Question

A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent enantiomeric excess?

Answer 1

Answer :  The % of (+) limonene isomer = 79%

                The % of (-) limonene isomer = 0%

                The % of enantiomeric excess = 58%

Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.

Given,

% of pure limonene enantiomer = The % of (+) limonene isomer = 79%

Therefore, The % of (-) limonene isomer = 0%

Formula used :  

$\%(+)\text{ isomer}=\frac{ee}{2}+50\%$

Where,         ee → enantiomeric excess

Now, put all the values in above formula, we get the value of enantiomeric excess (ee).

     ${ee}=\frac{\%(+)-50\%}{2}$

            $=\frac{79\%-50\%}{2}$

              = 58%

Answer 2

Answer:

The major enatiomer makes up 89.5% of the mixture

The minor enatiomer makes up 10.5% of the mixture

The percent enantiometric excess is 79%

Explanation:

% (+) = ee/2

79 % / 2 + 50%

= 39.5% + 50%

= 89.5%

and

% (-) = 100 % - 89.5 %

= 10.5%