Chapter 5, Problem 102AP

Interpretation Introduction

Interpretation:

Empirical formula and standard enthalpy of formation of a hydrocarbon is to be determined.

Concept introduction:

Empirical formula of a chemical compound is the simplest ratio of atoms present in it.

Molecular formula is whole number multiple of the empirical form given by the expression as follows:

$n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}$

Here, n is the whole number multiple.

The number of moles of a compound is determined as follows:

$m=\frac{\text{given weight}\left(wt\right)}{\text{molar mass}\left(M\right)}$

Answer to Problem 102AP

Solution:

(a)

Empirical formula of the compound is $\text{CH}$.

(b)

Enthalpy of formation of the hydrocarbon is $49\text{ kJ}/\text{mol}$.

Explanation of Solution

Given information: Combustion of $0.4196\text{ g}$ hydrocarbon releases $\text{1}\text{.419 g}$ of ${\text{CO}}_2$ and $0.290\text{ g}$ of ${\text{H}}_2\text{O}$.

a)

For determining the empirical formula of a compound, the number of moles of the product is determined by the expression as follows:

$m=\frac{\text{Given weight}\left(wt\right)}{\text{Molar mass}\left(M\right)}$

The number of moles of ${\text{CO}}_2$ is given as follows:

$m_{{\text{CO}}_2}=\frac{\left(w{t}_{{\text{CO}}_2}\right)}{\left(M_{{\text{CO}}_2}\right)}$

Substitute $\text{1}\text{.419 g}$ for $w{t}_{{\text{CO}}_2}$ and $\text{44}\text{.01 g}$ for $M_{{\text{CO}}_2}$:

$\begin{array}{c}{m}_{{\text{CO}}_2}=\left(\frac{1\times \text{1}\text{.419 }\cancel{\text{g}}}{\text{44}\text{.01 }\cancel{\text{g}}}\right)\\ =\text{0}\text{.032 mol}\end{array}$

The number of moles of ${\text{H}}_2\text{O}$ is given as follows:

$m_{{\text{H}}_2\text{O}}=\frac{\left(w{t}_{{\text{H}}_2\text{O}}\right)}{\left(M_{{\text{H}}_2\text{O}}\right)}$

Substitute $0.290\text{ g}$ for $w{t}_{{\text{H}}_2\text{O}}$ and $\text{18}\text{.016 g}$ for $M_{{\text{H}}_2\text{O}}$ as follows:

$\begin{array}{c}{m}_{{\text{H}}_2\text{O}}=\left(\frac{2\times 0.290\cancel{\text{g}}}{\text{18}\text{.016 }\cancel{\text{g}}}\right)\\ =\text{0}\text{.032 mol}\end{array}$

Therefore, molar ratio is calculated as follows:

$\begin{array}{c}{m}_{{\text{CO}}_2}:m_{{\text{H}}_2\text{O}}=\left(\frac{\text{0}\text{.032 }\cancel{\text{mol}}}{\text{0}\text{.032 }\cancel{\text{mol}}}\right):\left(\frac{\text{0}\text{.032 }\cancel{\text{mol}}}{\text{0}\text{.032 }\cancel{\text{mol}}}\right)\\ =1:1\end{array}$

Hence, the empirical formula is $\text{CH}$.

Given information: The molecular mass of the hydrocarbon is $76\text{ g}/\text{mol}$ and energy released in its combustion is $-17.55\text{ kJ}$.

b)

Molecular mass of lysine $=76\text{ g}/\text{mol}$

The empirical formula mass is calculated as follows:

$\begin{array}{c}\text{Empirical formula mass }=\left(\text{12}\text{.01 g}/\text{mol}+\text{1}\text{.008 g}/\text{mol}\right)\\ =\text{13}\text{.018 g}/\text{mol}\end{array}$

Now, whole number multiple $\left(n\right)$ can be calculated as follows:

$\begin{array}{c}n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}\\ =\left(\frac{76\cancel{\text{g}}/\cancel{\text{mol}}}{\text{13}\text{.018 }\cancel{\text{g}}/\cancel{\text{mol}}}\right)\\ =5.83\\ \approx 6\end{array}$

Molecular formula is as follows:

$\begin{array}{c}\text{Molecular formula}=n\times \text{Empirical formula}\\ =\text{6}\times \text{CH}\\ ={\text{C}}_6{\text{H}}_6\end{array}$

Therefore, molecular formula of the hydrocarbon is ${\text{C}}_6{\text{H}}_6$.

