In a chemical reaction, stoichiometry shows the mole-ratio of the reactants and products based on the coefficients which are written before them. It allows calculating the amount of any component of a chemical reaction if one of them is given using the mole method.

A chemical equation can be interpreted at molecular and molar levels. For example, the following equation shows the combustion reaction of ethanol (C₂H₅OH) producing carbon dioxide (CO₂) and water (H₂O):

Molar level: The coefficients in blue indicate, for example, that 1 mole of C₂H₅OH reacts with 3 moles of O₂ producing 2 moles of CO₂ and 3 moles of H₂O.

At the molecular level, it tells us that 1 molecule of C₂H₅OH reacts with 3 molecules of O₂ producing 2 molecules of CO₂ and 3 molecules of H₂O.

So, the coefficients show the ratio of the reaction components which is like the recipe of the reaction. However, we can have any number of moles of reactants, and, using stoichiometry (the ratio of the components), we can calculate how much products can be obtained.

For example, let’s determine how much CO₂ will be formed from 3.0 moles of C₂H₅OH by the following combustion reaction:

C₂H₅OH + 3O₂   →   2CO₂ + 3H₂O

The first thing is to set up the conversion factors correlating the two components – the one with the known amount and the target which we need to determine. This is done using the coefficients in the chemical equation which indicate that 1 mole of C₂H₅OH produces 2 moles of CO₂. Conversion factors are the two ratio fractions of these coefficients:

C₂H₅OH + 3O₂   →   2CO₂ + 3H₂O

Now, because you are asked to determine the amount of CO_2, pick the conversion factor such that C₂H₅OH is the denominator and can be canceled:

Next, write the moles of C₂H₅OH that are reacted (not the coefficient) and multiply it with this conversion factor:

Therefore, 3.0 mol C₂H₅OH will produce 6 mol CO_2, and this makes sense because the stoichiometry of the reaction tells us that there will always be two times more CO₂ produced from a given amount of C₂H₅OH. So, if it was, for example, 5 mol C₂H₅OH, 10 mol CO₂ would have been produced.

Let’s look at another example, where we need to determine the amount of a reactant required to produce certain moles of a product. Remember, as long as you know the quantity of one component in the reaction, whether it is a reactant or product, you can determine the amount of any other component.

For example: How many moles of N₂ must be used to produce 10 moles of NH₃ according to the following chemical equation?

3H₂ + N₂ → 2NH₃

Step 1. Write a conversion factor between N₂ and NH₃ such that NH₃ is on the bottom and can be canceled out:

Step 2. Write the moles of ammonia (10 mol) and multiply it by the conversion factor:

So, we will need 5 mol NH₃ to produce 10 mol NH₃ and this is consistent with the chemical equation which indicates that the amount of nitrogen required to produce ammonia is always two times less because of the 1:2 mole ratio.

Mass Calculations using Reaction Stoichiometry

Sometimes, or perhaps more often than not, the amounts of reaction components are given in grams or any other units. Remember, you will always need to convert them to moles first and then use the conversion factors to do the necessary calculations. This is, again, because chemical equations tell us the mole-ratio of the components by based on their coefficients.

For example: How many grams of Ba(NO₃)₂ can be prepared by reacting 16.5 g of HNO₃ with an excess of Ba(OH)₂?

Ba(OH)₂ + 2HNO₃  →  Ba(NO₃)₂ + 2H₂O

The plan would be to calculate the moles of HNO₃, then use the mole-ratio to determine the moles of Ba(NO₃)_2, and finally, convert the moles to the mass of Ba(NO₃)₂:

You can also do this step-by-step to better follow the entire calculation: