Stoichiometry

Stoichiometry is the numerical relationship between the reactants and products of a chemical reaction. In fact, the word ‘stoichiometry’ is derived from the Ancient Greek words stoicheion "element" and metron "measure". Stoichiometric techniques are used to calculate the required quantities of chemical reactants – or the substances in the reaction – needed to generate the desired amount of product using balanced chemical equations. A stoichiometric chemical reaction occurs when all reagents are consumed as a result of the reaction.

Solids, liquids, and gases react together to form chemical substances whose properties and behaviors shape practical applications. Understanding how the atoms and molecules making up these substances work together at the molecular level to create new compounds is fundamental to chemistry. At the basis of this work is stoichiometry.

Stoichiometry is based on some of the most fundamental principles in chemistry. The most central being the law of conservation of mass that stipulates that mass is neither created nor destroyed.

law of conservation of mass

Collect a cup of water from the faucet and find its mass. Now freeze the water and find the mass of the ice. The mass of liquid water is the same as the mass of the ice showing that mass is conserved.

The chemical properties of compounds are established by the assortment and arrangement of its atoms. When compounds and their elements rearrange in a reaction they do so in integer proportions. Based on these observations, other laws, the so-called law of definite proportions, law of multiple proportions, and law of reciprocal proportions were articulated. These laws all convey the fundamental idea that in a chemical reaction, the reactant molecules combine in definite ratios, e.g., one molecule of glucose always reacts with six molecules of oxygen. Matter cannot be created or destroyed; one element cannot be changed into another; and the number of atoms of each element must remain the same from the beginning to the end of the reaction (assuming no nuclear reactions such as fission, fusion, or radioactive decay are taking place).

You are familiar with the molecular description of water as \(\ce{H_{2}O}\); two atoms of hydrogen for every atom of oxygen. No matter what amount of water you have, whether it is the cup you are drinking from or a river you are rafting on, that ratio of hydrogen to oxygen remains the same; just as the ratio of eggs to butter to flour to sugar in a cake must always be the same to maintain taste and consistency.

Again, hydrogen and oxygen combine to form water \(\ce{H_{2}O}\) or hydrogen peroxide \(\ce{H_{2}O_{2}}\). In water, there are two atoms of hydrogen with one atom of oxygen and in hydrogen peroxide there are two atoms of hydrogen with two atoms of oxygen. The ratio of oxygen atoms to hydrogen atoms in these two compounds follows small, whole number ratios.

Consider hydrogen, oxygen, and carbon. These elements combine separately to form methane \(\ce{CH_{4}}\) and carbon dioxide \(\ce{CO_{2}}\). As we will see shortly when we discuss balancing chemical reactions, the ratios at which the elements combine remain simple multiples of each other.

\[\ce{carbon\, dioxide + hydrogen\, ->[{heat\, required}] methane + water}\]

\[\ce{methane + oxygen -> carbon\, dioxide + water}\]

Stoichiometry is used in chemistry to conveniently represent quantitative relationships between reactants and products in a chemical reaction. Chemical reactions are usually written in equation form with reactants/reagents on the left, products on the right, and an arrow in the middle indicating the direction of the reaction. Such as,

\[\ce{X + Y\, (reactants) -> XY\, (product)}\]

where XY is a new chemical entity distinct from the reactants. To ensure that the number of atoms of each type are conserved these equations must be balanced so that the number of atoms on the left is equivalent to the number of atoms on the right; the equation is balanced.

balancing chemical reactions

This reaction, the Haber process, makes ammonia \(\ce{NH_{3(g)}}\) by reacting nitrogen gas \(\ce{N_{2(g)}}\) with hydrogen gas \(\ce{H_{2(g)}}\) (where the subscript \(_{(g)}\) denotes the gaseous state):

\[N_{2(g)} + H_{2(g)} → NH_{3(g)}\]

This equation describes what you need to make ammonia but not how much – you’ve got to ensure that the number of atoms on the left is equal to the number on the right. In this reaction, \(N_{2}\) and \(H_{2}\) are both gases, there are always two atoms of nitrogen and two atoms of hydrogen, as such N—N and H—H; this can’t be changed as it would change the compounds. To balance the equation you can add coefficients, whole numbers in front of the reactants to tell you have many atoms or molecules you have.

This is a simple reaction that can be balanced by inspection. It is generally helpful to balance oxygen and hydrogen last as they are usually plentiful. So, consider nitrogen first. There are 2 nitrogen atoms on the left side, the reactant side, and 1 nitrogen atom on the right side, the product side. To balance the nitrogen atoms, place a coefficient of 2 in front of the ammonia on the right:

\[N_{2(g)} + H_{2(g)} → 2NH_{3(g)}\]

Now the number of nitrogen atoms on each side of the equation is equivalent. To balance hydrogen, you see there are 2 hydrogen atoms on the left and 6 hydrogen atoms on the right. So place a coefficient of 3 in front of the hydrogen atoms on the left and you have:

\[N_{2(g)} + 3H_{2(g)} → 2NH_{3(g)}\]

Final check: 2 nitrogen atoms on the left and 2 nitrogen atoms on the right 6 hydrogen atoms on the left and 6 hydrogen atoms on the right. This balanced equation can now be read as 1 nitrogen molecule reacting with 3 hydrogen molecules yields 2 ammonia molecules.

When elements are combined with correct stoichiometric ratios, all reagents are converted into product. When a reactant is consumed in a reaction, it is called a stoichiometric reactant. This is opposed to a catalytic reactant that is not consumed in the overall reaction but reacts in one step and is regenerated in another step.

