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Analytical chemistry

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A 5ml portion of table wine was diluted to 100 ml in a volumetric flask. The ethanol in a 20 ml aliquot was distilled into 100 ml of 0.05151M K2Cr2O7. heating completed oxidation of the alcohol to acetic acid:
Ch3Ch2OH + 2Cr2O7(2-) +16H(+) -> 3Ch3COOH + 4Cr(3+) + 11H2O following which the excess dichromate was titrated with 14.42mL of 0.02497M Fe(II). Calculate the weight-volume percentage of ethanol.
(RESULT: 11,72%)

So i tried doing this problem and again, i don't know how to continue (if i even started right)

n(K2Cr2O7)=0.1L*0.05151M=0,005151 mol
n(EtOH)=n(K2Cr2O7)/2

Cr2O7(2-) + 3Fe(2+) -> Cr(3+) + 3Fe(3+)
n(Fe2+)=0,01442L*0,02497M=0,0003601 mol
n(Fe2+)=n(Cr2O72-)*3
n(Cr2O72-)'=0,00012 mol

n(K2Cr2O7)=0,005151-0,00012=0,005031 mol
m(EtOH)=(0,005031/2)*46,07=0,11589 g

now i don't know how to continue :S

  • Analytical chemistry - ,

    You have made several errors. The way this works is that too much dichromate was used in the oxidation of ethanol to acetic acid. How much too much. That is what the back titration with Fe does--it tells you how much too much.
    1. mols Cr2O7^2- = ok
    2. You can't calculate mols EtOH from this (yet).
    3. The Fe/Cr2O78^2- is not balanced.
    6Fe^2+ + Cr2O7^2- ==> 2Cr^3+ + 6Fe^3+
    Then mols Fe = 0.01442*0.02497 = about 0.00036 (but you need to go through more accurately). mols Cr2O7^2- from the Fe titration is 1/6 thast which = about 0.00006.
    mols Cr2O7^2- used in the ethanol reaction = 0.005151-0.00006 = ?. Again, you need to go through this more accurately.

    Then net mols Cr2O7^2-/2 = mols EtOH.
    That times 46 g/mol converts to grams in the 20 mL portion you titrated.

    Multiply that by 5 to determine grams in the100 mL volumetric flask (which is also grams in the original 5 mL), then multily by 20 to find the amount in 100 mL of wine. I get 11.71.

  • Analytical chemistry - ,

    Thank you very much for your help :)

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