Stoichiometry // is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of positive integers. For example, in a reaction that forms ammonia (NH_{3}), exactly one molecule of nitrogen gas (N_{2}) reacts with three molecules of hydrogen gas (H_{2}) to produce two molecules of NH_{3}:
 Template:Chem/link + 3Template:Chem/link → 2Template:Chem/link
This particular kind of stoichiometry  describing the quantitative relationships among substances as they participate in chemical reactions  is known as reaction stoichiometry. In the example above, reaction stoichiometry describes the 1:3:2 ratio of molecules of nitrogen, hydrogen, and ammonia.
Stoichiometry can be used to determine quantities such as the amount of products (in mass, moles, volume, etc.) that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product). Stoichiometry calculations can predict how elements and components diluted in a standard solution react in experimental conditions. Stoichiometry is founded on the law of conservation of mass: the mass of the reactants equals the mass of the products.
Composition stoichiometry describes the quantitative (mass) relationships among elements in compounds. For example, composition stoichiometry describes the nitrogen to hydrogen ratio in the compound ammonia (NH_{3}): 1 mol of ammonia consists of 1 mol of nitrogen and 3 mol of hydrogen. As the nitrogen atom is about 14 times heavier than the hydrogen atom, the mass ratio is 14:3, thus 17 kg of ammonia contains 14 kg of nitrogen and 3 kg of hydrogen.
A stoichiometric amount or stoichiometric ratio of a reagent is the optimum amount or ratio where, assuming that the reaction proceeds to completion:
 All of the reagent is consumed
 There is no deficiency of the reagent
 There is no excess of the reagent.
A nonstoichiometric mixture, wherein reactions have gone to completion, will have only the limiting reagent consumed completely.
While almost all reactions have integerratio stoichiometry in amount of matter units (moles, number of particles), some nonstoichiometric compounds that cannot be represented by a ratio of welldefined natural numbers are known. These materials, therefore, violate the law of definite proportions that forms the basis of stoichiometry along with the law of multiple proportions.
Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume, and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio.
Etymology
The term stoichiometry was first used by Jeremias Benjamin Richter in 1792 when the first volume of Richter's Stoichiometry or the Art of Measuring the Chemical Elements was published. The term is derived from the Greek words στοιχεῖον stoicheion "element" and μέτρον metron "measure". In patristic Greek, the word Stoichiometria was used by Nicephorus to refer to the number of line counts of the canonical New Testament and some of the Apocrypha.
Definition
Stoichiometry rests upon the very basic laws that help to understand it better, i.e., law of conservation of mass, the law of definite proportions (i.e., the law of constant composition), and the law of multiple proportions. In general, chemical reactions combine in definite ratios of chemicals. Since chemical reactions can neither create nor destroy matter, nor transmute 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 actually involved in a reaction.
Chemical reactions, as macroscopic unit operations, consist of simply a very large number of elementary reactions, where a single molecule reacts with another molecule. As the reacting molecules (or moieties) consist of a definite set of atoms in an integer ratio, the ratio between reactants in a complete reaction is also in integer ratio. A reaction may consume more than one molecule, and the stoichiometric number counts this number, defined as positive for products (added) and negative for reactants (removed).^{[1]}
Different elements have a different atomic mass, and as collections of single atoms, molecules have a definite molar mass, measured with the unit mole (6.02 × 10^{23} individual molecules, Avogadro's constant). By definition, carbon12 has a molar mass of 12 g/mol. Thus, to calculate the stoichiometry by mass, 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.
Converting grams to moles
