Calculate the molar mass of ascorbic acid.

Question

A student crushed a vitamin C tablet that weighed 1.634g, and transferred 1.552g of the powder into a 100 mL volumetric flask. After diluting to the 100.00mL mark, they titrated a 20.00mL aliquot of this solution against 0.0106 M KIO3 ( the KIO2 was in the buret) with an excess HCl and KI. The initial buret reading was 0.53 mL; the final buret reading was 17.29 mL.

Answer 1

Work the titration first, then read what the numbers are telling you — because this data set has something instructive hiding in it.

Step 1 — titrant delivered

17.29 − 0.53 = 16.76 mL of 0.0106 M KIO₃

mol IO₃⁻ = (0.01676 L)(0.0106 M) = 1.777 × 10⁻⁴ mol

Step 2 — the stoichiometry (this is the part that decides everything)

The iodate does not touch the vitamin C directly. With excess KI and HCl it first generates iodine:

IO₃⁻ + 5 I⁻ + 6 H⁺ → 3 I₂ + 3 H₂O

and the iodine is what oxidises the ascorbic acid, 1:1:

C₆H₈O₆ + I₂ → C₆H₆O₆ + 2 I⁻ + 2 H⁺

So 1 mol IO₃⁻ ≡ 3 mol ascorbic acid. (Electron check: iodate goes I⁺⁵ → I⁻¹, a 6-electron reduction; ascorbic acid gives up 2 electrons. 6/2 = 3 ✓.)

mol ascorbic acid in the 20.00 mL aliquot = 3(1.777 × 10⁻⁴) = 5.330 × 10⁻⁴ mol

Step 3 — scale the aliquot up to the whole flask

The aliquot is 20.00 of 100.00 mL, so multiply by 5:

mol ascorbic acid in the flask = 2.665 × 10⁻³ mol

Step 4 — and here is the catch

Molar mass = mass ÷ moles requires the mass of pure ascorbic acid. If you use the 1.552 g of powder:

M = 1.552 g / 2.665 × 10⁻³ mol = 582 g/mol

Ascorbic acid is C₆H₈O₆, molar mass 176.12 g/mol. Your result is 3.3× too high — and that is not an arithmetic error. A vitamin C tablet is not pure ascorbic acid. Most of that 1.552 g is binder, filler, and coating.

So this experiment cannot determine the molar mass; it determines how much vitamin C is present. If your assignment really asks for molar mass, it is likely a wording slip — worth raising with your instructor.

What the data actually gives you

mass of ascorbic acid = (2.665 × 10⁻³ mol)(176.12 g/mol) = 0.469 g in the 1.552 g of powder → 30.2% ascorbic acid by mass

Scaling to the whole 1.634 g tablet:

0.469 g × (1.634 / 1.552) = 0.494 g ≈ 494 mg per tablet

That is a 500 mg vitamin C tablet, recovered to within 1.2%. Landing that close on a labelled value is strong evidence the titration, the 3:1 stoichiometry, and the aliquot scaling are all correct — which is exactly the confirmation the 582 g/mol figure could never give you.

Why the tablet was weighed twice: 1.634 g is the whole tablet, 1.552 g is what actually made it into the flask. Some powder is always lost in transfer, so the final scale-up by 1.634/1.552 is what converts your measured quantity back to a per-tablet basis. Skipping it under-reports the tablet by about 5%.