Hello, Chem 1210. Welcome back. Hopefully the exam went okay yesterday. Kind of like I mentioned to you in the announcement, I'll be going through those actually tomorrow on Friday.
And so watch out for that. By the end of Friday, you should see your final grade. All right. But anyway, so let's get back to, let's get back to stoichiometry. Last time, What we discussed was we went into detail on the concept of the mole and Avogadro's number and really the benefit and the utility in the quantity of the mole.
And really what we found there was that the mole in chemistry serves as the traffic interchange between really our scale and the atomic or particle scale. We had two really important conversion factors, if you recall there, where we went through the mole. The first was the molar mass of a substance, which gives us the grams, something we can measure, the mass of that substance in grams per mole. And then we also had Avogadro's number.
which was the number of particles, or we'll say atoms and molecules, per mole. And so we can use these two conversion factors to go in between mass and the total number of atoms and molecules we have in a substance. And we ended our discussion last time...
actually utilizing these conversion factors in a specific type of analysis. And that analysis was taking the mass percentage of a substance and going all the way to its empirical formula. And so we're going to do something similar today to start out. We're going to look at another type of analysis that's a little bit more specific, but the concept is very similar.
And that is, we're going to start today with... Combustion analysis. Combustion analysis. So remind me, what does a combustion reaction look like again?
I heard in the back for sure. Someone said combustion reaction is where we have a fuel in the presence of oxygen. We give it a little spark and we'll get carbon dioxide and water. This is what we call complete combustion. And so that's what every combustion reaction looks like.
And the thing is, we can actually back out what the chemical formula of our fuel is. I should say, yeah, anyway, we'll get there. We can back out the chemical formula of our fuel simply by measuring the masses of the products being carbon dioxide and water. So if we capture the CO2, we can measure the CO2 and water mass, total mass, then calculate the number of moles. of carbon and hydrogen that were in our original fuel sample.
And that part of our calculation is going to look just like last time. And the reason this works is if you take a look at our combustion reaction here, First of all, most fuels are hydrocarbons that we're going to run into on a regular basis. And so, what is a hydrocarbon again? It's carbons and hydrogen specifically. And if we take a look here, in our products, the only source of carbon here is in the carbon dioxide, and that had to come from the fuel because the only other reactant is the oxygen.
So whatever carbon we end up with in carbon dioxide... had to have come directly from the fuel. So those numbers should be the same because we have to conserve mass in our reaction.
Likewise, in water here, the hydrogen... is only bound in this product. And if we look on the reactant side, there's no hydrogen outside of the fuel. So all of the constant of the hydrogen has gone from the fuel and into the water. So if we measure the CO2 and water mass at the end of our reaction, we can actually get back to how much came from the reactants.
Oxygen is a little different story because we could have oxygen in our fuel and... We'll have to do one extra step because since we have O2 in the reactants, it's hard to tell whether the oxygen came from O2 or the fuel. Okay so let's take a look at an example real quick.
We have a hydrocarbon sample. And remember hydrocarbon means it consists of only carbon and hydrogen. And this sample has...
actually we don't need that part. So we're going to say hydrocarbon sample is combusted. To yield 299.4 grams of CO2 and 163.4 grams of water.
What is our sample's empirical formula? What is its empirical formula? Okay, so we're doing a combustion reaction and we have our sample here is a hydrocarbon so it consists of some number of carbon atoms, some number of hydrogen atoms of which we don't know yet.
If it's a combustion reaction, we're going to react it with oxygen and we're going to get carbon dioxide and water. Okay, so remember from what we just mentioned. How are we going to get to the number of carbon atoms here, the relative number of carbon atoms? Well, we need to evaluate the carbon dioxide mass because it contains all of the carbon that came from our original fuel sample.
So let's start with what we know here, which is 299.4 grams of CO2. And we need to get ourselves to moles of carbon atoms, okay? So we need moles of carbon at the end of this calculation.
That's where we should be. So how are we going to get there? First step, we've got to get into moles of CO2 first, okay? And how are we going to do that? How can we go between grams and moles?
