4. Warfarin (C19H16O4, 308.34 g/mol) converted to CHI3. One mole of CHI3 was pro

Question

4. Warfarin (C19H16O4, 308.34 g/mol) converted to CHI3. One mole of CHI3 was produced for each mole of warfarin that reacted. The collected CHI3 was reacted with silver to precipitate the iodine as AgI according to the following reaction.

CHI3 + 3Ag+ + H2O ? 3AgI(s) + 3H+ + CO(g)

The CHI3 produced from a 13.96 g sample was treated with 25.00 mL of 0.02979 M AgNO3. The excess Ag+ was then titrated with 2.38 mL of 0.05411 M KSCN. Calculate the concentration warfarin in the sample expressed as mg/kg.

Explanation / Answer

change values .....

moles Ag+ = 0.02500 L x 0.02964 M=0.0007410
moles KSCN = 3.82 x 10^-3 L x 0.05527 M = 0.000211
moles Ag+ that react with CH3I = 0.0007410 - 0.000211=0.0005299
moles CH3I = 0.0005299/3=0.0001766
mass CH3I = 0.0001766 x 393.7 g/mol= 0.06954 g
% = 0.06954 x 100/ 1.000 = 6.954

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