Chapter 19, Problem 84AE

Interpretation Introduction

Interpretation:

The molecular formula of the given compound has to be determined by using the ideal gas equation and molar mass of that compound.  The possible structures of the compound has to be drawn from the determined molecular formula.

Concept Introduction:

Ideal gas equation:

The ideal gas equation is the combined equation of gas laws, namely Boyle’s law, and Charles law. This equation describes the amount of gas, temperature, pressure, and volume.

$\text{PV}\text{=}\text{nRT}$

$\text{P}$ is the pressure.

$\text{V}$ is the volume.

$\text{T}$ is the temperature.

$\text{n}$ is the number of moles.

$\text{R}$ is the universal gas constant.

Explanation of Solution

The given information in the problem is recored as follows:

$\begin{array}{l}\text{V}\text{=140}\text{.3}\text{mL}\\ \text{T}\text{=100}{}^{\text{o}}\text{C}\\ \text{P}\text{=740}\text{mm}\text{Hg}\\ \text{m}\text{=0}\text{.442}\text{g}\end{array}$

The given temperature is converted to kelvin as shown here.

$\begin{array}{l}\text{T}\text{=100}{}^{\text{o}}\text{C}\\ \text{T}\text{=}\left(\text{100}\text{+273}\text{.15}\right)\text{K}\\ \text{T}\text{=373}\text{K}\end{array}$

The given volume in millilitres is converted into liters by the conversion factor as follows:

$\begin{array}{l}\text{V}\text{=140}\text{.3}\text{mL×}\frac{\text{1}\text{L}}{{\text{10}}^{\text{3}}\text{mL}}\\ \text{V}\text{=0}\text{.1403}\text{L}\end{array}$

Use the below ideal gas equation to find the moles of the vapour.

$\text{n}\text{=}\frac{\text{PV}}{\text{RT}}$

Substitute as follows.

$\begin{array}{l}\text{n}\text{=}\frac{\text{740}\text{mm}\text{Hg×0}\text{.1403}\text{L}}{\text{62}\text{.363}\text{mm}\text{Hg}\text{L}{\text{mol}}^{-1}{\text{K}}^{-1}\times 373\text{K}}\\ \text{n}\text{=}0.00446\text{mol}\end{array}$

Use the below equation to find the molar mass of the compound.

$\text{Molar mass}\text{=}\frac{\text{Mass}}{\text{Moles}}$

Substitute as follows.

$\begin{array}{c}\text{Molar mass}\text{=}\frac{\text{0}\text{.442}\text{g}}{\text{0}\text{.00446}\text{mol}}\\ \text{=}\text{99}\text{.1}\text{g/mol}\end{array}$

The empirical formula of the compound can be calculated as follows:

Assume that the mass of the sample is $\text{100}\text{.0}\text{g}$. The given percent composition is changed into grams for every element as shown below.

$\begin{array}{l}\text{C}\text{=}\text{24}\text{.3}\text{g}\\ \text{H}\text{=}\text{4}\text{.1}\text{g}\\ \text{Cl}\text{=71}\text{.7}\text{g}\end{array}$

The grams of each element is converted into moles by using the molar mass of the corresponding element.

$\begin{array}{l}\text{C:}\text{24}\text{.3}\text{g}\text{C×}\frac{\text{1}\text{mol}\text{C}\text{atoms}}{\text{12}\text{.01}\text{g}\text{C}}\text{=2}\text{.02}\text{mol}\text{C}\text{atoms}\\ \text{H:}\text{4}\text{.1}\text{g}\text{H×}\frac{\text{1}\text{mol}\text{H}\text{atoms}}{\text{1}\text{.008}\text{g}\text{H}}\text{=4}\text{.07}\text{mol}\text{H}\text{atoms}\\ \text{Cl:}\text{71}\text{.7}\text{g}\text{Cl×}\frac{\text{1}\text{mol}\text{Cl}\text{atoms}}{\text{35}\text{.45}\text{g}\text{Cl}}\text{=2}\text{.02}\text{mol}\text{Cl}\text{atoms}\end{array}$

Now the number of moles of each element is converted into whole numbers, dividing by the lowest mole.

$\begin{array}{l}\text{C}\text{=}\frac{\text{2}\text{.02}\text{mol}}{\text{2}\text{.02}\text{mol}}\text{=}\text{1}\\ \text{H}\text{=}\frac{\text{4}\text{.07}\text{mol}}{\text{2}\text{.02}\text{mol}}\text{=}\text{2}\\ \text{Cl}\text{=}\frac{\text{2}\text{.02}\text{mol}}{\text{2}\text{.02}\text{mol}}\text{=}\text{1}\end{array}$

Thus, the empirical formula of the compound is ${\text{CH}}_{\text{2}}\text{Cl}$.

The molecular formula of the compound can be determined from the empirical formula and the molar mass of that compound as follows.

$\begin{array}{c}{\text{CH}}_{\text{2}}\text{Cl}\text{molar}\text{mass}\text{=}\text{12}\text{.01}\text{g}\text{+}\text{2}\left(\text{1}\text{.008}\text{g}\right)\text{+}\text{35}\text{.45}\text{g}\\ \text{=}\text{49}\text{.48}\text{g}\end{array}$

Use the below equation to find the molecular formula.

$n=\frac{\text{Molar}\text{mass}}{\text{Mass}\text{of}\text{empirical}\text{formula}}$

Substitute as follows.

$\begin{array}{l}n=\frac{\text{99}\text{.1}\text{g}}{\text{49}\text{.48}\text{g}}\\ n=2\end{array}$

Hence, the molecular formula of the compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$.

The determined molecular formula for the compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ which has two isomers.

There are two carbon atoms in the given compound. So, place the two carbon atoms in a straight-line as the parent carbon chain as shown below.

Place the two chlorine atoms on the same carbon atom or for each carbon atom as shown below.

Now add four hydrogen atoms to each carbon atom in order to complete the two structures above because each carbon has four bonds around it. ]

Hence, the possible structures of the compound are show below.