Rate law of the reaction of propane with oxygen
The oxidation of hydrocarbons ($\ce{RH}$) such as this one proceed via a chain mechanism. The initiation step is 

$$\ce{RH + O2 -&gt; HO2^. + R^.}$$

and the propagation 

$$\ce{ R^.+O2 -&gt; RO2^. \\ RO2^. +RH -&gt; ROOH +R^.      }$$

But it does not stop there as the hydroperoxide decomposes $\ce{ ROOH -&gt; RO^. + OH^.}$ and there are other reactions such as 

$$\begin{align}
\ce{ R^. + O2 -&gt; RO^. + O^. \\ RH + OH^. -&gt; H2O + R^. \\ RCH2OO^. + O2 -&gt; RCH2O + O3  \\ RH + O3 -&gt; RO^. +HO2^.}\end{align}$$ 

and chain branching $\ce{ \quad O^. + RH -&gt; OH^. + R^.}$.

The exact mechanism depends on temperature and pressure, and in a plot of $T$ vs $p$ there are regions of slow reactions, explosion and hot and cold flames. (An original paper is by Newitt &amp; Thorne, J. Chem. Soc 1937, p1656). Thus you can appreciate that sorting out the mechanism is no simple task and will need much specialised equipment.