Chapter 5, Problem 5.153P

(a)

Interpretation Introduction

Interpretation:

The different molecular masses obtained for ${\text{PCl}}_{\text{3}}$ is to be determined.

Concept introduction:

The isotopes can be defined as the atom of the same element with same number different number of neutrons.

The molar mass is defined as the mass of the all the atoms are present in the molecules in grams per mole.

Answer to Problem 5.153P

The different molecular masses of ${\text{PCl}}_{\text{3}}$ are 136, 138,140 and 142.

Explanation of Solution

The molecular mass of ${\text{PCl}}_{\text{3}}$ vary due to various combinations of the two isotopes of $\text{Cl}$ as ${}^{35}\text{Cl}$ and ${}^{37}\text{Cl}$ with $\text{P}$. The molecule ${\text{PCl}}_{\text{3}}$ is formed by the one $\text{P}$ atom with three $\text{Cl}$ atoms.

The atomic mass of ${}^{35}\text{Cl}$ isotope of chlorine is 35. If all the three chlorine present in ${\text{PCl}}_{\text{3}}$ molecule are ${}^{35}\text{Cl}$ then molecular mass of ${\text{PCl}}_{\text{3}}$ is equal to the sum of atomic mass three ${}^{35}\text{Cl}$ atoms and atomic mass of $\text{P}$ atom. The molecular mass of ${\text{PCl}}_{\text{3}}$ is $136$.

The molecular mass of ${\text{PCl}}_{\text{3}}$ molecule formed by the combination of a phosphorous atom, 2 ${}^{35}\text{Cl}$ atoms, and one ${}^{37}\text{Cl}$ atom is equal to the sum of the atomic mass of 2 atoms of ${}^{35}\text{Cl}$ isotope, one atom of ${}^{37}\text{Cl}$ isotope and atomic mass of $\text{P}$ atom. The molecular mass of ${\text{PCl}}_{\text{3}}$ is $138$.

The molecular mass of ${\text{PCl}}_{\text{3}}$ molecule formed by the combination of a phosphorous atom, 1 ${}^{35}\text{Cl}$ atom, and two ${}^{37}\text{Cl}$ atoms is equal to the sum of the atomic mass of 1 atom of ${}^{35}\text{Cl}$ isotope, two atoms of ${}^{37}\text{Cl}$ isotope and atomic mass of $\text{P}$ atom. The molecular mass of ${\text{PCl}}_{\text{3}}$ is $140$.

The molecular mass of ${\text{PCl}}_{\text{3}}$ molecule formed by a combination of the phosphorous atom and three ${}^{37}\text{Cl}$ atoms is equal to the sum of atomic mass three atoms of ${}^{37}\text{Cl}$ isotope and atomic mass of $\text{P}$ atom. The molecular mass of ${\text{PCl}}_{\text{3}}$ is $142$.

The table is as follows:

$\begin{array}{ccccc}\text{P}& \text{first Cl}& \text{second Cl}& \text{third Cl}& \text{total}\\ 31& 35& 35& 35& 136\\ 31& 37& 35& 35& 138\\ 31& 37& 37& 35& 140\\ 31& 37& 37& 37& 142\end{array}$

Conclusion

The molecular mass of ${\text{PCl}}_{\text{3}}$ is different for a different combination of chlorine isotopes attached.

(b)

Interpretation Introduction

Interpretation:

Among the different combination of ${\text{PCl}}_{\text{3}}$ molecules the most abundant molecule is to be determined.

Concept introduction:

The expression to calculate the fraction abundance of each isotope is as follows:

  $\text{fractional abundance of the isotope}=\left(\frac{\text{Percent abundance of isotope}\left(\%\right)}{100\text{ \%}}\right)$

Answer to Problem 5.153P

The most abundant ${\text{PCl}}_{\text{3}}$ molecule that has molecular mass 136 is most abundant.

