8) How much of the S-enantiomer is present in 10 g of a mixture which has an enantiomeric excess of 20% of the R-enantiomer? A) 2 g B) 3 g C) 4 g D) 5 g E) 6 g
8) How much of the S-enantiomer is present in 10 g of a mixture which has an enantiomeric excess of 20% of the R-enantiomer? A) 2 g B) 3 g C) 4 g D) 5 g E) 6 g
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Enantiomer Calculations
#### Question 8
**Problem Statement:**
How much of the S-enantiomer is present in 10 g of a mixture which has an enantiomeric excess of 20% of the R-enantiomer?
- **Options:**
- A) 2 g
- B) 3 g
- C) 4 g
- D) 5 g
- E) 6 g
**Explanation:**
- Given data: Total mixture mass = 10 g, Enantiomeric excess (ee) = 20% of R-enantiomer
- Enantiomeric excess (ee) is defined as the difference between the percentages of the two enantiomers.
- If ee = 20%, then 80% of the mixture is composed of an equal mixture of R and S enantiomers, and 20% is the excess amount of the R-enantiomer.
- Let \( R \) be the amount of R-enantiomer and \( S \) be the amount of S-enantiomer.
- The amount of R-enantiomer in the 10 g mixture: \( R \) = \( 5g + \frac{(20/100) \times 10g}{2} \) = 6 g
- Therefore, the amount of S-enantiomer in the 10 g mixture: \( S \) = 10 - 6 = 4 g
- Hence, the correct answer is C) 4 g.
#### Question 9
**Problem Statement:**
A solution of 0.2 g/ml of a pure R-enantiomer in a 1 dm polarimeter sample tube rotates the plane of polarized light by +5°. What is the rotation observed on this solution in a 2 dm polarimeter sample tube?
- **Options:**
- A) 0
- B) -5
- C) 10
- D) -10
- E) 5
**Explanation:**
- Given data: concentration = 0.2 g/ml, tube length = 1 dm, observed rotation = +5°
- Optical rotation is directly proportional to the path length of the polarimeter tube.
- New tube length: 2 dm
- Rotation observed: \( +5° \times \frac{2 \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb987d8c1-664b-448b-b573-1ab7f66094f6%2F5affe4f0-445a-4554-904b-37d99af117e9%2Fafdhzkm_processed.png&w=3840&q=75)
Transcribed Image Text:### Enantiomer Calculations
#### Question 8
**Problem Statement:**
How much of the S-enantiomer is present in 10 g of a mixture which has an enantiomeric excess of 20% of the R-enantiomer?
- **Options:**
- A) 2 g
- B) 3 g
- C) 4 g
- D) 5 g
- E) 6 g
**Explanation:**
- Given data: Total mixture mass = 10 g, Enantiomeric excess (ee) = 20% of R-enantiomer
- Enantiomeric excess (ee) is defined as the difference between the percentages of the two enantiomers.
- If ee = 20%, then 80% of the mixture is composed of an equal mixture of R and S enantiomers, and 20% is the excess amount of the R-enantiomer.
- Let \( R \) be the amount of R-enantiomer and \( S \) be the amount of S-enantiomer.
- The amount of R-enantiomer in the 10 g mixture: \( R \) = \( 5g + \frac{(20/100) \times 10g}{2} \) = 6 g
- Therefore, the amount of S-enantiomer in the 10 g mixture: \( S \) = 10 - 6 = 4 g
- Hence, the correct answer is C) 4 g.
#### Question 9
**Problem Statement:**
A solution of 0.2 g/ml of a pure R-enantiomer in a 1 dm polarimeter sample tube rotates the plane of polarized light by +5°. What is the rotation observed on this solution in a 2 dm polarimeter sample tube?
- **Options:**
- A) 0
- B) -5
- C) 10
- D) -10
- E) 5
**Explanation:**
- Given data: concentration = 0.2 g/ml, tube length = 1 dm, observed rotation = +5°
- Optical rotation is directly proportional to the path length of the polarimeter tube.
- New tube length: 2 dm
- Rotation observed: \( +5° \times \frac{2 \,
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