Chemistry 6.3 Tips and Tricks for Balancing Equations - Welcome to tips and tricks for balancing chemical equations. The last video introduced what balancing equations is and why we do it. In this video, we're going to look at some helpful tips and tricks that you can use when you're balancing chemical equations. In this first example, we have a chemical equation showing the combustion of propane. If C3H8, that's propane gas, plus oxygen gas yields carbon dioxide and water. Your first step should always be to create a chart that keeps track of the number of atoms of each element on both sides of the reaction. In this particular reaction, we have the elements carbon, hydrogen, and oxygen. We're going to count the number of atoms that show up on the reaction side first. We have three carbons, eight hydrogens, and two oxygens. If we look at the product side next, you'll see that there's one carbon, two hydrogens in the water, and then a combined three oxygens, two from the carbon dioxide and one from the water. The first tip for balancing equations is to put a star next to any element that appears more than once on the same side of the reaction. So oxygen appears twice on the product side. So we're going to put a star next to it. And that's going to tell us we're going to balance this one last. Now that I know I'm going to do the oxygen last, I can choose to either balance that carbon or the hydrogen. And this is arbitrary. So I'm going to start with carbon. I have three on the reaction side and one on the product side. So I need to multiply the product side carbons over here by 3. I do that by writing a coefficient in front of the carbon dioxide. That changes my number of carbon to three, but it also changes my oxygens. I have three times two is six, plus the one from water gives me a total of seven oxygens on the product side. Now, I have hydrogen left, so I need to balance that. I have eight hydrogens on the left side, two on the right side, those are over here. To make both sides have eight atoms, I need to multiply this one by four. That will get me eight hydrogens. And now the hydrogens are balanced. Multiplying this water by four also changes a number of oxygens. I now have four oxygens from the water, and I have six oxygens from the carbon dioxide. So I have a total of ten oxygens on the product side. Now that carbon and hydrogen are both balanced, I can go ahead and take care of the oxygen, ten on the right side and two on the left side means I need to multiply the reactant side oxygens by five. And now this equation is balanced. The coefficients are one, five, three, and four, so I know that they are in the lowest whole numbers possible, and this is a correctly balanced equation. The key to balancing this example was recognizing that oxygen shows up twice on a single side of the reaction. So that was going to be the one that's balanced last. Let's take a look at our next example. In this example, we're going to make use of our second tip. And that's to recognize polyatomic groups as acting as a single unit. So I have a polyatomic here, that also shows up on the other side, my sulfate ion. I also have the hydroxide group, OH and OH. They both show up on both sides. What this means is that I'm not going to count the elements individually that are part of this polyatomic group when I make my chart of atoms. Now that I've identified the atomic groups, I'm going to list out the components that show up on the left and the right side. I have aluminum. I have the sulfate ion. I have calcium, and I have hydroxide. Those are my four different elements, or polyatomics, that show up on both sides of the equation. I'm going to count the number on each side. I have two aluminum's on the left and one on the right. I have three SO4 groups on the left, and only one on the right. For calcium, there's a one on the left and one on the right. And for hydroxide, I have two and three. It doesn't really matter where I start balancing, so I'll start with the aluminum. I need to multiply the product side aluminums by two, which is over here. That's going to change my aluminum count to two but it also changes my hydroxides on the product side. I will now have six hydroxide groups on the product side. I can keep going down my list here. I have SO4 next, three on the left, one on the right. So I'm going to multiply the product side sulfates by three, which is going to change sulfate to three, but also is going to change the calcium on the product side to three as well. Aluminums are balanced, SO4 is balanced. Let's do calcium next. Calcium is one on the reactant side, and three on the product side. I'm going to multiply the reactant side by three, which is going to change this to three. But it also changes the number of hydroxyl groups. Three times two gives me six on this side. That last step not only balance the calciums, but it also balanced the hydroxide groups. And now we have the full balance equation for this reaction. Example three shows a third trick that we can use to make balancing equations a little bit easier. If you find that you have hydrogen and hydroxide on the reaction side of your equation, but you have water on the product side, it's helpful to rewrite the water as HOH, a combination of hydrogen and hydroxide. This allows you to count hydrogen separate from hydroxide when you create your list, because now they are both on each side of the equation. So I have hydrogen on the reactant side. I have a PO4 group. Here's a polyatomic that shows up on both sides. I have sodium and I have hydroxide. If I go through and count now, there are three hydrogens on the left and one on the right, from this water that got broken up. It's one hydrogen and one hydroxide. For phosphate, PO4, there's one on the left and one on the right. For sodium, there's one of the left and three on the right, so that one needs to be balanced. And for hydroxide, I have one on the left and one on the right, hydroxide from the water. I'm going to start by balancing the hydrogens. I need to multiply the product side hydrogens by three. And those are over here where the water used to be. So I'm going to put a three in front of that. That's going to change my hydrogens on the product side to have three, but it also changes the hydroxides on the product side. There are now three hydroxides. So that means hydrogen is OK. Phosphate's already OK, so I'm going to skip that one. Let's look at sodium. Sodium has three on the product side. I need to have three on the reactant side. Sodium is right here, so I'm going to multiply this compound by three. That's going to give me three sodiums on the left, as well as three hydroxides that it also changes. Once I make this change, I could run through my list and see that everything is now balanced. So it's really only beneficial to break up this water into hydrogen and hydroxide, if you see hydrogen and hydroxide on the reaction side. That wraps up tips and tricks for balancing chemical equations, write down any questions you have in your notes, and bring them with you to class.