Balancing more complex chemical equations

- [Voiceover] Let's now see if we can balance a chemical equation with slightly more complex molecules. So, here we have a chemical equation, describing a chemical reaction. This is actually a combustion reaction. You have some ethylene right over here, in the presence of oxygen, and you need to get a little bit of energy to get this going, but then you're going to have this reaction that's actually going to release energy as well, but we're not accounting for the energy, at least the way we've written it. Right over here, you have some ethylene, and this little g in parentheses, says it's in the gas form or gaseous form, so gaseous ethylene plus some dioxygen molecule, which is the most prevalent form of oxygen molecule that you would find in the atmosphere. And so, that's also in the gas form. Put them together, you end up with some carbon dioxide gas and some liquid water. This is the classic combustion reaction. But now let's think about, how do we balance this thing? Let's make sure we have the same number of each atom on both sides. And when you see something more complicated like this, where, you know, here I have an oxygen and two different molecules over here, and a lot of these molecules have multiple elements in it. It might be very daunting. Where do I start? And this is where the art of balancing chemical equations starts to come into play. The general idea is, Try to balance the... try to balance the molecules that have multiple elements in them first, and leave the... molecules that only have one element in them for last. And the idea there is, is that these are harder. They're going to have all sorts of implications, and then, at the end of the day, you can just set a number here for the number of dioxygens. If you saved, say the ethylene for last, then every time, and you're trying to balance the carbons, you try to change the number of carbons, you're going to change the number of hydrogens, which is going to change the... You're going to have to balance over and over, and you're going to go into this really really really confusing circle. So, the best thing to do, try to balance the complex molecules first, and then save the single element molecules for last. So let's do that. So, let's start with the carbons. So, over here, I have two carbons. Over here, I only have one carbon. I only have one carbon. So, it seems like the best way to balance it is, I should have two molecules of carbon dioxide, and I haven't even thought about the oxygens yet. By putting that two there, that's going to change the number of oxygens I have on the righthand side. But at least it balances my carbons. I now have two carbons on the lefthand side, and I have two carbons on the righthand side. I’m no longer magically destroying a carbon atom, all right. Now, let's move on to the hydrogens, and remember, what I said is, let's wait to do the oxygens last, because we have a molecule that only contains oxygen right over here, so we'll save oxygen for last. So, let's do hydrogen next. So, hydrogen, right over here, we have four hydrogens. And on the righthand side, we have two hydrogens. So, it seems like the easiest thing to do to balance the hydrogens is to have two of these water molecules. Now I have four hydrogens here, and I have four hydrogens there. Now, let's do the oxygen. Now, let's do the oxygen. I've balanced the carbons and the hydrogens. And the reason why oxygen's going to be interesting, I can just count the amount of oxygen I now have here, after changing the amount of molecules I have. And then I can adjust this accordingly, because this is only going to affect the number of oxygens that I have on the lefthand side. Right now, on the lefthand side, I have two oxygens, and on the righthand side, let me count this, I have two O two's, really. So, this is going to be four oxygens here, and then I have, each of these water molecules has one oxygen, but I have two water molecules, so this is going to be two oxygens, two oxygens here. So, on the righthand side, I have four plus two oxygens. So, I have six oxygens on the righthand side. I need six oxygens on the lefthand side. I need this number to be six. So, how do I do that? Well, I just need three of these molecules. If I have three molecules, each of them have two oxygens, I'm going to have a total of six oxygens. And just like that, we have balanced this combustion reaction, this chemical equation.