How Do You Balance Chemical Equations?

Hi, and welcome to this video on balancing chemical equations!

In any reaction, the law of conservation of mass must be followed. In other words, when reactants recombine or decompose to form products, we can’t lose or create atoms- we have to end with the same number we started with. Chemists represent this in a balanced chemical equation, in which the number of each type of atom on the reactant side equals the number on the product side. Conversely, an unbalanced reaction has unequal numbers of each type of atom on the reactant and product side of the equation.

Balancing Equations Examples

For example, here’s the unbalanced reaction of the combustion of methane.

CH₄ + O₂ -> CO₂ + H₂O

Methane reacts with oxygen to produce carbon dioxide and water. While this equation accurately represents the chemical identity of the reactants and products, it’s unbalanced. If we count the atoms on each side of the reaction, we’ll discover that they are not equal!

It’s often helpful to keep track of our atom count in a table.

While counting, remember that the subscripts (the 4 in methane, 2 in oxygen, 2 in carbon dioxide, and 2 in water) they denote the number of atoms of the element preceding the subscript within the compound.

On the reactant side, we have 1 carbon, 4 hydrogen, and 2 oxygen atoms. On the product side, we have 1 carbon, 2 hydrogen, and 3 oxygen atoms. So, we need to balance this reaction.

Note that to balance equations, we have to honor the chemistry, meaning that you can’t simply add in extra compounds or atoms to achieve balance. This is because, of course, the equation must still reflect the chemistry that actually takes place. So for example, while this equation is technically balanced in that there is the same number of each element on either side of the reaction, it’s chemically inaccurate.

CH₄ + O₂ + O -> CO₂ + H₂O + H₂

This reaction doesn’t produce H₂ gas and certainly doesn’t use a lone oxygen atom as a reactant.

We also can’t adjust the subscripts of compounds to achieve balance, as this would change the chemical identity of the species.

CH₄ + O₃ -> CO₂ + H₄O

So again, while this is a balanced equation, we are no longer representing the combustion of methane. O₃ is ozone, and is chemically very different from O₂, and H₄O is hydrogen hydrate. By changing the subscripts, we change the chemical identity of the species and wind up with an entirely different and probably unviable reaction.

Instead, to balance a reaction, we adjust the number of each species in the chemical reaction. This is represented in the chemical equation by adding coefficients in front of the chemical species.

__CH₄ + __O₂ -> __CO₂ + __H₂O

For example, if we put a two in front of water, this means we have two water molecules, or 4 hydrogen and 2 oxygen atoms.

__CH₄ + __O₂ -> __CO₂ + 2H₂O

Notice that by doubling the water, hydrogen (in addition to carbon) is now balanced.

Now to finish we just need to balance oxygen. We have 2 on the reactant side and 4 on the product side, so by simply adding a second molecule of oxygen to the reactant side, we have ourselves a balanced reaction!

1CH₄ + 2O₂ -> 1CO₂ + 2H₂O

Note that it’s standard to drop the coefficient when it’s 1.

We can now read this equation as 1 molecule of methane reacts with 2 molecules of oxygen to produce 1 molecule of carbon dioxide and 2 molecules of water.

Notice that the equation is also balanced with these coefficients-

2CH₄ + 4O₂ -> 2CO₂ + 4H₂O

We’ve simply multiplied each coefficient by 2, which does give us a balanced chemical equation; however, we typically want the lowest whole number coefficients. So, similarly, while this

½ CH₄ + O₂ -> ½ CO₂ + H₂O

is also technically balanced, it uses fractions rather than whole numbers as coefficients. So once you’re finished balancing your reaction, make sure there’s no common denominator between coefficients or any fractional coefficients. If there are, either divide out by the common denominator or multiply to get the lowest whole number coefficients.

Now that we understand how to balance equations, let’s take a step back and think about why they are important. These coefficients not only ensure we follow the law of conservation of mass, but also dictate the calculation of reactants and products, or the stoichiometry, of the reaction (you’ll sometimes see the coefficients referred to as the stoichiometric coefficients).

For example, in the combustion of methane, to produce 1 mole of carbon dioxide, we need 2 moles of oxygen but only 1 mole of methane. Put another way, the mole ratio is 2:1 for oxygen and carbon dioxide but 1 to 1 for methane and carbon dioxide. Thus, these coefficients are necessary for calculating limiting reagents, yields, molarity, or other calculations that requires converting between moles of species within a reaction. Basically, it’s pretty important to have your chemical equation balanced correctly.

Though there is no one sure-fire way to balance chemical equations, there are a few useful techniques to apply. To summarize- it’s usually best to balance the elements in the more complex compounds (or multiple compounds) first and leave simple species for the latter steps. Second, for elements in only one species on either side, try using the subscript of the opposite side as a coefficient. Third, while neither of our examples dealt with charged ions, if you do encounter a chemical equation with ions, make sure that the charge is balanced as well! After all, you don’t want to lose electrons!

And lastly, some equations are simply going to be more challenging than others. It’s important not to get frustrated. Simply jump in and start by balancing one element and evaluate how that changed the entire chemical equation. Then move on to a second element, add a coefficient, and re-evaluate. Keep running through this process until you’ve got yourself a balanced equation- but remember to check that you’ve found the set of lowest whole-number coefficients.

Now that we’ve pointed out a few tips for balancing reactions, let’s practice some more. Pause the video and balance the following chemical equations: Oxidation of nitric oxide by oxygen, Roasting of the sulfide ore (pyrite-FeS₂), Combustion of ethane, and Decomposition of lead nitrate.

Oxidation of nitric oxide by oxygen
NO + O₂ -> NO₂

Roasting of the sulfide ore, pyrite- FeS₂
FeS₂ + O₂ -> Fe₂O₃ + SO₂

Combustion of ethane
C₂H₆ + O₂ -> CO₂ + H₂O

Decomposition of lead nitrate
Pb(NO₃)₂ -> PbO + NO₂ + O₂

Alright, hopefully that went well! Here are the four chemical equations balanced.

Oxidation of nitric oxide by oxygen
2NO + O₂ -> 2NO₂

Roasting of the sulfide ore, pyrite- FeS₂
4FeS₂ + 11O₂ -> 2Fe₂O₃ + 8SO₂

Combustion of ethane
2C₂H₆ + 7O₂ -> 4CO₂ + 6H₂O

Decomposition of lead nitrate
2Pb(NO₃)₂ -> 2PbO + 4NO₂ + O₂

The last three were pretty challenging, so don’t feel discouraged if they took you a while or you weren’t able to solve them. The best thing to do is to keep practicing and you’ll start to feel more comfortable with the process.

That’s it for today’s video. Thanks for watching, and happy studying!