Geometric Isomers Lab Calculations

[Erisdar]

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. :)

 

[ldc3]

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 R_product / R_reactant

where R is the number in the balanced chemical equation.

Mass of reagent = moles of reagent X molar mass (also called formula weight)

 

[Erisdar]

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. :)