Stoichiometry - Revision history

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	The two moles of oxygen required by the methane equation have a combined mass of approximately 64 g. Dividing 64 g by 16 g gives an oxygen-to-fuel mass ratio of approximately four. Thus, the simplified calculation requires about 4 kg of oxygen for 1 kg of methane. The factor of two in the reaction equation has become a factor of four in the mass comparison.<ref name="VR-3-3" />		The two moles of oxygen required by the methane equation have a combined mass of approximately 64 g. Dividing 64 g by 16 g gives an oxygen-to-fuel mass ratio of approximately four. Thus, the simplified calculation requires about 4 kg of oxygen for 1 kg of methane. The factor of two in the reaction equation has become a factor of four in the mass comparison.<ref name="VR-3-3" />
			[[File:fse-stoichiometry-v1.svg\|thumb\|upright=1.5\|alt=One mole of methane reacts with two moles of oxygen, corresponding to approximately 16 g and 64 g.\|Molar coefficients and mass proportions are different quantities in complete methane combustion. Based on Van Rossum (2027), §3.3.]]

	This difference is a common source of misunderstanding. The reaction consumes twice as many moles of oxygen as methane, but those moles have different masses. Applying the reaction coefficients directly to kilograms would therefore give the wrong oxygen requirement. The molar-mass conversion is the step that makes a chemical equation usable in a fuel-mass calculation.<ref name="VR-3-3" />		This difference is a common source of misunderstanding. The reaction consumes twice as many moles of oxygen as methane, but those moles have different masses. Applying the reaction coefficients directly to kilograms would therefore give the wrong oxygen requirement. The molar-mass conversion is the step that makes a chemical equation usable in a fuel-mass calculation.<ref name="VR-3-3" />
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	Using 28 g/mol for nitrogen, the accompanying nitrogen mass is 526.4 g. Adding the 160 g of oxygen gives 686.4 g of simplified air. Dividing this air mass by the 44 g propane mass gives 15.6 kg of air per kilogram of propane. The calculation shows why the oxygen requirement and the air requirement are numerically different quantities.<ref name="D-92" />		Using 28 g/mol for nitrogen, the accompanying nitrogen mass is 526.4 g. Adding the 160 g of oxygen gives 686.4 g of simplified air. Dividing this air mass by the 44 g propane mass gives 15.6 kg of air per kilogram of propane. The calculation shows why the oxygen requirement and the air requirement are numerically different quantities.<ref name="D-92" />
			[[File:fse-stoichiometry-v2.svg\|thumb\|upright=1.5\|alt=For 44 g propane, 160 g oxygen arrives with 526.4 g nitrogen in a total of 686.4 g simplified air.\|Oxygen and air requirements for the simplified complete combustion of propane. Based on Drysdale (2002), p. 1-92.]]

	<div class="fse-equation">Air-to-fuel mass ratio = (160 + 18.8 × 28) / 44 = 15.6 kg/kg</div>		<div class="fse-equation">Air-to-fuel mass ratio = (160 + 18.8 × 28) / 44 = 15.6 kg/kg</div>
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	This calculated concentration lies between the cited methane flammability limits, but it answers a different question. It identifies the ideal reactant balance for complete combustion. The limits describe flame propagation under the stated test conditions. Treating the calculated 9.5% as an ignition threshold would confuse a reaction-proportion calculation with an experimentally defined combustion boundary.<ref name="VR-3-3" /><ref name="B-172" />		This calculated concentration lies between the cited methane flammability limits, but it answers a different question. It identifies the ideal reactant balance for complete combustion. The limits describe flame propagation under the stated test conditions. Treating the calculated 9.5% as an ignition threshold would confuse a reaction-proportion calculation with an experimentally defined combustion boundary.<ref name="VR-3-3" /><ref name="B-172" />
			[[File:fse-stoichiometry-v3.svg\|thumb\|upright=1.5\|alt=Methane flammability range from 5 to 15 vol% with a separate calculated stoichiometric marker near 9.5 vol%.\|The calculated stoichiometric concentration is distinct from methane flammability limits. Based on Beyler (2002), p. 2-172, and Van Rossum (2027), §§3.3 and 4.2.]]

