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Stoichiometry

From Wikipedia, the free encyclopedia
Calculation of relative masses of reactants and products in chemical reactions
A stoichiometric diagram of thecombustion reaction ofmethane

Stoichiometry (/ˌstɔɪkiˈɒmɪtri/ ) is the relationships between the quantities ofreactants andproducts before, during, and followingchemical reactions.

Stoichiometry is based on thelaw of conservation of mass; the total mass of reactants must equal the total mass of products, so the relationship between reactants and products must form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of the products can be empirically determined, then the amount of the other reactants can also be calculated.

This is illustrated in the image here, where the unbalanced equation is:

CH4 (g) + O2 (g) → CO2 (g) + H2O (l)
However, the current equation is imbalanced. The reactants have 4 hydrogen and 2 oxygen atoms, while the product has 2 hydrogen and 3 oxygen. To balance the hydrogen, a coefficient of 2 is added to the productH2O, and to fix the imbalance of oxygen, it is also added toO2. Thus, we get:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

Here, onemolecule ofmethane reacts with two molecules ofoxygen gas to yield one molecule ofcarbon dioxide and two molecules of liquidwater. This particular chemical equation is an example ofcomplete combustion. The numbers in front of each quantity are a set of stoichiometric coefficients which directly reflect themolar ratios between the products and reactants. Stoichiometry measures these quantitative relationships, and is used to determine the amount of products and reactants that are produced or needed in a given reaction.

Describing the quantitative relationships among substances as they participate in chemical reactions is known asreaction stoichiometry. In the example above, reaction stoichiometry measures the relationship between the quantities of methane and oxygen that react to form carbon dioxide and water: for every mole ofmethane combusted, two moles of oxygen are consumed, one mole of carbon dioxide is produced, and two moles of water are produced.

Because of the well known relationship of moles toatomic weights, the ratios that are arrived at by stoichiometry can be used to determine quantities by weight in a reaction described by a balanced equation. This is calledcomposition stoichiometry.

Gas stoichiometry deals with reactions solely involving gases, where the gases are at a known temperature, pressure, and volume and can be assumed to beideal gases. For gases, the volume ratio is ideally the same by theideal gas law, but the mass ratio of a single reaction has to be calculated from themolecular masses of the reactants and products. In practice, because of the existence ofisotopes,molar masses are used instead in calculating the mass ratio.

Etymology

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The termstoichiometry was first used byJeremias Benjamin Richter in 1792 when the first volume of Richter'sAnfangsgründe der Stöchyometrie oder Meßkunst chymischer Elemente (German for 'Fundamentals of Stoichiometry, or the Art of Measuring the Chemical Elements') was published.[1] The term is derived from theAncient Greek wordsστοιχεῖον (stoikheîon), meaning 'element',[2] andμέτρον (métron), meaning 'measure'.Ludwig Darmstaedter andRalph E. Oesper have written a useful account on this.[3]

Definitions

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Main article:Amount of substance

Astoichiometric amount[4] orstoichiometric ratio of areagent is the optimumamount orratio where, assuming that the reaction proceeds to completion:

  1. All of the reagent is consumed
  2. There is no deficiency of the reagent
  3. There is no excess of the reagent.

Stoichiometry rests upon the very laws that help to understand it better, i.e.,law of conservation of mass, thelaw of definite proportions (i.e., thelaw of constant composition), thelaw of multiple proportions and thelaw of reciprocal proportions. In generalchemical reactions combine in definite ratios ofchemicals. Since chemical reactions can neither create nor destroy matter, nortransmute one element into another, the amount of each element must be the same throughout the overall reaction. For example, the number of atoms of a given element X on the reactant side must equal the number of atoms of that element on the product side, whether or not all of those atoms are involved in a reaction.[5]

Chemical reactions, asmacroscopic unit operations, consist of manyelementary reactions, where a single molecule reacts with another molecule. As the reactingmolecules (orformula units orion pairs) consist of a definite set ofatoms in aninteger ratio, the ratio between reactants in a complete reaction is also in an integer ratio. A reaction may consume more than one molecule, and thestoichiometric number counts this number, defined as positive forproducts (added) and negative forreactants (removed).[6] The unsigned coefficients are generally referred to as thestoichiometric coefficients.[7]

