[Haaaaaaanh]

The following question really has me stumped

"How much of the R-Enatiomer is present in 10g of a mixture which has an enatiomeric excess of 60% for the R-isomer"

Correct answer is 8g. I tried using the equation and got 6 grams which was incorrect, then I realized the r-enatiomer is basically an s isomer I think, and got 4grams so yeah I was just wondering if anyone can get 8grams as an answer or if the prof made a mistake.

[OC pro]

Simple calculation. Have you understood how ee % will be calculated?
60% ee means a 80:20 mixture of enantiomers. Therefore, 8g is the correct answer for R.
Enantiomeric excess is defined "as the absolute difference between the mole fraction of each enantiomer" (IUPAC)

[curiouscat]

A bottle and a cork together cost $1.10. The bottle costs $1.00 more than the cork. How much does the cork cost?

[Borek]

A bottle and a cork together cost $1.10. The bottle costs $1.00 more than the cork. How much does the cork cost?

It is not that simple, as definition of enantiomer excess is counterintuitive. 60% excess suggests there is 1.6 times the amount of the other.

[Haaaaaaanh]

Simple calculation. Have you understood how ee % will be calculated?
60% ee means a 80:20 mixture of enantiomers. Therefore, 8g is the correct answer for R.
Enantiomeric excess is defined "as the absolute difference between the mole fraction of each enantiomer" (IUPAC)

Can you explain how you got 80:20, my prof didn't really explain what enatiomeric excess was, just told us the formula for it.

[OC pro]

Have you already looked into a textbook or googled it? Wikipedia helps also...

[Haaaaaaanh]

Have you already looked into a textbook or googled it? Wikipedia helps also...

Ok I think I get it, so basically if the question states that there is an enatiomeric excess of say 70% for an R-enatiomer then the remaining 30% is racemic and therefore the total amount of R-enatiomer is 85%?

[Dan]

Correct.

The definition has already been posted:

as the absolute difference between the mole fraction of each enantiomer

So:

[tex]ee = 100(X_A - X_B)       (eq 1)[/tex]

Where X_A is the mole fraction of enantiomer A, X_B is the mole fraction of enantiomer B.

Since: [tex]X_A + X_B = 1[/tex]

         [tex]X_B = 1 - X_A        (eq 2)[/tex]

Sub (eq 2) into (eq 1)

[tex]ee = 100(2X_A - 1)[/tex]

[tex]X_A = \frac {\frac{ee}{100}+1}{2}[/tex]