Chapter 13, Problem 13.26P

Interpretation Introduction

Interpretation:

The molecular formula of hydrocarbon has to be determined.

Concept Introduction:

The expression to determine molecular weight as per ideal gas equation is given as follows:

  $M=\frac{mRT}{PV}$

Here,

$\text{R}$ denotes gas constant.

$V$ denotes the volume.

$n$ denotes the temperature.

$P$ denotes the pressure.

$m$ denotes mass.

$M$ denotes molecular mass.

In order to obtain the molecular formula of the compound, multiply the whole number with the subscript of each element present in the empirical formula.

Answer to Problem 13.26P

The molecular formula of hydrocarbon is ${\text{C}}_4{\text{H}}_6$.

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Substitute $\text{100 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=\text{100 }°\text{C}+273\text{ K}\\ =373\text{ K}\end{array}$

The conversion factor to convert $\text{mL}$ to $\text{L}$ is as follows:

  $1\text{ mL}={10}^{-3}\text{ L}$

Thus $\text{34}\text{.9 mL}$ is converted to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{34}\text{.9 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{\text{1 mL}}\right)\\ =0.0349\text{ L}\end{array}$

The conversion factor to convert $\text{mg}$ to $\text{g}$ is as follows:

  $1\text{ mg}={10}^{-3}\text{ g}$

Thus $\text{62}\text{.6 mg}$ is converted to $\text{g}$ as follows:

  $\begin{array}{c}\text{Mass}=\left(\text{62}\text{.6 mg}\right)\left(\frac{{10}^{-3}\text{ g}}{\text{1 mg}}\right)\\ =0.0626\text{ g}\end{array}$

The conversion factor to convert $\text{atm}$ to $\text{torr}$ is as follows:

  $1\text{ atm}=\text{760 torr}$

Thus, $\text{772 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{772 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =1.015\text{ atm}\end{array}$

The expression to determine molecular weight as per ideal gas equation is given as follows:

  $M=\frac{mRT}{PV}$        (2)

Substitute $0.0626\text{ g}$ for $m$, $0.0349\text{ L}$ for $V$, $1.015\text{ atm}$ for $P$, and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $373\text{ K}$ for $T$ in equation (2).

  $\begin{array}{c}M=\frac{\left(0.0626\text{ g}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(373\text{ K}\right)}{\left(1.015\text{ atm}\right)\left(0.0349\text{ L}\right)}\\ =54.09\text{ g/mol}\end{array}$

Molecular mass is $54.09\text{ g/mol}$.

The formula to convert mass in grams to moles is as follows:

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

$88.82\%$ by mass means $\text{0}\text{.8882 g}$ carbon in $\text{100 g}$ solution.

Substitute $\text{0}\text{.8882 g}$ for mass and $\text{12}\text{.01 g/mol}$ for molar mass of $\text{C}$ in equation (3) to calculate moles of $\text{C}$.

  $\begin{array}{c}\text{Moles of C}=\frac{\text{0}\text{.8882 g}}{\text{12}\text{.01 g/mol}}\\ =0.07395\text{ mol}\end{array}$

Substitute $\text{0}\text{.1118 g}$ for mass and $\text{1}\text{.008 g/mol}$ for molar mass of $\text{H}$ in equation (3) to calculate the moles of $\text{H}$.

  $\begin{array}{c}\text{Moles of H}=\frac{\text{0}\text{.1118 g}}{\text{1}\text{.008 g/mol}}\\ =0.11091\text{ mol}\end{array}$

Write moles of $\text{C}$ and $\text{H}$ as subscripts to obtain a preliminary formula as follows:

  ${\text{C}}_{0.07395}{\text{H}}_{0.11091}$

The smallest subscript is 0.07395. So, divide each subscript by 0.07395 as follows:

  ${\text{C}}_{\frac{0.07395}{0.07395}}{\text{H}}_{\frac{0.11091}{0.07395}}\to {\text{C}}_1{\text{H}}_{1.5}\to {\text{C}}_2{\text{H}}_3$

The expression to calculate the empirical formula mass is as follows:

  ${\text{Empirical formula mass of C}}_2{\text{H}}_3=\left(2\right)\left(\text{M of C}\right)+\left(3\right)\left(\text{M of H}\right)$        (4)

Substitute $12.01\text{ g/mol}$ for $\text{M}$ of $\text{C}$, $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ , in equation (4).

  $\begin{array}{c}{\text{Empirical formula mass of C}}_2{\text{H}}_3=\left(2\right)\left(12.01\text{ g/mol}\right)+\left(3\right)\left(1.008\text{ g/mol}\right)\\ =27.044\text{ g/mol}\end{array}$

The formula to calculate the whole number multiple is as follows:

  $\text{Whole-number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$        (5)

Substitute $54.09\text{ g/mol}$ for the molar mass of the compound and $27.044\text{ g/mol}$ for the empirical formula mass in equation (5) to calculate whole number multiple.

  $\begin{array}{c}\text{Whole}-\text{number multiple}=\frac{54.09\text{ g/mol}}{27.044\text{ g/mol}}\\ =2\end{array}$

Multiply the subscripts in ${\text{C}}_2{\text{H}}_3$ by 2 to obtain molecular formula.

  $\text{Molecular formula}=\left({\text{C}}_{2\left(2\right)}{\text{H}}_{3\left(2\right)}\right)\to \left({\text{C}}_4{\text{H}}_6\right)$

The molecular formula is ${\text{C}}_4{\text{H}}_6$.