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Is my expeirmental procedure sound?

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Using the apparatus that I've drawn, I have to deteremine the % composition H2O2 in an unknown solution.


    2. Relevant information
    H2O2 (aq) -enzyme-> H2O(l) + O2(g)
    or
    2H2O2 -Enzyme-> 2H2O(l) + O2(g) when balanced
    The test tube (1) contains H2O2 solution with an enzyme attached to a rxn. rod that will be lowered to start the rxn. The test tube is connected via tubing to a flask. The flask (2) is nearly filled with water. That flaks is connected via tubing to a beaker which has enough water to cover the tube's tip.

    As far as I understand, the rxn. catalyzed by the enzyme results in the above rxn. occurring. The rxn. in the testube produces O2 (g) which goes to the flask via tubing which forces H2O levels down due to pressure. This in turn forces H2O through the right side tubing and into the beaker (3).

    Of course, the process is simplified a bit.

    % by weight (mass) of element = (total mass of element present ÷ molecular mass) x 100

    3. The attempt at a solution
    i've already completed a laboratory exercise where I computed Pressure due to O2, Volume of O2 at STP, grams O2, moles O2, Molar Volume of O2 at STP.

    Using the unknown solution, I can repeat the steps for the first part of the experiment in which I determined the above values.

    I'm thinking that given an unknown solution of H2O2, I could put enough enzyme so that all the H2O2 has fully decomposed. By measuring the O2 produced, I can use stoichiometry to find the amount of H2O2 originally then using molar mass, calculate how much H2O2 there originally was. Using stiochiometry, I can calculate the amount of H2O producedby the rxn. and subtract this value from the volume of H2O in the test tube will give me the amount of H2O originally in the solution.

    Thus, to get % comp, I divided mass of H2O2 by mass of H2O2+mass of H2O x 100 to get % compositon.
     
  2. jcsd
  3. Oct 24, 2013 #2

    Borek

    User Avatar

    Staff: Mentor

    Hard to say anything not seeing the image. You have not mentioned partial pressure of water, you have not mentioned external pressure, you have not mentioned pressure exerted by the column of water present to list the most obvious things. Some of these can be eliminated by a good design of the apparatus, some can't.
     
  4. Oct 24, 2013 #3
    Thanks for the reply but I no longer need this thread as I was able to confirm from my TA that the procedure works with the apparatus. It was ore detailed there.
     
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