Chapter 3, Problem 3B.42E

Interpretation Introduction

Interpretation:

Molecular formula of the eugenol has to be determined.

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

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.42E

Molecular formula of the eugenol is ${\text{C}}_{10}{\text{H}}_{12}{\text{O}}_2$.

Explanation of Solution

The conversion factor to convert $280\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =280\text{°C}+273.15\\ =553.15\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$        (1)

Rearrange equation (1) to calculate $n$.

  $n=\frac{PV}{RT}$        (2)

Substitute $48.3\text{torr}$ for $P$, $62.3637\text{L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $500\text{mL}$ for $V$ and $553.15\text{K}$ for $T$ in equation (2) to calculate $n$.

  $\begin{array}{c}n=\frac{\left(48.3\text{torr}\right)\left(500\text{mL}\right)\left(\frac{0.001\text{L}}{1\text{mL}}\right)}{\left(62.3637\text{L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(553.15\text{K}\right)}\\ =0.0006957\text{mol}\end{array}$

The expression to relate number of moles, mass and molar mass is as follows:

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

Rearrange equation (3) to calculate molar mass of compound.

  $\text{molar mass of}\text{compound}=\frac{\text{mass of compound}}{\text{Number of moles}}$        (4)

Substitute $115\text{mg}$ for mass of compound and $0.0006957\text{mol}$ for number of moles in equation (4) to calculate molar mass of compound

  $\begin{array}{c}\text{molar mass of}\text{compound}=\frac{115\text{mg}\left(\frac{0.001\text{g}}{1\text{mg}}\right)}{0.0006957\text{mol}}\\ =165.3011\text{g/mol}\end{array}$

$\text{1}{\text{mol of CO}}_2$ has mass of $\text{44}\text{.01 g}$ then $50\text{mg}$ of ${\text{CO}}_2$ has moles calculated as follows:

  $\begin{array}{c}{\text{moles of CO}}_2=\left(50\text{mg}\right)\left(\frac{0.001\text{g}}{1\text{mg}}\right)\left(\frac{\text{1}{\text{mol of CO}}_2}{\text{44}\text{.01 g}}\right)\\ =0.001136\text{mol}\end{array}$

$\text{1}{\text{mol of H}}_2\text{O}$ has mass of $18.01528\text{ g}$ then $12.4\text{mg}$ of ${\text{H}}_2\text{O}$ has moles calculated as follows:

  $\begin{array}{c}{\text{moles of H}}_2\text{O}=\left(12.4\text{mg}\right)\left(\frac{0.001\text{g}}{1\text{mg}}\right)\left(\frac{\text{1}{\text{mol of H}}_2\text{O}}{18.01528\text{ g}}\right)\\ =0.0006883\text{mol }\end{array}$

In $0.001136\text{mol}$ of ${\text{CO}}_2$, $0.001136\text{mol}$ of $\text{C}$ is present.

In $0.0006883\text{mol}$ of ${\text{H}}_2\text{O}$, $2\left(0.0006883\text{mol}\right)$ of $\text{H}$ is present that is $0.001376\text{mol}$.

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

  $\text{Moles}\text{of}\text{C}=\frac{\text{mass of}\text{C}}{\text{molar mass of}\text{C}}$        (5)

Rearrange equation (5) to calculate mass of $\text{C}$.

  $\text{mass of}\text{C}=\left(\text{molar mass of}\text{C}\right)\left(\text{Moles}\text{of}\text{C}\right)$        (6)

Substitute $12.0107\text{g/mol}$ for molar mass of $\text{C}$ and $0.001136\text{mol}$ for moles of $\text{C}$ in equation (6) to calculate mass of $\text{C}$.

