Chapter 5, Problem 5.69QP

(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 5.69QP

Molecular mass of ${\text{CH}}_{\text{3}}\text{Cl}\text{=}\text{50}\text{.4874}\text{amu}$

Explanation of Solution

To calculate: Molecular mass of ${\text{CH}}_{\text{3}}\text{Cl}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{CH}}_{\text{3}}\text{Cl}\text{=}\text{1}\left(\text{mass}\text{of}\text{C}\text{in}\text{amu}\right)\text{+}\text{3}\left(\text{mass}\text{of}\text{H}\text{in}\text{amu}\right)\text{+}\text{1}\left(\text{mass}\text{of}\text{Cl}\text{in}\text{amu}\right)\\ \text{=}\text{1}\left(\text{12}\text{.0107}\text{amu}\right)\text{+}\text{3}\left(\text{1}\text{.0079}\text{amu}\right)\text{+}\text{1}\left(\text{35}\text{.453}\text{amu}\right)\\ \text{=}\left(\text{12}\text{.0107}\text{+}\text{3}\text{.0237}\text{+}\text{35}\text{.453}\right)\text{amu}\\ \text{=}\text{50}\text{.4874}\text{amu}\end{array}$

(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 5.69QP

Molecular mass of ${\text{N}}_{\text{2}}\text{O}\text{=}\text{44}\text{.0128}\text{amu}$

Explanation of Solution

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

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{N}}_{\text{2}}\text{O}\text{=}\text{2}\left(\text{mass}\text{of}\text{N}\text{in}\text{amu}\right)\text{+}\text{1}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{2}\left(\text{14}\text{.0067}\text{amu}\right)\text{+}\text{1}\left(\text{15}\text{.9994}\text{amu}\right)\\ \text{=}\left(\text{28}\text{.0134}\text{+}\text{15}\text{.9994}\right)\text{amu}\\ \text{=}\text{44}\text{.0128}\text{amu}\end{array}$

(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 5.69QP

Molecular mass of ${\text{SO}}_{\text{2}}\text{=}\text{64}\text{.0638}\text{amu}$

Explanation of Solution

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

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{SO}}_{\text{2}}\text{=1}\left(\text{mass}\text{of}\text{S}\text{in}\text{amu}\right)\text{+}\text{2}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{1}\left(\text{32}\text{.065}\text{amu}\right)\text{+}\text{2}\left(\text{15}\text{.9994}\text{amu}\right)\\ \text{=}\left(\text{32}\text{.065}\text{+}\text{31}\text{.9988}\right)\text{amu}\\ \text{=}\text{64}\text{.0638}\text{amu}\end{array}$

(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 5.69QP

Molecular mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}\text{=}\text{84}\text{.159}\text{amu}$

Explanation of Solution

To calculate: Molecular mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}\text{=}\text{6}\left(\text{mass}\text{of}\text{C}\text{in}\text{amu}\right)\text{+}\text{12}\left(\text{mass}\text{of}\text{H}\text{in}\text{amu}\right)\\ \text{=}\text{6}\left(\text{12}\text{.0107}\text{amu}\right)\text{+}\text{12}\left(\text{1}\text{.0079}\text{amu}\right)\\ \text{=}\left(\text{72}\text{.0642}\text{+}\text{12}\text{.0948}\right)\text{amu}\\ \text{=}\text{84}\text{.159}\text{amu}\end{array}$

(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 5.69QP

Formula mass of $\text{NaBr}\text{=}\text{102}\text{.8938}\text{amu}$

Explanation of Solution

To calculate: Formula mass of $\text{NaBr}$

$\begin{array}{l}\text{Formula}\text{mass}\text{of}\text{NaBr}\text{=}\text{1}\left(\text{mass}\text{of}\text{Na}\text{in}\text{amu}\right)\text{+}\text{1}\left(\text{mass}\text{of}\text{Br}\text{in}\text{amu}\right)\\ \text{=}\text{1}\left(\text{22}\text{.9898}\text{amu}\right)\text{+}\text{1}\left(\text{79}\text{.904}\text{amu}\right)\\ \text{=}\left(\text{22}\text{.9898}\text{+}\text{79}\text{.904}\right)\text{amu}\\ \text{=}\text{102}\text{.8938}\text{amu}\end{array}$

(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 5.69QP

Formula mass of ${\text{Cs}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\text{361}\text{.8736}\text{amu}$

Explanation of Solution

To calculate: Formula mass of ${\text{Cs}}_{\text{2}}{\text{SO}}_{\text{4}}$

$\begin{array}{l}\text{Formula}\text{mass}\text{of}{\text{Cs}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\text{2}\left(\text{mass}\text{of}\text{Cs}\text{in}\text{amu}\right)\text{+}\text{1}\left(\text{mass}\text{of}\text{S}\text{in}\text{amu}\right)\text{+}\text{4}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{2}\left(\text{132}\text{.9055}\text{amu}\right)\text{+}\text{1}\left(\text{32}\text{.065}\text{amu}\right)\text{+}\text{4}\left(\text{15}\text{.9994}\text{amu}\right)\\ \text{=}\left(\text{265}\text{.811}\text{+}\text{32}\text{.065}\text{+}\text{63}\text{.9976}\right)\text{amu}\\ \text{=}\text{361}\text{.8736}\text{amu}\end{array}$

(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 5.69QP

Formula mass of ${\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\text{=}\text{601}\text{.9238}\text{amu}$

Explanation of Solution

To calculate: Formula mass of ${\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$

$\begin{array}{l}\text{Formula}\text{mass}\text{of}{\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\text{=}\text{3}\left(\text{mass}\text{of}\text{Ba}\text{in}\text{amu}\right)\text{+}\text{2}\left(\text{1}\left(\text{mass}\text{of}\text{P}\text{in}\text{amu}\right)\text{+}\text{4}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\right)\\ \text{=}\text{3}\left(\text{137}\text{.327}\text{amu}\right)\text{+2}\left(\text{1}\left(\text{30}\text{.9738}\text{amu}\right)\text{+}\text{4}\left(\text{15}\text{.9994}\text{amu}\right)\right)\\ \text{=}\left(\text{411}\text{.981}\text{+}\text{2}\left(\text{30}\text{.9738}\text{+63}\text{.9976}\right)\right)\text{amu}\\ \text{=}\left(\text{411}\text{.981}\text{+}\text{2}\left(\text{94}\text{.9714}\right)\right)\text{amu}\\ \text{=}\left(\text{411}\text{.981}\text{+}\text{189}\text{.9428}\right)\text{amu}\\ \text{=601}\text{.9238}\text{amu}\end{array}$