Explanation of Solution

The expression to calculate the fraction abundance of ${}^{35}\text{Cl}$ isotope is as follows:

  ${\text{The fraction abundance of }}^{35}\text{Cl}=\frac{{\text{Perecentage abundace of }}^{35}\text{Cl}}{100\text{ \%}}$        (1)

Substitute $75\%$ for percentage abundace of ${}^{35}\text{Cl}$ in equation (1).

  $\begin{array}{c}{\text{The fraction abundance of }}^{35}\text{Cl}=\frac{75\text{ \%}}{100\text{ \%}}\\ =0.75\text{ \%}\end{array}$

The expression to calculate the fraction abundance of ${}^{37}\text{Cl}$ is as follows:

  ${\text{The fraction abundance of }}^{37}\text{Cl}=\frac{{\text{Perecentage abundance of }}^{37}\text{Cl}}{100\text{ \%}}$        (2)

Substitute $25\%$ for the percentage of ${}^{35}\text{Cl}$ in the equation (2).

  $\begin{array}{c}{\text{The fraction abundance of }}^{37}\text{Cl}=\frac{25\text{ \%}}{100\text{ \%}}\\ =0.25\end{array}$

The relative amount of each mass comes from the product of the relative abundance of each $\text{Cl}$ isotope.

The expression to calculate the relative mass of each $\text{Cl}$ isotope is as follows:

  $\text{Cl of mass 136}=\left[\begin{array}{l}\left({\text{fraction abundance of first }}^{35}\text{Cl }\right)\\ \left({\text{fraction abundance of second }}^{35}\text{Cl}\right)\\ \left({\text{fraction abundance of third }}^{35}\text{Cl}\right)\end{array}\right]$        (3)

Substitute $0.75$ for fraction abundance of first ${}^{35}\text{Cl}$, $0.75$ for fraction abundance of second ${}^{35}\text{Cl}$ and $0.75$ for fraction abundance of third ${}^{35}\text{Cl}$ in the equation (3).

  $\begin{array}{c}\text{Cl of mass 136}=\left[\left(\text{0}\text{.75 }\right)\left(\text{0}\text{.75 }\right)\left(\text{0}\text{.75 }\right)\right]\\ =0.421875\\ =0.42\end{array}$

The expression to calculate the relative mass of each $\text{Cl}$ isotope is as follows:

  $\text{Cl of mass 138}=\left[\begin{array}{l}\left({\text{fraction abundance of first }}^{37}\text{Cl }\right)\\ \left({\text{fraction abundance of second }}^{35}\text{Cl}\right)\\ \left({\text{fraction abundance of third }}^{35}\text{Cl}\right)\end{array}\right]$        (4)

Substitute $0.25$ for fraction abundance of first ${}^{37}\text{Cl}$, $0.75$ for fraction abundance of second ${}^{35}\text{Cl}$ and $0.75$ for fraction abundance of third ${}^{35}\text{Cl}$ in the equation (4).

  $\begin{array}{c}\text{Cl of mass 138}=\left[\left(\text{0}\text{.25 }\right)\left(\text{0}\text{.75 }\right)\left(\text{0}\text{.75 }\right)\right]\\ =0.140625\end{array}$

The expression to calculate the relative mass of each $\text{Cl}$ isotope is as follows:

  $\text{Cl of mass 140}=\left[\begin{array}{l}\left({\text{fraction abundance of first }}^{37}\text{Cl }\right)\\ \left({\text{fraction abundance of second }}^{37}\text{Cl}\right)\\ \left({\text{fraction abundance of third }}^{35}\text{Cl}\right)\end{array}\right]$        (5)

Substitute $0.25$ for fraction abundance of first ${}^{35}\text{Cl}$, $0.25$ for fraction abundance of second ${}^{35}\text{Cl}$ and $0.75$ for fraction abundance of third ${}^{35}\text{Cl}$ in the equation (5).

