“Stoichiometry” refers to the ratios between reactants and products in chemical reactions. For a typical chemical reaction in which generic reactants A and B combine to make products C and D -- i.e. A + B ---> C + D -- stoichiometric calculations allow the chemist to determine the number of grams of A she must add to the reaction mixture to react with compound B, as well as predict the number of grams of products C and D. Students, however, often find stoichiometry problems difficult because they involve calculations of the number of moles of substances. The key to making stoichiometry problems easy is to adopt and practice a methodical approach to the problems.
Balance the chemical reaction equation. A balanced reaction equation contains the same number of each type of atom on both sides of the reaction arrow. The reaction between hydrogen, H2, and oxygen, O2, to make water, H2O, for example, balances to 2 H2 + O2 ---> 2 H2O. This means that two molecules of hydrogen react with one molecule of oxygen to make 2 molecules of water.
Convert the mass of any reactant or product to moles by dividing the grams of material by its molecular weight. Moles simply represent another method of expressing the amount of substance. Note that carrying out a stoichiometric calculation only requires knowing the mass of a single reaction component. You can then calculate the masses of all other components. In the example from step 1, let’s assume that 1.0 grams of hydrogen will react. The molecular weight of hydrogen -- determined by adding together the atomic weights of all of the atoms in the molecular formula -- is 2.02 grams per mole. This means the reaction involves (1.0 grams) / (2.02 grams/mole) = 0.50 moles of hydrogen.
Multiply the moles of hydrogen by the appropriate stoichiometric ratio to determine the number of moles of any other substance involved in the reaction. The stoichiometric ratio simply represents the ratio of the coefficients from the balance chemical equation. Always place the coefficient of the compound whose mass you intend to calculate on top, and the coefficient of the compound whose mass you started with on bottom. In the example from step 1, we could calculate the moles of oxygen requires to react with hydrogen by multiplying by 1 / 2, or we could calculate the moles of water produced by multiplying by 2 / 2. Thus, 0.50 moles of H2 would require 0.25 moles of oxygen and produce 0.50 moles of water.
Finish the problem by converting moles of substance back to grams. Converting to moles required dividing by the compound molecular weight; converting back to grams therefore requires multiplying moles by molecular weight. In the case of hydrogen, this is unnecessary because we already know the reaction involve 1.0 grams of H2. In the case of oxygen, O2, the molecular weight is 32.00 grams/mole and 0.25 moles * 32.00 grams/mole = 8.0 grams of O2. In the case of water, the molecular weight is 18.02 grams/mole and 0.50 moles * 18.02 grams/mole = 9.0 grams of H2O.
Double-check your result by noting that the total grams of reactants must equal the total grams of products. In this case, the combined mass of H2 and O2 was 1.0 and 8.0 grams, respectively, for a total of 9.0 grams, and 9.0 grams of water was produced. This reflects the law of conservation of mass, which states that matter cannot be created or destroyed by a chemical reaction.