You weighed out 0.020 g of your crude aspirin product in order to determine the amount of salicylic acid impurity. Following the procedure outlined in the manual, you dissolved the solid and diluted the solution to a final volume of 10.0 mL. If the absorbance of your sample solution was 1.07, what was the percent salicylic acid in your product?
a. 6.4%.
b. 10%.
c. 7.8%.
d. 92%.
e. none of the above.

Answers

Answer 1

Answer:

7.8%

Explanation:

Given that:

The initial mass amount of aspirin = 0.020g

The standard molar mass of aspirin = 240 g/mol

Thus, the number of moles = mass/molar mass

= 0.020/240

= 0.0000833 moles

Now, the molarity of aspirin in the solution(diluted) [tex]C_1[/tex] = [tex]\dfrac{8.33 \times 10^{-5}}{0.01}[/tex]

= [tex]= 8.33 \times 10^{-3} \ M[/tex] (provided the volume v = 0.01 L)

The absorbance of the sample solution A =1.07

The path length (b) = 1 cm

From the standard value of salicylic acid, the coefficient (e)= 139.322 /M/cm

Now; according to Beer's law, the concentration of aspirin is:

A = e×b×c

[tex]c = \dfrac{A}{eb}[/tex]

[tex]c = \dfrac{1.07}{139.322 \times 1}[/tex]

c = 0.00768 M

Finally, relating the concentration of the aspirin, the percentage of salicylic acid the product  [tex]= \dfrac{0.00833-0.00768}{0.00833} \times 100[/tex]

= 7.8%


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