Eachelement has anatomic mass (usually given as an average in the form of thestandard atomic weight), and considering molecules as collections of atoms, every compound has amolecular mass (if molecular) orformula mass (if non-molecular), which when expressed indaltons is numerically equal to themolar mass ing/mol. By definition, the atomic mass ofcarbon-12 is exactly 12 Da, making its molar mass 12 g/mol. The number ofchemical entities per mole in a substance is given by theAvogadro constant, exactly6.02214076×1023 mol−1 since the2019 revision of the SI. Thus, to calculate the stoichiometry bymass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction. The mass ratios can be calculated by dividing each by the total in the whole reaction.

Elements in their natural state are mixtures ofisotopes of differing mass; thus, atomic masses and thus molar masses are not exactly integers. For instance, instead of an exact 14:3 proportion, 17.031 g ofammonia consists of 14.007 g ofnitrogen and 3 × 1.008 g ofhydrogen, because natural nitrogen includes a small amount ofnitrogen-15, and natural hydrogen includes hydrogen-2 (deuterium).

Astoichiometric reactant is a reactant that is consumed in a reaction, as opposed to acatalytic reactant, which is not consumed in the overall reaction because it reacts in one step and is regenerated in another step.

Converting grams to moles

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Stoichiometry is not only used to balance chemical equations but also used in "conversions" between quantities of a substance bydimensional analysis, e.g., converting fromgrams tomoles usingmolar mass as the "conversion factor", or from grams tomilliliters usingdensity. For example, to express 2.00 g of NaCl (sodium chloride) as anamount (in moles), one would do the following:

2.00 g NaCl58.44 g/mol=0.0342 mol NaCl{\displaystyle {\frac {2.00{\mbox{ g NaCl}}}{58.44{\mbox{ g/mol}}}}=0.0342\ {\text{mol NaCl}}}

In the above example, when written out in fraction form, the units of grams form amultiplicative identity, which is equivalent to one (g/g = 1), with the resulting amount in moles (the unit that was needed), as shown in the following equation,

(2.00 g NaCl1)(1 mol NaCl58.44 g NaCl)=0.0342 mol NaCl{\displaystyle \left({\frac {2.00{\mbox{ g NaCl}}}{1}}\right)\left({\frac {1{\mbox{ mol NaCl}}}{58.44{\mbox{ g NaCl}}}}\right)=0.0342\ {\text{mol NaCl}}}

Molar proportion

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Stoichiometry is often used to balance chemical equations (reaction stoichiometry). For example, the twodiatomic gases,hydrogen andoxygen, can combine to form a liquid, water, in anexothermic reaction, as described by the following equation:

2 H2 + O2 → 2 H2O

Reaction stoichiometry describes the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above equation.

The molar ratio allows for conversion between moles of one substance and moles of another. For example, in the reaction

2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

the amount of water that will be produced by the combustion of 0.27 moles ofCH
3
OH
is obtained using the molar ratio betweenCH
3
OH
andH
2
O
of 2 to 4.

(0.27 mol CH3OH1)(4 mol H2O2 mol CH3OH)=0.54 mol H2O{\displaystyle \left({\frac {0.27{\mbox{ mol }}\mathrm {CH_{3}OH} }{1}}\right)\left({\frac {4{\mbox{ mol }}\mathrm {H_{2}O} }{2{\mbox{ mol }}\mathrm {CH_{3}OH} }}\right)=0.54\ {\text{mol }}\mathrm {H_{2}O} }

The term stoichiometry is also often used for themolar proportions of elements in stoichiometric compounds (composition stoichiometry). For example, the stoichiometry of hydrogen and oxygen inH
2
O
is 2:1. In stoichiometric compounds, the molar proportions are whole numbers.

Determining amount of product

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Stoichiometry can also be used to find the quantity of a product yielded by a reaction. If a piece of solidcopper (Cu) were added to an aqueous solution ofsilver nitrate (AgNO3), thesilver (Ag) would be replaced in asingle displacement reaction forming aqueouscopper(II) nitrate (Cu(NO3)2) and solid silver. How much silver is produced if 16.00 grams of Cu is added to the solution of excess silver nitrate?

