oxygen to form one atom of carbon dioxide and two atoms of hydrogen, which will combine with one atom of oxygen to form a molecule of water. Thus, the total oxygen requirement is 2 moles for complete combustion under ideal conditions. Since the source of this oxygen is atmospheric air, the burner will have to handle 2/0.20946 = 9.548 moles of air, consisting of two moles of oxygen and 7.548 moles of atmospheric nitrogen.

The stoichiometric combustion equation for methane with air thus becomes:

Some combustion applications, however, require even a higher precision. This may be done by converting the components (such as those listed on a gas quality report) into elemental — or atomic — equivalents and expressing the chemical make-up using an equivalent formula, i.e., formula in the form CaHbOcNd. The subscript values are determined by multiplying the mole fraction of each component present in the natural gas mixture by the number of atoms of the four primary elements. This can be illustrated using the example of pipeline quality natural gas with the composition given in Table 5-9, which lists all hydrocarbons, as well as inert constituents.

Thus, this particular natural gas composition may be represented by C1.045

H4.034 O0.010 N0.036, with C1.040 H4.034 representing the hydrocarbon portion of the fuel. The mass of hydrocarbons is 16.557 lbm in each mole of fuel having a molar mass of 17.282 lbm (previously computed using A.G.A. Cat. #8806 computer program), easily computed from relative atomic masses components shown in Table 5-10.

Table 5-10 Relative Atomic Masses Components

Carbon, C 1.045 x 12.011 =

12.55 1.040 x 12.011

= 12.491

Hydrogen, H 4.034 x 1.00794 =

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