Chapter 7, Problem 7E.4P

Interpretation Introduction

Interpretation:

The vibrational frequencies (in $\text{Hz}$) of ${}^{\text{12}}{\text{C}}^{\text{16}}\text{O,}{}^{\text{13}}{\text{C}}^{\text{16}}\text{O},{}^{\text{12}}{\text{C}}^{\text{18}}\text{O},{}^{\text{13}}{\text{C}}^{\text{18}}\text{O}$ have to be calculated.

Concept introduction:

A harmonic oscillator consists of a particle of mass $m$ that undergoes a restoring force which is proportional to its amplitude $x$.  The restoring force is represented by $k_{\text{f}}$ which is known as the force constant.  As the force constant increases, the vibrational frequency increases.

Answer to Problem 7E.4P

The vibrational frequency of ${}^{\text{12}}{\text{C}}^{\text{16}}\text{O}$ is $\underset{\_}{6.42\times 10^{13}Hz}$.  The vibrational frequency of ${}^{\text{13}}{\text{C}}^{\text{16}}\text{O}$ is $\underset{\_}{6.29\times 10^{13}Hz}$.  The vibrational frequency of ${}^{\text{12}}{\text{C}}^{\text{18}}\text{O}$ is $\underset{\_}{6.27\times 10^{13}Hz}$.  The vibrational frequency of ${}^{\text{13}}{\text{C}}^{\text{18}}\text{O}$ is $\underset{\_}{6.13\times 10^{13}Hz}$.

Explanation of Solution

The formula to calculate vibrational frequency ($\nu$) is given below.

    $\nu =\frac{1}{2\pi }\sqrt{\frac{k_{\text{f}}}{\mu }}$                                                                                                   (1)

Where,

• $k_{\text{f}}$ is force constant.
• $\mu$ is the reduced mass of the molecule.

The reduced mass of ${}^{\text{12}}{\text{C}}^{\text{16}}\text{O}$ is calculated by the formula shown below.

    $\mu =\frac{m_{{}^{\text{12}}\text{C}}{m}_{{}^{\text{16}}\text{O}}}{m_{{}^{\text{12}}\text{C}}+m_{{}^{\text{16}}\text{O}}}$                                                                                              (2)

The mass of ${}^{\text{12}}\text{C}$ ($m_{{}^{\text{12}}\text{C}}$) is $\text{12}\text{amu}$.

The mass of ${}^{\text{16}}\text{O}$ ($m_{{}^{\text{16}}\text{O}}$) is $\text{16}\text{amu}$.

Substitute the value of $m_{{}^{\text{12}}\text{C}}$ and $m_{{}^{\text{16}}\text{O}}$ in equation (2).

    $\begin{array}{c}\mu =\frac{12\text{amu}\times 16\text{amu}}{12\text{amu}+16\text{amu}}\\ =6.857\text{amu}\end{array}$

The conversion of $\text{1}\text{amu}$ to $\text{kg}$ is shown below.

    $\text{1}\text{amu}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $6.857\text{amu}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}6.857\text{amu}=6.857\times \text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}\\ \text{=1}\text{.138}\times {\text{10}}^{-26}\text{kg}\end{array}$

The force constant ($k_{\text{f}}$) is $1857\text{N}{\text{m}}^{-1}$.

Substitute the value of $k_{\text{f}}$ and $\mu$ in equation (1).

    $\begin{array}{c}\nu =\frac{1}{2\times 3.14}\sqrt{\frac{1857\text{N}{\text{m}}^{\text{-1}}}{1.138\times {10}^{-26}\text{kg}}}\\ =\frac{1}{6.28}\sqrt{\frac{1857\text{N}{\text{m}}^{\text{-1}}}{1.138\times {10}^{-26}\text{kg}}\times \frac{\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}}{1\text{N}}}\\ =\frac{1}{6.28}\times 40.38\times {10}^{13}{\text{s}}^{-1}\\ =6.42\times {10}^{13}{\text{s}}^{-1}\end{array}$

The conversion of ${\text{s}}^{-1}$ to $\text{Hz}$ is shown below.

    $1{\text{s}}^{-1}=1\text{Hz}$

Therefore, the conversion of $6.42\times {10}^{13}{\text{s}}^{-1}$ to $\text{Hz}$ is shown below.

