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you have a mixture that is composed of 29% of the ( ) enantiomer and 71% the (-) enantiomer. what is the ee of the mixture?

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Final answer:

The enantiomeric excess (ee) of a mixture is 42%, which is the difference between the two enantiomers, calculated by subtracting the percent of the lesser enantiomer from the percent of the greater one.

Step-by-step explanation:

The enantiomeric excess (ee) of a mixture can be calculated by determining the difference in percentage between the two enantiomers. In this case, the mixture is composed of 29% of the (+) enantiomer and 71% of the (-) enantiomer. To find the enantiomeric excess (ee), one would subtract the percentage of the lesser enantiomer from the percentage of the greater enantiomer.

In the given mixture:

  • Percentage of (-) enantiomer = 71%
  • Percentage of (+) enantiomer = 29%

Calculating the enantiomeric excess:

  • ee = %(-) - %(+)
  • ee = 71% - 29%
  • ee = 42%

Therefore, the enantiomeric excess of the mixture is 42%

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