The following steps would be used:

  1. Write and balance the equation
  2. Mass to moles: Convert grams of Cu to moles of Cu
  3. Mole ratio: Convert moles of Cu to moles of Ag produced
  4. Mole to mass: Convert moles of Ag to grams of Ag produced

The complete balanced equation would be:

Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

For the mass to mole step, the mass of copper (16.00 g) would be converted to moles of copper by dividing the mass of copper by itsmolar mass: 63.55 g/mol.

(16.00 g Cu1)(1 mol Cu63.55 g Cu)=0.2518 mol Cu{\displaystyle \left({\frac {16.00{\mbox{ g Cu}}}{1}}\right)\left({\frac {1{\mbox{ mol Cu}}}{63.55{\mbox{ g Cu}}}}\right)=0.2518\ {\text{mol Cu}}}

Now that the amount of Cu in moles (0.2518) is found, we can set up the mole ratio. This is found by looking at the coefficients in the balanced equation: Cu and Ag are in a 1:2 ratio.

(0.2518 mol Cu1)(2 mol Ag1 mol Cu)=0.5036 mol Ag{\displaystyle \left({\frac {0.2518{\mbox{ mol Cu}}}{1}}\right)\left({\frac {2{\mbox{ mol Ag}}}{1{\mbox{ mol Cu}}}}\right)=0.5036\ {\text{mol Ag}}}

Now that the moles of Ag produced is known to be 0.5036 mol, we convert this amount to grams of Ag produced to come to the final answer:

(0.5036 mol Ag1)(107.87 g Ag1 mol Ag)=54.32 g Ag{\displaystyle \left({\frac {0.5036{\mbox{ mol Ag}}}{1}}\right)\left({\frac {107.87{\mbox{ g Ag}}}{1{\mbox{ mol Ag}}}}\right)=54.32\ {\text{g Ag}}}

This set of calculations can be further condensed into a single step:

mAg=(16.00 g Cu1)(1 mol Cu63.55 g Cu)(2 mol Ag1 mol Cu)(107.87 g Ag1 mol Ag)=54.32 g{\displaystyle m_{\mathrm {Ag} }=\left({\frac {16.00{\mbox{ g }}\mathrm {Cu} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Cu} }{63.55{\mbox{ g }}\mathrm {Cu} }}\right)\left({\frac {2{\mbox{ mol }}\mathrm {Ag} }{1{\mbox{ mol }}\mathrm {Cu} }}\right)\left({\frac {107.87{\mbox{ g }}\mathrm {Ag} }{1{\mbox{ mol Ag}}}}\right)=54.32{\mbox{ g}}}

Further examples

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Forpropane (C3H8) reacting withoxygen gas (O2), the balanced chemical equation is:

C3H8 + 5 O2 → 3 CO2 + 4 H2O

The mass of water formed if 120 g of propane (C3H8) is burned in excess oxygen is then

mH2O=(120. g C3H81)(1 mol C3H844.09 g C3H8)(4 mol H2O1 mol C3H8)(18.02 g H2O1 mol H2O)=196 g{\displaystyle m_{\mathrm {H_{2}O} }=\left({\frac {120.{\mbox{ g }}\mathrm {C_{3}H_{8}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {C_{3}H_{8}} }{44.09{\mbox{ g }}\mathrm {C_{3}H_{8}} }}\right)\left({\frac {4{\mbox{ mol }}\mathrm {H_{2}O} }{1{\mbox{ mol }}\mathrm {C_{3}H_{8}} }}\right)\left({\frac {18.02{\mbox{ g }}\mathrm {H_{2}O} }{1{\mbox{ mol }}\mathrm {H_{2}O} }}\right)=196{\mbox{ g}}}

Stoichiometric ratio

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Stoichiometry is also used to find the right amount of onereactant to "completely" react with the other reactant in achemical reaction – that is, the stoichiometric amounts that would result in no leftover reactants when the reaction takes place. An example is shown below using thethermite reaction,[citation needed]

