this is going to be an introduction to the topic of stochiometry and what that big complicated word means is just calculations involving amounts of reactants and products in reactions and that's covered in chapter 5 of the textbook and I included a picture from the beginning just above the table of contents from the textbook for chapter 5 shows a Rocket taking off and the point there is that often Rocket Fuel has various components and they need to be mixed in the exact correct ratio and so how you would calculate the amounts of each component you would need in the rocket fuel would be through sto geometry so we have seen that we can write equations for chemical reactions and we balance them so that the amount of each element is the same in the reactants and products but we didn't really say well what does that amount refer to what units are we talking about well as shown in this scheme from the textbook it could just be formula unit so it could be total number of molecules or something like that but as we've seen in chemistry we don't tend to deal with total number of atoms or molecules we use moles of course to count things so really this is what we're going to be doing is we are going to consider the coefficients in chemical reactions as referring to moles so stochiometry allows us to calculate how many moles of say a reactant or product B would either react with or be produced from a certain number of moles of another reactant or product a and so a reaction that we have studied already the combustion of propane and again here is the balanced equation for that so in sty geometry we would read that equation as saying okay if I have one mole of propane C3 h8 it would need to react with five moles of oxygen and then it would produce 3 moles of CO2 and four moles of water so that's what the St geometry tells us now this does not mean we need exactly these amounts but you know for whatever amount of propane reacted you would need five times as many moles of oxygen and you get three and four times the number of moles of CO2 in water so really what I'm saying here is that in stochiometry the balanced equation gives us the ratio of the moles of each reactant and product and so we can use that in calcul for example let's say 2 moles of propane reacted completely with oxygen so it says how many moles of CO2 would be produced so that's pretty easy because you can see that for every mole of CO2 every mole of propane that reacts you get three moles of CO2 so you just have to multiply the moles of CO2 by three but let's do this and use the principles of dimensional analysis to do the calculation like we've been doing all semester so if I have 2.0 moles of propane so the balanced equation tells us that we get 3 moles of CO2 for every one mole of propane that reacts so if I multiply by that ratio use that as the conversion factor notice the moles of propane cancel and that leaves me with moles of CO2 and of course 3 * 2 I would get 6.0 moles of CO2 now let's do another example same idea the math is just a little bit more complicated so now we have 2.0 moles of oxygen now let's say that amount of oxygen reacts with enough propane now in this case the ratio is that I get 3 moles of CO2 per 5 moles of o2 and so notice what I did is I just took the balanced equation or the coefficient I mean from the balanced equation for what we're trying to calculate and plug that in on top and then I divided by the coefficient for what we're given since O2 has a coefficient of five I plug that in there again now the moles of o2 are going to cancel and leave me moles of CO2 and so what I get is 2 * 3 / 5 which which works out to be 1.2 moles of CO2 so 2.0 moles of oxygen would produce 1.2 moles of CO2 so again you can use the balanced equation then and the coefficients to calculate the moles of any reactant or product related to moles of another reactant or product okay so that's fairly easy so far let's go one step further because we have seen that even though it's moles that matter in calculations it's how we count things in chemistry we can't measure moles directly usually but often you can measure grams you can way a sample and get its mass and we have seen that you can use in the molar mass to go between gr and moles so if we combine those two things that allows us to do something called gram to gam stochiometry so instead of just calculating how many moles of one thing you get from moles of another thing we can go from grams to grams so let's look at another example I'm going to change the reaction a little bit it's another combustion of a hydrocarbon this time is going to be octane which is C8 h18 and there is a balance equation for that reaction again it produces CO2 and water and other words we do have a fractional amount of o2 now of course if you don't like that or if you were asked to balance this equation using the simplest whole numbers of course you would just double all the coefficients and get the CO and get the equation I have below there I'm just going to use the top one because it makes the numbers a little easier and the amount of o2 is not going to come into my example so what does my example say so it's going going to ask how many G of CO2 could you get if 50 g of octane C8 h8 react completely so we're in this case we're going to do this in three steps so first of all again it's moles that matter so I need to convert the gr of C h18 to moles and so we have 50 g so of course we would need the molar mass so if you calculate the molar mass of C8 h18 it works out to be sorry 11 14.2 G per mole so if I divide 50 G by the molar mass I got 438 so that would be the moles of octane that we have okay now we can do just like I we did it in the previous examples and see how many moles of CO2 that would produce and you can see in this case it's an 8:1 ratio so I take the moles of octane that we have and the coefficients in the balanc equation say that I get 5 moles of C CO2 for every one mole of octane that reacts and that equals 3.50 so that's how many moles of CO2 we would get in this reaction but the question didn't ask for the moles it asked for the gram so I have one more step I have to convert the moles of CO2 to G CO2 1 Carbon 2 oxygen actually has a m mass of 44.0 G per mole so I'm going to take the 3.50 moles multipli by the molar mass 44.0 G per mole and I get a final answer of 154 G of CO2 so 50 g of octane would would produce 154 G of CO2 does that make sense I think so because essentially what you're doing is you're taking the octane you're taking away the hydrogen and adding two oxygens per carbon the oxygen weigh a lot more than hydrogens so it makes sense that the mass of CO2 is going to be more than the mass of octane that you started with okay let's just summarize what we've done here again it's three steps so and I did each step individually to show you what what we're doing here let's write this though just kind of as one uh long calculation in the format we've seen before for multi-step dimensional analysis type problems so I took the grams of octane and divided by the M mass that cancels the grams of octane gave me moles of octane and then I used the balanced equation to convert moles of octane to CO2 we saw the ratio is 8 to1 so I multiplied by eight that gave me the moles of CO2 and then I multiplied by the molar mass of CO2 the G per mole that cancels the moles of CO2 and I ended up with G of CO2 so so the pattern here is that I went from grams to moles then to moles of the other thing and then back to grams so we often refer to this as a grams to moles to moles to grams or you you could remember it as gmmg that is the process we're going to do for gram to gr stochiometry calculations that is also Illustrated in this another flowchart from the textbook and I cut off the bottom because I don't think we need to worry about total number of molecules you're probably not going to see that in calculations we're just going to be using moles as we've said so again you can start with the mass of a convert it to moles using the molar mass use the balance equation to go to the moles of B and then go back to the mass of B so in the examples we saw here a was a reactant and B was a product so we were calculating how much of a product you would get from a certain amount of reactant reacting but really A and B can be any reactant or product A and B could be both be reactants you could calculate you know how much of one reactant would you need to react with a certain amount of another reactant so both A and B can either be reactants or products it doesn't matter whether they're reactants or products the calculation is going to be the same