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The active ingredient in disulfiriam, a drug used for the treatment of chronic alcoholism,
is tetraethylthiurram disulfide. This drug has a molecular 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 SO2, which was absorbed in H2O2 to give H2SO4.The acid was treated with 22.13 mL of 0.03736 M NaOH. Calculate the
percentage tetraethylthiurram disulfide in the preparation.
Starting with what we know...
22.13 mL * 0.03736 mmol/mL = 0.8268 mmol NaOH
The reaction with sulfuric acid : H2SO4 + 2NaOH -> 2H2O + Na2SO4
0.8268 mmol NaOH / 2 = 0.4134 mmol H2SO4 = 0.4134 mmol S
There are 4 atoms of sulfur per molecule tetraethylthiurram disulfide (TET)
0.4134 mmol S / 4 = 0.1033 mmol TET
0.1033 mmol TET * 296.55 mg/mmol = 30.65 mg TET
(30.65 mg TET / 432.9 mg sample) x 100% = 7.080% TET
Answer: 7.080% TET