Lesson 8-1

Balancing Chemical Equations

     According to the law of conservation of mass, a chemical equation must be balanced.  This means that the total number of atoms on the reactant side must be equal to the total number of atoms on the product side.  This really involves three skills; interpreting a chemical formula, determining whether or not a chemical equation is balanced, and balancing the equation.

I. Interpreting a Chemical Formula.

     If you can read a chemical formula correctly, you can check the balance on a chemical equation.  One of the biggest problems for new Chemistry students is correctly reading the number of atoms inside parenthesis.  Let us practice this skill first, with a couple of examples.

Example 1.  How many atoms of each element are there in one formula unit of ammonium sulfide?

Ammonium Sulfide is   (NH₄)₂S

Remember that a subscript pertains only to the element that precedes it, unless it precedes parenthesis, in which case it is a multiplier for each element in the parenthesis.  In the example above, the subscript 4 only pertains to hydrogen, while the subscript 2 acts as a multiplier for both nitrogen and hydrogen, giving us as a final tally;

(NH₄)₂S

2 atoms of nitrogen;      8 atoms of hydrogen;    and 1 atom of sulfur.

Example 2.  How many atoms of each element are there in one formula unit of barium nitrate?

Barium Nitrate  = Ba(NO₃)₂

Now, the subscript 3 pertains only to the oxygen, but the subscript 2 becomes a multiplier for each element in the parenthesis.  Therefore;

Ba(NO₃)₂

1 atom of barium;      2 atoms of nitrogen;     and 6 atoms of oxygen.

II.  Checking the Balance of a Chemical Equation.

     When we write chemical equations for a chemical reaction, we use special numbers called coefficients to represent multiple molecules or formula units.  For example;

6H₂O

     As before, the subscript 2 pertains only to the hydrogen.  However, the coefficient 6 pertains to every element in the compound, whether or not they are found in parenthesis.  The 6 tells us that there are six molecules of water,  with a total of 12 atoms of hydrogen and 6 atoms of oxygen.  Once again, a coefficient pertains to every element in the compound, regardless of parenthesis.  You will need to keep this in mind when you check the balance of an equation.

Example 1.  Determine if the following reaction is balanced or not.

Ca(OH)_2(cr)  ---> CaO_(cr) + H₂O_(g)

Let us make an organized tally table and compare both sides of the equation;

    As you can see, this reaction is balanced, so no coefficients are necessary.

Example 2.  Check the balance on the following chemical reaction;

Ca(OH)_2(aq) + HCl_(aq) ---> CaCl_2(aq) + H₂O_(l)

     As you can see, this reaction is not balanced.  You are not allowed to change any subscripts, but coefficients may be added in order to obtain balance. 

III. Balancing Chemical Equations

Balancing chemical equations is a skill that only develops with practice, but for starters, look at the tally above.  Notice that you need more Cl on the reactant side.  What would the tally look like if we add a coefficient of 2 to the HCl on the reactant side?

Ca(OH)_2(aq) + 2HCl_(aq) ---> CaCl_2(aq) + H₂O_(l)

Now we need more oxygen and more hydrogen on the product side.  Let's add a coefficient of 2 to the H2O on the product side and check the balance again.

Ca(OH)_2(aq) + 2HCl_(aq) ---> CaCl_2(aq) + 2H₂O_(l)

Now the equation is balanced.

Example 2.  Write a balanced chemical equation for the reaction below;

Propane reacts with oxygen gas to yield carbon dioxide and water.

First, you need to be able to turn a word equation into a chemical equation.  The one above would become;

C₃H_8(g) + O_2(g) ----> CO_2(g) + H₂O_(g)

Now, let us tally the information in a table:

Well, a quick look shows us that we will need more hydrogen and more carbon on the right hand side.  Let us start by multiplying the number of hydrogen on the product side by four, giving us a total of 8 atoms of hydrogen.  Be aware that this will also change the number of oxygen atoms on the product side.  Let us look at how a coefficient of 4 in front of water changes things.

C₃H_8(g) + O_2(g) ----> CO_2(g) + 4H₂O_(g)

Now we have a match with the number of hydrogen atoms.  Let us balance the carbon atoms next, because in order to change the carbon atoms on the product side, it will also affect the number of oxygen atoms.  We need to multiply the number of carbon atoms on the product side by three, so we will place a coefficient of three in front of the carbon dioxide and check the tally again.

C₃H_8(g) + O_2(g) ----> 3CO_2(g) + 4H₂O_(g)

Now, we have matched the number of atoms for two of the elements.  A subscript of 5 in front of the oxygen on the reactant side should finish the job.

C₃H_8(g) + 5O_2(g) ----> 3CO_2(g) + 4H₂O_(g)

We have achieved proper balance!  In practice, the process is not nearly as long and tedious as this may have appeared.   Once you gain some experience, you will find that you can balance these equations quickly and painlessly.  Start practicing with the worksheets below, and be sure to browse the links for more information.

Please forward all questions, comments and criticisms to Gregory L. Curran.
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Last Modified March 24, 2004