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Gchem conceptual Question

Discussion in 'DAT Discussions' started by kkatranji, Jun 18, 2008.

  1. kkatranji

    kkatranji 2+ Year Member

    May 29, 2008
    Okay so I have a question about this equivalent mass/Normality stuff

    Since i'm really bad at Gchem I decided to start from the beginning with Schaums College Gchem and it's pretty good. Anyway, I got to the normality stuff, specifically question 13.12 in the Schaum text, and I think somethings not registering. So if someone would like to help it would be really great if you did.

    The question states...

    Calculate the mass (g) of FeSO4 that will be oxidized by 24.0mL of .250 N KMn04 in a solution acidified with sulfuric acid. The unbalanced equation for the reaction is below. The statement of normality of KMnO4 is with respect to the reaction (Mn changes from +7 to +2 during this reaction).

    MnO4- + Fe2+ + H+ ----> Fe3+ + Mn2+ + H20

    So in the beginning of the solution they solve for the Equivalent mass of FeSO4 (this I understand), Fe is oxidized with one electron therefore it equals molar mass/ oxidation change = 152/1

    Here is where I get messed up

    They set the equation up like this

    Number of eq KMnO4 = number of eq FeSO4

    (volume KMnO4) x (normality KMnO4) = (mass FeSO4)/(equivalent mass FeSO4)

    Maybe I don't understand the concept of equivalence but how are these two things equal to each other. can someone explain please?

    by the way the answer is w=.912g FeSO4 if someone was wondering.
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  3. harrygt

    harrygt 2+ Year Member

    Jun 16, 2008

    YOOOOP, I see what your problem is. As you feel, the tough part is the normality here. Once we convert normality to molarity, the rest of the question becomes a piece of cake.

    Mn goes from +7 to +2. a 5 degree difference in the oxidation number. That means you have to divide the normality by 5 to get the molarity .[Think of it this way: A 1 molar Mg(OH)2 is the same as a 2 normal Mg(OH)2, because it makes 2 OH- in solution. Over here, Mn+7 gets to +2, it means it's acting like 5 times more. So to get the molarity, you have to divide the normality by 5].
    The molarity is .25/5 = 0.05 M. The rest is a piece of cake after balanding the equation.
    I did the calculations and got the same answer as you provided. Hit me with a message if you don't get there.
    Ok, I feel like I should finish this before going home from the library.
    after balancing the equation, we can see that each mole of MnO4- react with 5 Fe2+. 0.024 L * 0.05 mol = 0.0012 moles of Mn04-
    multiply that by 5 [5 Fe2+ is oxidized by each mole MnO4-], you will get 0.006 moles of FeSO4 being oxidized. Finally, multiply this by the molar mass of FeSO4 which is 151.9, and you will get 0.9114.
    I hope this helped.
    I hate doing it using the equivalents. It can be really misleading. My suggestion is that you always convert the normality to molarity, and then solve the rest with a couple of simple stochiometry calculations.
    Last edited: Jun 18, 2008
  4. kkatranji

    kkatranji 2+ Year Member

    May 29, 2008
    First of all, thanks a lot for helping

    I think I definitely saw the problem better by converting from Normality back to Molarity.

    I think you did an extra step though, because you divided the N by 5, to get Molarity (.25/5 = 0.05 M), but then you multiplied by 5 later to return to Normality.

    I still have one question though.

    the book solves the problem by setting these two things equal

    (volume KMnO4) x (normality KMnO4) = (mass FeSO4)/(equivalent mass FeSO4)

    and they solve for mass.

    but I was wondering if you know how volume X normality is equal to the mass/equivalent mass.

    I guess I can always just memorize that relationship, but I'd rather understand it.

    Thanks again for your help, I really appreciate it.
  5. harrygt

    harrygt 2+ Year Member

    Jun 16, 2008
    OH no, That was no an extra step. each mole MnO4- reacts with 5 moles of Fe2+. That is why I multiplied the number of moles of MnO4- by 5 to get the number of moles of Fe2+. I did not mean to convert back to normality.
    I can see where that equation is coming from. But honestly, it would be really hard to explain it, because as I mentioned, this whole normality things is annoying for me too. I mean, I really can not explain that equation for you in words. It's kinda confusing. Guess what? I remember my high school chem teacher skipped the normality section of the chapter, and guess what again? my college chem teacher skipped the normality section too. Normality is not being taught in most courses these days. You don't even see it being used in text books in the problems a lot, except in the section where they skim through it.
    My suggestion is to convert normality to molarity and move on.
    Normality is kinda UNnormal for me!
  6. doc toothache

    doc toothache 10+ Year Member

    Jan 17, 2006
    By definition molarity is defined as the number of moles of a substance dissolved in a liter of solution. In oxidation reduction the normality is equal to the molarity divided by the change in the oxidation state. In the case of Fe the change in the oxidation state is one so for FeSO4, the equivalent wgt is 152/1).
    1. V1N1 = V2N2
    2. V x N = g/eq
    (Think of if it as the method used to calculate the amount of substance needed to to make 1 liter of 1 N solution.)
    The problem is solved as you showed it in your post.

    (volume KMnO4) x (normality KMnO4) = (mass FeSO4)/(equivalent mass FeSO4)
    ml x N x meq = mg
    24 x 0.25 x 152= mg FeSO4= 912=.912g
    Last edited: Jun 19, 2008

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