  $\begin{array}{c}\text{Cl of mass 140}=\left[\left(\text{0}\text{.25 }\right)\left(0.25\right)\left(\text{0}\text{.75 }\right)\right]\\ =0.046875\end{array}$

The expression to calculate the relative mass of each $\text{Cl}$ isotope is as follows:

  $\text{Cl of mass 142}=\left[\begin{array}{l}\left({\text{fraction abundance of first }}^{37}\text{Cl }\right)\\ \left({\text{fraction abundance of second }}^{37}\text{Cl}\right)\\ \left({\text{fraction abundance of third }}^{37}\text{Cl}\right)\end{array}\right]$        (6)

Substitute $0.25$ for fraction abundance of first ${}^{35}\text{Cl}$, $0.25$ for fraction abundance of second ${}^{35}\text{Cl}$ and $0.25$ for fraction abundance of third ${}^{35}\text{Cl}$ in the equation (6).

  $\begin{array}{c}\text{Cl of mass 142}=\left[\left(\text{0}\text{.25 }\right)\left(0.25\right)\left(\text{0}\text{.25 }\right)\right]\\ =\mathrm{0.015625.}\end{array}$

Conclusion

${\text{PCl}}_{\text{3}}$ molecule formed by the combination of three ${}^{35}\text{Cl}$ chlorine atoms is most abundant.

(c)

Interpretation Introduction

Interpretation:

The ratio of effusion rates of the heaviest and lightest ${\text{PCl}}_{\text{3}}$ molecule is to be calculated.

Concept introduction:

Effusion is explained as the movement of the gas molecule through a pinhole.

Diffusion can be explained as the mixing of one gas molecule with another gas molecule by random motion.

According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

  $\frac{\text{Rate of A}}{\text{Rate of B}}=\frac{u_{rms\text{ of A}}}{u_{rms\text{ of B}}}=\frac{\sqrt{M_{\text{B}}}}{\sqrt{M_{\text{A}}}}=\frac{\text{time B}}{\text{time A}}$

Here,

$M_{\text{B}}$ is the molar mass of gas $\text{B}$

$M_{\text{A}}$ is the molar mass of gas $\text{A}$.

Answer to Problem 5.153P

The ratio of effusion rates of the heaviest and lightest ${\text{PCl}}_{\text{3}}$ molecule is $0.979$.

Explanation of Solution

The heaviest ${\text{PCl}}_{\text{3}}$ molecule is ${\text{P}}^{37}{\text{Cl}}_{\text{3}}$ and the lightest ${\text{PCl}}_{\text{3}}$ molecule is ${\text{P}}^{35}{\text{Cl}}_{\text{3}}$.

The expression to calculate the ratio of the rate of the heaviest and lightest ${\text{PCl}}_{\text{3}}$ molecule is as follows:

  $\frac{{\text{Rate of P}}^{37}{\text{Cl}}_{\text{3}}}{{\text{Rate of P}}^{35}{\text{Cl}}_{\text{3}}}=\frac{\sqrt{{\text{Molar mass of P}}^{35}{\text{Cl}}_{\text{3}}}}{\sqrt{{\text{Molar mass of P}}^{37}{\text{Cl}}_{\text{3}}}}$        (7)

Substitute $136\text{g/mol}$ for ${\text{P}}^{35}{\text{Cl}}_{\text{3}}$ and $142\text{g/mol}$ for ${\text{P}}^{37}{\text{Cl}}_{\text{3}}$ in the equation (7).

  $\begin{array}{c}\frac{{\text{Rate of }}^{37}\text{Cl}}{{\text{Rate of }}^{35}\text{Cl}}=\frac{\sqrt{136\text{g/mol}}}{\sqrt{142\text{g/mol}}}\\ =0.978645\\ \approx \mathrm{0.979.}\end{array}$

Conclusion

The ratio of effusion rates of the heaviest and lightest ${\text{PCl}}_{\text{5}}$ molecule is less than 1 this suggests that the effusion rate of ${\text{P}}^{35}{\text{Cl}}_{\text{3}}$ is greater than ${\text{P}}^{37}{\text{Cl}}_{\text{3}}$.