	Temperature, pressure and mixture composition are part of the description of flammability limits. Beyler therefore states limits for a specified oxidant and diluent mixture, rather than presenting a fuel concentration in isolation. Stoichiometry retains its role as the balanced chemical reference. The additional combustion conditions determine how that reference relates to a particular flame or fire.<ref name="B-172" />		Temperature, pressure and mixture composition are part of the description of flammability limits. Beyler therefore states limits for a specified oxidant and diluent mixture, rather than presenting a fuel concentration in isolation. Stoichiometry retains its role as the balanced chemical reference. The additional combustion conditions determine how that reference relates to a particular flame or fire.<ref name="B-172" />

Peter: New FSE wiki article: Stoichiometry (written by GPT-6)

2026-11-01T21:35:26Z

New FSE wiki article: Stoichiometry (written by GPT-6)

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'''Stoichiometry''' describes the quantitative relationships between the substances consumed and produced in a chemical reaction. In fire safety engineering, it establishes how much oxygen is required for a specified amount of fuel to undergo complete [[Combustion|combustion]]. The relationship follows from a balanced reaction equation and the composition of the fuel.<ref name="VR-3-3">Van Rossum (2027), §3.3.</ref>

A stoichiometric mixture contains exactly the reactant proportions needed for the specified complete reaction. It is a chemical reference condition, rather than a complete description of a real fire. The reaction equation identifies the initial and final substances, but does not describe all intermediate reactions or establish how quickly combustion occurs.<ref name="D-92">Drysdale (2002), chapter 1-5, p. 1-92.</ref><ref name="VR-3-3" />

Stoichiometric calculations connect fuel mass, oxygen demand and air supply. They also provide the reference for distinguishing fuel-lean conditions, with excess oxidant, from fuel-rich conditions, with insufficient oxidant. In a compartment fire, that comparison helps describe the effect of restricted ventilation and the possibility of incomplete combustion products.<ref name="GL-58">Gottuk and Lattimer (2002), chapter 2-5, p. 2-58.</ref>

== Reading a balanced combustion equation ==

A balanced equation accounts for the atoms entering and leaving the specified reaction. The numbers before the chemical formulae express relative amounts of substance, measured in moles. They do not directly express kilograms of fuel, oxygen or products. Converting the equation into a mass requirement therefore needs a further step using the molar mass of each substance.<ref name="VR-3-3" />

For complete methane combustion, one mole of methane reacts with two moles of oxygen. The specified products are one mole of carbon dioxide and two moles of water. The equation therefore describes both the oxygen demand and the ideal product amounts. Van Rossum uses this reaction to explain the difference between reaction proportions and a detailed reaction mechanism.<ref name="VR-3-3" />

<div class="fse-equation">CH<sub>4</sub> + 2 O<sub>2</sub> → CO<sub>2</sub> + 2 H<sub>2</sub>O</div>

The coefficients preserve the carbon, hydrogen and oxygen totals. One carbon atom occurs in methane and in the resulting carbon dioxide. Four hydrogen atoms enter in methane and leave in two water molecules. The two oxygen molecules provide the four oxygen atoms divided between carbon dioxide and water. This is a direct reading of the balanced methane equation.<ref name="VR-3-3" />

The equation also contains assumptions about the products. Complete conversion to carbon dioxide and water is the specified reference reaction. A fire producing carbon monoxide, unburned hydrocarbons or soot has not followed that ideal product distribution. Gottuk and Lattimer explain why actual compartment-fire products cannot be predicted accurately by assuming ideal chemical behaviour alone.<ref name="GL-58" />

Physical states remain relevant when reaction equations are used for heat calculations. Drysdale distinguishes the chemical proportions from the thermochemical treatment of fuels and products. For example, a fuel can initially be liquid or gaseous even though its elemental composition is unchanged. Stoichiometry supplies the proportions; the associated energy calculation also needs the relevant initial and final states.<ref name="D-92" />

== Converting moles into mass ==

The mole is a measure of amount of substance. Molar mass connects that amount to an ordinary mass measurement. Dividing a substance's mass by its molar mass gives its amount in moles. Consistent units are necessary: a mass in grams pairs with grams per mole, while a mass in kilograms pairs with kilograms per mole.<ref name="VR-3-3" />