So we need grams. down here to cancel grams of CO2 out and we need mole CO2 up here. Well, how many grams of CO2 are in one mole?
All we have to do is refer to the periodic table to get the molecular weight or the molar mass of CO2. We've got 2 oxygens, so that's 2 times 16, which is 32, plus 12. And what we're going to end up with is then 44 grams of CO2. That's the molar mass, which is also the molecular weight of CO2.
So now if we finish our calculations, we'd have moles of CO2. But we don't need moles of CO2, we need moles of carbon. So here is going to be our next step.
We're going to ask how many moles of carbon are in one mole of carbon dioxide. Now remember, the mole is... numerically the same question as asking how many atoms of carbon are in one molecule of CO2 you're going to get the same number the answer will be the same every single time so all we need to do is look at the chemical formula and say in CO2 I only have one carbon for every one CO2 molecule.
That's the same thing as saying I have one mole of carbon atoms per one mole of CO2 molecules. And look at our units now. The way we've arranged this, mole CO2 cancels and guess what we've ended up with?
Moles of carbon. So let's plug that in. 299.4 divided by 44. We arrive at 6.80 moles of carbon.
Alrighty. So this is exactly the amount of carbon that was in our original fuel. That's the beauty of these mole calculations.
So the next thing we need for our hydrocarbons, we need hydrogen. Okay, and where is the hydrogen going to land in our reaction? Exclusively in the water. So we're going to use the mass of water.
to get to moles of hydrogen. And once we have the moles of hydrogen and carbon, we can get the empirical formula just like the last step in our last analysis. Okay, so our problem states that we have captured 163.4 grams of water in the reaction. And we're going to follow the exact same procedure here.
So we need the molar mass of water which is 18 grams per one mole of water. And then our second step is once again to ask how many moles of hydrogen atoms do I have for every one mole of water. What do you think the answer to that question is going to be?
If we just look at the chemical formula, how many hydrogen atoms do I have in one molecule of water? Two. So the molar relationship then is identical.
I've got two moles of hydrogen for every one mole of water now. Moles of water cancels, we're in moles of hydrogen. And so let's calculate 163.4 times 2 divided by 18 yields to us 18.155 moles of hydrogen. Okay, we'll say this is going to be step one, is to get to moles of the constituents of our fuel.
And then the next thing for the empirical formula is what are we going to do? We're going to take the mole. ratio.
What's the relative number of hydrogen atoms per carbon atoms or moles of hydrogen per moles of carbon? So we're gonna have 18.155 moles of hydrogen relative to our 6.8 moles of carbon. And again, I just like to put the smaller one on the bottom because I know I'll get something greater than one. And let's see, 18.155 divided by 6.8 gives us a ratio of 2.67 moles of hydrogen per one mole of carbon. Hmm.
That's not a whole number, which means that's not very nice. So what did we do last time to get to this? Because what we need here in our ratio has to be whole numbers, right?
We can't have 2.67 here. in our chemical formula. Just like before, we're going to multiply this ratio up and see when we land on a whole number.
So if I do 2.67 times two, I get 5.34. Okay, that's no good. That's not a whole number yet. 2.67 times three, though, gives us...
8 moles of hydrogen and then 1 times 3 is 3 moles of carbon. Alright, we're there! We've got whole numbers which means we have arrived at the empirical formula for this fuel.
So we would write then that our fuel is C3H8. Alright, there we have it. Isn't that cool?
The important part of this was to show you that we can relate the moles of one element in a compound to the total moles of that compound simply using the chemical formula. So now this always this procedure always works for hydrocarbons, right? That's all we need because we know it's only carbon and hydrogen. But sometimes we are burning a fuel that might contain oxygen.
If our sample contains oxygen, What we're going to do is we're going to find the total mass of our carbon and hydrogen and then subtract, excuse me, subtract that from the total mass. of the fuel compound. So we need, I'll just give you a second to write that, I guess, before I start talking. So if our sample contains oxygen, we can find the oxygen mass in our fuel by taking the original mass of the compound and subtracting out the contribution to the mass from carbon and hydrogen.
