What is the percentage of ASA in one tablet?

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AI Thread Summary
The discussion focuses on calculating the amount of acetylsalicylic acid (ASA) in a tablet and determining its percentage purity. The user outlines their calculations, starting with the molarity and volume of sodium hydroxide used in a titration, leading to the determination of moles of ASA. They calculate the mass of ASA as 0.324 grams from the moles derived and find the percentage purity to be 64.8% based on the original tablet mass of 0.500 grams. The calculations appear to be correct, confirming the user's approach to the problem. The thread emphasizes the importance of accurate stoichiometric calculations in determining the composition of pharmaceutical tablets.
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Homework Statement


Calculate the number of moles and the mass of ASA in one tablet. Determine the percentage of the original mass of the tablet that was actually ASA.

(this question was based off an experiment previously in the lesson)

Homework Equations


c = n/v
n = m/M

The Attempt at a Solution


Given:
HC_9H_7O_4 + NaOH --> C_9H-7O-4 + H_2O
I'm using the subscript b to represent the base, and subscript a to represent the acid.
c_b = 0.100 mol/L
V_b = 18.0 mL = 0.018 L
n_b = c_bV_b = 0.100mol/L x 0.018L = 1.8x10^{-3} mol

The mol ratio is 1:1, so the number of moles of the acid is also 1.8x10^{-3}
n =m/M
M_a = 180.17 g/mol
m = (1.8x10^{-3})(180.17 g/mol)
= 0.324 g
% purity = 0.324g/0.500g x 100% = 64.8%
(The original tablet was 0.500 g)

I was hoping someone could let me know if I'm doing this correctly.
Thank you!
 
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* % purity = 0.324g/0.500g = 64.8%
 
Looks OK to me.
 
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