Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
Question
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Chapter 16, Problem 16.3QAP
Interpretation Introduction

Interpretation:

The absorption frequency for CH and CD stretching vibration should be determined and compare the CH value with the range found in the charts.

Concept introduction:

An absorption band consists of a range of the energy and the wavelength in the electromagnetic spectrum. The absorption frequency can be found by the mass of atoms and the forces between atoms in a molecule.

Expert Solution & Answer
Check Mark

Answer to Problem 16.3QAP

The absorption frequency corresponding to the CH stretching vibration treating the group as a single diatomic CH molecule is 2984.65/cm.

The calculated value lies near to the range found in the correlation charts.

The absorption frequency for CD molecule is 2218.32/cm.

Explanation of Solution

The expression for the wave number is:

ν¯=12πckμ ...... (I)

Here, the velocity of light is c, the force constant is k, and the reduced mass is μ.

The expression for the reduced mass is:

μ=m1m2m1+m2

Here, mass of the carbon atom is m1 and the mass of the hydrogen atom is m2.

Substitute m1m2m1+m2 for μ in Equation (I).

ν¯=12πck(m1m2m1+m2) ...... (II)

Substitute 5.0×102N/m for k, 3×1010cm/s for c, 12g/mol for m1, 1g/mol for m2, and 3.14 for π in Equation (II).

ν¯=1(2)(3.14)(3×1010cm/s)(5.0×102N/m)((12g/mol)(1g/mol)(12g/mol)+(1g/mol))

=(5.307×1012s/cm)((5.0×102N/m)((12(g/mol)×(1kg1000g))(1g/mol)×(1kg1000g)(12(g/mol)×(1kg1000g))+(1(g/mol)×(1kg1000g))))

=(5.307×1012s/cm)((5.0×102N/m)(0.012kg/mol)(0.001kg/mol)(0.012kg/mol)+(0.001kg/mol))

=(5.307×1012s/cm)((5.0×102N/m)9.230×104kg/mol)

ν¯=(5.307×1012s/cm)((5.0×102N/m)9.230×104kg/mol×1mol6×1023atom×1atom)=(5.307×1012s/cm)((5.0×102kg/s2)1.538×1027kg)=(5.307×1012s/cm)(3.164×1029/s2)=(5.307×1012s/cm)(5.624×1014/s)

ν¯=2984.65/cm

Thus, the absorption frequency corresponding to the CH stretching vibration treating the group as a single diatomic CH molecule is 2984.65/cm.

The absorption frequency of CH lies within the range of 2850/cm1 to 2970/cm1 in the correlation charts. Therefore, the calculated value for the absorption frequency for CH molecule is near to the experimental value shown in Figure 176.

The expression for the wave number is:

ν¯=12πckμ ...... (III)

The expression for the reduced mass is:

μ=m1m3m1+m3

Here, mass the mass of the hydrogen atom is m3.

Substitute m1m3m1+m3 for μ in Equation (I).

ν¯=12πck(m1m3m1+m3) ...... (IV)

Substitute 5.0×102N/m for k, 3×1010cm/s for c, 12g/mol for m1, 1g/mol for m2, and 3.14 for π in Equation (II).

ν¯=1(2)(3.14)(3×1010cm/s)(5.0×102N/m)((12g/mol)(2g/mol)(12g/mol)+(2g/mol))

=(5.307×1012s/cm)((5.0×102N/m)((12(g/mol)×(1kg1000g))(2g/mol)×(1kg1000g)(12(g/mol)×(1kg1000g))+(2(g/mol)×(1kg1000g))))

=(5.307×1012s/cm)((5.0×102N/m)(0.012kg/mol)(0.002kg/mol)(0.012kg/mol)+(0.002kg/mol))

=(5.307×1012s/cm)((5.0×102N/m)1.714×103kg/mol)

ν¯=(5.307×1012s/cm)((5.0×102N/m)1.714×103(kg/mol)×(1mol6×1023atom×1atom))=(5.307×1012s/cm)((5.0×102kg/s2)2.856×1027kg)=(5.307×1012s/cm)(1.750×1029/s2)=(5.307×1012s/cm)(4.18×1014/s)

ν¯=2218.32/cm

Thus, the absorption frequency for CD molecule is 2218.32/cm.

Conclusion

The absorption frequency corresponding to the CH stretching vibration treating the group as a single diatomic CH molecule is 2984.65/cm.

The calculated value lies near to the range found in the correlation charts.

The absorption frequency for CD molecule is 2218.32/cm.

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