What Is the Theoretical Yield of Geometric Isomers in This Lab Experiment?

AI Thread Summary
The discussion focuses on calculating the theoretical yield of geometric isomers from a lab experiment involving maleic anhydride and water to produce maleic acid, followed by heating with hydrochloric acid to yield fumaric acid. The initial calculations mistakenly indicated a limiting reagent of maleic anhydride, with an incorrect theoretical mass of 505.37g for fumaric acid. Corrections were made to the mole ratio and equations, leading to a revised theoretical mass of 5.978g for fumaric acid. The experimental mass obtained was 3.23g, providing a more realistic basis for calculating the percentage yield. The conversation highlights the importance of accurate stoichiometric calculations in chemistry lab reports.
Erisdar
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Homework Statement



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
H2O:4.56mL=4.56g

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.

Homework Equations


The chemical equations I came up with are as follows:
C4H2O3+H2O→cis-C4H4O4
cis-C4H4O4+HCl→trans-C4H4O4+HCl

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

45.6gH2O×1molH2O/18.014gH2O=2.53molH2O

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. :)
 
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Erisdar said:
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

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)
 
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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