smurfhaven Posted August 4, 2010 Posted August 4, 2010 Please help to determine the (%w/w) benzoyl peroxide (C14H20O4) in a sample. 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.
smurfhaven Posted August 10, 2010 Author Posted August 10, 2010 Are the following steps correct? For sample weight 1 of 2.5011g, Benzoyl peroxide in 5mL of solution T = (8.58 mL - 0.80mL) x 12.11 mg/mL = 102.935mg Benzoyl peroxide in 100mL of solution T = 102.935mg x 20 = 2058.7mg %w/w benzoyl peroxide = (2058.7/1000)g /2.5011g x100% = 82.3 wt% Same steps are followed for sample weight 2, then the results are averaged out.
cypress Posted August 10, 2010 Posted August 10, 2010 Yes that appears to be correct. I don't see any errors at all.
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