The coefficients in the balanced equation put strict demands on the relative abundance of reactants. If the number of molecules of one far exceeds the number of molecules of the other (after taking their coefficients into account), we can have so-called limiting reagents that limit the amount of product that can be formed. If we mix reagents out of stoichiometric proportion, one of them will run out first and bring the reaction to a halt. An excess reagent is the reactant that is left over once the reaction has stopped because the limiting reactant has been consumed. If a reagent participates in, but isn’t changed by, a reaction it’s called a catalyst.

\(\underline{Here\, are\, a\, few\, steps\, to\, keep\, in\, mind\, when\, balancing\, chemical\, equations}\):

• Chemical equations can get complicated when there are many possible reactants and products. Choose an element to begin with – preferably an element in only one reactant and one product. Adjust the coefficients so the number of atoms of this element is the same on both sides of the equation.

• If polyatomic ions (ions comprised of many atoms) are on both sides of the equation, balance them as a unit.

• Continue balancing the remaining components in order of the most complicated substance remaining to the least complicated; using fractional coefficients, if necessary. Fractional coefficients can be made whole at the end by multiplying both sides of the equation by the denominator to obtain whole number coefficients.

• Do a final check on the numbers of atoms of each kind on both sides of the equation to make sure you have balanced the equation correctly.

try a non-trivial equation: photosynthesis

Photosynthesis is the process by which plants, and a few organisms, use energy from the sun to convert carbon dioxide and water into glucose (a sugar) and oxygen:

\[CO_{2(g)} + H_{2}O_{(l)} → sunlight → C_{6}H_{12}O_{6} + O_{2(g)}\]

Where \(\ce{CO_{2(g)}}\) is carbon dioxide, gas; \(\ce{H_{2}O_{(l)}}\) is water, liquid; \(\ce{C_{6}H_{12}O_{6}}\) is glucose; and \(\ce{O_{2(g)}}\) is oxygen, gas.

\(\underline{balance\, the\, equation}\):

It is evident that the equation above is not balanced. To do so, first, let’s choose carbon C to start. There is 1 C on the left and 6 on the right so we add the coefficient 6 to the left and we now have:

\[6 CO_{2(g)} + H_{2}O_{(l)} → sunlight → C_{6}H_{12}O_{6} + O_{2(g)}\]

Second, consider hydrogen H. There are 2 on the left and 12 on the right. If we put a coefficient of 6 in front of \(\ce{H_{2}O_{(l)}}\) we can balance the H as such:

\[6 CO_{2(g)} + 6 H_{2}O_{(l)} → sunlight → C_{6}H_{12}O_{6} + O_{2(g)}\]

Finally, only oxygen O remains to be done. We now have 18 on the left and 8 on the right. A coefficient of 6 in front of the \(\ce{O_{2(g)}}\) should do it. And the final overall balanced chemical equation for photosynthesis is:

\[6 CO_{2(g)} + 6 H_{2}O_{(l)} → sunlight → C_{6}H_{12}O_{6} + 6 O_{2(g)}\]

Each element is made up of different numbers of protons, neutrons, and electrons, and thus has a different atomic mass. Molecules, as collections of single atoms, have a molar mass (by definition, oxygen-16 has a molar mass of 16 g/mol). Molar mass is measured in terms of a unit mole which is \(6.02\,×\,10^{23}\) individual molecules. This number is known as Avogadro's constant.

To know stoichiometric relationships by mass, the number of molecules required for each reactant can be expressed in moles and multiplied by the molar mass of each reactant to give the mass of each reactant per mole of reaction. Stoichiometry is used to converting from units from grams to moles. This conversion is required because atoms and molecules are too small to count in a meaningful way so we use molar mass find the number of moles of atoms or molecules. This number can then be used in a chemical equation.

Converting grams to moles. How many moles in 100.0 grams of \(\ce{H_{2}O}\)?

First, lets determine how many grams are in 1 mole of \(\ce{H_{2}O}\). Look at the Periodic Table of Elements and you will see that the atomic mass of hydrogen = 1.01 g/mol and the atomic mass of oxygen is 16.0 g/mol.

Now, find the molecular mass of \(\ce{H_{2}O}\) to be:

= 2 (1.01 g/mol) + (16.0 g/mol)

= 2.02 g/mol + 16.0 g/mol

= 18.02 g/mol

The number of moles in 100.0 g of \(\ce{H_{2}O}\) is then found by:

100.0 g \(\ce{H_{2}O}\) × (1 mol/18.02 g) = 5.5 moles \(\ce{H_{2}}O\)

Out of curiosity, the number of \({H_{2}O}\) molecules would be:

5.5 moles × \(6.02\,×\,10^{23}\) = \(3.31\,×\,10^{24}\) \(\ce{H_{2}O}\) molecules – and that’s a number that’s hard to work with!

When dealing with liquids, other units are commonly employed. Molarity (M) is the number of moles of solute dissolved in one liter of a solution and is expressed in moles/L. or M. Molality (m) is the number of moles per kilogram of solvent and is expressed in moles/kg, or m.

While not used extensively, normality (N), or equivalent concentration, of a solution is defined as the molar concentration \(c_{i}\) divided by an equivalence factor \(f_{eq}\). To calculate normality, you would need a defined equivalence factor, which in turn depends on the definition of equivalents. This implies that the same solution can possess different normalities for different reactions. Due to normality’s ambiguity as a measure of concentration, molarity and molality are used for the majority of applications.