Find the weight percent of CHI_3 in an antiseptic.

Question

Iodoform (CHI_3) is used in some antiseptic preparations. Iodoform reacts with Ag ions to for AgI and CO gas:

CHI_3 3Ag H_2O ---> 3AgI (s) CO (g) 3H

Suppose that 1.000 g sample of an antiseptic preparation is treated with 25.00 mL of 0.02964 M AgNO_3 and then the excess Ag is titrated with 3.82 mL of 0.05527 M KSCN to a red FeSCN end point (Volhard method). What is the weight percent of CHI_3 (393.7 g/mol) in the antiseptic?

Answer 1

let's find total moles of AgNO3:

0.02500 litres @ 0.02964 mol/litre = 7.41e-4 moles of AgNO3

but then they back titrated to find the excess AgNO3:

0.00382 litres @ 0.05527mol/litre = 2.111e-4 moles of excess AgNO3

( 1 AgNO3 & 1 KSCN --> 1 AgSCN (s) )

so the moles of AgNO3 that were used to react with the CHI3 were:

0.0007410 total moles - 0.0002111 excess moles = 0.0005299 moles AgNO3

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so how many moles of CHI3 were there @ 1CHI3 & 3 AgNO3 --> :

0.0005299 mol AgNO3 @ 1 mol CHI3 / 3 mol AgNO3 = 1.766e-4 moles CHI3

so how many grams:

1.766e-4 moles CHI3 @ 393.7 g/mol = 0.06954 grams of CHI3

so what's the %:

0.06954 g CHI3 / 1.000g sample (100) = 6.954%

your answer is: 6.954% CHI3

( it actually became a three sig fig problem when they introduced the "3.82 ml")