A 10.0 mL sample of whiskey was diluted to 500.0 mL. A 4.00 mL aliquot of the diluted sample was removed and the ethanol, C2H5OH, was distilled into 50.00 mL of 0.02150 M K2Cr2O7 and oxidized to acetic acid.
The excess Cr2O72– was then back titrated with 17.1 mL of 0.1275 M Fe2+ , producing Cr3 and Fe3 . Calculate the weight per volume percent, %w/v, of ethanol in the original whiskey sample.
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I did
0.02150 M K2Cr2O7 * 0.05L = 0.001075 K2Cr2O7 TOTAL
then
0.01275 M Fe2 * 0.017 L = 0.00218025 mol Fe2
then multiplied that using the stoichiometric ratio from the back titration reaction of (1molK2Cr2O7/6molFe2+) which gives me = 0.00363375mol K2Cr2O7 reacted with Fe 2+.
Then I took the TOTAL- the reacted amount = 7.11625*10^(-4)mol of K2Cr2O7 reacted w/ ethanol
Then with that number of 7.11625*10^(-4)K2Cr2O7 * (3mol ethanol/2mol of K2Cr2O7)= 0.0010674375 mol C2H5OH (ethanol)
Then I used that number multiplied by the molecular formula of ethanol (46.0684g) to equal 0.049175g C2H5OH.
Then to determine the weight per volume of ethanol in the original whiskey sample, I took 500/4=125ml. Multiply this by 100. I got 0.03934%. This is wrong. What did I do wrong?
Help please!!