And all we're saying there is then if we know it only contains those three elements, whatever's left over has to be the oxygen content of the fuel. Now we have to do this because the oxygen um In CO2 and in our water products, that comes from both the fuel and the O2, right? The O2 atoms have to end up somewhere in these compounds as well.
So we can't do the exact same analysis we did before. But once we have... the mass of all three, then we proceed exactly the same.
We just find the moles of all three and take our mole ratios. So really it's no different from the last calculation. It's only one extra step at the very beginning. And so let's take a look at it. an example we're going to do half of the example and then for time's sake we can just say okay we're going to do the rest just like we did before let me show you how this step works so here's our example menthol or let's do it like this let's do combustion reaction of menthol combustion of menthol and this sample is 0.1005 grams specifically of menthol yields 0.2829 grams of CO2 and 0.1159 grams of water.
And we could ask, what is the empirical formula? Just like before. Alright, so step one, I'll say 1A here, is we need to find the carbon and hydrogen mass in grams. And we can get that from very similar calculation as to what we did on the last example. And here's the thing, if once we find the masses of carbon and hydrogen, if they add up to be the total mass of the sample, we're done.
We know that there's no oxygen there that way. So we automatically have a check. If the question doesn't mention if there's oxygen or not, you can always double check by finding the mass of carbon and hydrogen and comparing it to the total mass of the sample. If they're the same, it's a hydrocarbon. If this is a little bit smaller than the total mass, okay, there's something that we've left out, and it's probably oxygen.
Okay, so let's start with the carbon. The information given here is 0.2829 grams of CO2. How are we going to get to grams of carbon?
We have to go through the mole. You always have to go through moles. So I first need to get into moles of CO2 from grams.
We're going to do that just how we did before through the molar mass. Molar mass of CO2 is 44 grams. Gram CO2 cancels.
And we change over and get into carbon just like we did before by evaluating the molar content of carbon in one mole of CO2. And we know from the chemical formula, there's one mole of carbon for every one CO2. Okay, so this calculation looks just how it did in the last example. We're now in moles of carbon, but we need to get into grams of carbon.
How might we do that? Through the atomic weight, or the molar mass. So, one mole of carbon, we just go to the periodic table, right?
And we find it is 12. 12.01 grams of carbon. So we've basically gotten from grams to moles CO2, switched over at the mole. Okay, this is really the important step.
And then we can always get back to grams with the molar mass no matter what. So when we plug all this in, we should get about 0.07715 grams of carbon. Now, notice all the decimal places here.
For these calculations, take as many decimal places as you possibly can throughout the entire calculation. And the reason we want to do that is because at the very end... end of the day if we round off early our mole ratios may not look like a whole number even though they are so take as many decimal places as you can and really i should be doing better about decimal places in the molar masses as well. Take pretty much everything that you've got. Okay, so we have the total mass content of carbon in our fuel.
It's right here, 0.07715 grams. Now let's do the same thing for hydrogen. And we're told that we've got 0.1159 grams of water that we've collected after the combustion. So we're going to follow the exact same procedure. We're going to use the molar mass to go from grams of water to moles of water.
Now we can do the changeover where we go from moles water to moles of oxygen atoms. And remember we got two moles of hydrogen for every one mole of water. It's right there in the chemical formula. The answer is right there for us. And so what's the last step again?
Now we're in moles of hydrogen, we need to get back to grams, and we do that with a molar mass. So one mole of hydrogen is 1.00794 grams of hydrogen. Let's plug that into the calculator and what we get is 0.01298 grams of hydrogen.
Alright, now I'm going to say step 1B is, is there any oxygen there? And we're going to answer that by adding up the mass of the carbon and the mass of the hydrogen and then asking the question, does that equal the total mass? of the original fuel sample.
So let's see we had 0.07715 grams of carbon plus 0.01298 grams of hydrogen. What is that amount to? I got that that equals 0.09013 grams. Now, at the very beginning of our problem, we were told that the total was...