<div class="fse-equation">n = m / M</div>

{| class="wikitable"
! Symbol !! Description !! Unit
|-
| n || Amount of substance || mol
|-
| m || Mass of the substance || g
|-
| M || Molar mass, when mass is entered in grams || g/mol
|}

Van Rossum calculates methane's molar mass from its carbon and hydrogen atoms. The carbon contribution is 12.01 g/mol, and each of the four hydrogen contributions is 1.008 g/mol. Their sum is 16.042 g/mol, written as approximately 16.04 g/mol in the example. For a simple rounded mass comparison, the book uses about 16 g for one mole of methane.<ref name="VR-3-3" />

The two moles of oxygen required by the methane equation have a combined mass of approximately 64 g. Dividing 64 g by 16 g gives an oxygen-to-fuel mass ratio of approximately four. Thus, the simplified calculation requires about 4 kg of oxygen for 1 kg of methane. The factor of two in the reaction equation has become a factor of four in the mass comparison.<ref name="VR-3-3" />

This difference is a common source of misunderstanding. The reaction consumes twice as many moles of oxygen as methane, but those moles have different masses. Applying the reaction coefficients directly to kilograms would therefore give the wrong oxygen requirement. The molar-mass conversion is the step that makes a chemical equation usable in a fuel-mass calculation.<ref name="VR-3-3" />

The introductory book also gives approximate oxygen-demand factors for broader fuel groups. Its examples are about 1.25 kg of oxygen per kilogram of cellulosic fuel and 3.46 kg per kilogram of hydrocarbon fuel. These are simplified values for the stated groups. They do not replace a composition-specific calculation for every individual substance, as the propane calculation illustrates.<ref name="VR-3-3" /><ref name="D-92" />

== Worked example: oxygen and air for propane ==

Drysdale gives the complete combustion equation for propane as one mole of propane reacting with five moles of oxygen. The products are three moles of carbon dioxide and four moles of water. Using the rounded molar masses in that example, 44 g of propane requires 160 g of oxygen. The resulting oxygen-to-fuel mass ratio is approximately 3.64.<ref name="D-92" />

<div class="fse-equation">C<sub>3</sub>H<sub>8</sub> + 5 O<sub>2</sub> → 3 CO<sub>2</sub> + 4 H<sub>2</sub>O</div>

<div class="fse-equation">Oxygen-to-fuel mass ratio = 160 / 44 ≈ 3.64 kg/kg</div>

Air supplies nitrogen along with oxygen. Drysdale represents this using approximately 3.76 moles of nitrogen for each mole of oxygen, corresponding to the simplified oxygen-to-nitrogen proportion of 21:79. The five moles of oxygen therefore arrive with 18.8 moles of nitrogen. Nitrogen appears on both sides of this simplified complete-combustion equation.<ref name="D-92" />

<div class="fse-equation">C<sub>3</sub>H<sub>8</sub> + 5 O<sub>2</sub> + 18.8 N<sub>2</sub> → 3 CO<sub>2</sub> + 4 H<sub>2</sub>O + 18.8 N<sub>2</sub></div>

Using 28 g/mol for nitrogen, the accompanying nitrogen mass is 526.4 g. Adding the 160 g of oxygen gives 686.4 g of simplified air. Dividing this air mass by the 44 g propane mass gives 15.6 kg of air per kilogram of propane. The calculation shows why the oxygen requirement and the air requirement are numerically different quantities.<ref name="D-92" />

<div class="fse-equation">Air-to-fuel mass ratio = (160 + 18.8 × 28) / 44 = 15.6 kg/kg</div>

The inverse ratio is equally valid when clearly identified. Dividing the fuel mass by the air mass gives approximately 0.0641 kg of propane per kilogram of air. This is the stoichiometric fuel-to-air ratio for the same simplified reaction. It describes the same proportions as 15.6 kg/kg, but places fuel rather than air in the numerator.<ref name="D-92" />

The distinction between volume and mass fractions follows from the same calculation. Air's oxygen content is approximately 21% by volume in the introductory description. Gottuk and Lattimer use an oxygen mass fraction of approximately 0.23 in their fire calculations. These values have different bases and cannot be substituted for one another without changing the calculation.<ref name="VR-4-2">Van Rossum (2027), §4.2.</ref><ref name="GL-58" />