Stoichiometry is not only used to balance chemical equations but also used in conversions, i.e., converting from grams to moles, or from grams to milliliters. For example, to find the number of moles in 2.00 g of NaCl, one would do the following:
 $\backslash frac\{2.00\; \backslash mbox\{\; g\; NaCl\}\}\{58.44\; \backslash mbox\{\; g\; NaCl\; mol\}^\{1\}\}\; =\; 0.034\; \backslash \; \backslash text\{mol\}$
In the above example, when written out in fraction form, the units of grams form a multiplicative identity, which is equivalent to one (g/g=1), with the resulting amount of moles (the unit that was needed), is shown in the following equation,
 $\backslash left(\backslash frac\{2.00\; \backslash mbox\{\; g\; NaCl\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; NaCl\}\}\{58.44\; \backslash mbox\{\; g\; NaCl\}\}\backslash right)\; =\; 0.034\backslash \; \backslash text\{mol\}$
Molar proportions
Stoichiometry is often used to balance chemical equations (reaction stoichiometry). For example, the two diatomic gases, hydrogen and oxygen, can combine to form a liquid, water, in an exothermic reaction, as described by the following equation:
 2Template:Chem/link + Template:Chem/link → 2Template:Chem/link
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
2Template:Chem/link + 3Template:Chem/link → 2Template:Chem/link + 4Template:Chem/link
the number of moles of water that will be produced by the combustion of 0.27 moles of Template:Chem/link is obtained using the molar ratio between Template:Chem/link and Template:Chem/link of 2 to 4.
 $\backslash left(\backslash frac\{0.27\; \backslash mbox\{\; mol\; \}\backslash mathrm\{CH\_3OH\}\}\{1\}\backslash right)\backslash left(\backslash frac\{4\; \backslash mbox\{\; mol\; \}\backslash mathrm\{H\_2O\}\}\{2\; \backslash mbox\{\; mol\; \}\; \backslash mathrm\{CH\_3OH\}\}\backslash right)\; =\; 0.54\backslash \; \backslash text\{mol\}$
The term stoichiometry is also often used for the molar proportions of elements in stoichiometric compounds (composition stoichiometry). For example, the stoichiometry of hydrogen and oxygen in H_{2}O is 2:1. In stoichiometric compounds, the molar proportions are whole numbers.
Determining amount of product
Stoichiometry can also be used to find the quantity of a product yielded by a reaction. If a piece of solid copper (Cu) were added to an aqueous solution of silver nitrate (AgNO_{3}), the silver (Ag) would be replaced in a single displacement reaction forming aqueous copper(II) nitrate (Cu(NO_{3})_{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:
 Step 1  Write and Balance the Equation
 Step 2  Mass to Mole: Convert g Cu to moles Cu
 Step 3  Mole Ratio: Convert moles of Cu to moles of Ag produced
 Step 4  Mole to Mass: Convert moles Ag to grams of Ag produced
The complete balanced equation would be:
 Template:Chem/link + 2Template:Chem/link → Template:Chem/link + 2Template:Chem/link
For the mass to mole step, the amount of copper (16.00 g) would be converted to moles of copper by dividing the mass of copper by its molecular mass: 63.55 g/mol.
 $\backslash left(\backslash frac\{16.00\; \backslash mbox\{\; g\; Cu\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; Cu\}\}\{63.55\; \backslash mbox\{\; g\; Cu\}\}\backslash right)\; =\; 0.2518\backslash \; \backslash text\{mol\; Cu\}$
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.
 $\backslash left(\backslash frac\{0.2518\; \backslash mbox\{\; mol\; Cu\}\}\{1\}\backslash right)\backslash left(\backslash frac\{2\; \backslash mbox\{\; mol\; Ag\}\}\{1\; \backslash mbox\{\; mol\; Cu\}\}\backslash right)\; =\; 0.5036\backslash \; \backslash text\{mol\; Ag\}$
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:
 $\backslash left(\backslash frac\{0.5036\; \backslash mbox\{\; mol\; Ag\}\}\{1\}\backslash right)\backslash left(\backslash frac\{107.87\; \backslash mbox\{\; g\; Ag\}\}\{1\; \backslash mbox\{\; mol\; Ag\}\}\backslash right)\; =\; 54.32\; \backslash \; \backslash text\{g\; Ag\}$
This set of calculations can be further condensed into a single step:
 $m\_\backslash mathrm\{Ag\}\; =\; \backslash left(\backslash frac\{16.00\; \backslash mbox\{\; g\; \}\backslash mathrm\{Cu\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{Cu\}\}\{63.55\; \backslash mbox\{\; g\; \}\backslash mathrm\{Cu\}\}\backslash right)\backslash left(\backslash frac\{2\; \backslash mbox\{\; mol\; \}\backslash mathrm\{Ag\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{Cu\}\}\backslash right)\backslash left(\backslash frac\{107.87\; \backslash mbox\{\; g\; \}\backslash mathrm\{Ag\}\}\{1\; \backslash mbox\{\; mol\; Ag\}\}\backslash right)\; =\; 54.32\; \backslash mbox\{\; g\}$
Further examples
For propane (C_{3}H_{8}) reacting with oxygen gas (O_{2}), the balanced chemical equation is:
 Template:Chem/link + 5Template:Chem/link → 3Template:Chem/link + 4Template:Chem/link
The mass of water formed if 120 g of propane (C_{3}H_{8}) is burned in excess oxygen is then
 $m\_\backslash mathrm\{H\_2O\}\; =\; \backslash left(\backslash frac\{120.\; \backslash mbox\{\; g\; \}\backslash mathrm\{C\_3H\_8\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{C\_3H\_8\}\}\{44.09\; \backslash mbox\{\; g\; \}\backslash mathrm\{C\_3H\_8\}\}\backslash right)\backslash left(\backslash frac\{4\; \backslash mbox\{\; mol\; \}\backslash mathrm\{H\_2O\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{C\_3H\_8\}\}\backslash right)\backslash left(\backslash frac\{18.02\; \backslash mbox\{\; g\; \}\backslash mathrm\{H\_2O\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{H\_2O\}\}\backslash right)\; =\; 196\; \backslash mbox\{\; g\}$
Stoichiometric ratio
Stoichiometry is also used to find the right amount of one reactant to "completely" react with the other reactant in a chemical reaction  that is, the stoichiometric amounts that would result in no leftover reactants when the reaction takes place. An example is shown below using the thermite reaction,
 Template:Chem/link + 2Template:Chem/link → Template:Chem/link + 2Template:Chem/link
This equation shows that 1 mole of iron(III) oxide and 2 moles of aluminum will produce 1 mole of aluminium oxide and 2 moles of iron. So, to completely react with 85.0 g of iron(III) oxide (0.532 mol), 28.7 g (1.06 mol) of aluminium are needed.
 $m\_\backslash mathrm\{Al\}\; =\; \backslash left(\backslash frac\{85.0\; \backslash mbox\{\; g\; \}\backslash mathrm\{Fe\_2O\_3\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{Fe\_2\; O\_3\}\}\{159.7\; \backslash mbox\{\; g\; \}\backslash mathrm\{Fe\_2\; O\_3\}\}\backslash right)\backslash left(\backslash frac\{2\; \backslash mbox\{\; mol\; Al\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{Fe\_2\; O\_3\}\}\backslash right)\backslash left(\backslash frac\{26.98\; \backslash mbox\{\; g\; Al\}\}\{1\; \backslash mbox\{\; mol\; Al\}\}\backslash right)\; =\; 28.7\; \backslash mbox\{\; g\}$
Limiting reagent and percent yield
The limiting reagent is the reagent that limits the amount of product that can be formed and is completely consumed during the reaction. The excess reactant is the reactant that is left over once the reaction has stopped due to the limiting reactant.
Consider the equation of roasting lead(II) sulfide (PbS) in oxygen (O_{2}) to produce lead(II) oxide (PbO) and sulfur dioxide (SO_{2}):
 2Template:Chem/link + 3Template:Chem/link → 2Template:Chem/link + 2Template:Chem/link
To determine the theoretical yield of lead(II) oxide if 200.0 g of lead(II) sulfide and 200.0 grams of oxygen are heated in an open container:
 $m\_\backslash mathrm\{PbO\}\; =\; \backslash left(\backslash frac\{200.0\; \backslash mbox\{\; g\; \}\backslash mathrm\{PbS\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{PbS\}\}\{239.27\; \backslash mbox\{\; g\; \}\backslash mathrm\{PbS\}\}\backslash right)\backslash left(\backslash frac\{2\; \backslash mbox\{\; mol\; \}\backslash mathrm\{PbO\}\}\{2\; \backslash mbox\{\; mol\; \}\backslash mathrm\{PbS\}\}\backslash right)\backslash left(\backslash frac\{223.2\; \backslash mbox\{\; g\; \}\backslash mathrm\{PbO\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{PbO\}\}\backslash right)\; =\; 186.6\; \backslash mbox\{\; g\}$
 $m\_\backslash mathrm\{PbO\}\; =\; \backslash left(\backslash frac\{200.0\; \backslash mbox\{\; g\; \}\backslash mathrm\{O\_2\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{O\_2\}\}\{32.00\; \backslash mbox\{\; g\; \}\backslash mathrm\{O\_2\}\}\backslash right)\backslash left(\backslash frac\{2\; \backslash mbox\{\; mol\; \}\backslash mathrm\{PbO\}\}\{3\; \backslash mbox\{\; mol\; \}\backslash mathrm\{O\_2\}\}\backslash right)\backslash left(\backslash frac\{223.2\; \backslash mbox\{\; g\; \}\backslash mathrm\{PbO\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{PbO\}\}\backslash right)\; =\; 930.0\; \backslash mbox\{\; g\}$