0.1005 grams which is larger than the total mass of the carbon and hydrogen. So what is the remaining leftover mass? It must be oxygen. So then step two is that the mass of oxygen that came exclusively from our fuel, this is the way that we can extract the oxygen from the fuel away from the O2. That is going to be the total mass minus the carbon and hydrogen.
Just the leftover mass after carbon and hydrogen. And when we do that calculation, I found that the mass of oxygen in our sample is 0.0104 grams. Okay. Now at this point, we've got the masses of every single component of our fuel.
The mass of each element in our fuel. So the final step is going to be convert all those masses to moles. Then let's take...
And I'll let you try that out at this point. It's, we're just kind of going back, right? We can get into moles by just using the molar mass of each one. Then we've got moles of carbon, moles of hydrogen, moles of oxygen. We take the ratios to then arrive at the relationship between carbon, hydrogen, and oxygen.
What we're going to find is that the empirical formula for menthol is C10H20O. So you might give that a shot on your own, but we're going to move on and save a little bit of time here. But this was the new part.
Okay. I wanted to walk you through the new part compared to the last example. Everything else should be the same.
All righty. So let's go to the next topic. Our next topic is a really, really important one, and it's going to be one that we come across over and over and over again in this class and the next class. Pretty much your entire chemistry career, you're going to constantly run into these types of calculations and these relationships.
So we're going to explore the information that we get from balanced chemical equations. There is a lot of information in a balanced chemical equation and specifically A balanced equation, the balance part is important because it gives us the relative number of moles between each substance, between all of our reactants and all of our products. And remember that the relative number of moles are equivalent to the relative number of molecules that are actually reacting in our flask. So for example, let's go to one of our favorites. H2 gas.
Combustion. Presence of oxygen will give us 2 moles of water. So when we write out a chemical equation, remember the coefficients here tell us this information. This says that if we were to look down at the molecular level, if we only had 2 molecules of H2, One molecule of O2, we're going to get, in the end, two molecules of water. Now that's a really puny amount of water product, right?
Two molecules is not a whole lot of product. And that's going to be really probably tricky to track down because these molecules are going to be bouncing all over the place and it's really hard to see single molecules. But what we're saying with this first statement, the beauty of the mole, is that it's just a bigger counter.
So we could equivalently say that we've got two moles of hydrogen, which would be two grams of hydrogen, right? Because of its molar mass. Two moles of hydrogen could react with one mole of oxygen to generate two moles of water. this is what we're actually going to measure we're going to be looking at macroscopic scales here we're going to be looking at grams not single molecules but regardless the balanced chemical equation tells us that both are the same they're numerically the same thing And we say then, the terminology is that now that we balance our chemical equation, we can say 2 moles H2, 1 mole of O2, and 2 moles of water are called stoichiometrically equivalent, let's just say that, stoichiometric equivalence because this is the exact stoichiometry of the reaction. 2 moles H2 reacting with 1 mole O2 will give us 2 moles of water.
Having our stoichiometric equivalence from the balanced chemical equation enables us to do this really cool thing. We'll see. We can always calculate the amount of any substance and that amount is moles or grams, of course, we can do both, can always calculate the amount of any substance in our reaction so long as when we have the amount of one. So all we need to know is this is how much of the O2 is going to react to my reaction. And I can take that and calculate.
Exactly how much water we're going to make and how much H2 it's going to consume or vice versa. So let me show you what I mean by that. And how we use stoichiometric equivalence to do calculations on the amounts of substances in our reactions.
Our example will be how many moles of water... Here's how that's going to work. Let's start with what we know.
0.34 moles of O2. This is why, by the way, it's really important you always include the formula of the thing you're talking about. 0.34 moles of O2. Our chemical reaction was 2H2 plus O2 goes to two waters. Let's just put that in a little cloud.
So when I look at this, for every 1 mole of O2 that reacts in this reaction, how many moles of water are we going to produce? For every 1 mole here, we're going to produce 2 moles of water. This is directly grabbing the coefficient from our balanced chemical equation.
