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Snezhnost [94]
4 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]4 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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It is derived as shown below;

                        C                   H                O                  N

Mass          83.884             10.486        18.64            86.99

molar

mass                12                    1                  16                14

Moles       83.884/12         10.486/1       18.64/16        86.99/14

                   

                    6.99                 10.49              1.17                6.21

Divide

by

lowest      6.99/1.17         10.49/1.17           1.17/1.17            6.21/1.17

                       6                    9                         1                       5

Empirical formula  C₆H₉ON₅

learn more:

Empirical formula brainly.com/question/2790794

#learnwithBrainly

                 

                               

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