Calculating the heat of formation of propane given its enthalpy of combustion

Question

Given the equation $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$ and that enthalpies of formation for $\ce{H2O (l)}$ is $\pu{-285.3 kJ/mol}$ and $\ce{CO2 (g)}$ is $\pu{-393.5 kJ/mol}$, and the enthalpy of combustion for the reaction is $\pu{-2220.1 kJ/mol}$, I need to find the heat of formation of propane.

My initial idea was to use Hess's law, and I got $[3(-393.5) + 4(-285.3)] - [-2220.1] = \pu{-101.6 kJ}$

I then doubted myself because Hess's law gives the $\Delta H^o_\text{rxn}$ which is different from the $\Delta H^o_\mathrm{f}$ for $\ce{C3H8}$ that we are trying to find.

So I used another approach using the individual equations for the formation of $\ce{H2O, CO2}$, and $\ce{C3H8 + 5H2O -> 3CO2 + 4H2O}$ and combining the equations to get the equation for the formation of $\ce{C3H8}$ (which is $\ce{3C + 4H2 -> C3H8}$). After manipulating the equations, I got $+2220.1 - 1141.2 - 1180.5 = \pu{-101.6 kJ}$.

I got the same answer, and I want to understand why Hess's law still works even though the equation I used in Hess's law had nothing to do with the formation of $\pu{1 mol}$ of $\ce{C3H8}$. I understand my second method, but why does the first method work?

Answer 1

The equation I had earlier in my first method was wrong. Using Hess's law, we know the change in enthalpy of combustion to be $\pu{-2201.1 kJ/mol}$. Thus:

$$\Delta H^\circ_\text{rxn}=-22201.1 = [3(−393.5) + 4(−285.3)] − [5(0) + x]$$

where $x$ is the heat of formation of propane.

Solving the equation, we get $x = \pu{-101.6 kJ/mol}$.