Stoichiometry ( ) is the relationships between the quantities of reactants and products before, during and after chemical reactions. Stoichiometry is based on the law 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 product H2O, and to fix the imbalance of oxygen, it is also added to O2. Thus, we get: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) Here, one molecule of methane reacts with two molecules of oxygen gas to yield one molecule of carbon dioxide and two molecules of liquid water. This particular chemical equation is an example of complete combustion. The numbers in front of each quantity are a set of stoichiometric coefficients which directly reflect the molar 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 as reaction 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 of methane 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 to atomic 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 called composition 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 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, because of the existence of isotopes, molar masses are used instead in calculating the mass ratio.
Etymology The term stoichiometry was first used by Jeremias Benjamin Richter in 1792 when the first volume of Richter's Anfangsgründe der Stöchyometrie oder Meßkunst chymischer Elemente (German for 'Fundamentals of Stoichiometry, or the Art of Measuring the Chemical Elements') was published. The term is derived from the Ancient Greek words στοιχεῖον (stoikheîon), meaning 'element', and μέτρον (métron), meaning 'measure'. Ludwig Darmstaedter and Ralph E. Oesper have written a useful account on this.
Definitions
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. Stoichiometry rests upon the very 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), the law of multiple proportions and the law of reciprocal 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 involved in a reaction. Chemical reactions, as macroscopic unit operations, consist of many elementary reactions, where a single molecule reacts with another molecule. As the reacting molecules (or formula units or ion pairs) consist of a definite set of atoms in an integer ratio, the ratio between reactants in a complete reaction is also in an 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). The unsigned coefficients are generally referred to as the stoichiometric coefficients. Each element has an atomic mass (usually given as an average in the form of the standard atomic weight), and considering molecules as collections of atoms, every compound has a molecular mass (if molecular) or formula mass (if non-molecular), which when expressed in daltons is numerically equal to the molar mass in g/mol. By definition, the atomic mass of carbon-12 is exactly 12 Da, making its molar mass 12 g/mol. The number of chemical entities per mole in a substance is given by the Avogadro constant, exactly 6.02214076×1023 mol−1 since the 2019 revision of the SI. 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. Elements in their natural state are mixtures of isotopes 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 of ammonia consists of 14.007 g of nitrogen and 3 × 1.008 g of hydrogen, because natural nitrogen includes a small amount of nitrogen-15, and natural hydrogen includes hydrogen-2 (deuterium). 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.
Converting grams to moles Stoichiometry is not only used to balance chemical equations but also used in "conversions" between quantities of a substance by dimensional analysis, e.g., converting from grams to moles using molar mass as the "conversion factor", or from grams to milliliters using density. For example, to express 2.00 g of NaCl (sodium chloride) as an amount (in moles), one would do the following:
2.00 g NaCl 58.44 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 a multiplicative 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 NaCl 1 ) ( 1 mol NaCl 58.44 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 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:
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 of CH3OH is obtained using the molar ratio between CH3OH and H2O of 2 to 4.
( 0.27 mol C H 3 O H 1 ) ( 4 mol H 2 O 2 mol C H 3 O H ) = 0.54 mol H 2 O {\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 the molar proportions of elements in stoichiometric compounds (composition stoichiometry). For example, the stoichiometry of hydrogen and oxygen in H2O 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 (AgNO3), the silver (Ag) would be replaced in a single displacement reaction forming aqueous copper(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:
Write and balance the equation Mass to moles: Convert grams of Cu to moles of Cu Mole ratio: Convert moles of Cu to moles of Ag produced 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 its molar mass: 63.55 g/mol.
