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If a sample contains 63.0% of the r enantiomer and 37.0% of the s enantiomer,what is the enantiomeric excess of the mixture?

User Yudijohn
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1 Answer

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Final answer:

The enantiomeric excess of the mixture is 26.0%.

Step-by-step explanation:

The enantiomeric excess (ee) of a mixture is a measure of the excess of one enantiomer over the other in a sample. To calculate the enantiomeric excess, we subtract the percentage of the less abundant enantiomer from the percentage of the more abundant enantiomer. In this case, the enantiomeric excess can be calculated as follows:

Enantiomeric excess = % of r enantiomer - % of s enantiomer

Enantiomeric excess = 63.0% - 37.0% = 26.0%

User Anuith
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