Question

An analysis of generic antacid tablets labeled to contain 750 mg of CaCO3 (MM= 100.09 g/mol) per tablet of active ingredient was performed. The analysis was performed by dissolving a 0.985 g sample of the antacid tablet in 90.00 mL of 0.175 M HCl. The excess acid was back-titrated with exactly 33.15 mL of 0.155 M NaOH. The average weight of a tablet is 1.025 g. The tablet came from a bottle of 175 tablets that cost $4.99. Calculate the % relative error for the mg CaCO3 tablet found compared to the labeled content.

Answer 1

No of mol of HCl reacted = (90*0.175)-(33.15*0.155) = 10.61175 mmol

CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

1 mol CaCo3 = 2 mol HCl

No of mol of CaCo3 reacted = 10.61175/2 = 5.306 mmol

Amount of caCo3 in the sample = n*Mwt

        = 5.306*10^-3*100.09

        = 0.531 g

        = 531 mg

from the data , 0.985 g tablet = 531 mg CaCo3

                 1 tablet = 1.025 g ,

        one tablet containing Caco3 from analysis = 1.025*531/0.985

                              = 552.56 mg

% relative error = (750-552.56)/750*100 = 26.32%