Chapter 5, Problem 5.67QP

(a)

Interpretation Introduction

Interpretation: The molecular 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.

Some atomic weight of atoms are,

$\begin{array}{l}\text{C }\text{= 12}\text{.0107 amu}\\ \text{H }\text{= 1}\text{.0079 amu}\\ \text{Cl }\text{= 35}\text{.453 amu}\\ \text{N }\text{= 14}\text{.0067 amu}\\ \text{O }\text{= 15}\text{.9994 amu}\\ \text{S }\text{= 32}\text{.065 amu}\end{array}$

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

Answer to Problem 5.67QP

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

Explanation of Solution

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

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula. The molecular mass of ${\text{CH}}_{\text{3}}\text{Cl}$ was found to be $\text{50}\text{.4874}\text{amu}$.

(b)

Interpretation Introduction

Interpretation: The molecular 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.

Some atomic weight of atoms are,

$\begin{array}{l}\text{C }\text{= 12}\text{.0107 amu}\\ \text{H }\text{= 1}\text{.0079 amu}\\ \text{Cl }\text{= 35}\text{.453 amu}\\ \text{N }\text{= 14}\text{.0067 amu}\\ \text{O }\text{= 15}\text{.9994 amu}\\ \text{S }\text{= 32}\text{.065 amu}\end{array}$

To calculate: Molecular mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$

Answer to Problem 5.67QP

Molecular mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{=}\text{92}\text{.02894}\text{amu}$

Explanation of Solution

Molecular formula is given as ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{=}\text{2}\left(\text{mass}\text{of}\text{N}\text{in}\text{amu}\right)\text{+}\text{4}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{2}\left(\text{14}\text{.0067}\text{amu}\right)\text{+}\text{4}\left(\text{15}\text{.9994}\text{amu}\right)\\ \text{=}\left(\text{28}\text{.0134}\text{+}\text{63}\text{.9976}\right)\text{amu}\\ \text{=}\text{92}\text{.02894}\text{amu}\end{array}$

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ was found to be $\text{92}\text{.02894}\text{amu}$.

(C)

Interpretation Introduction

Interpretation: The molecular 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.

Some atomic weight of atoms are,

$\begin{array}{l}\text{C }\text{= 12}\text{.0107 amu}\\ \text{H }\text{= 1}\text{.0079 amu}\\ \text{Cl }\text{= 35}\text{.453 amu}\\ \text{N }\text{= 14}\text{.0067 amu}\\ \text{O }\text{= 15}\text{.9994 amu}\\ \text{S }\text{= 32}\text{.065 amu}\end{array}$

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

Answer to Problem 5.67QP

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

Explanation of Solution

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

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

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{SO}}_{\text{2}}$ was found to be $\text{64}\text{.0638}\text{amu}$.

(d)

Interpretation Introduction

Interpretation: The molecular 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.

Some atomic weight of atoms are,

$\begin{array}{l}\text{C }\text{= 12}\text{.0107 amu}\\ \text{H }\text{= 1}\text{.0079 amu}\\ \text{Cl }\text{= 35}\text{.453 amu}\\ \text{N }\text{= 14}\text{.0067 amu}\\ \text{O }\text{= 15}\text{.9994 amu}\\ \text{S }\text{= 32}\text{.065 amu}\end{array}$

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

Answer to Problem 5.67QP

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

Explanation of Solution

Molecular formula is given as ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$

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

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ was found to be $\text{84}\text{.159}\text{amu}$

(e)

Interpretation Introduction

Interpretation: The molecular 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.

