Question

Verifying the moles and mass of C₆O₆H₈ from KIO₃ titration and determing average % of vitamin C.

Experiment: Measured 50.0 mL of apple juice in an erlenmeyer flask. Add 1g KI and 5mL of 1 M HCI to the mixture. Add 2-3 mL of 0.5% starch solution. Titrate with 0.0100 M KIO3 through a burrette. Repeat 3 times.

Results:

	Juice 1	Juice 2	Juice 3
Juice Volume (mL)	50.0	50.0	50.0
mL of KIO3dispensed (mL)	5.15	5.92	4.98
moles of IO3-
moles of C6O6H8
Mass of C6O6H8 (mg)
Male % Vitamin C
Female % Vitamin C

What I have tried:

To find the number of IO₃^- moles:

[(mL of KIO₃ dispensed) * (0.001)] * (0.0100 M KIO₃) = moles IO₃^-

To find the number of C₆O₆H₈ moles:

[(mL of KIO₃ dispensed) * (0.001)] * (0.0100 M KIO₃) *3 = moles C₆O₆H₈

To find the mass of C₆O₆H₈:

([(mL of KIO₃ dispensed) * (0.001)] * (0.0100 M KIO₃) *3) * 176.124 mol C₆O₆H₈ = mass of C₆O₆H₈ (then convert to mg)

When I do this I get the following masses for C₆O₆H₈:

Juice 1	Juice 2	Juice 3
27.057 mg	31.280 mg	26.312 mg

The issue I'm having when calculating the mass of vitamin C in a single serving (240 mL) for each titration.

I'm assuming since 240/50 = 4.8, you would multiply the mass of C₆O₆H₈ by 4.8 for each titration. But I'm getting really big numbers when doing this. And the reason I'm would like help or clarification is because the next question states:

"The % Vitamin C is based on the recommended daily allowance which is 75 mg for females and 90 mg for males. Calculate the % of vitamin C based on your results from the mass of vitamin in a single serving (240mL) for the average male and average female"

But if 240 mL has 120% of the daily recommended allowance of Vitamin C, then 240 mL of apple juice should contain between 90 mg to 108 mg of Vitamin C which is much less than the results I am getting. I am not sure if I am reading the question wrong or if my calculations are wrong. Any help and explanations would be appreciated. Thank you!