Calculate Enantiomeric Excess (EE) of Penicillin G Mixture

[noxflos]

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?

 

[movies]

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.

 

[noxflos]

I got it...thanks for your help...It is 75% (+) enantiomer and 25% (-).