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Stoichometry Question

  1. Jul 12, 2004 #1
    I am taking chemistry 1A again just to refresh my skills. The teacher was telling us how to figure out the mass composition of oxalate in some crystals that we had just synthesized. The equation is written below

    2MnO4- + 5C2O42- + 16H+ ----> 2Mn2+ + 10CO2 + 8H2O

    The question is to find the Mass percent of oxalate in the green crystals. The molarity of MnO4- that was used was .0432M, and the amount that was used was 11.57ml. The mass of the green crystals was .225g.

    The teacher converted the Molarity to moles by multiplying it by .01157L. He then converted the moles of MnO4- to the moles of oxalate using the formula mentioned aboce.

    My question is since this is a titration dont the moles of the MnO4- added equal the moles of oxalate at the equivalce point. I dont understand why he used stoichometry to get the moles of oxalate. This is what i had learned in my previos class that the moles of acid or base added equals the moles of base being titrated at the equvialnce point. Isnt the fundamental concept of titration that the moles are equal at equivalance point. I may be missing something here so can some one explain it to me. And if my point of view is worng can some one tell me when that concept of "moles are equal at equivalnce point" is used.

  2. jcsd
  3. Jul 13, 2004 #2


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    If this is what they taught you before, they were wrong. However, I seriously doubt that any decent Chem teacher would teach the above wrong idea. I strongly suggest you revise the mole concept and understand the theory behind titrations before proceeding.

    Again, NO !

    This concept is never used .The correct statement is "the number of equivalents of the different species are equal at the equivalence point." In the above example, an equivalent of MnO4- has 2 moles and an equivalent of C2O42- has 5 moles.

    You can NOT relate the number of moles of different species WITHOUT using the balanced chemical equation !
  4. Jul 13, 2004 #3
    I think that was my problem, in the past i never had equations with different coefficients. Usually the equations were already balanced with every thing having a 1 to 1 ratio. I guess that is what messed me up. What you said at is what I meant to say.

  5. Jul 13, 2004 #4
    Would this be correct then, for example if you were doing a acid base titration

    nbase added = nacid initially

    correct or no
    What is the difference between this statement and the statement i made above.

    Last edited: Jul 13, 2004
  6. Jul 14, 2004 #5


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    It is the same as the previous statement, and hence, is incorrect for the same reason.
  7. Jul 15, 2004 #6
    This is not an acid/base titration. It is a redox titration.
    Last edited: Jul 15, 2004
  8. Jul 16, 2004 #7
    So it is only true for acid and base reactions.
  9. Jul 20, 2004 #8
    what is only true for acid/base titrations?
    the relation you gave above?
    that would only be true for strong, monoprotic acids (HCl, HNO3) reacting with a strong base (NaOH)
    if the acid were di or triprotic (H3PO4, H2SO4) then there would be 2-3 inflection points in your titration curve
    the pic at the bottom is an example
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