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 1.043g vitamin C tablet is crushed, treated, and diluted to 100.00mL in a volumetric flask with deionized water. Three 5.00mL samples of this solution are titrated with DCP that had a standardized concentration of 9.93x10^-4M. The three titrations took an average of 35.52mL 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)

Answer 1

Concentration of DCP = 9.93 x 10^-4 M

volume of DCP = 5.00 mL = 5.00 x 10^-3 L

moles of DCP = (Concentration of DCP) * (volume of DCP in Liter)

moles of DCP = (9.93 x 10^-4 M) * (5.00 x 10^-3 L)

moles of DCP = 4.965 x 10^-6 mol

Ascorbic acid reacts with DCP in a 1 : 1 stoichiometry

moles ascorbic acid present in 5.00 mL sample = moles of DCP

moles ascorbic acid present in 5.00 mL sample = 4.965 x 10^-6 mol

moles ascorbic acid present in 100.00 mL = (moles ascorbic acid present in 5.00 mL sample) * (100.00 mL / 5.00 mL)

moles ascorbic acid present in 100.00 mL = (4.965 x 10^-6 mol) * 20

moles ascorbic acid present in 100.00 mL = 9.93 x 10^-5 mol

mass ascorbic acid = (moles ascorbic acid present in 100.00 mL) * (molar mass ascorbic acid)

mass ascorbic acid = (9.93 x 10^-5 mol) * (176.124 g/mol)

mass ascorbic acid = 0.0175 g

mass ascorbic acid = 0.0175 g * (1000 mg / 1 g)

mass ascorbic acid = 17.5 mg