Stoichiometry

Overview: In this tutorial, the fundamentals of balancing chemical reactions are reviewed.

Skills:

Balancing Chemical Equations

New terms:

Stoichiometry
Stoichiometric Coefficients

In chemistry it is very important to understand the relationship between reactants and products in a reaction. Stoichiometry is exactly that. It is the quantitative relation between the number of moles (and therefore mass) of various products and reactants in a chemical reaction.
Chemical reactions must be balanced, or in other words, must have the same number of various atoms in the products as in the reactants.

If a chemical reaction is not balanced, no information about the relationship between products and reactants can be derived. So the first thing to do when you see a chemical reaction is to balance it. We balance reactions by adding coefficients in front of the reactants and products. These coefficients are the stoichiometric coefficients.

General Guidelines for Balancing Simple Equations

Assign a "1" as the coefficient for the most complex species (the one whose chemical formula has the greatest number of different elements).
Balance any single-element species last.
Eliminate fractional coefficients (although this is not necessary).
Add coefficients only; do not change the chemical formulas.
There must be the same number of atoms on the left and right sides of the chemical reaction.

These are just guidelines, not rules. Therefore, sometimes it may be necessary to deviate from these general guidelines.

Example 1.

Balance the chemical reaction.

This equation is not balanced since there are more N and O atoms on the left side of the equation. Let's start by using the guidelines. Assign a stoichiometric coefficient of 1 to the most complex compound, NO.

Now we can balance the remaining single-element compounds. In order to do this we will need to use fractional coefficients.

	(Balanced)
Note: Typically a stoichiometric coefficient of "1" is not explicitly included when writing the chemical equation.

We can get rid of the fractional coefficients by multiplying by 2 even though this is a perfectly acceptable balanced chemical equation.

(Balanced, but without fractional coefficients)

At the very beginning of this problem, perhaps you could see this was the answer. If you can see the balanced equation by sight, you don't need to go by the guidelines. Remember they are only guidelines to help if you run into trouble. You can see by simply adding a 2 in front of NO, we violate the first guideline even though it leads us to a balanced equation.

Example 2.

Balance the given chemical reaction.

This one may not be as easy to see the final answer so we will use the guidelines to balance the equation. N₂O₃ is the most complex species so we will add a 1 for its coefficient.

Now we can balance the remaining single element species. In order to balance the number of atoms we need 2 atoms of N and 3 atoms of oxygen on the left side of the equation. There are already 2 atoms of N so we can add 3/2 in front of oxygen to get 3 atoms of oxygen.

(Balanced)

The equation is now balanced. However, we can get rid of the fractional coefficient by again multiplying by 2.

(Balanced, without fractional coefficients)

Notice that in these two examples N₂ and O₂ react with a different stoichiometry to obtain different products. Is it necessary for the number of moles of the reactants to be equal to the number of moles of products? Answer

Not only does the stoichiometry tell us the mole relation between product and reactants but it will also tell us the mass relation.

Example 3.

How many grams of CO₂ are produced by the complete combustion of 1.00 g of glucose (C₆H₁₂O₆)? Remember that a combustion reaction is one in which the reactant (glucose in this case) reacts with O₂ to produce CO₂ and H₂O. So first let's write the skeleton equation (unbalanced equation).

Use the guidelines to balance the equation. We will assign a 1 to glucose.

We will balance the oxygen last because it is contained in a single-element species (O₂). This means we need to balance the C and H atoms next. So we need 6 atoms of C and 12 atoms of H on the right side of the reaction.

So now we have 6 atoms of C and 6 atoms of H₂ (which accounts for twelve H atoms). Now we can balance the oxygen. We have a total of 18 oxygen atoms on the right side. In order to have 18 atoms of oxygen on the left side we will need to assign a 6 to O₂.

	(Balanced)
or

What does this really say? We need one mole of glucose to react with 6 moles of O₂ to produce 6 moles of CO₂ and 6 moles of H₂O. But the question asks about the mass of CO₂ produced not the number of moles. We can use the molecular mass to convert.

The conversion factor of 6 moles of CO₂ per mole of glucose comes from the balanced chemical equation (or the correct stoichiometry).

Summary

Now you should be able to balance chemical equations and understand what relations a balanced equation can reveal.

Practice Quizzes: Stoichiometry/Limiting Reagents/Solutions

These two quizzes cover the three tutorial modules Stoichiometry, Limiting Reagents, and Solutions. You will probably want to review all three of these modules before trying the quiz.

Note: You will need a pencil, scratch paper, calculator, periodic table and equation sheet to work the practice quiz. Quiz questions are timed (4 minutes per question).

Open Periodic Table in a separate browser window.

Open Equations/Constants Page in a separate browser window.