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Snezhnost [94]
3 years ago
12

If the enantiomeric excess of a mixture is 75%, what are the % compositions of the major and minor enantiomer?

Chemistry
1 answer:
Maru [420]3 years ago
7 0

Let us say that R is the major enantiomer, while S is the minor enantiomer, therefore the formula for enantiomeric excess (ee) is:

ee = (R – S) * 100%

 

Let us further say that the fraction of R is x (R = x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:

75 = (x – (1 – x)) * 100

75 = 100 x – 100 + 100 x

200 x = 175

x = 0.875

 

Summary of answers:

R = major enantiomer = 0.875 or 87.5%

<span>S = minor enantiomer = (1 – 0.875) = 0.125 or 12.5%</span>

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The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

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<h3>Facts about the diluted solution</h3>

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

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brainly.com/question/2201903

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