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 stoichiometricallycalculated theoretical yield. Percent yield, then, is expressed in the following equation:
$\backslash mbox\{percent\; yield\}\; =\; \backslash frac\{\backslash mbox\{actual\; yield\}\}\{\backslash mbox\{theoretical\; yield\}\}\; \backslash times\; \backslash !\backslash ,\; 100$
If 170.0 g of lead(II) oxide is obtained, then the percent yield would be calculated as follows:
$\backslash mbox\{percent\; yield\}\; =\; \backslash frac\{\backslash mbox\{170.0\; g\; PbO\}\}\{\backslash mbox\{186.6\; g\; PbO\}\}\; \backslash times\; \backslash !\backslash ,\; 100\; =\; 91.12\backslash \%$
Example
Consider the following reaction, in which iron(III) chloride reacts with hydrogen sulfide to produce iron(III) sulfide and hydrogen chloride:
 2Template:Chem/link + 3Template:Chem/link → Template:Chem/link + 6Template:Chem/link
Suppose 90.0 g of FeCl_{3} reacts with 52.0 g of H_{2} S. To find the limiting reagent and the mass of HCl produced by the reaction, we could set up the following equations:
 $m\_\backslash mathrm\{HCl\}\; =\; \backslash left(\backslash frac\{90.0\; \backslash mbox\{\; g\; \}\backslash mathrm\{FeCl\_3\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{FeCl\_3\}\}\{162\; \backslash mbox\{\; g\; \}\backslash mathrm\{FeCl\_3\}\}\backslash right)\backslash left(\backslash frac\{6\; \backslash mbox\{\; mol\; \}\backslash mathrm\{HCl\}\}\{2\; \backslash mbox\{\; mol\; \}\backslash mathrm\{FeCl\_3\}\}\backslash right)\backslash left(\backslash frac\{36.5\; \backslash mbox\{\; g\; \}\backslash mathrm\{HCl\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{HCl\}\}\backslash right)\; =\; 60.8\; \backslash mbox\{\; g\}$
 $m\_\backslash mathrm\{HCl\}\; =\; \backslash left(\backslash frac\{52.0\; \backslash mbox\{\; g\; \}\backslash mathrm\{H\_2S\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{H\_2S\}\}\{34.1\; \backslash mbox\{\; g\; \}\backslash mathrm\{H\_2S\}\}\backslash right)\backslash left(\backslash frac\{6\; \backslash mbox\{\; mol\; \}\backslash mathrm\{HCl\}\}\{3\; \backslash mbox\{\; mol\; \}\backslash mathrm\{H\_2S\}\}\backslash right)\backslash left(\backslash frac\{36.5\; \backslash mbox\{\; g\; \}\backslash mathrm\{HCl\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{HCl\}\}\backslash right)\; =\; 111\; \backslash mbox\{\; g\}$
Thus, the limiting reagent is FeCl_{3} and the amount of HCl produced is 60.8 g.
To find what mass of excess reagent (H_{2}S) remains after the reaction, we would set up the calculation to find out how much H_{2}S reacts completely with the 90.0 g FeCl_{3}:
 $m\_\backslash mathrm\{H\_2S\}\; =\; \backslash left(\backslash frac\{90.0\; \backslash mbox\{\; g\; \}\backslash mathrm\{FeCl\_3\}\}\{1\}\backslash right)\backslash left(\backslash frac\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{FeCl\_3\}\}\{162\; \backslash mbox\{\; g\; \}\backslash mathrm\{FeCl\_3\}\}\backslash right)\backslash left(\backslash frac\{3\; \backslash mbox\{\; mol\; \}\backslash mathrm\{H\_2S\}\}\{2\; \backslash mbox\{\; mol\; \}\backslash mathrm\{FeCl\_3\}\}\backslash right)\backslash left(\backslash frac\{34.1\; \backslash mbox\{\; g\; \}\backslash mathrm\{H\_2S\}\}\{1\; \backslash mbox\{\; mol\; \}\backslash mathrm\{H\_2S\}\}\backslash right)\; =\; \{28.4\; \backslash mbox\{\; g\; \}\backslash mathrm\{reacted\}\}$
By subtracting this amount from the original amount of H_{2}S, we can come to the answer:
 $\{52.0\; \backslash mbox\{\; g\; \}\backslash mathrm\{H\_2S\}\}\; \; \{28.4\; \backslash mbox\{\; g\; \}\backslash mathrm\{H\_2S\}\}\; =\; \{23.6\; \backslash mbox\{\; g\; \}\backslash mathrm\{H\_2S\}\}$ $\{\backslash mathrm\{excess\}\}$
Different stoichiometries in competing reactions