Fe2O3 + 2 Al → Al2O3 + 2 Fe

This equation shows that 1 mole ofiron(III) oxide and 2 moles ofaluminium will produce 1 mole ofaluminium oxide and 2 moles ofiron. So, to completely react with 85.0 g ofiron(III) oxide (0.532 mol), 28.7 g (1.06 mol) of aluminium are needed.

mAl=(85.0 g Fe2O31)(1 mol Fe2O3159.7 g Fe2O3)(2 mol Al1 mol Fe2O3)(26.98 g Al1 mol Al)=28.7 g{\displaystyle m_{\mathrm {Al} }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}}

Limiting reagent and percent yield

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Main articles:Limiting reagent andYield (chemistry)

The limiting reagent is the reagent that limits the amount of product that can be formed and is completely consumed when the reaction is complete. An excess reactant is a reactant that is left over once the reaction has stopped due to the limiting reactant being exhausted.

Consider the equation of roastinglead(II) sulfide (PbS) in oxygen (O2) to producelead(II) oxide (PbO) andsulfur dioxide (SO2):

2 PbS + 3 O2 → 2 PbO + 2 SO2

To determine the theoretical yield of lead(II) oxide if 200.0 g of lead(II) sulfide and 200.0 g of oxygen are heated in an open container:

mPbO=(200.0 g PbS1)(1 mol PbS239.27 g PbS)(2 mol PbO2 mol PbS)(223.2 g PbO1 mol PbO)=186.6 g{\displaystyle m_{\mathrm {PbO} }=\left({\frac {200.0{\mbox{ g }}\mathrm {PbS} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {PbS} }{239.27{\mbox{ g }}\mathrm {PbS} }}\right)\left({\frac {2{\mbox{ mol }}\mathrm {PbO} }{2{\mbox{ mol }}\mathrm {PbS} }}\right)\left({\frac {223.2{\mbox{ g }}\mathrm {PbO} }{1{\mbox{ mol }}\mathrm {PbO} }}\right)=186.6{\mbox{ g}}}
mPbO=(200.0 g O21)(1 mol O232.00 g O2)(2 mol PbO3 mol O2)(223.2 g PbO1 mol PbO)=930.0 g{\displaystyle m_{\mathrm {PbO} }=\left({\frac {200.0{\mbox{ g }}\mathrm {O_{2}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {O_{2}} }{32.00{\mbox{ g }}\mathrm {O_{2}} }}\right)\left({\frac {2{\mbox{ mol }}\mathrm {PbO} }{3{\mbox{ mol }}\mathrm {O_{2}} }}\right)\left({\frac {223.2{\mbox{ g }}\mathrm {PbO} }{1{\mbox{ mol }}\mathrm {PbO} }}\right)=930.0{\mbox{ g}}}

Because a lesser amount of PbO is produced for the 200.0 g of PbS, it is clear that PbS is the limiting reagent.

In reality, the actual yield is not the same as the stoichiometrically-calculated theoretical yield. Percent yield, then, is expressed in the following equation:

percent yield=actual yieldtheoretical yield{\displaystyle {\mbox{percent yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}}

If 170.0 g of lead(II) oxide is obtained, then the percent yield would be calculated as follows:

percent yield=170.0 g PbO186.6 g PbO=91.12%{\displaystyle {\mbox{percent yield}}={\frac {\mbox{170.0 g PbO}}{\mbox{186.6 g PbO}}}=91.12\%}

Example

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Consider the following reaction, in whichiron(III) chloride reacts withhydrogen sulfide to produceiron(III) sulfide andhydrogen chloride:

2 FeCl3 + 3 H2S → Fe2S3 + 6 HCl

The stoichiometric masses for this reaction are:

324.41 gFeCl3, 102.25 gH2S, 207.89 gFe2S3, 218.77 g HCl

Suppose 90.0 g ofFeCl3 reacts with 52.0 g ofH2S. To find the limiting reagent and the mass of HCl produced by the reaction, we change the above amounts by a factor of 90/324.41 and obtain the following amounts:

90.00 gFeCl3, 28.37 gH2S, 57.67 gFe2S3, 60.69 g HCl

The limiting reactant (or reagent) isFeCl3, since all 90.00 g of it is used up while only 28.37 gH2S are consumed. Thus, 52.0 − 28.4 = 23.6 gH2S left in excess. The mass of HCl produced is 60.7 g.

