# Ammonia/oxygen reaction stoichiometry

#### mwm10620

2NH4 +3O2 >> 2NO2 +4H +2H2O
- 3.43g O2/g to convert

2NO2 + O2 >> 2NO3
- 1.14g O2/g NO2 oxidized

Total oxidation reaction:
NH4 + 2O2 >> NO3 +2H -H2O
- 4.57g O2/g N oxidized

Can someone please explain how the gram of O2 per gram of ammonia/nitrite is calculated? Thanks!

Last edited:

#### Woody

The mass of a single molecule of any substance can be determined,
however this is too tiny to be of practical use in the chemistry lab.
So the Molar mass is used instead, this is the mass of about 600,000,000,000,000,000,000,000 molecules of the substance
more precisely quoted in Wikipedia as 6.02214076×10^23. This is Avogadro's Number, also called a "mole".
(originally defined as the number of atoms in 12 grams of carbon-12, but now defined by ISO as precisely 6.02214076×10^23)

Tables giving the Molar Mass of a wide variety of substances are available, and the molar mass of most simple compounds is available on Wikipedia.
The measured mass of your substance divided by the molar mass gives the number of moles of molecules of that substance.
So, for example, the molar mass of oxygen molecules is 32 (note atomic mass is 16, but you are using O2 so...).
3.43/32=0.1071875 moles of oxygen molecules (0.1071875 * Avogadro's Number = about 0.6455x10^23 oxygen molecules)

You can see that using 0.6455x10^23 in your calculations will become tedious, which is why we use moles.
It is just the same concept as saying 0.1 billion, or 0.1 trillion or ... or 0.1 mole.

So given the number of moles of molecules of one reactant, and the reaction equation, you can work out the number of moles of molecules of the other reactants that will be needed.
given the number of moles of molecules of each reactant and the molar mass of each reactant you can work out the actual mass of each reactant needed.

Note that the phrase "the number of moles of molecules of each reactant" becomes tiresome to use repeatedly
so it is usually abbreviated to just "the number of moles of each reactant".

chip