This is a tutorial on Stoichiometry. Stoichiometry is basically defined as: So, what exactly does that mean? And how can we can we understand, mathematically, how everything written in the chemical equation relates to each other? Let’s take a look at a fairly simple reaction. Hydrogen [H₂]reacting with nitrogen [N₂] to produce ammonia [NH₃]. It is not balanced, so we’ll have to add some coefficients. Mastering stoichiometry requires an understanding of: So in other words: So we can look at coefficients as representing either a specific amount of something, an amount of particles, or as ratios of reactants and products in the reaction. To really understand how stoichiometry works, we really have to understand what those coefficients are actually telling us. If they are referring to a specific amount of particles, that would tell us how many particles are reacting and how many particles will be produced. For example, the way the reaction is written as we are looking at it right here, we would have three hydrogen molecules reacting with one nitrogen molecule and that produces two ammonia molecules. So let’s see how that works. The three hydrogens collide with one nitrogen, everything is broken apart and reformed as two ammonia molecules. In our everyday world, we are working with massive amounts of particles and so what the coefficients can actually tell us are simply the ratios of what is going on in the chemical reaction. For example, if we have six hydrogen molecules, they are going to react with two nitrogen molecules. Or if we have nine hydrogen molecules, they are going to react with three nitrogen molecules. So what I’d like you to be able to see here is that even with large amounts of molecules, this reaction still occurs with a ratio of three hydrogens to one nitrogen, producing two ammonia molecules. And with this particular reaction, it will always react in that three to one to two ratio no matter how many molecules you have. You can see we have produced, using nine hydrogens and three nitrogens, a total of six ammonia molecules. Let’s back up just a moment. So what’s really going on here? Three hydrogens reacting with one nitrogen, giving us two ammonias. But that’s really just a ratio. So no matter how many you have, it’s always going to be with this reaction in a three to one to two ratio. And so the coefficients of any chemical equation give the ratio of reactants to products in the reaction. The ratios can work for us mathematically, because we use very very large amounts of molecules at the everyday level. But the ratio remains constant. And now we can see how this all connects to molar amounts of particles. Take a look at the last line. We can actually see that the coefficients are molar ratios. We have three moles of hydrogen molecules reacting with one mole of nitrogen molecules. And that produces two moles of ammonia molecules. So the coefficients being ratios, they can represent any amounts, including molar amounts. And because we use such large amounts of particles on the level of molar amounts, we use the coefficients as molar ratios. Lets see how that ratio can work for us stoichiometrically. In other words, how can we use that ratio to relate an amount of one substance in a reaction to an amount of any other substance in the same reaction. Keep in mind that the mathematical relationship given by the coefficient ratios hold constant no matter how much of the substances are being reacted as we saw in the first part of the video. So in the rest of the video we will be looking at various examples to build up to what you might normally see in stoichiometry problems in a text book. We will look at: Let’s look at an example using this reaction of hydrogen and nitrogen. So if we have 7.5 moles on hydrogen, how many moles of nitrogen would react with that? We use the coefficient ratio of three hydrogens to one nitrogen to determine the answer. The amount given in the problem is 7.5 moles of hydrogen. And we multiply that by a fraction that represents the coefficient ratio, the one nitrogen to three hydrogen ratio, always putting on top what we want and on bottom what we have. It tells us that 2.5 moles of N₂ will react with 7.5 moles of H₂. And 7.5 to 2.5 is a 3:1 ratio. Let’s look at a second example: You want to find out how many moles of hydrogen are needed in the reaction to get .8 moles of ammonia, NH₃. We write down what has been given in the problem, 0.8 moles of NH₃, and multiply it by the ratio of three hydrogens for every 2NH₃. That tells us that 1.2 moles of hydrogen are needed to produce 0.8 moles of ammonia. In a problem like this where the other reactant, in this case Nitrogen, is not mentioned, we would just assume that there is enough nitrogen to react with that amount of hydrogen. So this is always the central part of the stoichiometric calculation, the mole – mole conversion. These mole – mole connections always follow the same format, where “A” is the amount given in the problem and “B” is the amount wanted in the problem. And multiplying by the coefficient ratio gets us from moles of A to moles of B, the moles given to the moles wanted. Again, this is always going to be the central calculation for any stoichiometry problem, keeping in mind that A, what you start with, goes on the bottom so that it cancels out, and B goes on top so that you end up with what you want. But there is a problem with this. We cannot really know, directly, the amount of moles because we cannot count out an amount of particles. But we can weigh amounts of substances, which we can then convert to moles. So that expands our map like this. Here is our ever present mole conversion, and given a mass of one substance in the reaction you can convert it to moles using molar mass and then convert to moles wanted and finally convert moles wanted to mass wanted using B’s