Chapter 5, Problem 5.68QP

(a)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

Answer to Problem 5.68QP

Molecular mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{=}\text{66}\text{.949}\text{amu}$

Explanation of Solution

To calculate: Molecular mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$

Molecular formula is given as ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{=}\text{2}\left(\text{mass}\text{of}\text{Li}\text{in}\text{amu}\right)\text{+}\text{1}\left(\text{mass}\text{of}\text{C}\text{in}\text{amu}\right)\text{+}\text{3}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{1}\left(\text{6}\text{.941}\text{amu}\right)\text{+}\text{1}\left(\text{12}\text{.0107}\text{amu}\right)\text{+}\text{3}\left(\text{15}\text{.9994}\right)\\ \text{=}\text{66}\text{.949amu}\end{array}$ The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ was found to be $\text{66}\text{.949amu}$.

(b)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

Answer to Problem 5.68QP

Molecular mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{=}\text{54}\text{.0902}\text{amu}$.

Explanation of Solution

To calculate: Molecular mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$

Molecular formula is given as ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{=}\text{2}\left(\text{mass}\text{of}\text{C}\text{in}\text{amu}\right)\text{+}\text{6}\left(\text{mass}\text{of}\text{H}\text{in}\text{amu}\right)\\ \text{=}\text{4}\left(\text{12}\text{.0107}\text{amu}\right)\text{+}\text{6}\left(\text{1}\text{.0079}\text{amu}\right)\\ \text{=}\text{54}\text{.0902}\text{amu}\end{array}$

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ was found to be $\text{54}\text{.0902}\text{amu}$.

(C)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

Answer to Problem 5.68QP

Molecular mass of ${\text{NF}}_{\text{2}}\text{=}\text{51}\text{.987}\text{amu}$

Explanation of Solution

To calculate: Molecular mass of ${\text{NF}}_{\text{2}}$

Molecular formula is given as ${\text{NF}}_{\text{2}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{NF}}_{\text{2}}\text{=}\text{1}\left(\text{mass}\text{of}\text{N}\text{in}\text{amu}\right)\text{+}\text{2}\left(\text{mass}\text{of}\text{F}\text{in}\text{amu}\right)\\ \text{=}\text{1}\left(\text{14}\text{.007}\text{amu}\right)\text{+}\text{2}\left(\text{18}\text{.99}\text{amu}\right)\\ \text{=}\text{51}\text{.987}\text{amu}\end{array}$

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{NF}}_{\text{2}}$ was found to be $\text{51}\text{.987}\text{amu}$.

(d)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

Answer to Problem 5.68QP

Formula mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{=}\text{101}\text{.961}\text{amu}$

Explanation of Solution

To calculate: Formula mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

Molecular formula is given as ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

$\begin{array}{l}\text{Formula}\text{mass}\text{of}{\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}\text{=}\text{2}\left(\text{mass}\text{of}\text{Al}\text{in}\text{amu}\right)\text{+}\text{3}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{2}\left(\text{26}\text{.9815}\text{amu}\right)\text{+}\text{3}\left(\text{15}\text{.9994}\text{amu}\right)\\ \text{=}\text{101}\text{.961}\text{amu}\end{array}$

The formula mass of the given substance was calculated using atomic weights of each elements present in its formula unit.  The molecular mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ was found to be $\text{101}\text{.961}\text{amu}$.

(e)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

Answer to Problem 5.68QP

Formula mass of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\text{=}\text{241}\text{.8597}\text{amu}$

Explanation of Solution

To calculate: Formula mass of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$

Molecular formula is given as $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$

$\begin{array}{l}\text{Formula}\text{mass}\text{of}\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\text{=}\text{1}\left(\text{mass}\text{of}\text{Fe}\right)\text{+}\text{3}\left(\text{1}\left(\text{mass}\text{of}\text{N}\text{in}\text{amu}\right)\text{+}\text{3}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\right)\\ \text{=}\text{55}\text{.845}\text{amu}\text{+3}\left(\text{14}\text{.0067}\text{+}\text{3}\left(\text{15}\text{.9994}\right)\right)\text{amu}\\ \text{=}\text{55}\text{.845}\text{amu}\text{+}\text{3}\left(\text{14}\text{.0067}\text{+}\text{47}\text{.9982}\right)\text{amu}\\ \text{=}\text{241}\text{.8597}\text{amu}\end{array}$

The formula mass of the given substance was calculated using atomic weights of each elements present in its formula unit.  The molecular mass of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ was found to be $\text{241}\text{.8597}\text{amu}$.

(f)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

Answer to Problem 5.68QP

Molecular mass of ${\text{PCl}}_{\text{5}}\text{=}\text{208}\text{.2388}\text{amu}$

To calculate: Molecular mass of ${\text{PCl}}_{\text{5}}$

Molecular formula is given as ${\text{PCl}}_{\text{5}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{PCl}}_{\text{5}}\text{=}\text{1}\left(\text{mass}\text{of}\text{P}\text{in}\text{amu}\right)\text{+}\text{5}\left(\text{mass}\text{of}\text{Cl}\text{in}\text{amu}\right)\\ \text{=}\text{1}\left(\text{30}\text{.9738}\text{amu}\right)\text{+}\text{5}\left(\text{35}\text{.453}\text{amu}\right)\\ \text{=}\text{208}\text{.2388}\text{amu}\end{array}$

Explanation of Solution

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{PCl}}_{\text{5}}$ was found to be $\text{208}\text{.2388}\text{amu}$.

(g)

Interpretation Introduction

Interpretation:

The molecular mass/formula mass of the given substances has to be calculated.

Concept introduction:

The sum of the atomic masses of all the atoms present in that molecule will give an overall mass of that molecule which is termed as Molecular mass.  The sum of the atomic masses of individual atoms present in a formula unit will be termed as formula mass, it may or may not be a formula of molecule.

Answer to Problem 5.68QP

Molecular mass of ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}\text{=}\text{100}\text{.9284}\text{amu}$.

Explanation of Solution

To calculate: Molecular mass of ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}$

Molecular formula is given as ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}\text{=}\text{3}\left(\text{mass}\text{of}\text{Mg}\text{in}\text{amu}\right)\text{+}\text{2}\left(\text{mass}\text{of}\text{N}\text{in}\text{amu}\right)\\ \text{=}\text{3}\left(\text{24}\text{.305}\right)\text{+}\text{2}\left(\text{14}\text{.0067}\right)\\ \text{=}\text{100}\text{.9284}\text{amu}\\ \end{array}$

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}$ was found to be $\text{100}\text{.9284}\text{amu}$.