And look at the units now. Moles of O2 cancels, and guess what? We're in moles of water now!
We're going to generate 0.68 moles of water in this reaction. So you plug in your stoichiometric equivalence here. And this is where we can convert from one substance to the next. So if we have one thing, we can always calculate anything else.
This question could have asked how many moles of hydrogen are going to be consumed if we use this much oxygen. We could do the same thing. We'd say one mole of oxygen is going to consume two moles of H2.
Isn't that cool? That is so cool. Now oftentimes, almost always, we're going to start with grams because that's something we measure, right?
There's no mole meter. There's only a mass balance. So we start with grams. And so if you have grams of one substance, we can get the others the mass or amounts of the others through going through moles first so for this example if we were starting with grams of O2 we couldn't just plug We couldn't just plug and chug and go straight to grams of water. You cannot do that.
Because we're not on the same scale mass-wise in a reaction relative to the balanced chemical equation. That is a representation of moles. That's a counter, right?
So we have to go first to moles. then calculate how we just did up here. So we go to moles of our substance and then convert to moles of another substance using our balanced equation and once we have moles there then we can use the molar mass of that substance to get to grams. So here's an example of that.
How many grams of water are produced from 10.2 grams of O2? How do we start this off? Always start with what you know. So we're given 10.2 grams. I'm running out of space but I hope we can make it.
Hopefully you can see that. Of course you can see that with a superior resolution on our camera. Crystal clear, right?
That was sarcasm if you couldn't tell. All right. So according to this and according to the way that the mole works, we need to get into moles of oxygen first. How can we do such a thing?
So I'll put moles of oxygen up here. So we need to convert grams to moles. The molar mass.
That's our... That's our turn style to go between grams and moles. O2 has two oxygen atoms, so 2 times 16 is 32 grams of oxygen. Oxygen is in one mole.
Grams cancels. Now what do we do next? Our traffic interchange, right?
Now we're going to moles of water. So we go refer back to our balanced equation which states that for every one mole of oxygen we're going to make two moles of water. Moles of oxygen cancels. We need grams of water.
So our next conversion factor will be the molar mass of water. And so we know that one mole of water is 18 grams. Moles waters cancel and when we actually do the unit analysis what unit are we left with? Grams of water. So I would calculate this by doing 10.2 times 2 times 18 and then dividing that by 32 times 1. We're going to get 11.5 grams of water.
Okay, I think this is a good time for a quick break. We'll see you in just a sec. All right, I hope you caught your breath a little bit. Got a breather. Maybe grabbed a snack or something.
I don't know about you guys, but what I found is during this kind of like quarantine era, my diet has turned into basically just intermittent snacking. Yeah, anyway, I'm sure most of you have experienced that as well. All right, so let me get organized real quick before we move on. So our next topic is going to be section 3.7 in your book.
And this topic is on limiting reactants. And let me start this section by presenting the following idea. In a chemical reaction involving at least two reactants, it can be more, our reaction, the progress of our reaction, can be limited And it will always be limited by... The reactant with a lower molar equivalent. So our reaction just a second ago in our last calculation, right, it said that we had 10.2 grams of O2.
And we just assumed that we had plenty of hydrogen to allow all of that 10.2 to react and go to products. Let me give you a real life example of eliminating reactant. And this really pertains more to like non-quarantine type stuff.
Remember when you had to keep up with your laundry? Well, you still do, of course, but it's a little bit different. So let's suppose that you have eight socks that are clean and six shirts.
To make it a week with clean clothes. Now, what's our balanced chemical reaction here? Well, if we think about it, if we think about one day, it's going to require two socks plus one shirt. And if they're clean, then we've accomplished... Achieving our social norm, okay?
For simplicity, we're leaving out like pants and shoes. Forget about all that, okay? In this society, only socks and a shirt, clean shirt matter.
Everything else can be dirty. Now, what's the problem here? How many days in a week do we have? We're talking about like a work week.
