# Biological Ecology

- Page ID
- 50953

Ecological (Biological) Stoichiometry

A new approach to studying the relationships in ecological or biological systems has been developed by R.W. Sterner, J.J. Elser, and others ^{[1]} ^{[2]}. It is called * ecological* or

*because it focuses on the chemical requirements of each trophic level, in addition to energy requirements. Proponents say that "Ecological stoichiometry recognizes that organisms themselves are outcomes of chemical reactions and thus their growth and reproduction can be constrained by supplies of key chemical elements [especially carbon (C), nitrogen (N) and phosphorus (P)]".*

**biological stoichiometry**^{[3]}

For example, by writing a chemical equation for photosynthesis in oceanic algae, we can predict which nutrients (nitrogen as nitrate, NO_{3}^{-}, phosphorus as hydrogen phosphate, HPO_{4}^{2}^{-}, etc.) are required for algae growth, and what products result from algal respiration.

\[\ce{106 CO2 (g) + 16NO3- (aq) + HPO4(2-) (aq) + 122 H2O (l) + 18 H+ (aq) <-> C106H263O110N16P1 (s) + 138 O2 (g)} \label{1}\]

The equation above does not include some minerals, like potassium ion (K^{+}), which may both consumed and produced in the processes, and the "formula" for algae (C_{106}H_{263}O_{110}N_{16}P_{1}) does not represent a single molecule, but just the overall composition of the algae (one might call it an "average" molecular formula)^{[4]}. It illustrates the * elemental homeostasis* exhibited by plants and animals--their elemental composition is fixed, regardless of the composition of their environment, so if the environment does not provide the correct elements in the correct ratios, the plants or animals do not thrive.

The double arrow indicates that the reaction may produce the **products** on the right from the **reactants** on the left during **photosynthesis**, or may proceed in the reverse direction during **respiration**. The abbreviations in parentheses indicate the state of the species; (*g*) indicates a gas, (*l*) a liquid, (*s*) a solid, and species which are dissolved in aqueous (water) solutions are denoted (*aq*).

This balanced chemical equation not only tells how many molecules of each kind are involved in a reaction, it also indicates the *amount* of each substance that is involved. Equation (1) says that 106 CO_{2} *molecules* can react with 122 H_{2}O *molecules*, 16 NO_{3}^{-} ions, 1 HPO_{4}^{2}^{-} and 18 H^{+} ions to give 1 C_{106}H_{263}O_{110}N_{16}P_{1}*"molecule"* and 138 O_{2} *molecules*. It also says that 106 *mol* of CO_{2} *molecules* can react with *mol* of 122 H_{2}O *molecules*, 16 *mol* of NO_{3}^{-} ions, 1 *mol* of HPO_{4}^{2}^{-} and 18*mol* of H^{+} ions to give 1 *mol* of C_{106}H_{263}O_{110}N_{16}P_{1}*"molecule"* and 138 *mol* of O_{2} *molecules*.

The balanced equation does more than this, though. It also tells us that 2 × 16 = 32 mol NO_{3}^{-} will react with 2 × 1 = 2 mol HPO_{4}^{2}^{-}, and that ½ × 16 = 8 mol NO_{3}^{-} requires only ½ × 1 = ½ mol HPO_{4}^{2}^{-}. In other words, the equation indicates that exactly 16 mol NO_{3}^{-} must react *for every* 1 mol HPO_{4}^{2}^{-} consumed. For the purpose of calculating how much HPO_{4}^{2}^{-} is required to react with a certain amount of NO_{3}^{-} therefore, the significant information contained in Eq. \(\ref{1}\) is the *ratio*

\[\dfrac{\text{1 mol HPO}_4^{2-}}{\text{16 mol NO}_3^-}\nonumber\]