Some atomic weight of atoms are,

$\begin{array}{l}\text{C }\text{= 12}\text{.0107 amu}\\ \text{H }\text{= 1}\text{.0079 amu}\\ \text{Cl }\text{= 35}\text{.453 amu}\\ \text{N }\text{= 14}\text{.0067 amu}\\ \text{O }\text{= 15}\text{.9994 amu}\\ \text{S }\text{= 32}\text{.065 amu}\end{array}$

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

Answer to Problem 5.67QP

Molecular mass of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{=}\text{34}\text{.0146}\text{amu}$

Explanation of Solution

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

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{=}\text{2}\left(\text{mass}\text{of}\text{H}\text{in}\text{amu}\right)\text{+}\text{2}\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{2}\left(\text{1}\text{.0079}\text{amu}\right)\text{+}\text{2}\left(\text{15}\text{.9994}\text{amu}\right)\\ \text{=}\left(\text{2}\text{.0158}\text{+}\text{31}\text{.9988}\right)\text{amu}\\ \text{=}\text{34}\text{.0146}\text{amu}\end{array}$

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ was found to be $\text{34}\text{.0146}\text{amu}$.

(f)

Interpretation Introduction

Interpretation: The molecular 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.

Some atomic weight of atoms are,

$\begin{array}{l}\text{C }\text{= 12}\text{.0107 amu}\\ \text{H }\text{= 1}\text{.0079 amu}\\ \text{Cl }\text{= 35}\text{.453 amu}\\ \text{N }\text{= 14}\text{.0067 amu}\\ \text{O }\text{= 15}\text{.9994 amu}\\ \text{S }\text{= 32}\text{.065 amu}\end{array}$

Answer to Problem 5.67QP

Molecular mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{=}\text{342}\text{.3 amu}$

To calculate: Molecular mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$

Explanation of Solution

Molecular formula is given as ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\text{=}\text{12}\left(\text{mass}\text{of}\text{C}\text{in}\text{amu}\right)\text{+}\text{22}\left(\text{mass}\text{of}\text{H}\text{in}\text{amu}\right)\text{+}11\left(\text{mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{12}\left(\text{12}\text{.0107}\text{amu}\right)\text{+}\text{22}\left(\text{1}\text{.0079}\right)\text{+}11\left(\text{15}\text{.9994}\text{amu}\right)\\ \text{=}\left(\text{144}\text{.1284}\text{+}\text{22}\text{.1738}\text{+}175.\text{9934}\right)\text{amu}\\ \text{=}\text{342}\text{.3 amu}\end{array}$

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ was found to be $\text{342}\text{.3 amu}$.

(g)

Interpretation Introduction

Interpretation: The molecular 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.

Some atomic weight of atoms are,

$\begin{array}{l}\text{C }\text{= 12}\text{.0107 amu}\\ \text{H }\text{= 1}\text{.0079 amu}\\ \text{Cl }\text{= 35}\text{.453 amu}\\ \text{N }\text{= 14}\text{.0067 amu}\\ \text{O }\text{= 15}\text{.9994 amu}\\ \text{S }\text{= 32}\text{.065 amu}\end{array}$

Answer to Problem 5.67QP

Molecular mass of ${\text{NH}}_{\text{3}}\text{=}\text{17}\text{.0304}\text{amu}$

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

Explanation of Solution

Molecular formula is given as ${\text{NH}}_{\text{3}}$

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{NH}}_{\text{3}}\text{=}\text{1}\left(\text{mass}\text{of}\text{N}\text{in}\text{amu}\right)\text{+}\text{3}\left(\text{mass}\text{of}\text{H}\text{in}\text{amu}\right)\\ \text{=}\text{1}\left(\text{14}\text{.0067}\text{amu}\right)\text{+}\text{3}\left(\text{1}\text{.0079}\text{amu}\right)\\ \text{=}\left(\text{14}\text{.0067}\text{+}\text{3}\text{.0237}\right)\text{amu}\\ \text{=}\text{17}\text{.0304}\text{amu}\end{array}$

The molecular mass of the given substance was calculated using atomic weights of each elements present in its molecular formula.  The molecular mass of ${\text{NH}}_{\text{3}}$ was found to be $\text{17}\text{.0304}\text{amu}$.