== Comparing an actual mixture with the reference ==

The [[Equivalence ratio|equivalence ratio]] compares an actual fuel-to-air ratio with the stoichiometric fuel-to-air ratio. Both ratios must use the same basis. Gottuk and Lattimer express the compartment-fire form using fuel mass-loss rate and incoming air mass-flow rate. Dividing their ratio by the stoichiometric fuel-to-air mass ratio gives a dimensionless measure, conventionally written as φ.<ref name="GL-58" />

<div class="fse-equation">φ = (ṁ<sub>f</sub> / ṁ<sub>a</sub>) / r</div>

{| class="wikitable"
! Symbol !! Description !! Unit
|-
| φ || Equivalence ratio || dimensionless
|-
| ṁ<sub>f</sub> || Fuel mass-loss rate || kg/s
|-
| ṁ<sub>a</sub> || Air mass-flow rate into the defined region || kg/s
|-
| r || Stoichiometric fuel-to-air mass ratio || kg/kg
|}

A value below one denotes fuel-lean or overventilated conditions. A value above one denotes fuel-rich or underventilated conditions. A value of one represents the stoichiometric proportion. In the ideal complete reaction, that proportion supplies exactly the oxygen needed to convert the fuel to the specified products without excess oxygen remaining.<ref name="GL-58" />

The definition of the denominator matters. Some sources use a fuel-to-air ratio, while others use its reciprocal, the air-to-fuel ratio. Gottuk and Lattimer explicitly identify this difference in notation. Substituting an air-to-fuel value into a formula defined with a fuel-to-air denominator changes the result, even if both source values are individually correct.<ref name="GL-58" />

The region being described also matters. A plume, an upper smoke layer and a whole compartment are different control volumes. A control volume is the selected region across whose boundaries fuel, air and products are counted. The SFPE treatment therefore requires the global equivalence ratio to be associated with a defined region, rather than treated as an unqualified property of the building.<ref name="GL-57">Gottuk and Lattimer (2002), chapter 2-5, pp. 2-56–2-57.</ref>

During changing conditions, a plume and the upper layer need not have the same ratio. The upper layer contains gases accumulated over time, while the plume can reflect a more recent fuel and air balance. Gottuk and Lattimer describe this difference through the residence time of gases in the layer. A stoichiometric comparison therefore also depends on how the flow history has been represented.<ref name="GL-57" />

== Heat release and ventilation-limited burning ==

Stoichiometry connects fuel consumption with oxygen consumption, but it does not assign the heat released by every reaction. Drysdale separately discusses heats of combustion expressed per mass of fuel, oxygen and air. For many listed fuels, the heat released per mass of oxygen consumed is approximately 13.1 kJ/g. This approximate relationship underlies the [[Oxygen consumption principle|oxygen consumption principle]].<ref name="D-93">Drysdale (2002), chapter 1-5, p. 1-93.</ref>

An energy value per kilogram of oxygen is different from an energy value per kilogram of fuel. Converting between them requires the corresponding oxygen-to-fuel mass relationship. This is one reason stoichiometry appears beside calorimetry in fire calculations. The chemical proportions identify the consumed quantities, while the energy relationship supplies the associated thermal output.<ref name="D-92" /><ref name="D-93" />

A [[Ventilation-controlled fire|ventilation-controlled fire]] develops when the available oxygen supply limits combustion. Van Rossum describes a growing compartment fire with openings that restrict the necessary air supply. Combustible gases can then remain unburned because the available oxygen is insufficient for complete combustion. The total fuel inventory alone no longer describes the combustion that can occur within the room.<ref name="VR-4-8">Van Rossum (2027), §4.8.</ref>

The ideal reaction equation remains a useful reference under these conditions, but does not become a prediction of actual products. Gottuk and Lattimer identify carbon monoxide, hydrogen and excess fuel among the products associated with underventilated burning. They also explain the limitations of representing all excess fuel simply as unburned hydrocarbons. Experimental correlations are needed to describe behaviour beyond that simplified chemical benchmark.<ref name="GL-58" />

