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

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 :  

Where,         ee → enantiomeric excess

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

              = 58%

answer: mercury

explanation: