Chapter 19, Problem 1P

(a)

To determine

The charge and the mass of an ionized hydrogen atom to three significant digits.

Answer to Problem 1P

The charge of ionized hydrogen atom is $\underset{\_}{+1.60\times {10}^{-19}\text{C}}$, and the mass of ionized hydrogen atom is $\underset{\_}{1.67\times {10}^{-27}\text{kg}}$.

Explanation of Solution

The ionized hydrogen atom ${\text{H}}^{+}$ is the loss of on electron from the hydrogen atom.

Write the expression for charge on ionized hydrogen atom ${\text{H}}^{+}$.

    $q=ne$        (I)

Here, $q$ is the charge on ionized hydrogen atom, $n$ is the number of electron lost, and $e$ is the charge of single electron.

Write the expression for mass of ionized hydrogen atom.

    $m=\left(1.0079u\right)m_{\text{p}}-m_e$        (II)

Here, $m$ is the mass of the ionized hydrogen atom, $m_{\text{p}}$ is the mass of proton, and $m_{\text{e}}$ is the mass of electron.

Conclusion:

Substitute $-1$ for $n$, and $-1.60\times {10}^{-19}\text{C}$ for $e$ in equation (I), to find $q$.

    $\begin{array}{c}q=\left(-1\right)\left(-1.60\times {10}^{-19}\text{C}\right)\\ =1.60\times {10}^{-19}\text{C}\end{array}$

Substitute $1.660\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$, and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation (II), to find $m$

    $\begin{array}{c}m=\left(1.0079u\right)\left(1.660\times {10}^{-27}\text{kg}\right)-\left(9.11\times {10}^{-31}\text{kg}\right)\\ =1.67\times {10}^{-27}\text{kg}\end{array}$

Therefore, the charge of ionized hydrogen atom is $\underset{\_}{+1.60\times {10}^{-19}\text{C}}$, and the mass of ionized hydrogen atom is $\underset{\_}{1.67\times {10}^{-27}\text{kg}}$.

(b)

To determine

The charge and the mass of an ionized sodium atom to three significant digits.

Answer to Problem 1P

The charge of ionized sodium atom is $\underset{\_}{+1.60\times {10}^{-19}\text{C}}$, and the mass of ionized hydrogen atom is $\underset{\_}{3.82\times {10}^{-26}\text{kg}}$.

Explanation of Solution

Write the expression for mass of ionized sodium atom.

    $m=\left(22.99u\right)m_{\text{p}}-m_e$        (III)

Here, $m$ is the mass of the ionized sodium atom, $m_{\text{p}}$ is the mass of proton, and $m_{\text{e}}$ is the mass of electron.

Conclusion:

Substitute $-1$ for $n$, and $-1.60\times {10}^{-19}\text{C}$ for $e$ in equation (I), to find $q$.

    $\begin{array}{c}q=\left(-1\right)\left(-1.60\times {10}^{-19}\text{C}\right)\\ =1.60\times {10}^{-19}\text{C}\end{array}$

Substitute $1.660\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$, and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation (III), to find $m$

    $\begin{array}{c}m=\left(22.99u\right)\left(1.660\times {10}^{-27}\text{kg}\right)-\left(9.11\times {10}^{-31}\text{kg}\right)\\ =3.82\times {10}^{-26}\text{kg}\end{array}$

Therefore, the charge of ionized sodium atom is $\underset{\_}{+1.60\times {10}^{-19}\text{C}}$, and the mass of ionized hydrogen atom is $\underset{\_}{3.82\times {10}^{-26}\text{kg}}$.

(c)

To determine

The charge and the mass of a chloride ion ${\text{cl}}^{-}$ to three significant digits.

Answer to Problem 1P

The charge of a chloride ion ${\text{cl}}^{-}$ is $\underset{\_}{1.60\times {10}^{-19}\text{C}}$, the mass of a chloride ion ${\text{cl}}^{-}$ is $\underset{\_}{5.89\times {10}^{-26}\text{kg}}$.

Explanation of Solution

Write the expression for mass of a chloride ion ${\text{cl}}^{-}$.

    $m=\left(35.453u\right)m_{\text{p}}-m_e$        (IV)

Here, $m$ is the mass of a chloride ion ${\text{cl}}^{-}$, $m_{\text{p}}$ is the mass of proton, and $m_{\text{e}}$ is the mass of electron.

Conclusion:

Substitute $-1$ for $n$, and $-1.60\times {10}^{-19}\text{C}$ for $e$ in equation (I), to find $q$.

