posted by .

Procedure: 2.500g of sample was dissolved in 75ml dimethylformamide and diluted to 100.0ml with the same solvent. (Solution T). To 5.0ml of Solution T, 20ml of acetone and 3ml of a 500g/l solution of potassium iodide were added and mixed .This was allowed to stand for 1min and then titrated with 0.1M sodium thiosulphate using 1ml of starch solution added towards the end of the titration as indicator. A blank titration was carried out.

1ml of 0.1M sodium thiosulphate is equivalent to 12.11mg of C14H20O4.

The following data were obtained based on 2 titre values: -
Sample Weight 1 = 2.5011g, titrant volume = 8.58ml
Sample Weight 2 = 2.5030g, titrant volume = 8.58ml
Blank titrant volume= 0.80ml

How do I calculate (%w/w) benzoyl peroxide in the sample?
Thanks.

8.58 mL - 0.80 blank = 8.50 mL used in the titration.
8.50 mL x 12.11 mg/mL = ??mg of the stuff in the 5 mL. That x 20 (100 mL/5 mL = 20x) = ?? mg stuff in the original 2.5011 g sample.
Change that (the mg peroxide) to grams, and convert to percent.
Percent = grams peroxide/2.5011)*100 = ??