# Why do enthalpy changes of combustion and formation not take into account oxygen?

I've seen many enthalpy changes in my studies and I don't understand why the oxygens on either side of reactions are not calculated (especially if the unreacted oxygens don't balance). For example:

$$\ce{2C(s, graphite) + 3H2(g) + 1/2O2(g) -> C2H5OH(l)}$$

• It would be much easier to discuss/answer the question if it contained a specific example. – airhuff Apr 2 '17 at 21:27
• @airhuff 2C (s, graphite) + 3H2 (g) + 1/2O2 (g) --> C2H5OH (l). – Yusuf Abukar Apr 2 '17 at 21:33

By convention elements in their most stable polymorph at standard conditions have at standard conditions (298.15 K) a standard enthalpies of formation $\Delta{}_fH^\circ$ of zero (one reference). This is why thermodynamics discerns C(graphite) from C(diamond), and C(fullerene), for example. Hence on first view it looks like zeros in your example (graphite, Hydrogen, Oxygen) got forgotten; but they are not.