Chapter 3, Problem 3B.40E

Interpretation Introduction

Interpretation:

Molecular formula of the hydrocarbon with $14.3\text{\% }\text{H}$ and $85.7\text{\% }\text{C}$ 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.

Answer to Problem 3B.40E

Molecular formula of the hydrocarbon that has $14.3\text{\% }\text{H}$ and $85.7\text{\% }\text{C}$ is ${\text{C}}_3{\text{H}}_6$.

Explanation of Solution

Replace all $"\%"$ from $"\text{g"}$. Thus $100\text{ g}$ of compound would contain $14.3\text{g H}$ and $85.7\text{g C}$.

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 $14.3\text{g}$ 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 $\text{H}$.

  $\begin{array}{c}\text{Number of moles}\text{of H}=\frac{14.3\text{g}}{\text{1}\text{.00784 g/mol}}\\ =14.1887\text{mol}\end{array}$

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 $85.7\text{g}$ 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 $\text{C}$.

  $\begin{array}{c}\text{Number of moles}\text{of H}=\frac{85.7\text{g}}{\text{12}\text{.0107 g/mol}}\\ =7.1353\text{mol}\end{array}$

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

  $\text{Preliminary formula}={\text{H}}_{14.1887}{\text{C}}_{7.1353}$

Each of subscript of $\text{H}$ and $\text{C}$ is divided by smallest value to determine empirical formula of compound. The smallest value is 7.1353. Therefore empirical formula of given compound is as follows:

  $\begin{array}{c}\text{Empirical formula}={\text{H}}_{\frac{14.1887}{7.1353}}{\text{C}}_{\frac{7.1353}{7.1353}}\\ ={\text{H}}_{1.98852}{\text{C}}_1\\ \approx {\text{H}}_2\text{C}\end{array}$

The conversion factor to convert $17\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =17\text{°C}+273.15\\ =290.15\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$        (3)

Rearrange equation (3) to calculate $n$.

  $n=\frac{PV}{RT}$        (4)

Substitute $508\text{torr}$ for $P$, $62.3637\text{L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $1.5\text{L}$ for $V$ and $290.15\text{K}$ for $T$ in equation (4) to calculate $P$.

  $\begin{array}{c}n=\frac{\left(508\text{torr}\right)\left(1.5\text{L}\right)}{\left(62.3637\text{L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(290.15\text{K}\right)}\\ =0.042111\text{mol}\end{array}$

The expression to relate number of moles, mass and molar mass of ${\text{CH}}_2$ is as follows:

  $\text{Number of moles}{\text{of CH}}_2=\frac{{\text{mass of CH}}_2}{\text{molar mass of}{\text{CH}}_2}$        (5)

Rearrange in equation (5) to calculate molar mass of ${\text{CH}}_2$.

  $\text{molar mass of}{\text{CH}}_2=\frac{{\text{mass of CH}}_2}{\text{Number of moles}{\text{of CH}}_2}$        (6)

Substitute $1.77\text{g}$ for mass of ${\text{CH}}_2$ and $0.042111\text{mol}$ for number of moles of ${\text{CH}}_2$ in equation (6) to calculate number of moles of ${\text{CH}}_2$.

  $\begin{array}{c}\text{molar mass of}{\text{CH}}_2=\frac{1.77\text{g}}{0.042111\text{mol}}\\ =42.0317\text{g/mol}\end{array}$

Mass from empirical formula can be calculated as follows:

  $\begin{array}{c}\text{emperical}\text{formula}=\left(\text{number}\text{of}\text{C}\right)\left(\text{mass}\text{of}\text{C}\right)+\left(\text{number}\text{of}\text{H}\right)\left(\text{mass}\text{of}\text{H}\right)\\ =\left(1\right)\left(\text{12}\text{.0107 g/mol}\right)+\left(2\right)\left(\text{1}\text{.00784 g/mol}\right)\\ =14.0263\text{g/mol}\end{array}$

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

  $\text{molar mass}=k\left(\text{emperical}\text{formula}\right)$        (7)

Rearrange equation (7) to calculate $k$.

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

Substitute $14.0263\text{g/mol}$ for empirical formula and $42.0317\text{g/mol}$ for molar mass in equation (8).

  $\begin{array}{c}k=\frac{42.0317\text{g/mol}}{14.0263\text{g/mol}}\\ =2.9954\\ \approx 3\end{array}$

Molecular formula of the hydrocarbon that has $14.3\text{\% }\text{H}$ and $85.7\text{\% }\text{C}$ is ${\text{C}}_3{\text{H}}_6$.