By looking at the stoichiometry of the reaction, one might have guessedFeCl3 being the limiting reactant; three times moreFeCl3 is used compared toH2S (324 g vs 102 g).

Different stoichiometries in competing reactions

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Often, more than one reaction is possible given the same starting materials. The reactions may differ in their stoichiometry. For example, themethylation ofbenzene (C6H6), through aFriedel–Crafts reaction usingAlCl3 as a catalyst, may produce singly methylated (C6H5CH3), doubly methylated (C6H4(CH3)2), or still more highly methylated (C6H6−n(CH3)n) products, as shown in the following example,

C6H6 + CH3Cl → C6H5CH3 + HCl
C6H6 + 2 CH3Cl → C6H4(CH3)2 + 2 HCl
C6H6 +n CH3Cl → C6H6−n(CH3)n +n HCl

In this example, which reaction takes place is controlled in part by the relativeconcentrations of the reactants.

Stoichiometric coefficient and stoichiometric number

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In lay terms, thestoichiometric coefficient of any given component is the number of molecules and/orformula units that participate in the reaction as written. A related concept is thestoichiometric number (using IUPAC nomenclature), wherein the stoichiometric coefficient is multiplied by +1 for all products and by −1 for all reactants.

For example, in the reactionCH4 + 2 O2CO2 + 2 H2O, the stoichiometric number ofCH4 is −1, the stoichiometric number ofO2 is −2, for CO2 it would be +1 and forH2O it is +2.

In more technically precise terms, the stoichiometric number in achemical reactionsystem of thei-th component is defined as

νi=ΔNiΔξ{\displaystyle \nu _{i}={\frac {\Delta N_{i}}{\Delta \xi }}\,}

or

ΔNi=νiΔξ{\displaystyle \Delta N_{i}=\nu _{i}\,\Delta \xi \,}

whereNi{\displaystyle N_{i}} is the number ofmolecules ofi, andξ{\displaystyle \xi } is the progress variable orextent of reaction.[8][9]

The stoichiometric number νi{\displaystyle \nu _{i}} represents the degree to which a chemical species participates in a reaction. The convention is to assign negative numbers toreactants (which are consumed) and positive ones toproducts, consistent with the convention that increasing the extent of reaction will correspond to shifting the composition from reactants towards products. However, any reaction may be viewed as going in the reverse direction, and in that point of view, would change in the negative direction in order to lower the system's Gibbs free energy. Whether a reaction actuallywill go in the arbitrarily selected forward direction or not depends on the amounts of thesubstances present at any given time, which determines thekinetics andthermodynamics, i.e., whetherequilibrium lies to theright or theleft of the initial state,

Inreaction mechanisms, stoichiometric coefficients for each step are alwaysintegers, since elementary reactions always involve whole molecules. If one uses a composite representation of an overall reaction, some may berationalfractions. There are often chemical species present that do not participate in a reaction; their stoichiometric coefficients are therefore zero. Any chemical species that is regenerated, such as acatalyst, also has a stoichiometric coefficient of zero.

The simplest possible case is anisomerization

A → B

in whichνB = 1 since one molecule of B is produced each time the reaction occurs, whileνA = −1 since one molecule of A is necessarily consumed. In any chemical reaction, not only is the totalmass conserved but also the numbers ofatoms of eachkind are conserved, and this imposes corresponding constraints on possible values for the stoichiometric coefficients.

There are usually multiple reactions proceeding simultaneously in anynatural reaction system, including those inbiology. Since any chemical component can participate in several reactions simultaneously, the stoichiometric number of thei-th component in thek-th reaction is defined as

νik=Niξk{\displaystyle \nu _{ik}={\frac {\partial N_{i}}{\partial \xi _{k}}}\,}

so that the total (differential) change in the amount of thei-th component is

dNi=kνikdξk.{\displaystyle dN_{i}=\sum _{k}\nu _{ik}\,d\xi _{k}.\,}

Extents of reaction provide the clearest and most explicit way of representing compositional change, although they are not yet widely used.

