Problem: A sample of a hydrocarbon is combusted completely in O2(g) to produce 24.01 g CO2(g), 4.917 g H2O(g), and 342 kJ of heat.Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

Determine the empirical formula of the hydrocarbon: (C and H only)

Recall that:

• 1 mole of C in 1 mole of CO₂. 
• 2 moles of H in 1 mole of H₂O.

Find moles of C and H:

$\mathrm{moles} \mathrm{of} \mathrm{C}=24.01 \cancel{\mathrm{g} {\mathrm{CO}}_2}\times \frac{1 \cancel{\mathrm{mol} {\mathrm{CO}}_2}}{44.01 \cancel{\mathrm{g} {\mathrm{CO}}_2}}\times \frac{1 \cancel{\mathrm{mol} \mathrm{C}}}{1 \cancel{\mathrm{mol} {\mathrm{CO}}_2}}$ 

moles C = 0.55 mol C

$\mathrm{moles} \mathrm{of} \mathrm{H}=4.917 \cancel{\mathrm{g} {\mathrm{H}}_2\mathrm{O}}\times \frac{1 \cancel{\mathrm{mol} {\mathrm{H}}_2\mathrm{O}}}{18.02 \cancel{\mathrm{g} {\mathrm{H}}_2\mathrm{O}}}\times \frac{2 \cancel{\mathrm{mol} \mathrm{H}}}{1 \cancel{\mathrm{mol} {\mathrm{H}}_2\mathrm{O}}}$

moles H = 0.545 mol H

Get smallest ratio of C, H (divide by 0.545):

$\mathrm{C}:\hspace{0.17em}\frac{0.55 \mathrm{mol} \mathrm{C}}{0.545}=1$           $\mathrm{H}:\hspace{0.17em}\frac{0.545 \mathrm{mol} \mathrm{H}}{0.545}=1$     

The empirical formula: CH

Balanced equation: 2CH + O₂ → CO₂ + H₂O

$$\begin{gathered} \mathbf{∆}{\mathbf{H}}_{\mathbf{rxn}}\mathbf{=}\mathbf{∆}\mathbf{H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}\mathbf{products}}\mathbf{-}\mathbf{∆}\mathbf{H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}\mathbf{reactants}} \\ -\mathbf{342}\mathbf{ }\mathbf{kJ}\mathbf{=}\mathbf{[}\mathbf{24}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{\left(}\frac{\mathbf{-}\mathbf{393}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{kJ}}{\cancel{\mathbf{mol}}}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{917}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}{\mathbf{18}\mathbf{.}\mathbf{02}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\left(}\frac{\mathbf{-}\mathbf{241}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{kJ}}{\cancel{\mathbf{mol}}}\mathbf{\right)}\mathbf{]}\mathbf{-}\mathbf{[}\mathbf{∆}\mathbf{H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}\mathbf{CH}}\mathbf{+}\mathbf{0}\mathbf{]} \end{gathered}$$