Theoretical Methane Yield (Buswell)
Computes the theoretical methane yield of a substrate with the Buswell equation, which closes the stoichiometric balance of anaerobic digestion of a CₙHₐO_bN_c compound into methane, carbon dioxide and ammonia: each mole of substrate yields (4n + a − 2b − 3c) ÷ 8 moles of methane, and dividing that by the molar mass and multiplying by the 22.414 L/mol molar volume gives the yield in litres of methane per gram at normal conditions, 0 °C and 1 atm. The less oxygen the molecule already carries, the more reduced it is and the more methane it yields: cellulose and glucose land at 0.41 and 0.37 L/g with 50% methane in the biogas, while a fat such as tristearin exceeds 1.02 L/g and reaches 71% methane — the methane fraction of the biogas is exactly (4n + a − 2b − 3c) ÷ 8n, since all the substrate carbon leaves either as methane or as carbon dioxide. The value is a thermodynamic ceiling, not a design forecast: in practice a digester delivers 60% to 80% of it, because part of the substrate becomes bacterial biomass and part never becomes accessible to the enzymes within the available retention time. Enter the number of carbon, hydrogen, oxygen and nitrogen atoms in the substrate's empirical formula.
Result
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The methane ceiling a substrate can reach
Before shipping a sample off for a BMP assay that ties up the lab for thirty to sixty days, it helps to know how much methane the material could yield at best. Buswell's equation gives that ceiling in minutes, from the empirical formula alone. Two uses pay for themselves right away: catching a supplier who promises yields above stoichiometry, which nobody delivers, and grading a working digester by dividing what it produces by what the chemistry allowed.
For a CₙHₐO_bN_c substrate each mole yields (4n + a − 2b − 3c) ÷ 8 moles of methane; the remaining carbon leaves as carbon dioxide and the nitrogen as ammonia. Divide by molar mass, multiply by 22.414 L/mol and you get yield per gram at STP — 0 °C and 1 atm, the convention this screen adopts and the one most digestion literature quotes. The less oxygen the molecule already carries, the more it yields: the defaults (C₆H₁₀O₅, the cellulose repeat unit) return 0.4147 L/g with 50% methane in the biogas, glucose lands at 0.373 and a fat clears 1.0 L/g.
Treat the figure as a thermodynamic ceiling, never a forecast. A working digester returns 60% to 80% of it, because some carbon builds bacterial biomass and some material — lignin, crystalline cellulose, plastics — never gets hydrolysed within the retention time. The equation knows nothing about ammonia inhibition, which bites hardest on nitrogen-rich substrates, nor about sulphur stealing electrons to leave as H₂S. And the molar volume belongs to STP: if your meter logs gas at 35 °C on the reactor outlet, multiply by 1.13 before comparing.
Frequently asked questions
Where does the 0.4147 L/g from the defaults come from?
Why may nitrogen sit at zero while carbon may not?
How do I turn this into cubic metres of biogas per day?
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The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.