Chemical equations do not come already balanced. This must be done before the equation can be used in a chemically meaningful way.

All chemical calculations you will see in other units must be done with a balanced equation.

A balanced equation has equal numbers of each type of atom on each side of the equation.

The The Law of Conservation of Mass is the rationale for balancing a chemical equation. The law was discovered by Antoine Laurent Lavoisier (1743-94) and this is his formulation of it, translated into English in 1790 from the *Traité élémentaire de Chimie* (which was published in 1789):

"We may lay it down as an incontestible axiom, that, in all the operations of art and nature, nothing is created; an equal quantity of matter exists both before and after the experiment; the quality and quantity of the elements remain precisely the same; and nothing takes place beyond changes and modifications in the combination of these elements."

A less wordy way to say it might be:

"Matter is neither created nor destroyed."

Therefore, we must finish our chemical reaction with as many atoms of each element as when we started.

**Problem #1:** Balance the following equation:

H_{2}+ O_{2}---> H_{2}O

It is an unbalanced equation (sometimes also called a skeleton equation). This means that there are UNEQUAL numbers at least one atom on each side of the arrow. By the way, a skeleton equation is not wrong, it just hasn't been balanced yet. Presenting it as being balanced would be wrong.

In the example equation, there are two atoms of hydrogen on each side, BUT there are two atoms of oxygen on the left side and only one on the right side.

Remember this: A balanced equation MUST have EQUAL numbers of EACH type of atom on BOTH sides of the arrow.

An equation is balanced by changing coefficients in a somewhat trial-and-error fashion. It is important to note that only the coefficients can be changed, NEVER a subscript.

The coefficient times the subscript gives the total number of atoms.

Three quick examples before balancing the equation.

(a) 2 H_{2}- there are 2 x 2 atoms of hydrogen (a total of 4).(b) 2 H

_{2}O - there are 2 x 2 atoms of hydrogen (a total of 4) and 2 x 1 atoms of oxygen (a total of 2).(c) 2 (NH

_{4})_{2}S - there are 2 x 1 x 2 atoms of nitrogen (a total of 4), there are 2 x 4 x 2 atoms of hydrogen (a total of 16), and 2 x 1 atoms of sulfur (a total of 2).

So, now to balancing the example equation:

H_{2}+ O_{2}---> H_{2}O

The hydrogen are balanced, but the oxygens are not. We have to get both balanced. We put a two in front of the water and this balances the oxygen.

H_{2}+ O_{2}---> 2 H_{2}O

However, this causes the hydrogen to become unbalanced. To fix this, we place a two in front of the hydrogen on the left side.

2 H_{2}+ O_{2}---> 2 H_{2}O

This balances the equation.

Two things you CANNOT do when balancing an equation.

1) You cannot change a subscript.

You cannot change the oxygen's subscript in water from one to two, as in:

H_{2}+ O_{2}---> H_{2}O_{2}

True, this is a balanced equation, but you have changed the substances in it. H_{2}O_{2} is a completely different substance from H_{2}O. So, it's not the answer to the question that was asked.

2) You cannot place a coefficient in the middle of a formula.

The coefficient goes at the beginning of a formula, not in the middle, as in:

H_{2}+ O_{2}---> H_{2}2O

Water only comes as H_{2}O and you can only use whole formula units of it.

Two more points:

1) Make sure that your final set of coefficients are all whole numbers with no common factors other than one. For example, this equation is balanced:

4 H_{2}+ 2 O_{2}---> 4 H_{2}O

However, all the coefficients have the common factor of two. Divide through to eliminate common factors like this.

Technically, the equation just above is balanced, but only if you ignore the "no common factors other than one" rule. The correct answer has all common factors greater than one removed. If you were to answer a test question balanced as above, you will probably only get partial credit, if that.

2) NO fractions allowed in the final answer, only whole numbers. For example:

H_{2}+ Cl_{2}---> (1/2)HCl

is an allowable step along the way to the answer, but it is not the final answer.

The use of fractions in balancing is a powerful tool. Look for it in the solved examples.

**Problem #2:** Balance the following equation: H_{2} + Cl_{2} ---> HCl

Remember that the rule is: A balanced equation MUST have EQUAL numbers of EACH type of atom on BOTH sides of the arrow.

The correctly balanced equation is:

H_{2}+ Cl_{2}---> 2 HCl

Placement of a two in front of the HCl balances the hydrogen and chlorine at the same time.

**Problem #3:** Balance the following equation: O_{2} ---> O_{3}

Hint: think about what the least common multiple is between 2 and 3. That's right - six.

The LCM tells you how many of each atom will be needed. Your job is to pick coefficients that get you to the LCM.

The correctly balanced equation is:

3 O_{2}---> 2 O_{3}

**Problem #4:** Balance the following equation:

Na + H_{2}O ---> NaOH + H_{2}

In the skeleton equation as written, the Na and the O are already balanced. So, we look only at the H.

Notice that the H must come in twos on the left-hand side. That means we must have an even number of hydrogen on the right-hand side. We do this:

Na + H_{2}O ---> 2NaOH + H_{2}

to make an even number of hydrogens on the right. We then balance the H like this:

Na + 2H_{2}O ---> 2NaOH + H_{2}

Notice that the first placing of a 2 messed up the balance of the Na and the O. In addition, notice that the second placing of a 2 corrected the imbalance in the oxygen.

The last step is to balance the Na:

2Na + 2H_{2}O ---> 2NaOH + H_{2}

and it's done.

Go back to the skeleton equation and balance it like this:

Na + H_{2}O ---> NaOH + (1/2)H_{2}

Multiply through by 2 to get the final answer (which should not have any fractions in it).

How many oxygens are indicated: 3 Ca(NO_{3})_{2}

Balance these equations:

Zn + HCl ---> ZnCl_{2} + H_{2}

KClO_{3} ---> KCl + O_{2}

S_{8} + F_{2} ---> SF_{6}

Fe + O_{2} ---> Fe_{2}O_{3}

C_{2}H_{6} + O_{2} ---> CO_{2} + H_{2}O

The last problem above involved the use of fractional coefficients. Balance these three equations using ONLY fractional coefficients:

S_{8} + F_{2} ---> SF_{6}

C_{4}H_{10} + O_{2} ---> CO_{2} + H_{2}O

S_{8} + O_{3} ---> SO_{2}

Answers to the three fractional coefficient problems

Be careful on using fractions. For example, (1/2)H_{2}O is not a correct use of fractions. Why not? (1/2)H_{2} = one H atom, but (1/2)O = one half of one atom. You cannot split atoms in chemical reactions.

Generally speaking, fractions are mostly used with diatomics (with O_{2} is the most common). However, as you delve into this, you will sometimes see something like (1/2)H_{2}O ussed in a balancing step, but you will never see it as the answer.

Want more balancing practice? Here's Balancing Worksheet #1 with 50 problems and answers.

Want still more? Here's Balancing Worksheet #2 with 60 more problems and answers.