Chapter 2, Problem 2.29P

(a)

Interpretation Introduction

Interpretation:

Molecular mass of ${\text{BrN}}_{\text{3}}$ has to be calculated to five significant figures.

Concept Introduction:

In a molecule, the sum of atomic mass of all the atoms is known as molecular mass.  Atomic mass of an atom is the average of all the isotope of the particular element.  Molecular mass can be used to calculate the mass percentage of the atom present in the molecule.

Explanation of Solution

Given halogen compound is ${\text{BrN}}_{\text{3}}$.  Atomic mass of $\text{Br}$ is $\text{79}\text{.904}\text{g/mol}$.  Atomic mass of $\text{N}$ is $\text{14}\text{.0067}\text{g/mol}$.  Molecular mass of ${\text{BrN}}_{\text{3}}$ is calculated as shown below.

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{BrN}}_{\text{3}}=\text{atomic}\text{mass}\text{of}\text{Br}+3(\text{atomic}\text{mass}\text{of}\text{N})\\ =79.904\text{g/mol + 3(14}\text{.0067}\text{g/mol)}\\ =79.904\text{g/mol +42}\text{.0201}\text{g/mol}\\ =\text{121}\text{.92}\text{g/mol}\end{array}$

Therefore, molecular mass of ${\text{BrN}}_{\text{3}}$ is $\text{121}\text{.92}\text{g/mol}$.

(b)

Interpretation Introduction

Interpretation:

Molecular mass of ${\text{NaIO}}_{\text{3}}$ has to be calculated to five significant figures.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given halogen compound is ${\text{NaIO}}_{\text{3}}$.  Atomic mass of $\text{Na}$ is $\text{22}\text{.989770}\text{g/mol}$.  Atomic mass of iodine is $\text{126}\text{.90447}\text{g/mol}$.  Atomic mass of oxygen is $\text{15}\text{.9994}\text{g/mol}$.  Molecular mass of ${\text{NaIO}}_{\text{3}}$ is calculated as shown below.

$\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{NaIO}}_{\text{3}}=\text{atomic}\text{mass}\text{of}\text{Na}+\text{atomic}\text{mass}\text{of}\text{I+}3(\text{atomic}\text{mass}\text{of}\text{O})\\ =22.989770\text{g/mol + 126}\text{.90447}\text{g/mol + 3(15}\text{.9994}\text{g/mol)}\\ =22.989770\text{g/mol + 126}\text{.90447}\text{g/mol + 47}\text{.9982}\text{g/mol}\\ =\text{197}\text{.89}\text{g/mol}\end{array}$

Therefore, molecular mass of ${\text{NaIO}}_{\text{3}}$ is $\text{197}\text{.89 g/mol}$.

(c)

Interpretation Introduction

Interpretation:

Molecular mass of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ has to be calculated to five significant figures.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given halogen compound is ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$.  Atomic mass of $\text{C}$ is $\text{12}\text{.0107}\text{g/mol}$.  Atomic mass of $\text{Cl}$ is $\text{35}\text{.453}\text{g/mol}$.  Atomic mass of fluorine is $\text{18}\text{.9984032}\text{g/mol}$.  Molecular mass of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ is calculated as shown below.

$\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}=1(\text{atomic}\text{mass}\text{of}\text{C)}+2(\text{atomic}\text{mass}\text{of}\text{Cl)}+2(\text{atomic}\text{mass}\text{of}\text{F})\\ =12.0107\text{g/mol + 2(35}\text{.453}\text{g/mol) + 2(18}\text{.9984032}\text{g/mol)}\\ =\text{12}\text{.0107}\text{g/mol + 70}\text{.906}\text{g/mol + 37}\text{.9968064}\text{g/mol}\\ =\text{120}\text{.91}\text{g/mol}\end{array}$

Therefore, molecular mass of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ is $\text{120}\text{.91 g/mol}$.

(d)

Interpretation Introduction

Interpretation:

Molecular mass of ${\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{6}}$ has to be calculated to five significant figures.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given halogen compound is ${\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{6}}$.  Atomic mass of $\text{C}$ is $\text{12}\text{.0107}\text{g/mol}$.  Atomic mass of $\text{Cl}$ is $\text{35}\text{.453}\text{g/mol}$.  Atomic mass of hydrogen is $\text{1}\text{.008}\text{g/mol}$.  Molecular mass of ${\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{6}}$ is calculated as shown below.

$\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{6}}=14(\text{atomic}\text{mass}\text{of}\text{C)}+9(\text{atomic}\text{mass}\text{of}\text{H)}+6(\text{atomic}\text{mass}\text{of}\text{Cl})\\ =14(12.0107\text{g/mol) + 9(1}\text{.008}\text{g/mol) + 6(35}\text{.453}\text{g/mol)}\\ =\text{168}\text{.1498}\text{g/mol + 9}\text{.072}\text{g/mol + 212}\text{.718}\text{g/mol}\\ =\text{389}\text{.94}\text{g/mol}\end{array}$

Therefore, molecular mass of ${\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{6}}$ is $\text{389}\text{.94 g/mol}$.