    $6.42\times {10}^{13}{\text{s}}^{-1}=6.42\times {10}^{13}\text{Hz}$

Therefore the vibrational frequency of ${}^{\text{12}}{\text{C}}^{\text{16}}\text{O}$ is $\underset{\_}{6.42\times 10^{13}Hz}$.

The reduced mass of ${}^{\text{13}}{\text{C}}^{\text{16}}\text{O}$ is calculated by the formula shown below.

    $\mu =\frac{m_{{}^{\text{13}}\text{C}}{m}_{{}^{\text{16}}\text{O}}}{m_{{}^{\text{13}}\text{C}}+m_{{}^{\text{16}}\text{O}}}$                                                                                              (3)

The mass of ${}^{\text{13}}\text{C}$ ($m_{{}^{\text{13}}\text{C}}$) is $\text{13}\text{amu}$.

The mass of ${}^{\text{16}}\text{O}$ ($m_{{}^{\text{16}}\text{O}}$) is $\text{16}\text{amu}$.

Substitute the value of $m_{{}^{\text{13}}\text{C}}$ and $m_{{}^{\text{16}}\text{O}}$ in equation (3).

    $\begin{array}{c}\mu =\frac{13\text{amu}\times 16\text{amu}}{13\text{amu}+16\text{amu}}\\ =7.17\text{amu}\end{array}$

The conversion of $\text{1}\text{amu}$ to $\text{kg}$ is shown below.

    $\text{1}\text{amu}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $7.17\text{amu}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}7.17\text{amu}=7.17\times \text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}\\ =\text{1}\text{.19}\times {\text{10}}^{-26}\text{kg}\end{array}$

The force constant ($k_{\text{f}}$) is $1857\text{N}{\text{m}}^{-1}$.

Substitute the value of $k_{\text{f}}$ and $\mu$ in equation (1).

    $\begin{array}{c}\nu =\frac{1}{2\times 3.14}\sqrt{\frac{1857\text{N}{\text{m}}^{\text{-1}}}{\text{1}\text{.19}\times {\text{10}}^{-26}\text{kg}}}\\ =\frac{1}{6.28}\sqrt{\frac{1857\text{N}{\text{m}}^{\text{-1}}}{\text{1}\text{.19}\times {\text{10}}^{-26}\text{kg}}\times \frac{\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}}{1\text{N}}}\\ =\frac{1}{6.28}\times 39.5\times {10}^{13}{\text{s}}^{-1}\\ =6.29\times {10}^{13}{\text{s}}^{-1}\end{array}$

The conversion of ${\text{s}}^{-1}$ to $\text{Hz}$ is shown below.

    $1{\text{s}}^{-1}=1\text{Hz}$

Therefore, the conversion of $6.29\times {10}^{13}{\text{s}}^{-1}$ to $\text{Hz}$ is shown below.

    $6.29\times {10}^{13}{\text{s}}^{-1}=6.29\times {10}^{13}\text{Hz}$

Therefore the vibrational frequency of ${}^{\text{13}}{\text{C}}^{\text{16}}\text{O}$ is $\underset{\_}{6.29\times 10^{13}Hz}$.

The reduced mass of ${}^{\text{12}}{\text{C}}^{\text{18}}\text{O}$ is calculated by the formula shown below.

    $\mu =\frac{m_{{}^{\text{12}}\text{C}}{m}_{{}^{\text{18}}\text{O}}}{m_{{}^{\text{12}}\text{C}}+m_{{}^{\text{18}}\text{O}}}$                                                                                              (4)

The mass of ${}^{\text{12}}\text{C}$ ($m_{{}^{\text{12}}\text{C}}$) is $\text{12}\text{amu}$.

The mass of ${}^{\text{18}}\text{O}$ ($m_{{}^{\text{18}}\text{O}}$) is $\text{18}\text{amu}$.

Substitute the value of $m_{{}^{\text{12}}\text{C}}$ and $m_{{}^{\text{18}}\text{O}}$ in equation (4).

    $\begin{array}{c}\mu =\frac{12\text{amu}\times 18\text{amu}}{12\text{amu}+18\text{amu}}\\ =7.2\text{amu}\end{array}$

The conversion of $\text{1}\text{amu}$ to $\text{kg}$ is shown below.

    $\text{1}\text{amu}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $7.2\text{amu}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}7.2\text{amu}=7.2\times \text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}\\ =\text{1}\text{.195}\times {\text{10}}^{-26}\text{kg}\end{array}$

The force constant ($k_{\text{f}}$) is $1857\text{N}{\text{m}}^{-1}$.