Often, more than one reaction is possible given the same starting materials. The reactions may differ in their stoichiometry. For example, the methylation of benzene ($\backslash mathrm\{C\_6H\_6\}$), through a FriedelCrafts reaction using $\backslash mathrm\{AlCl\_3\}$ as a catalyst, may produce singly methylated $(\backslash mathrm\{C\_6H\_5CH\_3\})$, doubly methylated $(\backslash mathrm\{C\_6H\_4(CH\_3)\_2\})$, or still more highly methylated $(\backslash mathrm\{C\_6H\}\_\{6n\}(\backslash mathrm\{CH\_3\})\_n)$ products, as shown in the following example,
 $\backslash mathrm\{C\_6H\_6\; +\; CH\_3Cl\; \backslash rightarrow\; C\_6H\_5CH\_3\; +\; HCl\}\backslash ,$
 $\backslash mathrm\{C\_6H\_6\; +\; \backslash ,2\backslash \; CH\_3Cl\; \backslash rightarrow\; C\_6H\_4(CH\_3)\_2\; +\; 2HCl\}\backslash ,$
 $\backslash mathrm\{C\_6H\_6\}\; +\; \backslash ,n\backslash \; \backslash mathrm\{CH\_3Cl\}\; \backslash rightarrow\; \backslash mathrm\{C\_6H\}\_\{6n\}(\backslash mathrm\{CH\_3\})\_n\; +\; n\backslash ,\backslash mathrm\{HCl\}\backslash ,$
In this example, which reaction takes place is controlled in part by the relative concentrations of the reactants.
Stoichiometric coefficient
In lay terms, the stoichiometric coefficient (or stoichiometric number in the IUPAC nomenclature^{[2]}) of any given component is the number of molecules that participate in the reaction as written.
For example, in the reaction CH_{4} + 2 O_{2} → CO_{2} + 2 H_{2}O, the stoichiometric coefficient of CH_{4} would be 1 and the stoichiometric coefficient of O_{2} would be 2.
In more technically precise terms, the stoichiometric coefficient in a chemical reaction system of the i–th component is defined as
 $\backslash nu\_i\; =\; \backslash frac\{dN\_i\}\{d\backslash xi\}\; \backslash ,$
or
 $dN\_i\; =\; \backslash nu\_i\; d\backslash xi\; \backslash ,$
where N_{i} is the number of molecules of i, and ξ is the progress variable or extent of reaction (Prigogine & Defay, p. 18; Prigogine, pp. 4–7; Guggenheim, p. 37 & 62).
The extent of reaction ξ can be regarded as a real (or hypothetical) product, one molecule of which produced each time the reaction event occurs. It is the extensive quantity describing the progress of a chemical reaction equal to the number of chemical transformations, as indicated by the reaction equation on a molecular scale, divided by the Avogadro constant (in essence, it is the amount of chemical transformations). The change in the extent of reaction is given by dξ = dn_{B}/ν_{B}, where ν_{B} is the stoichiometric number of any reaction entity B (reactant or product) an dn_{B} is the corresponding amount.^{[3]}
The stoichiometric coefficient ν_{i} represents the degree to which a chemical species participates in a reaction. The convention is to assign negative coefficients to reactants (which are consumed) and positive ones to products. However, any reaction may be viewed as "going" in the reverse direction, and all the coefficients then change sign (as does the free energy). Whether a reaction actually will go in the arbitrarily selected forward direction or not depends on the amounts of the substances present at any given time, which determines the kinetics and thermodynamics, i.e., whether equilibrium lies to the right or the left.
If one contemplates actual reaction mechanisms, stoichiometric coefficients will always be integers, since elementary reactions always involve whole molecules. If one uses a composite representation of an "overall" reaction, some may be rational fractions. 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 a catalyst, also has a stoichiometric coefficient of zero.
The simplest possible case is an isomerism
 $A\; \backslash iff\; B$
in which ν_{B} = 1 since one molecule of B is produced each time the reaction occurs, while ν_{A} = −1 since one molecule of A is necessarily consumed. In any chemical reaction, not only is the total mass conserved but also the numbers of atoms of each kind are conserved, and this imposes corresponding constraints on possible values for the stoichiometric coefficients.
There are usually multiple reactions proceeding simultaneously in any natural reaction system, including those in biology. Since any chemical component can participate in several reactions simultaneously, the stoichiometric coefficient of the i–th component in the k–th reaction is defined as
 $\backslash nu\_\{ik\}\; =\; \backslash frac\{\backslash partial\; N\_i\}\{\backslash partial\; \backslash xi\_k\}\; \backslash ,$