Equation for combustion of the hydrocarbon is represented below.

${\text{2C}}_6{\text{H}}_6+15{\text{O}}_2\to 12{\text{CO}}_2+6{\text{H}}_2\text{O}$

Energy produced by $0.419{\text{ g C}}_6{\text{H}}_6=-17.55\text{ kJ}$

Mass of ${\text{C}}_6{\text{H}}_6$ used is calculated as follows:

$\begin{array}{c}{\text{m}}_{C_6{H}_6}=\text{2}\times \left[\text{6}\times \text{12}\text{.01 g}/\text{mol}+\text{6}\times \text{1}\text{.008 g}/\text{mol}\right]\\ =\text{2}\times \left[72.060+6.048\right]\text{ g}/\text{mol}\\ =\text{156}\text{.216 g}/\text{mol}\end{array}$

Energy produced by $\text{ 156}\text{.216 g}/\text{mol}$ is calculated as follows:

$\begin{array}{c}\Delta H_{rxn}^{\circ }=\left(\frac{-17.55\text{ kJ}}{0.419\text{ g}}\right)\times \text{156}\text{.216 g}/\text{mol}\\ =\left(\text{41}\text{.88 kJ}/\cancel{\text{g}}\right)\left(\text{156}\text{.216 }\cancel{\text{g}}/\text{mol}\right)\\ =-6543\text{ kJ}/\text{mol }\end{array}$

Also, total enthalpy of the reaction is given by the equation as follows:

$\Delta H_{rxn}^{\circ }=12\left(\Delta {\text{H}}_{\text{f}}^{\circ }{\text{CO}}_2\right)+6\left(\Delta {\text{H}}_{\text{f}}^{\circ }{\text{H}}_2\text{O}\right)-2\left(\Delta {\text{H}}_{\text{f}}^{\circ }{\text{C}}_6{\text{H}}_6\right)$

Substitute $-6543\text{ kJ}/\text{mol }$ for $\Delta H_{rxn}^{\circ }$, $-393.5\text{ kJ}/\text{mol }$ for $\Delta H_f^{\circ }{}_{C{O}_2}$ and $-285.8\text{ kJ}/\text{mol }$ for $\Delta H_f^{\circ }{}_{H_{2}O}$ in the above equation to determine the enthalpy of formation.

$\begin{array}{c}-6543\text{ kJ}/\text{mol }=12\left(-393.5\text{ kJ}/\text{mol }\right)+6\left(-285.8\text{ kJ}/\text{mol }\right)-2\left(\Delta H_f^{\circ }{}_{C_6{H}_6}\right)\\ -6543\text{ kJ}/\text{mol}=\left(-4722\text{ kJ}/\text{mol}-1714.8\text{ kJ}/\text{mol}-2\left(\Delta H_f^{\circ }{}_{C_6{H}_6}\right)\right)\\ -6543\text{ kJ}/\text{mol}=-6436.8\text{ kJ}/\text{mol}-2\left(\Delta H_f^{\circ }{}_{C_6{H}_6}\right)\\ 2\left(\Delta H_f^{\circ }{}_{C_6{H}_6}\right)=-6436.8\text{ kJ}/\text{mol}+6543\text{ kJ}/\text{mol}\end{array}$

Therefore, $\Delta H_f^{\circ }{}_{C_6{H}_6}$ can be calculated as:

$\begin{array}{c}2\left(\Delta H_f^{\circ }{}_{C_6{H}_6}\right)=102\text{ kJ}/\text{mol}\\ \Delta H_f^{\circ }{}_{C_6{H}_6}=\left(\frac{\text{102 kJ}/\text{mol}}{2}\right)\\ \Delta H_f^{\circ }{}_{C_6{H}_6}=49\text{ kJ}/\text{mol}\end{array}$

Hence, the standard enthalpy of the formation of ${\text{C}}_6{\text{H}}_6$ is $49\text{ kJ}/\text{mol}$.