So on the weekend, right, you just hang out in pajamas. But for a work week, we have five days, okay? So the problem here is that our socks will run out.
On day four. But, we still have plenty of shirts left over. But that doesn't matter.
The shirts don't matter because the socks are going to make us run out of all of our clean outfits. So let's write that. Doesn't matter about our excess shirts because no socks. The more I say socks to think about this example, I'm realizing how stupid of an example this is. But it's a relatable one, right?
We all hate laundry. So, the same concept pertains to chemical reactions. Let's go back to our...
One of our favorite reactions. 2H2 plus O2 will give us water. So the coefficients here tell us that we need twice as many moles of hydrogen as we do oxygen in order to make this reaction go to completion. So we need two moles of hydrogen for every one mole of oxygen.
What if we have... 10 moles of hydrogen and 7 moles of oxygen in actuality. This is just what we've got. Well, we can ask the question using our balanced chemical equation, if I consume all of this much, how much of the other thing will it consume? So let's ask that.
We're starting with 10 moles of H2. Let's find out how many moles of oxygen will be consumed in a reaction if we use all 10 moles. What we're going to do then is take that number and compare with the actual number of moles of oxygen that we've got.
So how do we do this conversion? Coefficients in our balanced chemical equation holds the key. So this says that for every two moles of H2 We're going to consume one mole of O2, and that's going to be then that five moles of O2 will be consumed in a reaction if all of the hydrogen is used.
Now, do we have 5 moles of oxygen? We sure do. In fact, we've got more.
We've got 7 moles. So what this means is that O2 is in excess by 2 moles. Okay, 7 minus 5. And that, we're going to have two moles, that excess is left over when the hydrogen runs out.
Because when we lose one reactant or the other, it doesn't matter which one, no matter what, the reaction stops. Because our equation tells us we need both to make our product. So when one runs out, the reaction is over. So we would call then H2 the limiting reactant.
Because the limiting reactant is always the one that is totally consumed in the reaction. and then the reaction stops on us. Does that make sense? We almost never have exactly the right molar equivalence in a reaction. But if we know which one the limiting reactant is, then we can use the moles of that one to calculate the moles of product that we're going to make.
Another example of this is combustion in your car. What runs out? The gasoline or the oxygen?
It's always the gasoline. We're always having to refill the fuel because O2 is in excess. So in your car or truck, your fuel is the limiting reactant. And O2 is always the excess because there's plenty of that going around in the air to enable the combustion.
Alright, let's do another example of that one. And usually how these problems look is you're given, you've got this much of one reactant and this much of the other reactant. How much product are you going to make? And so what we need to do is firstly identify which one is the limiting reactant and which one is an excess.
And then we're going to use the limiting reactant to calculate the quantity of the product. So here's example 2. We're going to look at ammonia synthesis. So N2 plus 3H2 goes to 2NH3s.
And check my balancing on all this, by the way, guys. Alright, we have 1.40 moles of N2 and 2.10 moles of H2. We're going to ask how much ammonia, which is NH3, how much ammonia is produced And what is the limiting reactant?
Should be a question mark, but it's the world's tiniest question mark there. Okay, now we can do this one of two ways. I'll tell you the way I generally do this because you get kind of a two-in-one calculation.
but it's one extra step. We could either do what we just did on the last one and take the moles of one and calculate how many moles of the other reactant will we consume if this is all gone and compare it to our number. Or what I'll show you on this example is we can calculate how many moles of the product will make if we consume all of one reactant versus if we consume all of the other reactant. And how do we know which one of those will be the limiting reactant? Because that amount of product will be lower.
Does that make sense? It's like saying, you know, if you've got a five-gallon jug of gasoline versus a one-gallon jug of gasoline, we can calculate how many miles we're going to go on one gallon versus five gallons. But we know the one gallon is going to be less. And anyway, that's a bad example because it's the same thing.
But you kind of get where I'm going. So let's do that. Let's go to moles of ammonia from each. So, okay. So I'll say step one, what are the theoretical amounts of product in the scenario where we consume all of one and then the next scenario where we consume all the other.