    $\begin{array}{c}q=\left(-1\right)\left(-1.60\times {10}^{-19}\text{C}\right)\\ =1.60\times {10}^{-19}\text{C}\end{array}$

Substitute $1.660\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$, and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation (IV), to find $m$

    $\begin{array}{c}m=\left(35.453u\right)\left(1.660\times {10}^{-27}\text{kg}\right)-\left(9.11\times {10}^{-31}\text{kg}\right)\\ =5.89\times {10}^{-26}\text{kg}\end{array}$

Therefore, The charge of a chloride ion ${\text{cl}}^{-}$ is $\underset{\_}{1.60\times {10}^{-19}\text{C}}$, the mass of a chloride ion ${\text{cl}}^{-}$ is $\underset{\_}{5.89\times {10}^{-26}\text{kg}}$.

(d)

To determine

The charge and the mass of a doubly ionized calcium atom ${\text{Ca}}^{++}$ to three significant digits.

Answer to Problem 1P

The charge of a doubly ionized calcium atom ${\text{Ca}}^{++}$ is $\underset{\_}{3.20\times {10}^{-19}\text{C}}$, and the mass of a doubly ionized calcium atom ${\text{Ca}}^{++}$ is $\underset{\_}{6.65\times {10}^{-26}\text{kg}}$.

Explanation of Solution

Write the expression for a doubly ionized calcium atom ${\text{Ca}}^{++}$.

    $m=\left(40.078u\right)m_{\text{p}}-n{m}_e$        (V)

Here, $m$ is the mass of a doubly ionized calcium atom ${\text{Ca}}^{++}$, $m_{\text{p}}$ is the mass of proton, $n$ is the number of electron lost, and $m_{\text{e}}$ is the mass of electron.

Conclusion:

Substitute $-2$ for $n$, and $-1.60\times {10}^{-19}\text{C}$ for $e$ in equation (I), to find $q$.

    $\begin{array}{c}q=\left(-2\right)\left(-1.60\times {10}^{-19}\text{C}\right)\\ =+3.20\times {10}^{-19}\text{C}\end{array}$

Substitute $1.660\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$, $-2$ for $n$, and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation (V), to find $m$

    $\begin{array}{c}m=\left(40.078u\right)\left(1.660\times {10}^{-27}\text{kg}\right)-2\left(9.11\times {10}^{-31}\text{kg}\right)\\ =6.65\times {10}^{-26}\text{kg}\end{array}$

Therefore, the charge of a doubly ionized calcium atom ${\text{Ca}}^{++}$ is $\underset{\_}{3.20\times {10}^{-19}\text{C}}$, and the mass of a doubly ionized calcium atom ${\text{Ca}}^{++}$ is $\underset{\_}{6.65\times {10}^{-26}\text{kg}}$.

(e)

To determine

The charge and the mass of a ${\text{N}}^{3-}$ ion to three significant digits.

Answer to Problem 1P

The charge of a ${\text{N}}^{3-}$ ion is $\underset{\_}{-4.80\times {10}^{-19}\text{C}}$, and the mass of a ${\text{N}}^{3-}$ ion is $\underset{\_}{2.33\times {10}^{-26}\text{kg}}$.

Explanation of Solution

Write the expression for a ${\text{N}}^{3-}$ ion

    $m=\left(14.007u\right)m_{\text{p}}+n{m}_e$        (VI)

Here, $m$ is the mass of a ${\text{N}}^{3-}$ ion, $m_{\text{p}}$ is the mass of proton, $n$ is the number of electron gained, and $m_{\text{e}}$ is the mass of electron.

Conclusion:

Substitute $3$ for $n$, and $-1.60\times {10}^{-19}\text{C}$ for $e$ in equation (I), to find $q$.

    $\begin{array}{c}q=\left(3\right)\left(-1.60\times {10}^{-19}\text{C}\right)\\ =-4.80\times {10}^{-19}\text{C}\end{array}$

Substitute $1.660\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$, $3$ for $n$, and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation (VI), to find $m$

    $\begin{array}{c}m=\left(14.007u\right)\left(1.660\times {10}^{-27}\text{kg}\right)+3\left(9.11\times {10}^{-31}\text{kg}\right)\\ =2.33\times {10}^{-26}\text{kg}\end{array}$

Therefore, the charge of a ${\text{N}}^{3-}$ ion is $\underset{\_}{-4.80\times {10}^{-19}\text{C}}$, and the mass of a ${\text{N}}^{3-}$ ion is $\underset{\_}{2.33\times {10}^{-26}\text{kg}}$.

(f)

To determine

The charge and the mass of a ${\text{N}}^{4+}$ ion to three significant digits.

Answer to Problem 1P

The charge of a ${\text{N}}^{4+}$ ion is $\underset{\_}{6.40\times {10}^{-19}\text{C}}$, and the mass of a ${\text{N}}^{4+}$ ion is $\underset{\_}{2.32\times {10}^{-26}\text{kg}}$.

Explanation of Solution

Write the expression for a ${\text{N}}^{4+}$ ion

    $m=\left(14.007u\right)m_{\text{p}}-n{m}_e$        (VII)

Here, $m$ is the mass of a ${\text{N}}^{4+}$ ion, $m_{\text{p}}$ is the mass of proton, $n$ is the number of electron lost, and $m_{\text{e}}$ is the mass of electron.

