Chapter 3, Problem 3B.39E

Interpretation Introduction

Interpretation:

Molecular formula of the compound with $2.1\text{\% }\text{H}$, $24.7\text{\% }\text{C}$ and $73.2\text{\% }\text{Cl}$ has to be determined.

Concept Introduction:

An ideal gas contains a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is a theoretical concept. Gases that show perfect elastic collision are practically not possible. At higher $T$ and lower $P$ gases behave almost ideally.

Ideal gas law is an equation of state for any hypothetical gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

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.

Empirical formula represents simplest positive integer ratio of atoms in the compound. It only gives proportions of elements in compound. Molecular formula consists of chemical symbols for the respective elements followed by numeric subscript that denotes the number of atom of each element present in molecule.

Answer to Problem 3B.39E

Chemical formula of hydrocarbon is ${\text{C}}_2{\text{H}}_2{\text{Cl}}_2$.

Explanation of Solution

Replace all $"\%"$ from $"\text{g"}$. Thus $100\text{ g}$ of compound would contain $2.1\text{g H}$, $24.7\text{g C}$, and $73.2\text{g Cl}$.

The expression to relate number of moles, mass and molar mass of $\text{H}$ is as follows:

  $\text{Number of moles}\text{of H}=\frac{\text{mass of H}}{\text{molar mass of}\text{H}}$        (1)

Substitute $2.1\text{g}$ for mass of $\text{H}$ and $\text{1}\text{.00784 g/mol}$ for molar mass of $\text{H}$ in equation (1) to calculate number of moles of $\text{H}$.

  $\begin{array}{c}\text{Number of moles}\text{of H}=\frac{2.1\text{g}}{\text{1}\text{.00784 g/mol}}\\ =2.08366\text{mol}\end{array}$

The expression to relate number of moles, mass and molar mass of $\text{C}$ is as follows:

  $\text{Number of moles}\text{of C}=\frac{\text{mass of C}}{\text{molar mass of}\text{C}}$        (2)

Substitute $24.7\text{g}$ for mass of $\text{C}$ and $\text{12}\text{.0107 g/mol}$ for molar mass of $\text{C}$ in equation (2) to calculate number of moles of $\text{C}$.

  $\begin{array}{c}\text{Number of moles}\text{of C}=\frac{24.7\text{g}}{\text{12}\text{.0107 g/mol}}\\ =2.05649\text{mol}\end{array}$

The expression to relate number of moles, mass and molar mass of $\text{Cl}$ is as follows:

  $\text{Number of moles}\text{of Cl}=\frac{\text{mass of Cl}}{\text{molar mass of}\text{Cl}}$        (3)

Substitute $73.2\text{g}$ for mass of $\text{Cl}$ and $35.453\text{ g/mol}$ for molar mass of $\text{Cl}$ in equation (3) to calculate number of moles of $\text{Cl}$.

  $\begin{array}{c}\text{Number of moles}\text{of C}=\frac{73.2\text{g}}{35.453\text{ g/mol}}\\ =2.0647\text{mol}\end{array}$

Preliminary formula for compound is formed with moles of $\text{H}$, $\text{Cl}$ and $\text{C}$ written in subscripts. Therefore it can be written as follows:

  $\text{Preliminary formula}={\text{H}}_{2.08366}{\text{C}}_{2.05649}{\text{Cl}}_{2.0647}$

Each of subscript of $\text{H}$, $\text{Cl}$ and $\text{C}$ is divided by smallest value to determine empirical formula of compound. The smallest value is 2.05649. Therefore, empirical formula of given compound is as follows:

  $\begin{array}{c}\text{Empirical formula}={\text{H}}_{\frac{2.08366}{2.05649}}{\text{C}}_{\frac{2.05649}{2.05649}}{\text{Cl}}_{\frac{2.0647}{2.05649}}\\ ={\text{H}}_{1.01321}{\text{C}}_1{\text{Cl}}_{1.0039}\\ \approx \text{HCCl}\end{array}$

The conversion factor to convert $0\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =0\text{°C}+273.15\\ =273.15\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$        (4)

Rearrange equation (4) to calculate $n$.

  $n=\frac{PV}{RT}$        (5)

Substitute $1.1\text{atm}$ for $P$, $8.20574\times {10}^{-2}\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $755\text{mL}$ for $V$ and $273.15\text{K}$ for $T$ in equation (5) to calculate $P$.

  $\begin{array}{c}n=\frac{\left(1.1\text{atm}\right)\left(755\text{mL}\right)\left(\frac{0.001\text{L}}{\text{1}\text{mL}}\right)}{\left(8.20574\times {10}^{-2}\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(273.15\text{K}\right)}\\ =0.037052\text{mol}\end{array}$

The expression to relate number of moles, mass and molar mass of $\text{CHCl}$ is as follows:

  $\text{Number of moles}\text{of CHCl}=\frac{\text{mass of CHCl}}{\text{molar mass of}\text{CHCl}}$        (6)

Rearrange in equation (6) to calculate molar mass of $\text{CHCl}$.

  $\text{molar mass of}\text{CHCl}=\frac{\text{mass of CHCl}}{\text{Number of moles}\text{of CHCl}}$        (7)

Substitute $3.557\text{g}$ for mass of $\text{CHCl}$ and $0.037052\text{mol}$ for number of moles of $\text{CHCl}$ in equation (7) to calculate number of moles of $\text{CHCl}$.

  $\begin{array}{c}\text{molar mass of}\text{CHCl}=\frac{3.557\text{g}}{0.037052\text{mol}}\\ =96.0002\text{g/mol}\end{array}$

Mass from empirical formula can be calculated as follows:

  $\begin{array}{c}\text{emperical}\text{formula}=\left(\begin{array}{c}\left(\text{number}\text{of}\text{C}\right)\left(\text{mass}\text{of}\text{C}\right)+\left(\text{number}\text{of}\text{H}\right)\left(\text{mass}\text{of}\text{H}\right)\\ +\left(\text{number}\text{of}\text{Cl}\right)\left(\text{mass}\text{of}\text{Cl}\right)\end{array}\right)\\ =\left(1\right)\left(\text{12}\text{.0107 g/mol}\right)+\left(1\right)\left(\text{1}\text{.00784 g/mol}\right)+\left(1\right)\left(35.453\text{ g/mol}\right)\\ =48.47154\text{g/mol}\end{array}$

Molar mass and empirical formula are related by formula as follows:

  $\text{molar mass}=k\left(\text{emperical}\text{formula}\right)$        (8)

Rearrange equation (8) to calculate $k$.

  $k=\frac{\text{molar mass}}{\text{emperical}\text{formula}}$        (9)

Substitute $48.47154\text{g/mol}$ for empirical formula and $96.0002\text{g/mol}$ for molar mass in equation (9).

  $\begin{array}{c}k=\frac{96.0002\text{g/mol}}{48.47154\text{g/mol}}\\ =1.9805\\ \approx 2\end{array}$

Molecular formula of the compound that has $2.1\text{\% }\text{H}$, $24.7\text{\% }\text{C}$ and $73.2\text{\% }\text{Cl}$ is ${\text{C}}_2{\text{H}}_2{\text{Cl}}_2$.