Answer to Question #110342 in Organic Chemistry for NBOR NATHANIEL

Question

Tablet powder containing ca. 0.25 g of furosemide (frusemide) is shaken with 300ml of 0.1 M NaOH to extract the acidic furosemide (frusemide). The extract is then made up to 500 ml with 0.1 M NaOH. A portion of the extract is filtered and 5 ml of the filtrate is made up to 250 ml with 0.1M NaOH. The absorbance of the diluted extract is measured at 271 nm. The A (1%, 1cm) value at 271 nm is 580 in basic solution. From the data below calculate the % of stated content in a sample of furosemide tablets: Stated content per tablet; 40 mg of furosemide (frusemide) Weight of 20 tablets = 1.656 g Weight of tablet powder taken for assay = 0.5195 g Absorbance reading = 0.596

Answer 1

Expected content in tablet powder taken: (0.5195 / 1.656) × 40 × 20 = 251.0 mg

Dilution factor: 5 - 250 mL = 50

Concentration in diluted tablet extract: 0.596 / 580 = 0.001028 g/100 mL = 1.028 mg / 100 ml

Concentration in original tablet extract: 1.028 × 50 = 51.40 mg / 100 mL

Volume of original extract: 500 mL

Therefore, amount of furosemide in original extract: 51.40 × 5 = 257.0

Percentage of stated content: (257.0 / 251.0) × 100.0 = 102.4 %