The distinction is relevant to modelling fuel consumption and smoke production. A stoichiometric oxygen requirement can help establish whether the specified air flow can support complete burning. It does not independently establish the actual concentrations of every combustion product. The SFPE discussion relates those concentrations to combustion conditions, transport and the selected description of the fire compartment.<ref name="GL-54">Gottuk and Lattimer (2002), chapter 2-5, p. 2-54.</ref><ref name="GL-58" />

The quantity called a species yield expresses the mass of a particular product per mass of fuel burned. Oxygen is treated similarly through its depletion, meaning the oxygen consumed per mass of fuel. Gottuk and Lattimer compare these quantities with their theoretical maxima to describe how closely actual burning approaches complete combustion. This separates the ideal amount allowed by the fuel composition from the amount produced under the actual fire conditions.<ref name="GL-57" />

A gas concentration is another distinct measurement. It describes the composition of the sampled gas, which includes more than the product mass alone. The SFPE treatment distinguishes wet measurements, which include water vapour, from dry measurements taken after water removal. Removing water changes the reported fraction of the remaining gases. Consequently, a product amount calculated from stoichiometry cannot be compared directly with an unspecified measured concentration without accounting for the basis of that measurement.<ref name="GL-57" />

== Stoichiometric concentration and flammability limits ==

A [[Premixed flame|premixed flame]] burns a mixture in which fuel and oxidant have mixed before combustion. Van Rossum contrasts this with flames where fuel and surrounding air mix at the flame. Stoichiometric proportions can describe the premixed composition, but they do not imply that only one composition can burn. Flame propagation is possible over a range of appropriate mixture compositions.<ref name="VR-4-3">Van Rossum (2027), §4.3.</ref><ref name="B-172">Beyler (2002), chapter 2-7, p. 2-172.</ref>

[[Flammability limits]] define the boundaries of that range for specified conditions. Beyler gives methane limits of 5% and 15% by volume in air at normal temperature and pressure. Those values concern whether a flame can propagate through the mixture. Neither value is the stoichiometric oxygen requirement, and the interval between them is not a range of complete-reaction coefficients.<ref name="B-172" />

The methane equation provides a separate concentration calculation. One mole of methane needs two moles of oxygen. With the simplified air composition of 21% oxygen, that oxygen is contained in approximately 9.52 moles of air. Methane is therefore one part of a total mixture containing approximately 10.52 parts, giving a calculated stoichiometric concentration of about 9.5% by volume.<ref name="VR-3-3" /><ref name="VR-4-2" />

This calculated concentration lies between the cited methane flammability limits, but it answers a different question. It identifies the ideal reactant balance for complete combustion. The limits describe flame propagation under the stated test conditions. Treating the calculated 9.5% as an ignition threshold would confuse a reaction-proportion calculation with an experimentally defined combustion boundary.<ref name="VR-3-3" /><ref name="B-172" />

Temperature, pressure and mixture composition are part of the description of flammability limits. Beyler therefore states limits for a specified oxidant and diluent mixture, rather than presenting a fuel concentration in isolation. Stoichiometry retains its role as the balanced chemical reference. The additional combustion conditions determine how that reference relates to a particular flame or fire.<ref name="B-172" />

== See also ==
* [[Adiabatic flame temperature]]

== Notes ==
<references />

== References ==
* Beyler, C. L. (2002). Flammability limits of premixed and diffusion flames. In P. J. DiNenno (Ed.), ''SFPE handbook of fire protection engineering'' (3rd ed., chapter 2-7, pp. 2-172–2-187). National Fire Protection Association.
* Drysdale, D. D. (2002). Thermochemistry. In P. J. DiNenno (Ed.), ''SFPE handbook of fire protection engineering'' (3rd ed., chapter 1-5, pp. 1-90–1-98). National Fire Protection Association.
* Gottuk, D. T., & Lattimer, B. Y. (2002). Effect of combustion conditions on species production. In P. J. DiNenno (Ed.), ''SFPE handbook of fire protection engineering'' (3rd ed., chapter 2-5, pp. 2-54–2-82). National Fire Protection Association.
* Van Rossum, E. (2027). ''Introduction to Fire Safety Engineering''. Fire Safety Academy.

[[Category:Combustion]]