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If a sample is said to contain a 98.0% enantiomeric excess of the compound (+)-camphor, what would be the observed specific rotation if the pure (+)-camphor was found to have a rotation of 44.1?

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nuhulawal20

Answer:

the observed specific rotation is 43.218

Explanation:

Given the data in the question;

percentage of enantiomeric excess = 98.0%

observed specific rotation = ? { represented by x }

specific rotation of pure compound = 44.1

Now, we know that;

% of enantiomeric excess = ( observed specific rotation / specific rotation of pure compound ) × 100%

so we substitute

98.0 % = ( x / 44.1 ) × 100%

0.98 = x / 44.1

x = 0.98 × 44.1

x = 43.218

Therefore, the observed specific rotation is 43.218

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