The sulfur in a 0.575 g sample of an organic thiamine was converted to H2S, whic

Question

The sulfur in a 0.575 g sample of an organic thiamine was converted to H2S, which was then absorbed in 15.00 mL of 0.0320 M I2. This amount of I2 was in excess of that required to oxidize the H2S to elemental sulfur; I- was the reduction product.

H2S + I2 --> S + 2I- + 2H+

The excess I2 was determined by titration with thiosulfate S2O32-, the reaction products being tetrathionate S4O62- and I- (see below). 26.29 mL of 0.0175 M Na2S2O3 was required to reduce the excess I2. How much sulfur (mass) was in the original sample?

I2 + 2S2O32- --> 2I- + S4O62-

This is the way my teacher solved it:

Mass of S = moles of Sulfur = moles of H2S = moles of I2

Mole of I2 in 1st rxn = mol of S in 1st rxn + mol of I2 excess
                                = mol of S in 1st rxn + mol of (S2O3)^-2 in 2nd rxn

Mol of I2 = 0.0320M(15mL) = 0.48 mmol of I2

Mol of (S2O3)^-2 in 2nd rxn = 1/2(0.0175M)(26.29mL) = 0.23003 mmol

0.48mmol of I2 in 1st rxn - 0.23003 mmol in 2nd rxn =0.24996 mmol of S

0.24996 mmol of S * (32.066 mg/0.1 mmol) = 8.02 mg of S

I don't understand this procedure. Any help or suggestions will be greatly appreciated.

Explanation / Answer

I2 will be consumed in rexn with H2S forming S.

the no. of moles of S formed will be equal to the no. of moles of I2 that will react with H2S.

but here H2S is limiting, so there will be some I2 which will remain unreacted.

the unreacted I2 will then react with Na2S2O3 as per the rexn. one mole of I2 will be consumed per 2 moles of Na2S2O3.

therefore, moles of I2= moles of S formed in 1st rexn+moles of I2 remaining

=moles of S formed in 1st rexn+1/2 moles of Na2S2O3 ( as 1 mole of I2 reacts with 2 moles of

Na2S2O3)

moles of Na2S2O3 = 0.0175*26.96= 0.46 mmol

therefore moes of I2 used in second rexn= 0.5*0.46=0.23

and total moles of I2= 0.032*15=0.48mmol

therefore moles of I2 used in 1st rexn = 0.48-0.23=0.25mmol

hence moles of S = moles of I2 used in 1st rexn=0.25 mmol

hence weight of S= 0.25*32= 8 mg

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