We shall call such a ratio derived from a balanced chemical equation a **stoichiometric ratio** and give it the symbol *S*. Thus, for Eq. \(\ref{1}\),

\[\text{S}(\dfrac{\text{HPO}_4^{2-}}{\text{NO}_3^-}) = \dfrac{\text{1 mol HPO}_4^{2-}}{\text{16 mol NO}_3^-}\nonumber\]

Thus for algae to grow, 1 HPO_{4}^{2}^{-} ion must be present for every 16 NO_{3}^{-} ions. Often, this amount of HPO_{4}^{2}^{-} is not present, so algae cannot grow. But if a good source of HPO_{4}^{2}^{-} is added (for example, as modern detergent runoff), the stoichiometric ratio is established and an * algae bloom* results.

The word *stoichiometric* comes from the Greek words *stoicheion*, “element,“ and *metron*, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.

Example \(\PageIndex{1}\): Stoichiometric Ratios

Derive all possible stoichiometric ratios from Eq. \(\ref{1}\)

**Solution** Any ratio of amounts of substance given by coefficients in the equation may be used:

\[\text{S}(\dfrac{\text{NO}_3^{-}}{\text{O}_2}) = \dfrac{\text{16 mol NO}_3^{-}}{\text{138 mol O}_2}\nonumber\]

\[\text{S}(\dfrac{\text{CO}_2}{\text{NO}_3^-}) = \dfrac{\text{106 mol CO}_2}{\text{16 mol NO}_3^-}\nonumber\]

\[\text{S}(\dfrac{\text{O}_2}{\text{H}_2\text{O}}) = \dfrac{\text{138 mol O}_2}{\text{122 mol H}_2\text{O}}\nonumber\]

\[\text{S}(\dfrac{\text{CO}_2}{\text{H}_2\text{O}}) = \dfrac{\text{106 mol CO}_2}{\text{122 mol H}_2\text{O}}\nonumber\]

There are many more stoichiometric ratios, each of which relates any two of the reactants or products. Eq. (2) gives one of them.

*When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios*. Using Eq. (1) as an example, this means that the ratio of the amount of CO_{2} consumed to the amount of N_{3} ^{-} consumed must be the stoichiometric ratio S(CO_{2}/NO_{3}^{-}):

\(\dfrac{n_{CO_2} (consumed)}{n_{NO_3-} (consumed)} = \text{S}(\dfrac{\text{CO}_2}{\text{NO}_3^-}) = \dfrac{\text{106 mol CO}_2}{\text{16 mol NO}_3^-}\nonumber\)

Similarly, the ratio of the amount of O_{2} produced to the amount of CO_{2} consumed must be

S(O_{2}/CO_{2}):

\(\dfrac{n_{O_2} (produced)}{n_{CO_2} (consumed)} = \text{S}(\dfrac{\text{O}_2}{\text{CO}_2}) = \dfrac{\text{138 mol O}_2}{\text{106 mol CO}_2}\nonumber\)

In general we can say that

\[\large\text{Stoichiometric ratio }\left( \dfrac{\text{X}}{\text{Y}} \right)=\Large\dfrac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}} \label{11}\]

or, in symbols,

\[\large\text{S}\left( \dfrac{\text{X}}{\text{Y}} \right)= \Large\dfrac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}} \label{12}\]

Note

Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent *any* reactant or *any* product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants *consumed* and the amounts of products *produced* will be in appropriate stoichiometric ratios.

Example \(\PageIndex{2}\): Oxygen Produced

Find the amount of oxygen produced when 3.68 mol CO_{2} is consumed according to Eq. (1).