Substitute the value of $k_{\text{f}}$ and $\mu$ in equation (1).

    $\begin{array}{c}\nu =\frac{1}{2\times 3.14}\sqrt{\frac{1857\text{N}{\text{m}}^{\text{-1}}}{\text{1}\text{.195}\times {\text{10}}^{-26}\text{kg}}}\\ =\frac{1}{6.28}\sqrt{\frac{1857\text{N}{\text{m}}^{\text{-1}}}{\text{1}\text{.195}\times {\text{10}}^{-26}\text{kg}}\times \frac{\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}}{1\text{N}}}\\ =\frac{1}{6.28}\times 39.42\times {10}^{13}{\text{s}}^{-1}\\ =6.27\times {10}^{13}{\text{s}}^{-1}\end{array}$

The conversion of ${\text{s}}^{-1}$ to $\text{Hz}$ is shown below.

    $1{\text{s}}^{-1}=1\text{Hz}$

Therefore, the conversion of $6.27\times {10}^{13}{\text{s}}^{-1}$ to $\text{Hz}$ is shown below.

    $6.27\times {10}^{13}{\text{s}}^{-1}=6.27\times {10}^{13}\text{Hz}$

Therefore the vibrational frequency of ${}^{\text{12}}{\text{C}}^{\text{18}}\text{O}$ is $\underset{\_}{6.27\times 10^{13}Hz}$.

The reduced mass of ${}^{\text{13}}{\text{C}}^{\text{18}}\text{O}$ is calculated by the formula shown below.

    $\mu =\frac{m_{{}^{\text{13}}\text{C}}{m}_{{}^{\text{18}}\text{O}}}{m_{{}^{\text{13}}\text{C}}+m_{{}^{\text{18}}\text{O}}}$                                                                                              (5)

The mass of ${}^{\text{13}}\text{C}$ ($m_{{}^{\text{13}}\text{C}}$) is $\text{13}\text{amu}$.

The mass of ${}^{\text{18}}\text{O}$ ($m_{{}^{\text{18}}\text{O}}$) is $\text{18}\text{amu}$.

Substitute the value of $m_{{}^{\text{13}}\text{C}}$ and $m_{{}^{\text{18}}\text{O}}$ in equation (5).

    $\begin{array}{c}\mu =\frac{13\text{amu}\times 18\text{amu}}{13\text{amu}+18\text{amu}}\\ =7.548\text{amu}\end{array}$

The conversion of $\text{1}\text{amu}$ to $\text{kg}$ is shown below.

    $\text{1}\text{amu}=\text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}$

Therefore, the conversion of $7.548\text{amu}$ to $\text{kg}$ is shown below.

    $\begin{array}{c}7.548\text{amu}=7.548\times \text{1}\text{.66}\times {\text{10}}^{-27}\text{kg}\\ \text{=1}\text{.252}\times {\text{10}}^{-26}\text{kg}\end{array}$

The force constant ($k_{\text{f}}$) is $1857\text{N}{\text{m}}^{-1}$.

Substitute the value of $k_{\text{f}}$ and $\mu$ in equation (1).

    $\begin{array}{c}\nu =\frac{1}{2\times 3.14}\sqrt{\frac{1857\text{N}{\text{m}}^{\text{-1}}}{\text{1}\text{.252}\times {\text{10}}^{-26}\text{kg}}}\\ =\frac{1}{6.28}\sqrt{\frac{1857\text{N}{\text{m}}^{\text{-1}}}{\text{1}\text{.252}\times {\text{10}}^{-26}\text{kg}}\times \frac{\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}}{1\text{N}}}\\ =\frac{1}{6.28}\times 38.51\times {10}^{13}{\text{s}}^{-1}\\ =6.13\times {10}^{13}{\text{s}}^{-1}\end{array}$

The conversion of ${\text{s}}^{-1}$ to $\text{Hz}$ is shown below.

    $1{\text{s}}^{-1}=1\text{Hz}$

Therefore, the conversion of $6.13\times {10}^{13}{\text{s}}^{-1}$ to $\text{Hz}$ is shown below.

    $6.13\times {10}^{13}{\text{s}}^{-1}=6.13\times {10}^{13}\text{Hz}$

Therefore the vibrational frequency of ${}^{\text{13}}{\text{C}}^{\text{18}}\text{O}$ is $\underset{\_}{6.13\times 10^{13}Hz}$.