Stoichiometry Tutorial: Step by Step Video + review problems explained | Crash Chemistry Academy

This is a

tutorial

on

stoichiometry

. Stoichiometry is basically defined as: So what exactly does that mean? And how 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 reacts with nitrogen to produce ammonia. It's not balanced, so we'll have to add some coefficients. Mastering

stoichiometry

requires an understanding of: In other words: we can think of coefficients as representing a specific amount of something, a number of particles, or as proportions of reactants and products in the reaction. To really understand how stoichiometry works, we really have to understand what those coefficients really tell us.

If they mean a specific number of particles, that would tell us how many particles are reacting and how many particles will be produced. For example, writing the reaction as we are seeing it here, we would have three molecules of hydrogen reacting with one molecule of nitrogen and that produces two molecules of ammonia. So let's see how it works. The three hydrogens collide with one nitrogen, everything breaks down and reforms into two ammonia molecules. In our everyday world, we work with massive quantities of particles and therefore what the coefficients can really tell us are simply the proportions of what happens in the chemical reaction.

For example, if we have six hydrogen molecules, they will react with two nitrogen molecules. Or if we have nine molecules of hydrogen, they will react with three molecules of nitrogen. So what I would like you to see here is that even with large amounts of molecules, this reaction still occurs at a ratio of three hydrogens to one nitrogen, producing two molecules of ammonia. And with this particular reaction, you're always going to react in that three to one to two ratio, no matter how many molecules you have. You can see that we have produced, using nine hydrogens and three nitrogens, a total of six ammonia molecules.

Let's back up for a moment. So what's really going on here? Three hydrogens react with one nitrogen and give us two ammonias. But in reality that is just a proportion. So no matter how many you have, it will always be with this reaction in a ratio of three to one to two. And so, the coefficients of any chemical equation give the ratio of reactants to products in the reaction. Ratios can work for us mathematically, because we use very, very large quantities of molecules on the everyday level. But the proportion remains constant. And now we can see how all of this connects to molar quantities of particles.

Take a look at the last line. In fact, we can see that the coefficients are mole ratios. We have three moles of hydrogen molecules reacting with one mole of nitrogen molecules. And that produces two moles of ammonia molecules. So, since they are ratio coefficients, they can represent any quantity, including molar quantities. And since we use such large quantities of particles at the level of molar quantities, we use the coefficients as mole ratios. Let's see how that relationship can work stoichiometrically for us. 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?

Note that the mathematical relationship given by the ratios of the coefficients remains constant no matter how many substances react, as we saw in the first part of the

video

. So in the rest of the

video

we'll look at several examples to flesh out what you might typically see in stoichiometry

problems

in a textbook. We'll see: Let's look at an example using this reaction of hydrogen and nitrogen. So if we have 7.5 moles of hydrogen, how many moles of nitrogen would react with that? We use the ratio of coefficients of three hydrogens to one nitrogen to determine the answer.

The quantity given in the problem is 7.5 moles of hydrogen. And we multiply that by a fraction that represents the proportion of coefficients, the proportion of one nitrogen to three hydrogen, always putting what we want at the top and what we have at the bottom. 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 know how many moles of hydrogen are needed in the reaction to obtain 0.8 moles of ammonia, NH₃. We write down what was given in the problem, 0.8 moles of NH₃, and multiply it by the proportion of three hydrogens for every 2NH₃.

That tells us that it takes 1.2 moles of hydrogen 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 simply assume that there is enough nitrogen to react with that amount of hydrogen. This is always the central part of the stoichiometric calculation, the mol-mol conversion. These mol-mol connections always follow the same format, where "A" is the quantity given in the problem and "B" is the desired quantity in the problem. And multiplying by the ratio of coefficients we obtain from moles of A to moles of B, the given moles to the desired moles.