  $\begin{array}{c}\text{mass of C}=\left(12.0107\text{g/mol}\right)\left(0.001136\text{mol}\right)\\ =0.013644\text{g}\end{array}$

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

  $\text{Moles}\text{of}\text{H}=\frac{\text{mass of}\text{H}}{\text{molar mass of}\text{H}}$        (7)

Substitute $1.00784\text{g/mol}$ for molar mass of $\text{H}$ and $0.001136\text{mol}$ for moles of $\text{H}$ in equation (7) to calculate mass of $\text{H}$.

  $\begin{array}{c}\text{mass of}\text{H}=\left(1.00784\text{g/mol}\right)\left(0.001376\text{mol}\right)\\ =0.001386\text{g}\end{array}$

$18.8\text{mg}$ equal to $0.0188\text{g}$ of eugenol contain $0.013644\text{g}$ of $\text{C}$ and $0.001386\text{g}$ of $\text{H}$ then

Mass of $\text{O}$ present is calculated as follows:

  $\begin{array}{c}\text{mass}\text{of}\text{O}=\left(\text{mass of eugenol}\right)-\left(\text{mass of C}+\text{mass of H}\right)\\ =\left(\text{0}\text{.0188}\text{g}\right)-\left(\text{0}\text{.013644}\text{g}+\text{0}\text{.001386}\text{g}\right)\\ =\text{0}\text{.00377}\text{g}\end{array}$

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

  $\text{Moles}\text{of}\text{O}=\frac{\text{mass of}\text{O}}{\text{molar mass of}\text{O}}$        (8)

Substitute $\text{0}\text{.00377}\text{g}$ for mass of $\text{O}$ and $\text{12}\text{.0107 g/mol}$ for molar mass of $\text{O}$ in equation (8) to calculate moles of $\text{O}$.

  $\begin{array}{c}\text{Moles}\text{of O}=\frac{0.00377\text{g}}{15.999\text{ g/mol}}\\ =0.000235\text{mol}\end{array}$

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

  $\text{Preliminary formula}={\text{H}}_{0.001376}{\text{C}}_{0.001136}{\text{O}}_{0.000235}$

Each of subscript of $\text{H}$, $\text{O}$ and $\text{C}$ is divided by smallest value to determine empirical formula of compound. The smallest value is 0.000235. Therefore empirical formula of given compound is as follows:

  $\begin{array}{c}\text{Empirical formula}={\text{H}}_{\frac{0.001376}{0.000235}}{\text{C}}_{\frac{0.001136}{0.000235}}{\text{O}}_{\frac{0.000235}{0.000235}}\\ ={\text{H}}_{5.85531}{\text{C}}_{4.83404}\text{O}\\ \approx {\text{H}}_6{\text{C}}_5\text{O}\end{array}$

Empirical mass from empirical formula can be calculated as follows:

  $\begin{array}{c}\text{Empirical mass}=\left(\begin{array}{l}\left(\text{number}\text{of}\text{C}\right)\left(\text{mass}\text{of}\text{C}\right)+\\ \left(\text{number}\text{of}\text{H}\right)\left(\text{mass}\text{of}\text{H}\right)+\left(\text{number}\text{of}\text{O}\right)\left(\text{mass}\text{of}\text{O}\right)\end{array}\right)\\ =\left(5\right)\left(\text{12}\text{.0107 g/mol}\right)+\left(6\right)\left(\text{1}\text{.00784 g/mol}\right)+\left(\text{15}\text{.999 g/mol}\right)\\ =82.09954\text{g/mol}\end{array}$

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

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

Rearrange equation (9) to calculate $k$.

  $k=\frac{\text{molar mass}}{\text{emperical}\text{formula}}$        (10)

Substitute $82.09954\text{g/mol}$ for empirical formula and $165.3011\text{g/mol}$ for molar mass in equation (10).

  $\begin{array}{c}k=\frac{165.3011\text{g/mol}}{82.09954\text{g/mol}}\\ =1.94244\\ \approx 2\end{array}$

Molecular formula of the eugenol is ${\text{C}}_{10}{\text{H}}_{12}{\text{O}}_2$.