Conclusion:

Substitute $-4$ for $n$, and $-1.60\times {10}^{-19}\text{C}$ for $e$ in equation (I), to find $q$.

    $\begin{array}{c}q=\left(-4\right)\left(-1.60\times {10}^{-19}\text{C}\right)\\ =+6.40\times {10}^{-19}\text{C}\end{array}$

Substitute $1.660\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$, $-4$ for $n$, and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation (VII), to find $m$

    $\begin{array}{c}m=\left(14.007u\right)\left(1.660\times {10}^{-27}\text{kg}\right)-4\left(9.11\times {10}^{-31}\text{kg}\right)\\ =2.32\times {10}^{-26}\text{kg}\end{array}$

Therefore, The charge of a ${\text{N}}^{4+}$ ion is $\underset{\_}{6.40\times {10}^{-19}\text{C}}$, and the mass of a ${\text{N}}^{4+}$ ion is $\underset{\_}{2.32\times {10}^{-26}\text{kg}}$.

(g)

To determine

The charge and the mass of the nucleus of nitrogen atom to three significant digits.

Answer to Problem 1P

The charge of the nucleus of nitrogen atom is $\underset{\_}{1.12\times {10}^{-18}\text{C}}$, and the mass of the nucleus of nitrogen atom is $\underset{\_}{2.32\times {10}^{-26}\text{kg}}$.

Explanation of Solution

Consider nitrogen atom as seven times ionized nitrogen atom ${\text{N}}^{7+}$.

Write the expression for a ${\text{N}}^{7+}$ ion

    $m=\left(14.007u\right)m_{\text{p}}-n{m}_e$        (VIII)

Here, $m$ is the mass of a ionized nitrogen atom ${\text{N}}^{7+}$, $m_{\text{p}}$ is the mass of proton, $n$ is the number of electron lost, and $m_{\text{e}}$ is the mass of electron.

Conclusion:

Substitute $-7$ for $n$, and $-1.60\times {10}^{-19}\text{C}$ for $e$ in equation (I), to find $q$.

    $\begin{array}{c}q=\left(-7\right)\left(-1.60\times {10}^{-19}\text{C}\right)\\ =+1.12\times {10}^{-18}\text{C}\end{array}$

Substitute $1.660\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$, $-7$ for $n$, and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation (VIII), to find $m$

    $\begin{array}{c}m=\left(14.007u\right)\left(1.660\times {10}^{-27}\text{kg}\right)-7\left(9.11\times {10}^{-31}\text{kg}\right)\\ =2.32\times {10}^{-26}\text{kg}\end{array}$

Therefore, the charge of the nucleus of nitrogen atom is $\underset{\_}{1.12\times {10}^{-18}\text{C}}$, and the mass of the nucleus of nitrogen atom is $\underset{\_}{2.32\times {10}^{-26}\text{kg}}$.

(h)

To determine

The charge and the mass of molecular ion ${\text{H}}_2{\text{O}}^{-}$ to three significant digits.

Answer to Problem 1P

The charge of molecular ion ${\text{H}}_2{\text{O}}^{-}$ is $\underset{\_}{-1.60\times {10}^{-19}\text{C}}$, and the mass of molecular ion ${\text{H}}_2{\text{O}}^{-}$ is $\underset{\_}{2.99\times {10}^{-26}\text{kg}}$.

Explanation of Solution

Write the expression for charge of ${\text{H}}_2{\text{O}}^{-}$ molecular ion

    $m=\left[2\left(1.0079u+15.999\right)\right]m_{\text{p}}+n{m}_e$        (IX)

Here, $m$ is the mass of molecular ion ${\text{H}}_2{\text{O}}^{-}$, $m_{\text{p}}$ is the mass of proton, $n$ is the number of electron lost, and $m_{\text{e}}$ is the mass of electron.

Conclusion:

Substitute $1$ for $n$, and $-1.60\times {10}^{-19}\text{C}$ for $e$ in equation (I), to find $q$.

    $\begin{array}{c}q=\left(1\right)\left(-1.60\times {10}^{-19}\text{C}\right)\\ =-1.60\times {10}^{-19}\text{C}\end{array}$

Substitute $1.660\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$, $1$ for $n$, and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation ((IX), to find $m$

    $\begin{array}{c}m=\left[2\left(1.0079u+15.999\right)\right]\left(1.660\times {10}^{-27}\text{kg}\right)+1\left(9.11\times {10}^{-31}\text{kg}\right)\\ =2.99\times {10}^{-26}\text{kg}\end{array}$

Therefore, The charge of molecular ion ${\text{H}}_2{\text{O}}^{-}$ is $\underset{\_}{-1.60\times {10}^{-19}\text{C}}$, and the mass of molecular ion ${\text{H}}_2{\text{O}}^{-}$ is $\underset{\_}{2.99\times {10}^{-26}\text{kg}}$.