Why are there different values for enthalpy of combustion, depending on the calculation method? Take for example the combustion of ethanol: $$\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O (l)}$$
If I calculate this using enthalpy of formation:
\begin{array}{c|c} \text{Substance} & \Delta H^\circ /\pu{(kJ/mol)} \\ \hline \ce{H2O(l)} & -285.8 \\ \ce{CO2(g)} & -393.5 \\ \ce{C2H5OH(l)} & -277.0 \end{array}
\begin{align} \Delta H &= \left[2\cdot(\pu{-393.5 kJ/mol}) + 3\cdot(\pu{-285.8 kJ/mol})\right] -\left[\pu{-277.0 kJ/mol} + 3\cdot(\pu{0 kJ/mol})\right]\\ &= \pu{-1367.4 kJ/mol} \end{align}
If I calculate this using bond energy:
\begin{array}{c|c} \text{Bonds} & \Delta H^\circ /\pu{(kJ/mol)} \\ \hline \ce{C-C} & 347 \\ \ce{C-H} & 414 \\ \ce{O-H} & 464 \\ \ce{C-O} & 351 \\ \ce{C=O(CO2)} & 803 \\ \ce{O=O} & 498 \end{array}
\begin{align} \Delta H &= \pu{347 kJ/mol} + \pu{351kJ/mol} + \pu{464 kJ/mol} + 5\cdot(\pu{414 kJ/mol}) \\ &\quad + 3\cdot(\pu{498 kJ/mol}) + 4\cdot(\pu{-803 kJ/mol}) + 6\cdot(\pu{-464 kJ/mol})\\ &= \pu{-1270 kJ/mol} \end{align}
Why is there such a big difference between them?