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?

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

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter