Chapter 2, Problem 2.30E

Interpretation Introduction

(a)

Interpretation:

The molecular weight of the given molecule in $\text{u}$ is to be stated.

Concept introduction:

The total number of protons and neutron which are present in an atom is collectively known as atomic mass of that atom. The total number of protons present in the nucleus of an atom is known as atomic number of that atom.

Answer to Problem 2.30E

The molecular weight of the given molecule, nitrogen dioxide in $\text{u}$ is $46.01\text{u}$.

Explanation of Solution

The given molecule is nitrogen dioxide, ${\text{NO}}_2$. This molecule contains one nitrogen atom and two oxygen atoms.

According to the periodic table, the atomic weight of one nitrogen atom is $14.01\text{u}$ and atomic weight of one oxygen atom is $16.00\text{u}$. Thus, the molecular weight of nitrogen dioxide molecule is calculated as,

$\text{Molecular}\text{weight}\text{of}{\text{NO}}_2=\left(\text{atomic}\text{weight}\text{of}\text{N}\right)+\left(2\times \text{atomic}\text{weight}\text{of}\text{O}\right)$

Substitute the value of atomic weight of one nitrogen atom and oxygen atom in the above expression.

$\begin{array}{rll}\text{Molecular}\text{weight}\text{of}{\text{NO}}_2& =& \left(14.01\text{u}\right)+\left(2\times 16.00\text{u}\right)\\ & =& 46.01\text{u}\end{array}$

Thus, the molecular weight of the given molecule, nitrogen dioxide in $\text{u}$ is $46.01\text{u}$.

Conclusion

The molecular weight of the given molecule, nitrogen dioxide in $\text{u}$ is $46.01\text{u}$.

Interpretation Introduction

(b)

Interpretation:

The molecular weight of the given molecule in $\text{u}$ is to be stated.

Concept introduction:

The total number of protons and neutron which are present in an atom is collectively known as atomic mass of that atom. The total number of protons present in the nucleus of an atom is known as atomic number of that atom.

Answer to Problem 2.30E

The molecular weight of the given molecule, ammonia in $\text{u}$ is $17.034\text{u}$.

Explanation of Solution

The given molecule is ammonia, ${\text{NH}}_{\text{3}}$. This molecule of ammonia contains one nitrogen atom and three hydrogen atoms.

According to the periodic table, the atomic weight of one nitrogen atom is $14.01\text{u}$ and the atomic weight of one hydrogen atom is $1.008\text{u}$. Thus, the molecular weight ammonia molecule is calculated as,

$\text{Molecular}\text{weight}\text{of}{\text{NH}}_{\text{3}}=\left(\text{atomic}\text{weight}\text{of}\text{N}\right)+\left(3\times \text{atomic}\text{weight}\text{of}\text{H}\right)$

Substitute the value of atomic weights of nitrogen and hydrogen atoms in the above expression.

$\begin{array}{rll}\text{Molecular}\text{weight}\text{of}{\text{NH}}_{\text{3}}& =& \left(14.01\text{u}\right)+\left(3\times 1.008\text{u}\right)\\ & =& 17.034\text{u}\end{array}$

Thus, the molecular weight of the given molecule, ammonia in $\text{u}$ is $17.034\text{u}$.

Conclusion

The molecular weight of the given molecule, ammonia in $\text{u}$ is $17.034\text{u}$.

Interpretation Introduction

(c)

Interpretation:

The molecular weight of the given molecule in $\text{u}$ is to be stated.

Concept introduction:

The total number of protons and neutron which are present in an atom is collectively known as atomic mass of that atom. The total number of protons present in the nucleus of an atom is known as atomic number of that atom.

Answer to Problem 2.30E

The molecular weight of the given molecule, glucose in $\text{u}$ is $180.156\text{u}$.

Explanation of Solution

The given molecule is glucose, ${\text{C}}_6{\text{H}}_{12}{\text{O}}_{\text{6}}$. This molecule of glucose contains six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.

According to the periodic table, the atomic weight of one hydrogen atom is $1.008\text{u}$, the atomic weight of one oxygen atom is $16.00\text{u}$ and the atomic weight of one carbon atom is $12.01\text{u}$. Thus, the molecular weight of glucose molecule is calculated as,

$\text{Molecular}\text{weight}\text{of}{\text{C}}_6{\text{H}}_{12}{\text{O}}_{\text{6}}=\left[\begin{array}{c}\left(6\times \text{atomic}\text{weight}\text{of}\text{C}\right)+\\ \left(12\times \text{atomic}\text{weight}\text{of}\text{H}\right)+\\ \left(6\times \text{atomic}\text{weight}\text{of}\text{O}\right)\end{array}\right]$

Substitute the value of atomic weights of hydrogen, oxygen and carbon atoms in the above expression.

