Students performed a procedure similar to Part III of this experiment (Analyzing
ID: 554634 • Letter: S
Question
Students performed a procedure similar to Part III of this experiment (Analyzing a Vitamin Supplement for Vitamin C Content) as described in the procedure section. A 1.044 g vitamin C tablet is crushed, treated, and diluted to 100.00 mL in a volumetric flask with deionized water. Three 5.00 mL samples of this solution are titrated with DCP that had a standardized concentration of 9.92x10-4 M. The three titrations took an average of 42.87 mL of DCP. (MM Ascorbic Acid = 176.124 g/mol) Calculate the mass (in mg) in the vitamin C tablet. (MM Ascorbic Acid = 176.124 g/mol)
Explanation / Answer
Ans. Moles of DCP consumed = [DCP] x Volume of solution in liters
= (9.92 x 10-4 M) x 0.04287 L
= 4.2527 x 10-5 mol
# 1 mol DCP neutralizes 1 mol ascorbic acid (Vitamin C).
During titration 5.0 mL ascorbic acid aliquot is neutralized by 42.87 mL DCP. So, the moles of ascorbic acid in 5.0 mL aliquot must be equal to the number of moles consumed.
So,
Moles of Ascorbic acid in 5.0 mL aliquot = 4.2527 x 10-5 mol
Moles of Ascorbic acid in 100.0 mL soln. = (100.0 mL / 5.0 mL) x moles on 5 mL aliquot
= (100.0 mL / 5.0 mL) x 4.2527 x 10-5 mol
= 8.5054 x 10-4 mol
Mass of Ascorbic acid in 100.0 mL soln. = Moles x Molar mas
= (8.5054 x 10-4 mol) x (176.124 g/ mol)
= 0.150 g
= 150.0 mg
# Since the 100.0 mL of ascorbic acid sample is prepared from 1.044 g tablet, the total mass of ascorbic acid in 100 mL ascorbic acid solution must be equal to its amount in the tablet.
Hence, mass of ascorbic acid in 1,044 g tablet = 150.0 mg