so that the total (differential) change in the amount of the i–th component is
 $dN\_i\; =\; \backslash sum\_k\; \backslash nu\_\{ik\}\; d\backslash xi\_k.\; \backslash ,$
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 present { N_{i} } (state variables), and the representation in terms of the actual compositional degrees of freedom, as expressed by the extents of reaction { ξ_{k} }. The transformation from a vector expressing the extents to a vector expressing the amounts uses a rectangular matrix whose elements are the stoichiometric coefficients [ ν_{i k} ].
The maximum and minimum for any ξ_{k} 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 purely kinematic restriction on the reaction simplex, a hyperplane in composition space, or N‑space, whose dimensionality equals the number of linearlyindependent 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 of chemical kinetics and thermodynamic equilibrium, every chemical reaction is reversible, at least to some degree, so that each equilibrium point must be an interior point of the simplex. As a consequence, extrema for the ξ's will not occur unless an experimental system is prepared with zero initial amounts of some products.
The number of physicallyindependent 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) reaction paths 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 be molecules or moles. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules. The N's and ξ's are reduced to molar units by dividing by Avogadro's number. While dimensional mass units may be used, the comments about integers are then no longer applicable.
Stoichiometry matrix
In complex reactions, stoichiometries are often represented in a more compact form called the stoichiometry matrix. The stoichiometry matrix is denoted by the symbol, $\backslash mathbf\{N\}$.
If a reaction network has $\backslash mathit\{n\}$ reactions and $\backslash mathit\{m\}$ participating molecular species then the stoichiometry matrix will have corresponding $\backslash mathit\{m\}$ rows and $\backslash mathit\{n\}$ columns.
For example, consider the system of reactions shown below:
 S_{1} → S_{2}
 5S_{3} + S_{2} → 4S_{3} + 2S_{2}
 S_{3} → S_{4}
 S_{4} → S_{5}.
This systems comprises four reactions and five different molecular species. The stoichiometry matrix for this system can be written as:
 $$
\mathbf{N} = \begin{bmatrix}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
where the rows correspond to S_{1}, S_{2}, S_{3}, S_{4} and S_{5}, respectively. Note that the process of converting a reaction scheme into a stoichiometry matrix can be a lossy transformation, for example, the stoichiometries in the second reaction simplify when included in the matrix. This means that it is not always possible to recover the original reaction scheme from a stoichiometry matrix.
Often the stoichiometry matrix is combined with the rate vector, v to form a compact equation describing the rates of change of the molecular species:
 $$
\frac{d\mathbf{S}}{dt} = \mathbf{N} \cdot \mathbf{v}.
Gas stoichiometry
Gas stoichiometry is the quantitative relationship (ratio) between reactants and products in a chemical reaction with reactions that produce gases. Gas stoichiometry applies when the gases produced are assumed to be ideal, 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, the standard 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 unknown volume or mass of a gaseous product or reactant. For example, if we wanted to calculate the volume of gaseous NO_{2} produced from the combustion of 100 g of NH_{3}, by the reaction:
 4NH_{3} (g) + 7O_{2} (g) → 4NO_{2} (g) + 6H_{2}O (l)
we would carry out the following calculations:
 $100\; \backslash \; \backslash mbox\{g\}\backslash ,NH\_3\; \backslash cdot\; \backslash frac\{1\; \backslash \; \backslash mbox\{mol\}\backslash ,NH\_3\}\{17.034\; \backslash \; \backslash mbox\{g\}\backslash ,NH\_3\}\; =\; 5.871\; \backslash \; \backslash mbox\{mol\}\backslash ,NH\_3\backslash $
There is a 1:1 molar ratio of NH_{3} to NO_{2} in the above balanced combustion reaction, so 5.871 mol of NO_{2} will be formed. We will employ the ideal gas law to solve for the volume at 0 °C (273.15 K) and 1 atmosphere using the gas law constant of R = 0.08206 L · atm · K^{−1} · mol^{−1} :
$PV$

