide, so

Moles of tetraethylthiurram disulfide present = 0.0004133884*0.25 = 0.0001033471

Mass of tetraethylthiurram disulfide present = mass*MW = 0.0001033471*296.55 = 0.030647582505

Mass of sample = 0.4329

So % tetraethylthiurram disulfide in sample = (0.030647582505/0.4329) * 100 = 7.079%

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