Chapter 2, Problem 135QRT

(a)

Interpretation Introduction

Interpretation:

The diatomic compound $\text{BrCl}$ is a reddish-brown gas.  The naturally occurring isotope of elements are given below:

  $\begin{array}{l}\text{Isotope}\text{Natural}\text{abundance,\%}\\ \text{-----------------------------------------------}\\ \text{Br-79}\text{50}\text{.69}\\ \text{Br-81}\text{49}\text{.31}\\ \text{Cl-35}\text{75}\text{.77}\\ \text{Cl-37}\text{24}\text{.23}\end{array}$

The compound has to be named.

Explanation of Solution

The given $\text{BrCl}$ is a binary compound.  Then the name of the compound is bromine monochloride.

(b)

Interpretation Introduction

Interpretation:

The diatomic compound $\text{BrCl}$ is a reddish-brown gas.  The naturally occurring isotope of elements are given below:

  $\begin{array}{l}\text{Isotope}\text{Natural}\text{abundance,\%}\\ \text{-----------------------------------------------}\\ \text{Br-79}\text{50}\text{.69}\\ \text{Br-81}\text{49}\text{.31}\\ \text{Cl-35}\text{75}\text{.77}\\ \text{Cl-37}\text{24}\text{.23}\end{array}$

The number of different types of $\text{BrCl}$ molecules are possible by using the sum of mass numbers as the criteria for type has to be determined.

Explanation of Solution

There are four isotopic combinations: ${}^{\text{79}}{\text{Br-}}^{\text{35}}{\text{Cl,}}^{\text{79}}{\text{Br-}}^{\text{37}}{\text{Cl,}}^{\text{81}}{\text{Br-}}^{\text{35}}\text{Cl}\text{and}{}^{\text{81}}{\text{Br-}}^{\text{37}}\text{Cl}$.

The sum of the mass numbers is:

    $\text{79+35=114,79+37=116,81+35=116}\text{and}\text{87+37=118}$.

Using the sum of the mass numbers as a criterion, there are three types: 114, 116 and 118.

(c)

Interpretation Introduction

Interpretation:

The diatomic compound $\text{BrCl}$ is a reddish-brown gas.  The naturally occurring isotope of elements are given below:

  $\begin{array}{l}\text{Isotope}\text{Natural}\text{abundance,\%}\\ \text{-----------------------------------------------}\\ \text{Br-79}\text{50}\text{.69}\\ \text{Br-81}\text{49}\text{.31}\\ \text{Cl-35}\text{75}\text{.77}\\ \text{Cl-37}\text{24}\text{.23}\end{array}$

The most abundant type in (b) has to be determined.

Explanation of Solution

Abundance of the diatomic molecule is determined by combining the percents.

Because there are two different isotopic combinations composing the 116 type, the abundance of this type comes from two combinations ${}^{\text{79}}{\text{Br-}}^{\text{37}}\text{Cl}\text{and}{}^{81}{\text{Br-}}^{\text{35}}\text{Cl}$.

    ${}^{\text{79}}{\text{Br-}}^{\text{37}}\text{Cl}$ has an abundance of $\text{(0}\text{.5069)}\times \text{(0}\text{.2423)=0}\text{.1228,}\text{or}\text{12}\text{.28\%}$

    ${}^{\text{81}}{\text{Br-}}^{\text{35}}\text{Cl}$ has an abundance of $\text{(0}\text{.4931)}\times \text{(0}\text{.7577)=0}\text{.3736}\text{or}\text{37}\text{.36\%}$

    ${}^{\text{79}}{\text{Br-}}^{\text{35}}\text{Cl}$ has an abundance of $\text{(0}\text{.5069)}\times \text{(0}\text{.7577)=0}\text{.3841,}\text{or}\text{38}\text{.41\%}$

    ${}^{\text{81}}{\text{Br-}}^{\text{37}}\text{Cl}$ has an abundance of $\text{(0}\text{.4931)}\times \text{(0}\text{.2423)=0}\text{.1195,}\text{or}\text{11}\text{.95\%}$

The total abundance of the type is $\text{37}\text{.36\%}\text{+}\text{12}\text{.28}\text{=}\text{49}\text{.64\%}$.

The most abundant type is 116 type.

(d)

Interpretation Introduction

Interpretation:

The diatomic compound $\text{BrCl}$ is a reddish-brown gas.  The naturally occurring isotope of elements are given below:

  $\begin{array}{l}\text{Isotope}\text{Natural}\text{abundance,\%}\\ \text{-----------------------------------------------}\\ \text{Br-79}\text{50}\text{.69}\\ \text{Br-81}\text{49}\text{.31}\\ \text{Cl-35}\text{75}\text{.77}\\ \text{Cl-37}\text{24}\text{.23}\end{array}$

The second most abundant has to be determined.

Explanation of Solution

Abundance of the diatomic molecule is determined by combining the percents.

Because there are two different isotopic combinations composing the 116 type, the abundance of this type comes from two combinations ${}^{\text{79}}{\text{Br-}}^{\text{37}}\text{Cl}\text{and}{}^{81}{\text{Br-}}^{\text{35}}\text{Cl}$.

    ${}^{\text{79}}{\text{Br-}}^{\text{37}}\text{Cl}$ has an abundance of $\text{(0}\text{.5069)}\times \text{(0}\text{.2423)=0}\text{.1228,}\text{or}\text{12}\text{.28\%}$

    ${}^{\text{81}}{\text{Br-}}^{\text{35}}\text{Cl}$ has an abundance of $\text{(0}\text{.4931)}\times \text{(0}\text{.7577)=0}\text{.3736}\text{or}\text{37}\text{.36\%}$

    ${}^{\text{79}}{\text{Br-}}^{\text{35}}\text{Cl}$ has an abundance of $\text{(0}\text{.5069)}\times \text{(0}\text{.7577)=0}\text{.3841,}\text{or}\text{38}\text{.41\%}$

    ${}^{\text{81}}{\text{Br-}}^{\text{37}}\text{Cl}$ has an abundance of $\text{(0}\text{.4931)}\times \text{(0}\text{.2423)=0}\text{.1195,}\text{or}\text{11}\text{.95\%}$

The abundance of the 1174 type is $\text{38}\text{.41\%}$.

The second most abundant type is 114 type.