What is the percentage of ASA in one tablet?

  • Thread starter Thread starter pbonnie
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the percentage of acetylsalicylic acid (ASA) in a tablet using a titration method. The participant correctly determined the number of moles of ASA as 1.8x10-3 mol based on a sodium hydroxide (NaOH) concentration of 0.100 mol/L and a volume of 18.0 mL. The mass of ASA was calculated to be 0.324 g, leading to a purity percentage of 64.8% when compared to the original tablet mass of 0.500 g. The calculations and methodology were confirmed as accurate by other forum members.

PREREQUISITES
  • Understanding of stoichiometry and molarity calculations
  • Familiarity with titration techniques and acid-base reactions
  • Knowledge of molecular weight calculations, specifically for acetylsalicylic acid (C9H8O4)
  • Basic skills in percentage calculations and purity determination
NEXT STEPS
  • Review titration methods for determining concentrations in acid-base reactions
  • Study the properties and reactions of acetylsalicylic acid (ASA)
  • Learn about different methods for calculating purity in pharmaceutical tablets
  • Explore advanced stoichiometry problems involving multiple reactants
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in pharmaceutical analysis or quality control of medications will benefit from this discussion.

pbonnie
Messages
92
Reaction score
0

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!
 
Physics news on Phys.org
* % purity = 0.324g/0.500g = 64.8%
 
Looks OK to me.
 

Similar threads

Determining Volume of Base for ASA Experiment
  • · Replies 2 ·
Replies
2
Views
5K
Finding mass of acetyl salicylic acid in aspirin tablet
  • · Replies 2 ·
Replies
2
Views
22K
What is the Correct Calculation for Percentage Yield of Alkene from Alcohol?
  • · Replies 6 ·
Replies
6
Views
22K