With complex reaction systems, it is often useful to consider both the representation of a reaction system in terms of the amounts of the chemicals presentNi } (state variables), and the representation in terms of the actual compositionaldegrees of freedom, as expressed by the extents of reactionξk }. The transformation from avector expressing the extents to a vector expressing the amounts uses a rectangularmatrix whose elements are the stoichiometric numbersνi k ].

Themaximum and minimum for anyξk occur whenever the first of the reactants is depleted for the forward reaction; or the first of the "products" is depleted if the reaction as viewed as being pushed in the reverse direction. This is a purelykinematic restriction on the reactionsimplex, ahyperplane in composition space, orN‑space, whosedimensionality equals the number oflinearly-independent chemical reactions. This is necessarily less than the number of chemical components, since each reaction manifests a relation between at least two chemicals. The accessible region of the hyperplane depends on the amounts of each chemical species actually present, a contingent fact. Different such amounts can even generate different hyperplanes, all sharing the same algebraic stoichiometry.

In accord with the principles ofchemical kinetics andthermodynamic equilibrium, every chemical reaction isreversible, at least to some degree, so that each equilibrium point must be aninterior point of the simplex. As a consequence, extrema for theξs will not occur unless an experimental system is prepared with zero initial amounts of some products.

The number ofphysically-independent reactions can be even greater than the number of chemical components, and depends on the various reaction mechanisms. For example, there may be two (or more) reactionpaths for the isomerism above. The reaction may occur by itself, but faster and with different intermediates, in the presence of a catalyst.

The (dimensionless) "units" may be taken to bemolecules ormoles. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules. TheNs andξs are reduced to molar units by dividing by theAvogadro constant. While dimensionalmass units may be used, the comments about integers are then no longer applicable.

Stoichiometry matrix

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Main article:Chemical reaction network theory

In complex reactions, stoichiometries are often represented in a more compact form called the stoichiometry matrix. The stoichiometry matrix is denoted by the symbolN.[10][11][12]

If a reaction network hasn reactions andm participating molecular species, then the stoichiometry matrix will have correspondinglym rows andn columns.

For example, consider the system of reactions shown below:

S1 → S2
5 S3 + S2 → 4 S3 + 2 S2
S3 → S4
S4 → S5

This system comprises four reactions and five different molecular species. The stoichiometry matrix for this system can be written as:

N=[10001100011000110001]{\displaystyle \mathbf {N} ={\begin{bmatrix}-1&0&0&0\\1&1&0&0\\0&-1&-1&0\\0&0&1&-1\\0&0&0&1\\\end{bmatrix}}}

where the rows correspond toS1, S2, S3, S4 and S5, respectively. The process of converting a reaction scheme into a stoichiometry matrix can be a lossy transformation: for example, the stoichiometries in the second reaction simplify when included in the matrix. This means that it is not always possible to recover the original reaction scheme from a stoichiometry matrix.

Often the stoichiometry matrix is combined with the rate vector,v, and the species vector,x to form a compact equation, thebiochemical systems equation, describing the rates of change of the molecular species:

dxdt=Nv.{\displaystyle {\frac {d\mathbf {x} }{dt}}=\mathbf {N} \cdot \mathbf {v} .}

Gas stoichiometry

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Gas stoichiometry is the quantitative relationship (ratio) between reactants and products in achemical reaction with reactions that producegases. Gas stoichiometry applies when the gases produced are assumed to beideal, and the temperature, pressure, and volume of the gases are all known. The ideal gas law is used for these calculations. Often, but not always, thestandard temperature and pressure (STP) are taken as 0 °C and 1 bar and used as the conditions for gas stoichiometric calculations.

