Calculate Enantiomeric Excess (EE) of Penicillin G Mixture

  • Thread starter Thread starter noxflos
  • Start date Start date
  • Tags Tags
    Ee Mixture
Click For Summary
SUMMARY

The enantiomeric excess (EE) of a Penicillin G mixture was calculated based on its optical rotation. Given that pure Penicillin G has an optical rotation of +206 degrees and the sample measured +103 degrees, the EE was determined to be 50%. Further calculations revealed that the composition of the mixture consists of 75% (+) enantiomer and 25% (-) enantiomer. This calculation is crucial for understanding the purity and effectiveness of the Penicillin G mixture.

PREREQUISITES
  • Understanding of optical rotation and its significance in stereochemistry.
  • Knowledge of enantiomers and enantiomeric excess (EE) calculations.
  • Familiarity with mole fractions and their application in composition analysis.
  • Basic principles of organic chemistry, particularly related to chiral compounds.
NEXT STEPS
  • Study the calculation methods for enantiomeric excess in chiral mixtures.
  • Learn about the significance of optical rotation in determining the purity of chiral substances.
  • Explore the concept of mole fractions and their role in chemical composition analysis.
  • Investigate the properties and applications of Penicillin G in pharmaceutical chemistry.
USEFUL FOR

Chemists, pharmaceutical researchers, and students studying organic chemistry, particularly those focused on stereochemistry and the analysis of chiral compounds.

noxflos
Messages
10
Reaction score
0
The question reads: The optical rotation of pure pencicillin G is +206 degrees. A sample of pencillin G was found to have an optical rotation of +103 degrees. Calculate the enantiomeric excess (ee) in the mixture.

I did so by: 103/203 x 100 = 50%

Here is where I am lost:

Next calculate the precent composition of the (+) enantiomer.

What the heck is that? I know it is a mole fraction but how do I know what to plug in. Any ideas? :rolleyes:
 
Chemistry news on Phys.org
My best guess would be they want to know what percentage of the supposed mixture is the (+) enantiomer. Remember that the ee refers to the excess of one enantiomer, not the amount of one enantiomer. Think of the number of moles of each enantiomer that would be necessary to form a mixture with 50% ee and then calculate the mole fraction from there.

I hope that is clear, I don't want to give it away though.
 
I got it...thanks for your help...It is 75% (+) enantiomer and 25% (-).
 

Similar threads

Chemistry Understanding chemical potential of gas in mixture versus pure gas, used in derivation of equilibrium constant
  • · Replies 1 ·
Replies
1
Views
2K
How is the Calorimetry Mixture Problem Calculated?
  • · Replies 1 ·
Replies
1
Views
4K
Calculate the final temperature of the mixture
  • · Replies 7 ·
Replies
7
Views
3K
Optical Isomerism: Enantiomeric Excess Calculation from -13.9 Degrees
  • · Replies 3 ·
Replies
3
Views
2K
Chemistry What is the percentage of KClO3 in a mixture after removing oxygen?
  • · Replies 12 ·
Replies
12
Views
3K
Percentage yield in a reaction problem
  • · Replies 4 ·
Replies
4
Views
3K
Calculating the Optical Rotation Value
  • · Replies 1 ·
Replies
1
Views
2K
How to Determine Mole Fraction and Mass Percent from Osmotic Pressure Data?
  • · Replies 3 ·
Replies
3
Views
2K
Unknown specific heat of a mixture of several substances of known specific heat
  • · Replies 6 ·
Replies
6
Views
16K
Chemistry Calculating Mole Fraction from Mass Fraction: Benzene-Toluene Mixture Example
  • · Replies 4 ·
Replies
4
Views
4K