Video: Balancing Equations

In this video we will learn how to balance chemical equations and we will discover some shortcuts we can use to balance multi atom groups

17:16

Video Transcript

In this video, we will learn how to balance chemical equations. And we will discover some shortcuts that we can use to balance multiatom groups.

Let’s start by looking at what we mean by the word “balanced” when we’re talking about reaction equations. Here we have two reaction equations. Both of these equations have N2 and H2 on the reactant side. And they have NH3, ammonia, on the right-hand side. The difference between these two reaction equations is that the one on the left is not balanced, while the one on the right is balanced.

A balanced equation follows the law of conservation of mass. The law of conservation of mass says that, in an isolated system, mass is neither created nor destroyed by a chemical reaction. What this means is that the mass of our reactants must exactly equal the mass of our products. But when writing reaction equations, we don’t really talk about grams, which is what we would usually associate with mass. So how does this apply?

When writing a balanced reaction equation, we need to have the exact same number of atoms of each element on the reactant side as we do on the product side. When we look back at our unbalanced reaction equation, we can see that we have two nitrogen atoms on the left, but only one nitrogen atom on the right. Likewise, we have two hydrogen atoms on the left, but three hydrogen atoms on the right. So this is why our first equation here is unbalanced.

In the balanced example, we can see that we’ve added coefficients in front of one of the reactants and the product. This means that we now have two nitrogen atoms on both sides and six hydrogen atoms on both sides. So this is why our second equation is balanced. So now that we understand what a balanced equation is and why it’s important, let’s have a look at how we go about balancing an equation.

The process of balancing a reaction equation can sometimes be a case of trial and error. Luckily, with practice, this will become a faster and more efficient process. But we’ll go through some key steps to get us started.

We’ll use the reaction of sodium metal with water as an example to go through the process. It can be useful to begin with the key reactant. This could be the most complex reactant or the one we’re most interested in. In this example, let’s begin with sodium. We can set up a little table, a bit like this, and count the number of sodium atoms in each of our reactants and then again for each of our products. In this case, we have one sodium atom on our reactants and one sodium atom on our products. So the sodium atoms are balanced for now.

Let’s move on to the next element. If we look at hydrogen, we have two hydrogen atoms in our water. So that’s a total of two hydrogen atoms in the reactants. But we need to be careful when looking at our products since two of our products contain hydrogen atoms. There is one hydrogen atom in our sodium hydroxide and two hydrogen atoms in our hydrogen gas. This gives us a total of three on the product side versus only two on the reactant side. So this is clearly not balanced yet.

Now, let’s go through and do our oxygen atoms. We can see that we have one oxygen atom in our reactants and one on our products. So the oxygen atoms are balanced for now. The next step to create a balanced equation is to add coefficients in front of some of the reactants and all products in order to change the number of the hydrogen atoms. Be mindful though that if you add a coefficient in front of water, for example, you’re going to change the number of hydrogen atoms but also change the number of oxygen atoms. So after adjusting any coefficients, you should always go back and recheck the number of all of your atoms.

The only unbalanced element in this equation at the moment is hydrogen. We have one extra hydrogen atom in our products than we do in our reactants. In an ideal world, we would place coefficients so that we only alter the number of hydrogen atoms, though of course this may not always be possible.

Our last product is the only molecule in our reaction which only contains hydrogen. But can we add a coefficient to this H2 to make our reaction balanced? Technically, we could add a coefficient of one-half. This would reduce the number of hydrogen atoms on the product side and, therefore, leave us with two hydrogens, both sides, making it balanced. At the same time, it doesn’t affect the numbers of any of the other elements.

While this is technically a balanced equation, it’s always nicer if we can have whole-number coefficients wherever possible. A really easy way to do this is to simply multiply the entire reaction by two. This results in a coefficient of two being placed in front of sodium, water, and sodium hydroxide. And the half coefficient in front of our hydrogen has disappeared. When we recheck the number of each element on both sides of our equation, we can see that we are still balanced.

Now that we know the basics of how to balance an equation, let’s see if there are any tricks or tips for making this process easier. It turns out that there is a little shortcut for reaction equations which involve polyatomic ions. Remember that polyatomic ions are made up of more than one atom. Examples of polyatomic ions are nitrate, NO3−; sulfate, SO42−; and ammonium, NH4+.

We can modify our method for balancing an equation when it involves polyatomic ions. We are able to treat polyatomic ions as a single unit and balance them as such. But note that this only works if the polyatomic ion is present both in the products and the reactants. To understand this more clearly, let’s have a go at an example. Let’s have a look at this example reaction.

Aqueous solutions of Pb(NO3)2 and NaCl are mixed, producing aqueous NaNO3 and a PbCl2 precipitate. Write a balanced equation with state symbols.