$=\; nRT$

$V$

$=\; \backslash frac\{nRT\}\{P\}\; =\; \backslash frac\{5.871\; \backslash cdot\; 0.08206\; \backslash cdot\; 273.15\}\{1\}\; =\; 131.597\; \backslash \; \backslash mbox\{L\}\backslash ,NO\_2$

Gas stoichiometry often involves having to know the molar mass of a gas, given the density of that gas. The ideal gas law can be rearranged to obtain a relation between the density and the molar mass of an ideal gas:
 $\backslash rho\; =\; \backslash frac\{m\}\{V\}$ and $n\; =\; \backslash frac\{m\}\{M\}$
and thus:
 $\backslash rho\; =\; \backslash frac\; \{M\; P\}\{R\backslash ,T\}$
where:


$P$

= absolute gas pressure

$V$

= gas volume

$n$

= number of moles

$R$

= universal ideal gas law constant

$T$

= absolute gas temperature

$\backslash rho$

= gas density at $T$ and $P$

$m$

= mass of gas

$M$

= molar mass of gas

Stoichiometry of combustion
In the combustion 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 because of slow combustion or insufficient mixing of fuel and oxygen  this is not due to stoichiometry.) Different hydrocarbon fuels have a different contents of carbon, hydrogen and other elements, thus their stoichiometry varies.
Fuel

By mass ^{[4]}

By volume ^{[5]}

Percent fuel by mass

Gasoline

14.7 : 1

—

6.8%

Natural gas

17.2 : 1

9.7 : 1

5.8%

Propane (LP)

15.67 : 1

23.9 : 1

6.45%

Ethanol

9 : 1

—

11.1%

Methanol

6.47 : 1

—

15.6%

Hydrogen

34.3 : 1

2.39 : 1

2.9%

Diesel

14.5 : 1

0.094 : 1

6.8%

Gasoline engines can run at stoichiometric airtofuel 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, leaving less time for evaporation and mixing. Thus, it would form soot (black smoke) at stoichiometric ratio.
Stoichiometric reactants and catalytic reactants
A stoichiometric reactant is a reactant that is consumed in a reaction, as opposed to a catalytic reactant, which is not consumed in the overall reaction because it reacts in one step and is regenerated in another step.
References

 Library of Congress Catalog No. 6729540
 Library of Congress Catalog No. 6720003
 Zumdahl, Steven S. Chemical Principles. Houghton Mifflin, New York, 2005, pp 148–150.
 Internal Combustion Engine Fundamentals, John B. Heywood
External links
 Engine Combustion primer from the University of Plymouth
 Free Stoichiometry Tutorials from Carnegie Mellon's ChemCollective
 Stoichiometry AddIn for Microsoft Excel for calculation of molecular weights, reaction coëfficients and stoichiometry.
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