( 16.00 g Cu 1 ) ( 1 mol Cu 63.55 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 Cu 1 ) ( 2 mol Ag 1 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 Ag 1 ) ( 107.87 g Ag 1 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:
m A g = ( 16.00 g C u 1 ) ( 1 mol C u 63.55 g C u ) ( 2 mol A g 1 mol C u ) ( 107.87 g A g 1 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 For propane (C3H8) reacting with oxygen 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
m H 2 O = ( 120. g C 3 H 8 1 ) ( 1 mol C 3 H 8 44.09 g C 3 H 8 ) ( 4 mol H 2 O 1 mol C 3 H 8 ) ( 18.02 g H 2 O 1 mol H 2 O ) = 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 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,
Fe2O3 + 2 Al → Al2O3 + 2 Fe This equation shows that 1 mole of iron(III) oxide and 2 moles of aluminium 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 A l = ( 85.0 g F e 2 O 3 1 ) ( 1 mol F e 2 O 3 159.7 g F e 2 O 3 ) ( 2 mol Al 1 mol F e 2 O 3 ) ( 26.98 g Al 1 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
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 roasting lead(II) sulfide (PbS) in oxygen (O2) to produce lead(II) oxide (PbO) and sulfur 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:
m P b O = ( 200.0 g P b S 1 ) ( 1 mol P b S 239.27 g P b S ) ( 2 mol P b O 2 mol P b S ) ( 223.2 g P b O 1 mol P b O ) = 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}}}
m P b O = ( 200.0 g O 2 1 ) ( 1 mol O 2 32.00 g O 2 ) ( 2 mol P b O 3 mol O 2 ) ( 223.2 g P b O 1 mol P b O ) = 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 yield theoretical 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 PbO 186.6 g PbO = 91.12 % {\displaystyle {\mbox{percent yield}}={\frac {\mbox{170.0 g PbO}}{\mbox{186.6 g PbO}}}=91.12\%}
Example Consider the following reaction, in which iron(III) chloride reacts with hydrogen sulfide to produce iron(III) sulfide and hydrogen chloride:
2 FeCl3 + 3 H2S → Fe2S3 + 6 HCl The stoichiometric masses for this reaction are:
324.41 g FeCl3, 102.25 g H2S, 207.89 g Fe2S3, 218.77 g HCl Suppose 90.0 g of FeCl3 reacts with 52.0 g of H2S. 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 g FeCl3, 28.37 g H2S, 57.67 g Fe2S3, 60.69 g HCl The limiting reactant (or reagent) is FeCl3, since all 90.00 g of it is used up while only 28.37 g H2S are consumed. Thus, 52.0 − 28.4 = 23.6 g H2S left in excess. The mass of HCl produced is 60.7 g. By looking at the stoichiometry of the reaction, one might have guessed FeCl3 being the limiting reactant; three times more FeCl3 is used compared to H2S (324 g vs 102 g).
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 (C6H6), through a Friedel–Crafts reaction using AlCl3 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 relative concentrations of the reactants.
Stoichiometric coefficient and stoichiometric number In lay terms, the stoichiometric coefficient of any given component is the number of molecules and/or formula units that participate in the reaction as written. A related concept is the stoichiometric 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 reaction CH4 + 2 O2 → CO2 + 2 H2O, the stoichiometric number of CH4 is −1, the stoichiometric number of O2 is −2, for CO2 it would be +1 and for H2O it is +2. In more technically precise terms, the stoichiometric number in a chemical reaction system of the i-th component is defined as
ν i = Δ N i Δ ξ {\displaystyle \nu _{i}={\frac {\Delta N_{i}}{\Delta \xi }}\,}
or
Δ N i = ν i Δ ξ {\displaystyle \Delta N_{i}=\nu _{i}\,\Delta \xi \,}
where N i {\displaystyle N_{i}} is the number of molecules of i, and ξ {\displaystyle \xi } is the progress variable or extent of reaction. The stoichiometric number ν i {\displaystyle \nu _{i}} represents the degree to which a chemical species participates in a reaction. The convention is to assign negative numbers to reactants (which are consumed) and positive ones to products, 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 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 of the initial state, In reaction mechanisms, stoichiometric coefficients for each step are always 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 isomerization
A → B in which νB = 1 since one molecule of B is produced each time the reaction occurs, while νA = −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 number of the i-th component in the k-th reaction is defined as
ν i k = ∂ N i ∂ ξ k {\displaystyle \nu _{ik}={\frac {\partial N_{i}}{\partial \xi _{k}}}\,}
so that the total (differential) change in the amount of the i-th component is
d N i = ∑ k ν i k d ξ k . {\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 present { Ni } (state variables), and the representation in terms of the actual compositional degrees of freedom, as expressed by the extents of reaction { ξk }. The transformation from a vector expressing the extents to a vector expressing the amounts uses a rectangular matrix whose elements are the stoichiometric numbers [ νi k ]. The maximum and minimum for any ξk occur whenever the first of the reactants is depleted for the forward reaction; or the first of