Let’s begin writing our balanced equation by simply putting the reactants and the products down. We have lead nitrate and sodium chloride as our reactants. And we have sodium nitrate and lead chloride as our products. We can already add our state symbols as well. We are told that both of our reactants are aqueous, as well as our sodium nitrate, but our lead chloride is a precipitate, making it a solid.

Now we just need to balance our equation. So let’s set up our table for counting the atoms. Our most complex reactant is, of course, our lead nitrate. So let’s begin there. We have one lead atom on the left and one lead atom on the right. So for now, our lead atoms are balanced.

Next, our shortcut can come in handy. When we look closely, we can see that we have the nitrate ion on both sides of our equation. So rather than counting this nitrate ion as nitrogen atoms and oxygen atoms, we can simply count it as a single unit. When we do this, we see that we have two nitrate ions in our reactant and only one in our product. So this is not yet balanced.

Next, let’s finish counting the rest of our elements. We can see that our sodium atoms are balanced, but our chlorine atoms are not. So here we have two elements which need to become balanced. Let’s begin by balancing our nitrate ions. By adding two as a coefficient in front of our sodium nitrate, we have balanced our nitrate ions. However, don’t forget this also changes the number of sodium atoms.

Now we need to rebalance our sodium atoms. We have two on the products and only one on the reactants. So let’s add a coefficient of two in front of our sodium chloride reactant. Remember that this also alters the number of chlorine atoms. So now our sodium atoms are balanced, let’s move on to our chlorines.

When we come to recheck our chlorines, we can see that actually we’ve already balanced them. And a quick recheck of all of our numbers shows that this is now our final balanced equation with state symbols.

We now have a process by which we can try to balance equations and a handy little shortcut for when there are polyatomic ions involved. So let’s use this on some questions.

Gallium(III) bromide can be produced by the reaction of gallium with bromine. Write a balanced molecular equation for this reaction including state symbols.

This question is asking us to write a chemical equation from the information about a reaction given. More importantly, it’s asking for a balanced molecular equation. When an equation is balanced, it obeys the law of conservation of mass. This means in practice that we need to have the same number of atoms of each element on both the reactant side and the product side.

Let’s begin writing our equation by extracting the reactants and the products from the information given. We are told that this reaction produces gallium(III) bromide. So this must be our product. Now we just need the reactants. We are told in the question that these are gallium and bromine. The chemical symbol for gallium is Ga. It can be found on the periodic table just below aluminum. The symbol for bromine is Br. However, don’t forget that bromine exists as a diatomic. So this is really Br2.

Now we just need to work out the formula for gallium(III) bromide. Remember that bromine forms a negative anion. This Br− anion is called bromide. We are told that we have gallium(III) in our gallium bromide. So this means that we have gallium(III) plus as a cation. The formula for gallium bromide can’t be GaBr. This is because the charges don’t balance. We need two more negative charges from the bromide to balance out the three plus charge on the gallium. We can do this by adding two extra bromide anions. So our formula for gallium(III) bromide is GaBr3.

Next, we need to balance our equation. To balance our equation, let’s start with the main reactant. In this case, let’s go with gallium. We have one gallium on the reactants and one on the products. So for now, the gallium atoms are balanced. Next, we come to bromine. We can see that we have two bromine atoms on the left but three on the right. So this is not yet balanced.

We need to alter our coefficients to balance our bromines. Ideally, what we want is one more bromine atom on our reactant side. So we could place a coefficient of three over two in front of our Br2. This would give us three bromine atoms on both sides of our reaction, making it balanced.

However, it’s not always nice to have fractional coefficients. So let’s get rid of our fraction by multiplying the whole reaction by two. This simply doubles the number of all of the atoms on both sides of our equation. So it’s still balanced. But we can always recheck just to be sure.

Finally, we need to add our state symbols. State symbols tell us whether our reactants or products are a solid, liquid, gas, or an aqueous solution. Bromine is, of course, a liquid at room temperature. Gallium is technically a solid at room temperature. But for this reaction, it’s going to need to be heated. Interestingly, the melting point of gallium is only slightly above room temperature. So it’s a metal that can melt in your hand. So we’ll label gallium as a liquid. Gallium bromide is a white powder, so we can label that as a solid. So here is our balanced molecular equation.

Let’s summarize our key points. Balanced reaction equations obey the law of conservation of mass. This means that a balanced reaction will have the same number of atoms of each element on the reactant side as it does on the product side. The process of balancing equations is sometimes a trial-and-error approach. But with practice, you’ll get better at it. The general method is to start with your main reactant, adjust the coefficients, and recheck, repeating this process until your equation is balanced.

We’ve also seen that there is a shortcut you can use when dealing with polyatomic ions. We can balance polyatomic ions like nitrate and sulfate as a single unit, making life much simpler. Remember though that this only works if the polyatomic ion is present on both sides of our reaction equation.

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