Gas stoichiometry calculations solve for the unknownvolume ormass of a gaseous product or reactant. For example, if we wanted to calculate the volume of gaseousNO2 produced from the combustion of 100 g ofNH3, by the reaction:

4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (l)

we would carry out the following calculations:

100gNH31molNH317.034gNH3=5.871molNH3{\displaystyle 100\,\mathrm {g\,NH_{3}} \cdot {\frac {1\,\mathrm {mol\,NH_{3}} }{17.034\,\mathrm {g\,NH_{3}} }}=5.871\,\mathrm {mol\,NH_{3}} }

There is a 1:1 molar ratio ofNH3 toNO2 in the above balanced combustion reaction, so 5.871 mol ofNO2 will be formed. We will employ theideal gas law to solve for the volume at 0 °C (273.15 K) and 1 atmosphere using thegas law constant ofR = 0.08206 L·atm·K−1·mol−1:

PV=nRTV=nRTP=5.871 mol0.08206LatmmolK273.15 K1 atm=131.597LNO2{\displaystyle {\begin{aligned}PV&=nRT\\V&={\frac {nRT}{P}}\\&={\frac {5.871{\text{ mol}}\cdot 0.08206\,{\frac {\mathrm {L\cdot atm} }{\mathrm {mol\cdot K} }}\cdot 273.15{\text{ K}}}{1{\text{ atm}}}}\\&=131.597\,\mathrm {L\,NO_{2}} \end{aligned}}}

Gas stoichiometry often involves having to know themolar mass of a gas, given thedensity of that gas. The ideal gas law can be re-arranged to obtain a relation between thedensity and themolar mass of an ideal gas:

ρ=mV{\displaystyle \rho ={\frac {m}{V}}}    and   n=mM{\displaystyle n={\frac {m}{M}}}

and thus:

ρ=MPRT{\displaystyle \rho ={\frac {MP}{R\,T}}}

where:

  • P = absolute gaspressure
  • V = gasvolume
  • n = amount (measured inmoles)
  • R = universal ideal gas law constant
  • T = absolute gastemperature
  • ρ = gas density atT andP
  • m = mass of gas
  • M = molar mass of gas

Stoichiometric air-to-fuel ratios of common fuels

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See also:Air–fuel ratio andCombustion

In thecombustion reaction, oxygen reacts with the fuel, and the point where exactly all oxygen is consumed and all fuel burned is defined as the stoichiometric point. With more oxygen (overstoichiometric combustion), some of it stays unreacted. Likewise, if the combustion is incomplete due to lack of sufficient oxygen, fuel remains unreacted. (Unreacted fuel may also remain due to physical rather than chemical factors, such as slow combustion or insufficient mixing of fuel and oxygen – this is not due to stoichiometry.) Different hydrocarbon fuels have different contents of carbon, hydrogen and other elements, thus their stoichiometry varies.

Oxygen makes up only 20.95% of the volume of air, and only 23.20% of its mass.[13] The air-fuel ratios listed below are much higher than the equivalent oxygen-fuel ratios, due to the high proportion of inert gasses in the air.(The ratio of 14.7 as listed for gasoline only applies for combustion in combination with a catalytic converter.)

FuelRatio by mass[14]Ratio by volume[15][full citation needed]Percent fuel by massMain reaction
Gasoline14.7 : 16.9%2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
Natural gas14.5 : 19.7  : 16.9%CH4 + 2 O2 → CO2 + 2 H2O
Propane (LP)15.67 : 123.9 : 16.45%C3H8 + 5 O2 → 3 CO2 + 4 H2O
Ethanol9 : 111.1%C2H6O + 3 O2 → 2 CO2 + 3 H2O
Methanol6.47 : 115.6%2 CH4O + 3 O2 → 2 CO2 + 4 H2O
n-Butanol11.2 : 18.2%C4H10O + 6 O2 → 4 CO2 + 5 H2O
Hydrogen34.3 : 12.39 : 12.9%2 H2 + O2 → 2 H2O
Diesel14.5 : 1[citation needed]6.8%2 C12H26 + 37 O2 → 24 CO2 + 26 H2O
Methane17.23 : 19.52 : 15.5%CH4 + 2 O2 → CO2 + 2 H2O
Acetylene13.26 : 1[citation needed]11.92 : 17.0%2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
Ethane16.07 : 1[citation needed]16.68 : 15.9%2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
Butane15.44 : 1[citation needed]30.98 : 16.1%2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
Pentane15.31 : 1[citation needed]38.13 : 16.1%C5H12 + 8 O2 → 5 CO2 + 6 H2O

