Question

A 0.300 g sample of an aspirin tablet is used to prepare 100.0 mL of a stock solution by dissolving the tablet in 1 M NaOH then diluting with distilled water to the 100 mL mark of a volumetric flask.  A 0.30 mL sample of the stock solution is then transferred to a volumetric flask and diluted with 0.02 M iron(III) chloride to the 10.00 mL mark.  The tetraaquasalicylatoiron(III) complex concentration in this 10.00 mL sample is determined to be 4.3x10^-4M.  What is the percentage of acetylsalicylic acid in the original aspirin tablet?  Do not include the % sign in your answer.  For example, for a value of 100% enter 100.

Would really like a step by step answer. So i can understand the process.

Answer 1

Just go through my explanation, Iam sure u would be satisfied.

Ans:

The concentration of aspirin (C9H8O4) in the stock solution can be determined by using aspirin's molecular weight and the total volume of the solution:

0.300 g C9H8O4 x (1 mol C9H8O4 / 180.157 g C9H8O4) = 0.001665 mol C9H8O4
Concentration of C9H8O4 in stock solution: 0.001665 mol C9H8O4 / 0.1 L = 0.01665 M C9H8O4

A sample of the stock solution is further diluted with V1 = 0.3 mL and V2 = 10 mL
Concentration of C9H8O4 after dilution:
M1V1 = M2V2
(0.01665 M)(0.3 mL) = M2(10 mL), M2 = 0.0004995 M C9H8O4
Number of moles of C9H8O4 in diluted solution: 0.0004995 M C9H8O4 x 0.01 L = 0.000004995 mol C9H8O4

Complex concentration = 0.00043 M
Number of moles of complex = 0.00043 M x 0.01 L = 0.0000043 mol
Number of moles of complex must equal to number of moles of acetylsalicylic acid = 0.0000043 mol

Percentage = (0.0000043 mol / 0.000004995 mol) x 100% = 86.9%