Chapter 3, Problem 3.9E

(a)

Interpretation Introduction

Interpretation:

Molecular formula for unknown compound has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  $PV=nRT$        (1)

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 3.9E

Mass of ${\text{NaN}}_3$ required to fill air bag with nitrogen is $138\text{ g}$.

Explanation of Solution

Conversion factor to convert $25°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =25°\text{C}+273.15\\ =298.15\text{K}\end{array}$

Rearrange equation (1) to determine value of $n$ for unknown gas as follows:

  $n=\frac{PV}{RT}$        (2)

Substitute $1.81\text{ atm}$ for $P$, $200\text{ mL}$ for $V$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $298.15\text{K}$ for $T$ in equation (2).

  $\begin{array}{c}n=\left(\frac{\left(1.81\text{ atm}\right)\left(200\text{ mL}\right)}{\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298.15\text{K}\right)}\right)\left(\frac{{10}^{-3}\text{ L }}{1\text{ mL}}\right)\\ =\text{1}\text{.48}\times {\text{10}}^{-2}\text{ mol}\end{array}$

Number of moles of nitrogen present in $0.414\text{ g}$ can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\frac{0.414\text{ g}}{14\text{ g}\cdot {\text{mol}}^{-1}}\\ =0.02957\text{ mol}\end{array}$

Number of moles of hydrogen gas present in $0.0591\text{ g}$ can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\frac{0.0591\text{ g}}{1\text{ g}\cdot {\text{mol}}^{-1}}\\ =0.0591\text{ mol}\end{array}$

Ratio of nitrogen to hydrogen can be calculated as follows:

  $\begin{array}{c}\text{Ratio}\left(\text{N:H}\right)=\frac{0.02957\text{ mol}}{0.0591\text{ mol}}\\ =\frac{1}{2}\end{array}$

Therefore formula unit is ${\text{NH}}_2$. Molar mass of formula unit can be calculated as follows:

  $\begin{array}{c}\text{Molar mass}=1\left(14\text{ g}\cdot {\text{mol}}^{-1}\right)+2\left(1\text{ g}\cdot {\text{mol}}^{-1}\right)\\ =16\text{ g}\cdot {\text{mol}}^{-1}\end{array}$

Total molar mass of gas can be calculated as follows:

  $\begin{array}{c}\text{Molar mass}=\frac{\text{Mass}}{\text{Moles}}\\ =\frac{0.473\text{ g}}{\text{1}\text{.48}\times {\text{10}}^{-2}\text{ mol}}\\ \simeq 32\text{ g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, number of formula unit of ${\text{NH}}_2$ present in unknown gas is 2. Hence molecular formula for unknown gas is ${\text{N}}_2{\text{H}}_4$.

(b)

Interpretation Introduction

Interpretation:

Lewis structure of ${\text{N}}_2{\text{H}}_4$ has to be determined.

Concept Introduction:

Lewis structure represents covalent bonds and describes valence electrons configuration of atoms. The covalent bonds are depicted by lines and unshared electron pairs by pairs of dots. The sequence to write Lewis structure of some molecule is given as follows:

• The central atom is identified and various other atoms are arranged around it. This central atom so chosen is often the least electronegative.
• Total valence electrons are estimated for each atom.
• single bond is first placed between each atom pair.
• The electrons left can be allocated as unshared electron pairs or as multiple bonds around symbol of element to satisfy the octet (or duplet) for each atom.
• Add charge on overall structure in case of polyatomic cation or anion.

Explanation of Solution

The molecule ${\text{N}}_2{\text{H}}_4$ consists of two $\text{N}$ and four $\text{H}$ atoms.

Electronic configuration of $\text{N}$ is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^3$. Thus, it possesses 5 valence electrons.

Electronic configuration of $\text{H}$ is ${\text{1s}}^1$. Thus, it possesses 1 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{N}}_2{\text{H}}_4$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=2\left(5\right)+4\left(1\right)\\ =14\left(7\text{ pairs}\right)\end{array}$

Skeleton structure ${\text{N}}_2{\text{H}}_4$ has four single bonds between hydrogen and nitrogen that comprise 8 electrons. Also, one single bond between nitrogen and nitrogen that comprise 2 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=14-10\\ =4\left(\text{2 pairs}\right)\end{array}$

Hence, 4 electrons are allocated as 2 lone pairs on each nitrogen atom. The Lewis structure of ${\text{N}}_2{\text{H}}_4$ is as follows:

(c)

Interpretation Introduction

Interpretation:

Amount of nitrogen gas effused through small hole in $\text{25}\text{.0 min}$ has to be determined.

Concept Introduction:

Effusion is explained as movement of the gas molecule through small hole. Diffusion can be explained as one gas molecule mix with another gas molecule by random motion.

Answer to Problem 3.9E

Amount of nitrogen gas effused through small hole in $\text{25}\text{.0 min}$ is $0.42\text{ mmol}$.

Explanation of Solution

The expression for molar masses of ${\text{N}}_2{\text{H}}_4$, ${\text{NH}}_3$ and their effusion rate is as follows:

  $\frac{\frac{{\text{Amount of N}}_2{\text{H}}_4\text{ passed}}{{\text{Time taken by N}}_2{\text{H}}_4}}{\frac{{\text{Amount of NH}}_3\text{ passed}}{{\text{Time taken by NH}}_3}}=\sqrt{\frac{M_{{\text{NH}}_3}}{M_{{\text{N}}_2{\text{H}}_4}}}$        (3)

Rearrange equation (3) to calculate amount of ${\text{NH}}_3$ as follows:

  ${\text{Amount of N}}_2{\text{H}}_4\text{ passed}=\left(\begin{array}{l}\left({\text{Time taken by N}}_2{\text{H}}_4\right)\left(\sqrt{\frac{M_{{\text{NH}}_3}}{M_{{\text{N}}_2{\text{H}}_4}}}\right)\\ \left(\frac{{\text{Amount of NH}}_3\text{ passed}}{{\text{Time taken by NH}}_3}\right)\end{array}\right)$        (4)

Substitute $25.0\text{ min}$ time taken by ${\text{N}}_2{\text{H}}_4$, $15.0\text{ min}$ time taken by ${\text{NH}}_3$, $0.35\text{ mmol}$ amount of ${\text{NH}}_3$, $17.03\text{ g/mol}$ for molar mass of ${\text{NH}}_3$, and $32.05\text{ g/mol}$ for molar mass of ${\text{N}}_2{\text{H}}_4$ in equation (4).

  $\begin{array}{c}{\text{Amount of N}}_2{\text{H}}_4\text{ passed}=\left(\begin{array}{l}\left(25.0\text{ min}\right)\left(\sqrt{\frac{17.03\text{ g/mol}}{32.05\text{ g/mol}}}\right)\\ \left(\frac{0.35\text{ mmol}}{15.0\text{ min}}\right)\end{array}\right)\\ =0.42\text{ mmol}\end{array}$

Hence amount of nitrogen gas effused through small hole in $\text{25}\text{.0 min}$ is $0.42\text{ mmol}$.