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!!
To find the weight per volume percent (%w/v) of ethanol in the original whiskey sample, you need to correct your calculations.
First, calculate the number of moles of K2Cr2O7 used in the back titration.
0.02150 M K2Cr2O7 * 0.05 L = 0.001075 mol K2Cr2O7
Next, calculate the number of moles of Fe2+ used in the back titration.
0.1275 M Fe2+ * 0.0171 L = 0.00218025 mol Fe2+
Now, find the moles of K2Cr2O7 reacted with the ethanol.
Since the stoichiometric ratio between K2Cr2O7 and ethanol is 2:3, you can use the mole ratio to find the moles of ethanol.
0.001075 mol K2Cr2O7 * (3 mol ethanol / 2 mol K2Cr2O7) = 0.0016125 mol ethanol
Next, convert the moles of ethanol to grams.
0.0016125 mol C2H5OH * 46.0684 g/mol = 0.07429 g C2H5OH
Now, calculate the weight per volume percent of ethanol in the original whiskey sample.
The 4.00 mL aliquot was diluted to 500.0 mL. Therefore, you need to use the diluted volume and the diluted weight (0.07429 g) to calculate the %w/v.
(0.07429 g / 500.0 mL) * 100 = 0.01486% w/v
So, the weight per volume percent of ethanol in the original whiskey sample is 0.01486% w/v.
Please note that this calculation assumes that the density of the whiskey sample is equal to the density of water. If the density of the whiskey is different, you would need to incorporate that into your calculations.