Question

Determine the mg of aspirin in a commercial aspirin tablet labeled to contain 5 grains using the following data: 800.0 mg of pure aspirin was treated as the experiment called for and diluted to exactly 1000.0 mL. A reagent blank was prepared and set to read 0.00 absorbance. Exactly 10.00 and 5.00 mL of the standard aspirin solution were each treated with 10 mL of Fe 3+ solution, diluted to exactly 100.0 mL and found to have absorbances of 0.837 and 0.410 respectively. A 374.5 mg commercial aspirin tablet was treated as the experiment called for and diluted to exactly 500.0 mL. Exactly 10.00 and 5.00 mL of the sample solution were each treated with 10 mL of Fe 3+ solution, diluted to exactly 100.0 mL and found to have absorbances of 0.694 and 0.336 respectively. Comment

Answer 1

1 mole of aspirin reacts with 1 mole of Fe3+

concentration of pure aspirin stock solution = 0.8g/180.157 g/mol x 0.1 L = 0.044 M

10 ml (A) and 5 ml (B) of stock taken + 10 ml of Fe3+ added to each and both diluted to 100 ml

final concentration of aspirin in A = 0.044 x 10/100 = 4.4 x 10^-3 M

final concentration of Aspirin in B = 0.044 x 5/100 = 2.2 x 10^-3 M

Absorbance for A = 0.837

Absorbance for B = 0.410

concentration of aspitin in commercial sample

A1/A2 = C1/C2

with,

A1 = 0.837

A2 = 0.694

C1 = 4.4 x 10^-3 M

C2 = ?

feed values,

C2 = 4.4 x 10^-3 x 0.692/0.837 = 3.64 x 10^-3 M

So the concentration of aspirin in the commercial 100 ml diluted solution = 3.64 x 10^-3 M

moles of aspirin = 3.64 x 10^-3 M x 0.01 = 3.64 x 10^-5 mols in 10 ml solution

moles of aspirin in 100 ml solution = 3.64 x 10^-4 mols

mass of aspiring = 3.64 x 10^-4 x 180.157 = 0.0656 g = 65.6 mg of aspirin in the commercial sample