**Solution** The amount of oxygen produced must be in the stoichiometric ratio S(O_{2}/CO_{2}) to the amount of carbon dioxide consumed:

\(\text{S}(\dfrac{\text{O}_2}{\text{CO}_2}) = \dfrac{n_{O_2} (produced)}{n_{CO_2}(consumed)}\)

Multiplying both sides by*n*_{CO2}_{ consumed}, we have

\(n_{O_2} (produced) = n_{CO_2}(consumed) \times \text{S}(\dfrac{\text{O}_2}{\text{CO}_2}) = \text{3.68 mol CO}_2 \times \dfrac{\text{138 mol O}_2}{\text{106 mol CO}_2}= \text{4.79 mol O}_2\)

This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form

\[\large \text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \\ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}\]

or symbolically.

\[\large n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}}\]

When using stoichiometric ratios, be sure you *always* indicate moles *of what*. You can only cancel moles of the same substance. In other words, 1 mol CO_{2} cancels 1 mol CO_{2} but does not cancel 1 mol O_{2}.

The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.

Example \(\PageIndex{3}\): Algae Growth

Equation (1) may be used to calculate the ratio of nitrate to phosphate in a fertilizer that is optimal for algal growth. Phosphate fertilizers may be in liquid form, containing phosphoric acid^{[5]}, while nitrate fertilizers may contain solid KNO_{3}. If 100 g of KNO_{3} is applied to a small pond, what is the stoichiometric mass of H_{3}PO_{4} required for optimal algae growth?

**Solution** The problem asks that we calculate the mass of H_{3}PO_{4} from a mass of KNO_{3}. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the mass of KNO_{3} to the amount of KNO_{3}. We can thus calculate the amount of KNO_{3}. The amount of NO_{3}^{-} will be the same (there is 1 mol of NO_{3}^{-} in 1 mol of KNO_{3}). We can then use a stoichiometric ratio to calculate the amount of HPO_{4}^{2}^{-} that would stoichiometrically react with the given mass of NO_{3}^{-}. This is the same as the amount of H_{3}PO_{4}, and we can use the molar mass to convert the amount to the mass.

This requires the stoichiometric ratio

\[\text{S}(\dfrac{\text{HPO}_4^{2-}}{\text{NO}_3^-}) = \dfrac{\text{1 mol HPO}_4^{2-}}{\text{16 mol NO}_3^-}\nonumber\]

The *amount* of KNO_{3} in 100 g is

\(\text{n} (mol) = \dfrac{\text{m} (g)}{\text{M} (g/mol)} = \dfrac{\text{100 g}}{\text{101.1} g/mol} = \text{0.989 mol}\)

The *amount* of H_{3}PO_{4} required is then

\(n_{HPO_4(2-)} = n_{NO_3}(consumed) \times \text{conversion factor} = \text{0.989 mol NO}_3^- \times \dfrac{\text{1 mol HPO}_4^{2-}}{\text{16 mol NO}_3^-} = \text{0.0618 mol HPO}_4^{2-}\)

The molar mass of H_{3}PO_{4} (98.0 g/mol) is used to calculate the *mass*:

\(\text{m} (g) = \text{n} (mol) \times \text{M} (g/mol) = \text{0.0618 mol} \times \text{98.0} \dfrac{g}{mol} = \text{6.06 g}\)

Very little phosphate is required compared to nitrate, so the phosphate in discharged washwater may well cause the eutrophication shown in the figure above when (as is often the case) plenty of nitrate is available.

With practice this kind of problem can be solved in one step by concentrating on the units.

From ChemPRIME: 3.1: Equations and Mass Relationships

## References

- ↑ Ecological Stoichiometry: The biolgy of elements from Molecules to Biosphere; Robert W. Sterner and James J. Elser; Princeton University Press, Princeton, NJ, 2002.
- ↑ http://en.Wikipedia.org/wiki/Ecological_stoichiometry
- ↑ http://www.plosbiology.org/article/info:doi%2F10.1371%2Fjournal.pbio.0050181
- ↑ Ecological Stoichiometry: The biolgy of elements from Molecules to Biosphere; Robert W. Sterner and James J. Elser; Princeton University Press, Princeton, NJ, 2002, p. 30.
- ↑ www.extension.umn.edu/distrib...ms/dc6288.html

## Contributors and Attributions

Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.