Asked by Adrianna
A pure sample of (R)-limonene has a specific rotation of +125.6. If a sample had a specific rotation of +35.6, how much of each enantiomer is present in the sample? Show your work.
I don't even know where to begin....please help :(
I don't even know where to begin....please help :(
Answers
Answered by
DrBob222
I suggest X = + enantiomer
and 1-X = - enantiomer.
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125.6X + (1-X)*-125.6 = 35.6 and solve for X and 1-X. These will be fractions and you can convert to percent by multiplying by 100.
Note: If the + enantiomer has +125.6 then the - enantiomer will have -125.6.
I have seen this worked a quickie way but I've forgotten how to do it the short way. This should give you the correct answers.
and 1-X = - enantiomer.
---------------------------
125.6X + (1-X)*-125.6 = 35.6 and solve for X and 1-X. These will be fractions and you can convert to percent by multiplying by 100.
Note: If the + enantiomer has +125.6 then the - enantiomer will have -125.6.
I have seen this worked a quickie way but I've forgotten how to do it the short way. This should give you the correct answers.
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