Again, this will always be the central calculation for any stoichiometry problem, keeping in mind that A, what you start with, goes down so it cancels, and B goes up so you end up with what you want. want. But there is a problem with this. We can't really know the number of moles directly because we can't count the number of particles. But we can weigh quantities of substances that we can then convert into moles. That expands our map like this. Here is our ever-present mole conversion, and given the mass of a substance in the reaction, you can convert it to moles using the molar mass and then convert it to the desired moles and finally convert the desired moles to the desired mass using the molar mass of B.

We can now see how this map can guide us in a mass-to-mass stoichiometry problem involving 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 complete the map to fit our problem. We start with a mass of nitrogen that they give us and we want to end with a mass of ammonia, which is the desired substance. We can see that we must first convert the mass of N₂ to moles of N₂ using the molar mass of nitrogen from the periodic table, which is 28 grams of N₂ in one mole of N₂.

Then moles of N₂ to moles of NH₃ using the coefficient relationship and finally 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 supplied is 42 grams. So we multiply that by one mole of nitrogen times 28 grams of nitrogen. This calculation gives us 1.5 moles of N₂. The next

step

is to convert moles of N₂ to moles of NH₃. So here's the core part of any stoichiometric calculation: converting moles of one substance to moles of another substance using the ratio of coefficients.

The last

step

converts moles of ammonia to mass of ammonia, using the molar mass of ammonia. Then 51 grams of NH₃ will be produced from the complete reaction of 42 grams of N₂. Typically, with stoichiometry, as in dimensional analysis, we can generally more efficiently write chain calculations. And you can see, as in any properly written dimensional analysis, all the unwanted units cancel out and the desired unit is the one you are left with. If the units don't cancel out, you're setting it up incorrectly and will get the wrong response, so you'll have to go back and check 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 set up correctly, what you are putting into your calculator is simply multiplying all the numbers in the numerator and dividing by everything that is in the nominator. I hope you can see the usefulness of stoichiometry here. A simple way for the chemist to determine how much product will be formed in the reaction or how much of one reactant is needed to react with a certain amount 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're doing now. Here is the following equation. We can't do stoichiometry without coefficients, so we will need to perform a balance. Again, these coefficients, as in any reaction, are ratios that we can use as a central part of the stoichiometric conversion of 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 given quantity, 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 a coefficient ratio of 7 to 4, and then to mass of O₂ using the molar mass of O₂. Make sure everything cancels out so that we are left with a mass of oxygen, which amounts to 19.6 grams. The calculation is to multiply the numbers above and divide by the numbers below.

Let's do one more mass - mass practice, using the combustion of ethane, C₂H₆. First we will add coefficients to balance. Why don't you pause the video to figure it out and then come back to see how it went? If you're having trouble getting started, let's make a map first and follow the map. Here is our map. We start with what we are given, convert it to moles, then the mol-mol conversion, and then convert it to mass of carbon dioxide. We start with what we are given, convert it to moles, then the mol - mole conversion and then convert it to mass of carbon dioxide.

So the reaction of 37.5 g of C₂H₆ produces 110 g of carbon dioxide. Let us now move on to other types of stoichiometric calculations. The core calculation is always provided by the mol - mol calculation. And so any given quantity that can be converted to moles can expand the repertoire of calculations. For example, suppose the problem provides a certain number of particles in a reaction instead of mass, or you want 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 gases in the equation at standard temperature and pressure, you can use the fact that one mole of gas particles occupies 22.4 L at STP. Now we can see the map that shows us the route to convert to or from particles, mass or volume of gas. For example, a problem that gives us the volume of gas and asks for particles, would first convert the volume of gas to given moles and the desired moles to desired particles, using these conversions. A problem that gives us mass and requests gas volume would use this route in the conversions.

You can also use the map to end at the moles, like here, or end at the given moles, like here. Using the last equation, the combustion of C₂H₆, let's use the map to find the mass of water produced from a given number of particles, in this case, C₂H₆ molecules. The map gives us the path from particles to mass. We first convert the molecules to moles of C₂H₆, then to moles of water, and then to mass of water. By reacting 2.8 x 10²⁴ molecules of C₂H₆, 251 grams of water are obtained. So that's the basic stoichiometry.