It can be predicted from the stoiciometric oxygen required for complete combustion and the LFL.

Write a balanced oxidation reaction equation to CO2 and H2O.

Example: C3H8 + 5O2 --> 3CO2 + 4H2O

LFL of propane = 2.2% in air

2.2 x 5 = 11.0%

MOC of propane = 11.5% by lab determination.

Good luck,
Latexman

I learned it as the Minimum Oxygen Concentration (MOC).  LOC = MOC.

Good luck,
Latexman

The minimum concentration of oxygen to propagate a flame by the energy generated in the reaction is the basis for inerting and is called the MOC. Inerting is used to bring the oxygen content to at least 4% points below the MOC.

In the absence of experimental data the procedure explained by Latexman can be used as an estimate that has been confirmed to work for many hydrocarbons.

Published LFL (low flammable limit) values for vapor-air mixtures are generally based on the fuel gas in air at 25^oC and 760 mm Hg.

For many organic compounds it was found that the LFL can be approximated by taking 55% of the stoichiometric molar concentration of the fuel in the vapor-air mixture.

Visiting the following threads may add more information:

thread135-73870
thread798-47819
thread816-58335
 

As noted in the threads 25362 has referenced, be sure to look at NFPA 69, Standard on Explosion Prevention Systems.  They have listings of LOC for a number of chemicals but unfortunately I did not see cyclohexane and acrylic acid.

I would point out that LOC is affected by the material used as a diluent.  For example, according to NFPA 69 the LOC for propane is 11.5% for a N2/air environment, the LOC for propane is 14.5% for a CO2/air environment.