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Homework Help: Geometric Isomers Lab Calculations

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the theoretical yield of the isomers from your data. Note: Find the limiting reagent.

    List of Masses I obtained during lab:
    Maleic Anhydride (C4H2O3): 5.05g, molar mass=98.1g/mol
    Impure Fumaric Acid: 2.28g
    Pure Fumaric Acid(trans-C4H4O4): 1.45g
    Pure Maleic Acid(cis-C4H4O4): 1.78g

    In the lab maleic anhydride is mixed with water and filtrated to produce maleic acid.
    The filtrate is mixed with hydrochloric acid and heated to produce fumaric acid.

    2. Relevant equations
    The chemical equations I came up with are as follows:

    3. The attempt at a solution
    As hinted I started by trying to find the limiting reagent (I'll use / for divide):

    Converting grams to moles here.
    5.05gC4H2O3 × 1molC4H2O3/9.81g=0.0515molC4H2O3


    mol ratio =2.53/98.1=49.15 >1 ∴C4H2O3 is the limiting reagent

    theoretical mass=5.05gC4H2O3×1molC4H4O4/1molC4H2O3×100.07gC4H4O4/1molC4H4O4=505.37gC4H4O4

    This seems very wrong as my mass obtained of maleic acid and fumaric acid are 1 to 2 grams only.
    This is my first and last chemistry course required for my degree so the above, I realize, may be awfully wrong.
    Any help is appreciated. :)
  2. jcsd
  3. Mar 30, 2013 #2
    I think you have the mol ratio wrong, but at least you have the correct limiting reagent.

    The next equation is definitely incorrect, the units don't work out properly.
    You want the following general equations:

    Moles of product = Moles of reactant X Rproduct / Rreactant

    where R is the number in the balanced chemical equation.

    Mass of reagent = moles of reagent X molar mass (also called formula weight)
  4. Mar 30, 2013 #3
    For the mol ratio it was supposed to be 2.53/0.0515 thanks for catching my error.

    Your two equations above helped immensely.
    For anyone checking this over:
    I used the first equation and the molar mass of C4H2O3 to get 0.0515mol C4H4O4.
    From this I used the second equation to obtain a theoretical mass of 5.978g C4H4O4.
    Adding the masses of the pure maleic and fumaric acids I get an experimental mass of 3.23g.
    For calculating % yield, 3.23g and 5.98g makes much more sense than my initial 505.4g.

    Thanks a bunch to ldc3 for the help and now I can finish writing my lab report. :)
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