molar mass. We can now see how this map can guide us in a mass to mass stoichiometry problem that involves multiple unit conversions. The setup uses the same structure as any dimensional analysis problem. Let’s go back to the ammonia reaction. What mass of NH₃ will be produced from 42 grams of N₂ reacting completely? Let’s fill in the map to fit our problem. We start with a mass of nitrogen that we’re given and want to end up with some mass of ammonia, which is the substance wanted. We can see that we must first convert mass of N₂ to moles of N₂ using nitrogen’s molar mass from the periodic table, which is 28 grams of N₂ in one mole of N₂. Then [we must convert] moles of N₂ to moles of NH₃ using the coefficient ratio, and finally [convert] from moles of NH₃ to mass of NH₃ using the molar mass of NH₃ from the periodic table, which is 17 grams of NH₃ in one mole of NH₃. The amount of nitrogen given is 42 grams. So we multiply that by one mole of nitrogen per 28 grams of nitrogen. This calculation gives us 1.5 moles of N₂. The next step is converting moles of N₂ to moles of NH₃. So here’s the central part of any stoichiometric calculation: converting moles of one substance to moles of another substance using the coefficient ratio. The last step converts moles of ammonia to mass of ammonia, using the molar mass of ammonia. So 51 grams of NH₃ will be produced from the complete reaction of 42 grams of N₂. Normally, with stoichiometry, as in dimensional analysis, in general we can more efficiently write the calculations in a chain. And you can see, as in any correctly written dimensional analysis, all unwanted units cancel, and the wanted unit is the one you are left with. If the units don’t cancel, you are setting it up incorrectly and will get an incorrect answer, so you will need to go back and review the conversions to determine how to fix it. So the steps here are in this chain. This calculation gives us moles of nitrogen, the next gives us moles of NH₃, and finally the mass of NH₃. Once it is set up correctly, what you are putting in your calculator is simply multiplying all the numbers in the numerator, and dividing by everything in the nominator. Hope you can see the usefulness of stoichiometry here. A straightforward way for the chemist how to determine how much product will form in the reaction, or how much of one reactant is needed to react with some quantity of the other reactant. We’ll do a couple more practice problems and then see how we can extend stoichiometry beyond these mass to mass calculations that we are doing now. Here is the next equation. We cannot do stoichiometry without coefficients, and so we will need to balance. Again, these coefficients, like in any reaction, are ratios that we can use as the central part of the stoichiometric conversion for moles of one substance in the reaction to moles of some other substance in the reaction. So what mass of O₂ is needed to react with 5.95 grams of NH₃? Identify the amount given, which is 5.95 grams of NH₃, and identify what is wanted, which is mass of oxygen. Let’s draw a map to guide our chain of equations. Mass of NH₃ to moles of NH₃, to moles of oxygen, to mass of oxygen. We have the coefficients from the equation and the molar masses from the periodic table. Starting with 5.95 grams of NH₃, convert to moles of NH₃, then to moles of O₂ with the 7 to 4 coefficient ratio, and then to mass of O₂ using the molar mass of O₂. Make sure everything cancels out so that we are left with mass of oxygen, which comes out to 19.6 grams. The calculation is multiplying the numbers on top and dividing by the numbers on bottom. Lets do one more mass – mass practice, using the combustion of ethane, C₂H₆. First we’ll add coefficients to balance. Why don’t you pause the video to work it out and then come back to see how you did? If you have trouble getting started, let’s first make a map and follow the map. Here is our map. We start with what is given, convert to moles, then the mole – mole conversion,then convert to mass of carbon dioxide [CO₂]. We start with what is given, convert to moles, then the mole – mole conversion, then convert to mass of carbon dioxide. So the reaction of 37.5 g of C₂H₆, produces 110 g of carbon dioxide. Now lets move on to other types of stoichiometric calculations. The central calculation is always provided by the mole – mole calculation. And so any quantity given that can be converted to moles can expand the repertoire of calculations. For example, suppose the problem gives some amount of particles in a reaction rather than mass, or wants to end up with particles. Particles refers to atoms, ions, molecules, or formula units. These can be converted to moles and from moles using Avogadro’s number, 6.02 x 10²³, because 1 mole of particles is equal to 6.02 x 10²³ particles. Similarly, if you are working with volumes of gasses in the equation at standard temperature and pressure, you can use the fact that a mole of gas particles occupies 22.4 L at STP. Now we can se the map giving us the path to convert to or from particles, mass, or gas volume. For example, a problem that gives us gas volume and asks for particles, would first convert gas volume to moles given, moles wanted to particles wanted, using these conversions. A problem giving us mass and asking for gas volume would use this pathway in conversions. You can also use the map to end at moles, such as here, or end at moles given, such as here. Using the last equation, the combustion of C₂H₆, lets use the map to find the mass of water produced from a given amount of particles, in this case, molecules of C₂H₆. The map gives us the path from particles to mass. We first convert molecules to moles of C₂H₆, then to moles of water, then to mass of water, reacting 2.8 x 10²⁴ C₂H₆ molecules produces 251 grams of water. So that is basic stoichiometry.