Chemical equations, like mathematical equations, must be balanced.

Chemical equations and mathematical equations have a lot in common. The equal sign in a math equation means that both sides, however they're written, are equivalent expressions of the same thing:

2(3) - 4 = 5(2) - 8 ← Try it for yourself. The arrow or arrows in a chemical equation mean the same thing. Later, we'll also do some algebra with chemical equations that will help us solve real problems.

Balancing chemical equations

Chemical equations must be mass balanced and charge balanced across the or ⇌ sign.

  • Mass balance means the same number and kind of atoms on each side
  • Charge balance means the same number of unpaired + and - charges on each side.

In this section we'll be entirely concerned with mass balance, and we'll defer charge balance until we discuss redox reactions (reactions in which electrons are transferred between reactants and products) in another section.

From time to time, you will have to balance a chemical equation before going on to do more useful or meaningful things, like figure out how much of a reactant to weigh out to carry out a reaction.

A properly-balanced chemical equation is the absolute basis for most of the calculations we need to do in chemistry. You really need to learn it, and the better you become at it, the easier you'll learn what comes next.

Mass balance is nothing more than another statement of one of the most important "laws" in science:

Conservation of mass

The requirement of mass balance – that there be the same number of each type of atom on each side of a chemical equation – follows from the law of conservation of mass: Mass can neither be created from nothing, nor destroyed*.

* Einstein, of course showed that matter with mass can be directly converted into energy, but we don't normally encounter that in our study of chemistry. It will be covered in the physics section.

Examples of balanced equations

The best way to illustrate mass balance is to take a look at some balanced equations, then we'll look at unbalanced reactions and see how we can bring them into balance.

The decomposition of hydrogen peroxide (H2O2 - you can buy a 2-3% aqueous solution in the pharmacy) is shown below.

The reaction is balanced. On the left (reactant side) there are two hydrogen atoms and two oxygen atoms. They're arranged differently on the right (product side), but the numbers are the same. This equation is mass balanced. Mass does not appear out of nowhere on the right, nor does it disappear on the left . . . ahh, the universe is spinning in greased grooves.

Now let's look at a slightly more complicated reaction (below), one with numerical coefficients in front of the reactants and products. Ignore those for a moment and check for balance: Without those coefficients, there are 8 carbon, 18 hydrogen and 2 oxygen atoms on the reactant side; and 1 carbon, 2 hydrogen and 3 oxygen atoms on the product side ... clearly not balanced. That's why we need the coefficients.

Forget about how to assign those coefficients for now and let's just check the balance again, this time including the coefficients, which tell us how many of each reactant or product must participate in the reaction.

On the left, 2C8H18 means two molecules of C8H18 (octane - like in your gas tank). That sums to 16 total carbons atoms and 36 total hydrogen atoms on the left side. Applying the same logic, 25 O2 molecules means 50 oxygen atoms.

I've written out all 16 CO2 molecules on the right, just to help count the atoms. 16 CO2's is 16 carbons and 32 oxygens. Add the oxygens to the 18 from H2O and the oxygens match, 50 on each side. The carbons and hydrogens match, too. The equation is balanced; equal numbers of each atom appear on either side.

The equation above means that in order for octane to react fully with oxygen, there must be 12-1/2 oxygen molecules for every octane. It's a lot more pleasing of course not to split up O2 molecules, so we settle on the whole-number ratio 25 O2 for every two C8H18.

The reaction will produce 16 molecules of CO2 and 18 molecules of water. With fewer than 25 O2 molecules for those two octanes, not all of the octane will react.

Starting with an unbalanced equation

Our job before doing anything meaningful with a chemical equation will always be to make sure it's balanced. Here's our first example of an unbalanced equation and how to balance it:

Clearly this equation is unbalanced: There are two hydrogen atoms on each side, but the oxygens are unbalanced, two on the left, one on the right. What to do?

Balancing chemical equations involves a bit of trial-and-error and you have to practice enough to develop a little intuition about it.

Start this one by looking at the oxygens. Notice that there's an odd number (1) on the right and an even number (2) on the left.

If we multiply the right side by a coefficient of two, the oxygens will balance, 2 and 2:

H2   +   O2   ⇌   2H2O

That even-odd thing is a key observation. There is no whole-number coefficient you can multiply O2 by to get an odd number of oxygens, so you'll need to multiply H2O (with an odd number of oxygens) by an even number. Now there are four hydrogens on the right, but that's easy to fix by multiplying the H2 on the left by 2:

2H2   +   O2   ⇌   2H2O

That's it. all balanced, and we now know that it takes two molecules of H2 to react with one molecule of O2 and produce 2 molecules of water in order for our reaction to obey the law of conservation of mass. These ratios of reactants and products will be hugely important in work we'll do later, so keep on moving through this section.

Another example

Now let's look at a more complicated example. This is the unbalanced equation for the combustion of phenol. The structure of phenol is shown, just in case you're curious. The equation in the yellow box is the important part.

The reaction is clearly unbalanced. We just need to notice that the number of carbons is different, 6 on the left, 1 on the right to know that. There are many possible approaches to balancing this equation; what follows is just one, and I encourage you to develop your own style of balancing after following a few examples.

What follows is just my train of thought as I write this. So far as I can recall, I've never balanced this equation before now:

Both sides have a moeity with an odd number of oxygens, so I'm not overly concerned about multiplying everything by two right now, though I might have to later, who knows?. I'll start with the carbons (a random choice, but there's probably some "intuiting" going on in my brain because I've done this many times):

C6H6O   +   O2   ⇌   6CO2   +   H2O

Now the hydrogens: I've got six on the left, so I'll have to multiply H2O by 3:

C6H6O   +   O2   ⇌   6CO2   +   3H2O

OK, so that's 6 carbons and 6 hydrogens on each side. Now I'll do the oxygens. This last step could spoil one of the other elements I've just balanced, so be aware of that — you might have to start over, but not this time:

C6H6O   +   7O2   ⇌   6CO2   +   3H2O

That's got it. When you're done balancing, take a look at your coefficients. If they're all divisible by a number, say 2, go ahead and do that to reduce the coefficients to their lowest-possible whole-number (integer) values. These are fine.

Pro tips:

Learning to balance an equation is part practice and part intuition that builds over much practice. Keep going and don't be afraid to try new ideas to get an equation to balance. You'll develop your own intuition and skill.

2. Beware of how coefficients and subscripts multiply to give you the number of atoms present in an equation. Here are a few examples:

2 SiCl4 Each molecule contains one Si and 4 Cl atoms, so two of them are 2 Si and 8 Cl.
2 Al2(OH)3 Each molecule contains 2 Al and 3 copies of (OH), or 3 O atoms and 3 H atoms. Two of these yields: 4 Al, 6 O and 6 H
3 Fe2(SO4)3 Each molecule contains 2 Fe and 3 copies of the SO4 group. Three molecules contain: 6 Fe, 9 S, 36 O. To get the number of oxygens, it's (4)(3)(3) = 36.
2 Al2(SO4)3 Each molecule contains 2 Al and three copies of the SO4 group. Two of these contain: 4 Al, 6 S, 24 O.

Practice problems

Balance the following reactions (some may already be balanced). Over over or click on each equation to check your answer.

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