Gasoline engines can run at stoichiometric air-to-fuel ratio, because gasoline is quite volatile and is mixed (sprayed or carburetted) with the air prior to ignition. Diesel engines, in contrast, run lean, with more air available than simple stoichiometry would require. Diesel fuel is less volatile and is effectively burned as it is injected.[16]

See also

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References

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  1. ^Richter, J. B. (1792).Anfangsgründe der Stöchyometrie ... (in 3 vol.s) [Rudiments of Stoichiometry ...] (in German). Vol. 1. Breslau and Hirschberg, (Germany): Johann Friedrich Korn der Aeltere. p. 121.Die Stöchyometrie (Stöchyometria) ist die Wissenschaft die quantitativen oder Massenverhältnisse ... zu messen, in welchen die chemischen Elemente ... gegen einander stehen.transl. "Stoichiometry (stoichiometria) is the science of measuring the quantitative or mass relations in which the chemical "elements" exist in relation to each other." [On pp. 3–7, Richter explains that an "element" is a pure substance, and that a "chemical element" (chymisches Element (Elementum chymicum)) is a substance that cannot be resolved into dissimilar substances by known physical or chemical means. Thus, for example,aluminium oxide was a "chemical element" because in Richter's time, it couldn't be resolved further into its component elements.]
  2. ^Sinnott, R. K. (2005).Coulson and Richardson's Chemical Engineering (4th ed.). Amsterdam, Paris:Elsevier,Butterworth-Heinemann. p. 36.ISBN 978-0-7506-6538-4.
  3. ^L. Darmstaedter; R. E. Oesper (1928). "Jeremias Benjamin Richter".Journal of Chemical Education.5 (7):785–790.Bibcode:1928JChEd...5..785D.doi:10.1021/ed005p785.
  4. ^What's in a Name? Amount of Substance, Chemical Amount, and Stoichiometric Amount Carmen J. Giunta Journal of Chemical Education 2016 93 (4), 583-586doi:10.1021/acs.jchemed.5b00690
  5. ^"Stoichiometry of Chemical Reactions"(PDF).
  6. ^IUPAC,Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "stoichiometric number,ν".doi:10.1351/goldbook.S06025
  7. ^Nijmeh, Joseph; Tye, Mark (2 October 2013)."Stoichiometry and Balancing Reactions".LibreTexts. Retrieved5 May 2021.
  8. ^Prigogine & Defay, p. 18; Prigogine, pp. 4–7; Guggenheim, p. 37 & 62
  9. ^IUPAC,Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "extent of reaction,ξ".doi:10.1351/goldbook.E02283
  10. ^Ghaderi, Susan; Haraldsdóttir, Hulda S.; Ahookhosh, Masoud; Arreckx, Sylvain; Fleming, Ronan M.T. (August 2020)."Structural conserved moiety splitting of a stoichiometric matrix".Journal of Theoretical Biology.499 110276.Bibcode:2020JThBi.49910276G.doi:10.1016/j.jtbi.2020.110276.hdl:1887/3134882.PMID 32333975.
  11. ^Hofmeyr, Jan-hendrik S. (2001). "Metabolic control analysis in a nutshell".In Proceedings of the 2 Nd International Conference on Systems Biology:291–300.CiteSeerX 10.1.1.324.922.
  12. ^Reder, Christine (21 November 1988). "Metabolic control theory: A structural approach".Journal of Theoretical Biology.135 (2):175–201.Bibcode:1988JThBi.135..175R.doi:10.1016/s0022-5193(88)80073-0.PMID 3267767.
  13. ^"Universal Industrial Gases, Inc: Composition of Air - Components & Properties of Air - Answers to "What is air?" - "What is air made up of?" -" What are air products and what are they used for?"".
  14. ^John B. Heywood: "Internal Combustion Engine Fundamentals page 915", 1988
  15. ^North American Mfg. Co.: "North American Combustion Handbook", 1952
  16. ^"Air-fuel ratio, lambda and engine performance". Retrieved2019-05-31.
  • Zumdahl, Steven S.Chemical Principles. Houghton Mifflin, New York, 2005, pp 148–150.
  • Internal Combustion Engine Fundamentals, John B. Heywood

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