$\begin{array}{rll}\text{Molecular}\text{weight}\text{of}{\text{C}}_6{\text{H}}_{12}{\text{O}}_{\text{6}}& =& \left[\begin{array}{c}\left(6\times 12.01\text{u}\right)+\left(12\times 1.008\text{u}\right)+\\ \left(6\times 16.00\text{u}\right)\end{array}\right]\\ =180.156\text{u}\end{array}$

Thus, the molecular weight of the given molecule, glucose in $\text{u}$ is $180.156\text{u}$.

Conclusion

The molecular weight of the given molecule, glucose in $\text{u}$ is $180.156\text{u}$.

Interpretation Introduction

(d)

Interpretation:

The molecular weight of the given molecule in $\text{u}$ is to be stated.

Concept introduction:

The total number of protons and neutron which are present in an atom is collectively known as atomic mass of that atom. The total number of protons present in the nucleus of an atom is known as atomic number of that atom.

Answer to Problem 2.30E

The molecular weight of the given molecule, ozone in $\text{u}$ is $4\text{8}\text{.00}\text{u}$.

Explanation of Solution

The given molecule is ozone, ${\text{O}}_3$. It is a triatomic molecule and possesses three atoms of oxygen.

According to the periodic table, the atomic weight of one oxygen atom is $16.00\text{u}$. Thus, the molecular weight of ozone molecule is calculated as,

$\text{Molecular}\text{weight}\text{of}{\text{O}}_{\text{3}}=3\times \left(\text{atomic}\text{weight}\text{of}\text{O}\right)$

Substitute the value of atomic weight of one oxygen atom in the above expression.

$\begin{array}{rll}\text{Molecular}\text{weight}\text{of}{\text{O}}_{\text{3}}& =& 3\times \left(16.00\text{u}\right)\\ & =& 4\text{8}\text{.00}\text{u}\end{array}$

Thus, the molecular weight of the given molecule, ozone in $\text{u}$ is $4\text{8}\text{.00}\text{u}$.

Conclusion

The molecular weight of the given molecule, ozone in $\text{u}$ is $4\text{8}\text{.00}\text{u}$.

Interpretation Introduction

(e)

Interpretation:

The molecular weight of the given molecule in $\text{u}$ is to be stated.

Concept introduction:

The total number of protons and neutron which are present in an atom is collectively known as atomic mass of that atom. The total number of protons present in the nucleus of an atom is known as atomic number of that atom.

Answer to Problem 2.30E

The molecular weight of the given molecule, ethylene glycol in $\text{u}$ is $62.068\text{u}$.

Explanation of Solution

The given molecule is ethylene glycol, ${\text{C}}_2{\text{H}}_6{\text{O}}_2$. This molecule of ethyl alcohol contains two carbon atoms, six hydrogen atoms and two oxygen atoms.

According to the periodic table, the atomic weight of one hydrogen atom is $1.008\text{u}$, the atomic weight of one oxygen atom is $16.00\text{u}$ and the atomic weight of one carbon atom is $12.01\text{u}$. Thus, the molecular weight of ethylene glycol molecule is calculated as,

$\text{Molecular}\text{weight}\text{of}{\text{C}}_2{\text{H}}_6{\text{O}}_2=\left[\begin{array}{c}\left(2\times \text{atomic}\text{weight}\text{of}\text{C}\right)+\\ \left(6\times \text{atomic}\text{weight}\text{of}\text{H}\right)+\\ \left(2\times \text{atomic}\text{weight}\text{of}\text{O}\right)\end{array}\right]$

Substitute the value of atomic weights of hydrogen, oxygen and carbon atoms in the above expression.

$\begin{array}{rll}\text{Molecular}\text{weight}\text{of}{\text{C}}_2{\text{H}}_6{\text{O}}_2& =& \left[\begin{array}{c}2\times \left(12.01\text{u}\right)+6\times \left(1.008\text{u}\right)+\\ 2\times \left(16.00\text{u}\right)\end{array}\right]\\ =62.068\text{u}\end{array}$

Thus, the molecular weight of the given molecule, ethylene glycol in $\text{u}$ is $62.068\text{u}$.

Conclusion

The molecular weight of the given molecule, ethylene glycol in $\text{u}$ is $62.068\text{u}$.