Chapter 16, Problem 16.3QAP

Interpretation Introduction

Interpretation:

The absorption frequency for $-\text{C}-\text{H}$ and $-\text{C}-D$ stretching vibration should be determined and compare the $-\text{C}-\text{H}$ 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.

Answer to Problem 16.3QAP

The absorption frequency corresponding to the $-\text{C}-\text{H}$ stretching vibration treating the group as a single diatomic $\text{C}-\text{H}$ molecule is $2984.65/\text{cm}$.

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

The absorption frequency for $\text{C}-\text{D}$ molecule is $2218.32/\text{cm}$.

Explanation of Solution

The expression for the wave number is:

$\overline{\nu }=\frac{1}{2\pi c}\sqrt{\frac{k}{\mu }}$ ...... (I)

Here, the velocity of light is $c$, the force constant is $k$, and the reduced mass is $\mu$.

The expression for the reduced mass is:

$\mu =\frac{m_1{m}_2}{m_1+m_2}$

Here, mass of the carbon atom is $m_1$ and the mass of the hydrogen atom is $m_2$.

Substitute $\frac{m_1{m}_2}{m_1+m_2}$ for $\mu$ in Equation (I).

$\overline{\nu }=\frac{1}{2\pi c}\sqrt{\frac{k}{\left(\frac{m_1{m}_2}{m_1+m_2}\right)}}$ ...... (II)

Substitute $5.0\times {10}^2\text{N}/\text{m}$ for $k$, $3\times {10}^{10}\text{cm}/\text{s}$ for $c$, $12\text{g}/\text{mol}$ for $m_1$, $1\text{g}/\text{mol}$ for $m_2$, and $3.14$ for $\pi$ in Equation (II).

$\overline{\nu }=\frac{1}{\left(2\right)\left(3.14\right)\left(3\times {10}^{10}\text{cm}/\text{s}\right)}\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{\left(\frac{\left(12\text{g}/\text{mol}\right)\left(1\text{g}/\text{mol}\right)}{\left(12\text{g}/\text{mol}\right)+\left(1\text{g}/\text{mol}\right)}\right)}}$

$=\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{\left(\frac{\left(12\left(\text{g}/\text{mol}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\right)\left(1\text{g}/\text{mol}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)}{\left(12\left(\text{g}/\text{mol}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\right)+\left(1\left(\text{g}/\text{mol}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\right)}\right)}}\right)$

$=\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{\frac{\left(0.012\text{kg}/\text{mol}\right)\left(0.001\text{kg}/\text{mol}\right)}{\left(0.012\text{kg}/\text{mol}\right)+\left(0.001\text{kg}/\text{mol}\right)}}}\right)$

$=\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{9.230\times {10}^{-4}\text{kg}/\text{mol}}}\right)$

$\begin{array}{c}\overline{\nu }=\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{9.230\times {10}^{-4}\text{kg}/\text{mol}\times \frac{1\text{mol}}{6\times {10}^{23}\text{atom}}\times 1\text{atom}}}\right)\\ =\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{kg}/{\text{s}}^2\right)}{1.538\times {10}^{-27}\text{kg}}}\right)\\ =\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{3.164\times {10}^{29}/{\text{s}}^2}\right)\\ =\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(5.624\times {10}^{14}/\text{s}\right)\end{array}$

$\overline{\nu }=2984.65/\text{cm}$

Thus, the absorption frequency corresponding to the $-\text{C}-\text{H}$ stretching vibration treating the group as a single diatomic $\text{C}-\text{H}$ molecule is $2984.65/\text{cm}$.

The absorption frequency of $-\text{C}-\text{H}$ lies within the range of $2850/{\text{cm}}^{-1}$ to $2970/{\text{cm}}^{-1}$ in the correlation charts. Therefore, the calculated value for the absorption frequency for $-\text{C}-\text{H}$ molecule is near to the experimental value shown in Figure $17-6$.

The expression for the wave number is:

$\overline{\nu }=\frac{1}{2\pi c}\sqrt{\frac{k}{\mu }}$ ...... (III)

The expression for the reduced mass is:

$\mu =\frac{m_1{m}_3}{m_1+m_3}$

Here, mass the mass of the hydrogen atom is $m_3$.

Substitute $\frac{m_1{m}_3}{m_1+m_3}$ for $\mu$ in Equation (I).

$\overline{\nu }=\frac{1}{2\pi c}\sqrt{\frac{k}{\left(\frac{m_1{m}_3}{m_1+m_3}\right)}}$ ...... (IV)

Substitute $5.0\times {10}^2\text{N}/\text{m}$ for $k$, $3\times {10}^{10}\text{cm}/\text{s}$ for $c$, $12\text{g}/\text{mol}$ for $m_1$, $1\text{g}/\text{mol}$ for $m_2$, and $3.14$ for $\pi$ in Equation (II).

$\overline{\nu }=\frac{1}{\left(2\right)\left(3.14\right)\left(3\times {10}^{10}\text{cm}/\text{s}\right)}\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{\left(\frac{\left(12\text{g}/\text{mol}\right)\left(2\text{g}/\text{mol}\right)}{\left(12\text{g}/\text{mol}\right)+\left(2\text{g}/\text{mol}\right)}\right)}}$

$=\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{\left(\frac{\left(12\left(\text{g}/\text{mol}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\right)\left(2\text{g}/\text{mol}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)}{\left(12\left(\text{g}/\text{mol}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\right)+\left(2\left(\text{g}/\text{mol}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\right)}\right)}}\right)$

$=\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{\frac{\left(0.012\text{kg}/\text{mol}\right)\left(0.002\text{kg}/\text{mol}\right)}{\left(0.012\text{kg}/\text{mol}\right)+\left(0.002\text{kg}/\text{mol}\right)}}}\right)$

$=\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{1.714\times {10}^{-3}\text{kg}/\text{mol}}}\right)$

$\begin{array}{c}\overline{\nu }=\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{N}/\text{m}\right)}{1.714\times {10}^{-3}\left(\text{kg}/\text{mol}\right)\times \left(\frac{1\text{mol}}{6\times {10}^{23}\text{atom}}\times 1\text{atom}\right)}}\right)\\ =\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{\frac{\left(5.0\times {10}^2\text{kg}/{\text{s}}^2\right)}{2.856\times {10}^{-27}\text{kg}}}\right)\\ =\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(\sqrt{1.750\times {10}^{29}/{\text{s}}^2}\right)\\ =\left(5.307\times {10}^{-12}\text{s}/\text{cm}\right)\left(4.18\times {10}^{14}/\text{s}\right)\end{array}$

$\overline{\nu }=2218.32/\text{cm}$

Thus, the absorption frequency for $\text{C}-\text{D}$ molecule is $2218.32/\text{cm}$.

Conclusion

The absorption frequency corresponding to the $-\text{C}-\text{H}$ stretching vibration treating the group as a single diatomic $\text{C}-\text{H}$ molecule is $2984.65/\text{cm}$.

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

The absorption frequency for $\text{C}-\text{D}$ molecule is $2218.32/\text{cm}$.