Chapter 5, Problem 5.106QE

A compound is 82.7% carbon and 17.3% hydrogen, and has a molar mass of approximately 60 g/mol. When 1.000 g of this compound burns in excess oxygen, the enthalpy change is −49.53 kJ.

1. (a) What is the empirical formula of this compound?
2. (b) What is the molecular formula of this compound?
3. (c) What is the standard enthalpy of formation of this compound?
4. (d) Two compounds that have this molecular formula appear in Appendix G. Which one was used in this exercise?

Interpretation Introduction

Interpretation:

Empirical formula of the compound gas has to be determined.

Answer to Problem 5.106QE

Empirical formula of the compound is ${\text{C}}_2{\text{H}}_5$.

Explanation of Solution

Compound is 82.7% carbon and 17.3% hydrogen. Hence, $\text{100 g}$ compound contains $\text{82}\text{.7 g}$ Carbon and $\text{17}\text{.3 g}$ Hydrogen.

Number of moles of Carbon and Hydrogen is determined as follows,

  $\begin{array}{c}No.\text{ of moles of Carbon =}\frac{Mass}{Molar\text{ mass}}\\ \\ \text{= }\frac{\text{82}\text{.7 g C}}{\text{12}\text{.01 g/mol}}\text{ = 6}\text{.88 mol}\\ \\ \text{No}\text{.of}\text{}\text{ moles of Hydrogen = }\frac{\text{17}\text{.3 g H}}{1.01\text{ g/mol}}\text{ = 17}\text{.1 mol}\end{array}$

Empirical formula of the compound is determined as follows,

  $\begin{array}{cccc}& C& :& H\\ No.\text{ of moles:}& \text{6}\text{.88}& :& 17.1\\ Divide\text{ with least no}\text{.:}& \frac{\text{6}\text{.88}}{\text{6}\text{.88}}& :& \frac{17.1}{\text{6}\text{.88}}\\ & 1& :& 2.5\\ Convert\text{ to whole no}\text{.:}& 2& :& 5\end{array}$

Therefore, empirical formula of the compound is ${\text{C}}_2{\text{H}}_5.$

Interpretation Introduction

Interpretation:

Molecular formula of the compound gas has to be determined.

Answer to Problem 5.106QE

Molecular formula of the compound is ${\text{C}}_4{\text{H}}_{10}$.

Explanation of Solution

Empirical formula mass is determined as follows,

  $\begin{array}{l}Empirical\text{ formula mass = }\left(\text{2}\times \text{12}\text{.01 g/mol C}\right)\text{ +}\left(5\times 1.01\text{ g/mol H}\right)\\ \\ \text{ = 29}{\text{.07 g/mol C}}_2{\text{H}}_5\end{array}$

Molecular formula of the compound is $n\left(C_2{H}_5\right)$.

  $\begin{array}{l}Molecular\text{ formula = n}\times \text{ Empirical formula}\\ \\ \text{n = }\frac{Molar\text{ mass}}{Empirical\text{ formula mass}}\text{ = }\frac{ \text{60 g/mol}}{\text{29}\text{.06 g/mol}}\text{ = 2}\end{array}$

Therefore, molecular formula of the compound is ${\text{C}}_4{\text{H}}_{10}.$

Interpretation Introduction

Interpretation:

Standard enthalpy of formation of the compound has to be determined.

Answer to Problem 5.106QE

Standard enthalpy of formation of the given compound is $-123.19\text{ kJ/mol}$.

Explanation of Solution

Number of moles in $1.00{\text{ g C}}_4{\text{H}}_{10}$ is determined as follows,

  $\begin{array}{c}n\text{ = }\frac{Mass}{Molar\text{}\text{ mass}}\\ \\ =\frac{\text{1}.00\text{ g}}{\text{58}\text{.12} \text{g/mol}}\\ \\ =0.0172\text{ mol}\end{array}$

Enthalpy change when $0.0172\text{ mol}$ of ${\text{C}}_4{\text{H}}_{10}$ is burned is $-49.53\text{ kJ}$. Hence enthalpy change when ${\text{1 mol C}}_4{\text{H}}_{10}$ is burned can be determined as follows,

  $\begin{array}{c}\Delta H\text{ = }\frac{-49.53\text{ kJ}\times \text{ 1 mol}}{0.0172\text{ mol}}\\ \\ =-2880\text{ kJ}\end{array}$

The balanced thermochemical equation for the combustion of cyclopropane is given as:

  $C_5{H}_{10}\left(g\right) + \frac{13}{2}{O}_2\left(g\right) \underset{}{\to } 4C{O}_2\left(g\right) + 5H_{2}O\left(l\right)\Delta H=-\text{2091 kJ}$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of CO}}_{\text{2}}\left(g\right)\text{=}-\text{393}\text{.51}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of H}}_{2}O\left(l\right)\text{=}-\text{285}\text{.83}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of}{\text{O}}_2\left(g\right)\text{=}0\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of }{C}_4{H}_{10}\left(g\right)\text{=}\text{?}\\ \\ {\text{ΔH}}_{rxn}^o=-\text{2880 kJ}\end{array}$

Standard enthalpy of formation of the given compound can be calculated by the equation:

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Substitute the values as follows,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{4 mol}\times -\text{393}\text{.51}\text{kJ}/\text{mol}\right)\text{+}\left(\text{5 mol×}-\text{285}\text{.83}\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{1 mol}\times {\text{ΔH}}_f^o\text{of }{C}_4{H}_{10}\right)\text{+}\left(\frac{13}{2}\text{ mol×}0\text{kJ}/\text{mol}\right)\right]\\ \\ -\text{2880}\text{kJ =}\left(-3003.19\text{ kJ}\right)-\left(\text{1 mol}\times {\text{ΔH}}_f^o\text{of }{C}_4{H}_{10}\right)\\ \\ {\text{ΔH}}_f^o\text{of }{C}_4{H}_{10}\text{= }-123.19\text{ kJ/mol}\end{array}$

Therefore, standard enthalpy of formation of the given compound is $-123.19\text{ kJ/mol}\text{.}$

Interpretation Introduction

Interpretation:

Compound used for the exercise has to be determined.

Answer to Problem 5.106QE

The compound used must be ${\text{n-C}}_{\text{4}}{\text{H}}_{\text{10}}$.

Explanation of Solution

For ${\text{n-C}}_{\text{4}}{\text{H}}_{\text{10}}$, ${\text{ΔH}}_f^o=\text{}-124.73\text{ kJ/mol}$.

For ${\text{i-C}}_{\text{4}}{\text{H}}_{\text{10}}$, ${\text{ΔH}}_f^o=\text{}-131.60\text{ kJ/mol}$.

Standard enthalpy of formation of the given compound is $-123.19\text{ kJ/mol}$.

Standard enthalpy of formation of the given compound is nearly equal to the standard enthalpy of formation of normal butane. Hence, the compound used must be ${\text{n-C}}_{\text{4}}{\text{H}}_{\text{10}}.$