Question

Government and industry analysts routinely measure the quantity of ascorbic acid (vitamin C, C6H8O6) in commercial products such as fruit juices and vitamin tablets. The ascorbic acid reacts with excess iodine, and the amount of I2 remaining is determined with sodium thiosulfate (Na2S2O3). It is suspected by the Montreal Fraud Division that a prominent natural foods importer is defrauding his customers by claiming that his vitamin C tablets contain 45% of the active ingredient ascorbic acid. In one analysis, a 2.55- gram vitamin tablet is mixed with water and treated with 53.20 mL of 0.1030 M I2 according to the equation:

C6H8O6(s) + I2(aq)  C6H6O6(aq) + 2 HI(aq)

After all the ascorbic acid reacts, the excess I2 reacts with 27.54 mL of 0.1153 M Na2S2O3 according to the equation:

I2(aq) + 2 Na2S2O3(aq)  2 NaI(aq) + Na2S4O6(aq)

a) How many grams of ascorbic acid are in the vitamin tablet?

b) Should the Montreal police arrest the importer? Why?

Answer is 26.7%.

Answer 1

Answer-

Moles of Thosulphate used = vol of thiosulphate in L * molarity

                                        = 0.02754 L * 0.115 M =3.17*10^-3 moles

2 moles Na2s2O3 reacts with 1 mol I2.

Moles of I2 = 1.584*10^-3 moles

Moles of ascorbic acid present = 1.584*10^-3 moles

Mass of ascorbic acid = 1.584*10^-3 moles * molar mass of ascorbic acid

                                 = 1.584*10^-3 moles *176.12g/mol =0.28 g

% of ascorbic acid = 0.28/2.55gm *100% =10.9 %

% od ascorbic acid is only 10% when it claims to be 45%. Police should arrest the imposter.