Jump to content
APC Forum

Oxygen Balance Equation


Recommended Posts

Hello all,


I was reading up on oxygen balance and I think I understand the concept but not well enough.

The oxygen balance equation is as follows per Wikipedia






X = number of atoms of carbon, Y = number of atoms of hydrogen, Z = number of atoms of oxygen, and M = number of atoms of metal (metallic oxide produced).



Why do we include the molar molecular mass and the -1600? This makes it different from straightforward stoichiometric balancing, but why? Why does a heavier element/molecule have a more positive oxygen balance? Does this imply that it would need a higher oxygen content in the overall formula?


For example, Alany says in this thread (http://www.pyrosociety.org.uk/forum/topic/1955-sodium-nitrate-compositions/) that you need less Sodium nitrate than Potassium nitrate for the same application, I don't understand why that is if they both consist of 1 metal, 1 nitrogen, and 3 oxygen.


To maintain the same oxygen balance you should use 0.84 times as much NaNO3 as KNO3.

That would make "BP" approximately:

72 NaNO3
17 C
11 S


In order to balance the stoichiometric equation or predict the products of reaction I would only look at the elements present, not their weights. In order to predict the products I would follow something like this: http://en.wikipedia.org/wiki/Chemical_explosive#Balancing_chemical_explosion_equations.


Thank you,




edit: "molar mass" corrected to "molecular mass"

Edited by AzoMittle
Link to comment
Share on other sites

Think in these terms: K is heavier then Na, thus the same amount of Na nitrate (in weight!) contains more oxygen.


M KNO3 = 40+14+(3*16)=102

M NaNO3= 23+14+(3*16)=85


You can clearly see that both molecules contain the same amount of oxygen, but weight for weight, NaNO3 has more.


Let's make the percentages, and find out how much NaNO3 will provide the SAME oxygen amount.

Since the oxygen is the same in both molecules:


102g KNO3 (1 mole)------------------------- 100%

85g NaNO3 (1 mole)------------------------- X



x=82*100./102 = 83.3% (your 0.84 is rounded)


So you need 83.3 percent of the required amont of KNO3 to provide the SAME amount of oxygen.


The formulation "0.84 of the required KNO3 amount" is perfectly fine. We like to work with weights rather then atoms, molecules and enthalpies.


While the burning equations will be similar, the weights of the reagents will not be.

Moreover, that Na may provide different salts then K since it's less reactive but these are details.

Edited by a_bab
Link to comment
Share on other sites

I wondered about the 1600 thing for a very long time. When I found out, I basically slapped myself in the forehead since it's so stupid. 1600 is 16*100. AKA molecular mass of oxygen * 100. The 100 converts from a fraction to a percentage.


As far as the molecular mass thing. If you follow though the units, knowing about the 1600 thing, you'll see that it's needed to cancel all the units. By making it overall unitless, it actually makes it easier to balance formulas. The formula is basically mass of oxygen used/excess, divided by the mass of the total compound. Then you can go from there.


I've generally had an issue with the formula when metals are included. There really should be a factor based on oxygen atoms it requires. Maybe I'm missing something, but it works if you have a lead salt. Lead (II) being the thermodynamic product. However, if you had a silver salt, I'd think you should need a 1/2 factor.


Anyway, the OB% formula really only works for high explosives.

Link to comment
Share on other sites

Wow okay, that makes a lot more sense, thank you.


Why does it only work for HE? Is it because of BP being a physical mixture instead of chemically bonded; or because of reaction rate or what?

Link to comment
Share on other sites

OB works for HEs but they change the rules often.

For pyro OB doesn't mean a lot as the burning usually happens in air, often stars have a low OB -sometimes to cause certain intermediates to form in the flame, and sometimes to let the star use atmospheric oxygen as it travels (see Tiger Tail)

  • Like 1
Link to comment
Share on other sites

They just function, in general, by different mechanisms in my opinion. High explosives are designed to convert as much chemical energy into kinetic energy as possible. They do this by forming the most thermodynamically stable molecules in the shortest amount of time. The reaction will never be perfect, but the simple approximation of CO2, H2O, and N2 being the products isn't too far off for a balanced mixture.


Short of binary flash powders, most low explosives will probably have dozens to hundreds of reaction pathways. The amount of energy produced isn't always enough to allow everything to yield the biggest thermodynamic sinks. This usually works out in our favor. Burn times are manageable, a variety of colors are able to be produced, and complex effects like strobes and glitters are possible. This is a simplification of course, but even for relatively simple mixtures such as whistle mix, I'd bet the reaction produces easily 10-15 measurable reaction products.


The OB formula probably would work acceptably for simple binary or trinary fully inorganic low explosive mixtures, but little more.

  • Like 1
Link to comment
Share on other sites

  • Create New...