Mass of Vitamin C sample analyzed = .123 +/- .001 g Mass of KIO 3 used for solut
ID: 1007907 • Letter: M
Question
Mass of Vitamin C sample analyzed = .123 +/- .001 g
Mass of KIO3 used for solution = .534 +/- .0014 g
Moles of KIO3 = .00249 +/- .2622 %
Molar concentration of KIO3 solution = .00249 +/- .2622 %
Buret reading (initial) = 2.650 +/- .05 mL
Buret reading (final, after adding KIO3 solution) = 21.90 +/- .05 mL
With the above laboratory data, please find with absolute error: (1) Volume of KIO3 added, (2) Moles of IO3- dispensed, (3) Moles of I3- produced, (4) Moles of C6H8O6 in sample, and (5) Mass (in grams) of C6H8O6 in sample. Please see attached photo for reference.
Explanation / Answer
1) Volume of KIO3 added = Initial burette reading - Final reading = 26.50 -21.90 = 4.6 ml = 0.0046L
NOTE: initial burette reading is given as 2.650 ml which is less than the final reading. This does not make sense, I am assuming this must be 26.50 instead of 2.650 ml
2) Moles of IO3- = Molarity of IO3- * volume added = 0.00249 * 0.0046 = 1.15*10^-5 moles
3) As per the reaction:
1 mole of IO3- produces 3 moles of I3- = 3 * 1.15 * 10^-5 = 3.45*10^-5 moles
4) Since I3- and C6H8O6 react in the ratio 1:1, the # moles of C6H8O6 consumed = 3.45 * 10^-5 which is the moles present in the sample.
5) Mass of C6H6O8 in the sample = moles of C6H8O6 * mol wt C6H6O8 = 3.45 * 10^-5 * 176 = 0.00607 g