So we've got 1.4 moles of N2 we need to get to moles of ammonia. How do we do that? Through our balanced chemical equation, which states that for every one mole of N2, we're going to produce two moles of ammonia.
And that's going to give us 2.80 moles of our product. Now remember we're asking a question by doing this calculation and that question is if We consume all of our nitrogen, we would theoretically produce 2.8 moles of the product. Now we're going to ask the exact same question, but for the other reactive. If we consume 2.1 moles of H2, how many moles of product will we make? Well, let's go back to our balanced chemical equation.
This states that for every 3 moles now of hydrogen, Every 3 moles of H2 will yield 2 moles of the product. Our unit is moles of ammonia. So let's calculate what this number would be. 1.40 moles of product. So the question is, which one is the limiting reactant?
We've got these two scenarios. Always, if you do this strategy in answering a question like this, the lower, the lowest, I should say, product yield is going to be the correct answer because that suggests then that hydrogen has to be the limiting reactant and our final product yield is going to be 1.4 moles because these are the actual amounts of our things that we have whichever answer gives us a lower product yield is going to be the reality because that means that this 2.1 moles of hydrogen is going to be completely diminished once we produce 1.4 moles of the product and the reaction will stop so we have nitrogen in excess and h2 is the limiting reactant And our final product yield is 1.4 moles. Does that make sense?
So we'll say step two is determining the limiting product by simply comparing the theoretical yields if we consume completely either reactant. Now how would we get, let's say maybe the question asked for us to calculate the amount of ammonia in grams. What would we do? We would take our 1.4 moles and we would multiply it by the molar mass, which would get us in grams. Okay, now we could also sometimes these questions might have another part that says out of whichever reactant is the excess how much excess will you have and What do you think think on that for a second?
How could we calculate how much leftover nitrogen we've got? in Fact let's do it. Let's do the calculation why not what the heck How much excess nitrogen will be left over after a reaction is complete?
Well, we're going to work backwards because we know exactly how much product we're going to produce. It's 1.4 moles. So let's start there. We know the reaction is complete at 1.4 moles.
So why don't we calculate how much nitrogen... was consumed once we hit our final product yield. And then we'll compare that to our original number.
So we need to get to moles of N2. So we're going to, once again, go to our balanced chemical equation for that, which says that for every 2 moles of ammonia, it will consume 1 mole of N2. moles of NH3 goes away and we're left with the answer is 0.70 moles of N2 And that, remember, is how much was actually consumed in the reaction.
Because we know that the reaction is complete once NH3 has reached 1.4. So if that's the case, the excess is going to be the total that we started with minus how much was consumed, right? And that is... I said that really weird.
That is 2.1... I'm sorry, nitrogen was 1.4. 1.4 moles minus 0.7 moles.
So we've only used half of the nitrogen because we have 0.7 moles left in excess. So we just work our way backwards. Once we've established what's the final product yield limited by the limiting reactant, then we can use that to find how much of the other reactant will be absorbed or consumed in that process and simply subtract that from what we started with. Isn't this cool, y'all?
We could calculate then grams too in either of these. So we would know everything about the amounts of our reactants and products in our reaction. And hopefully, I hope you see the value in that from an industrial standpoint.
In that, if we have this much of a reactant and this much of a reactant, we might be talking about tons by the way, we want to be able to predict, well how much of my product am I going to make? You know, granted, this is like the maximum amount you'll make. This assumes nothing, you know, will happen that our reaction is 100% efficient, which of course is not realistic, but at least gives us kind of a number as to what we can hope for in a chemical reaction.
And so let's get to that point. Let's talk about product yield for a second. And what we've been calculating here is what we just said. This is the amount of our product that we'll make assuming nothing goes wrong.
Now what's the problem with that philosophy? Because the freaking real world, things go wrong, right? We can almost never have, well, we can never have perfect efficiency in really any procedure. There's just going to be some error, there's going to be some losses. So I want to introduce to you two terms that get at this.
The first is what we're going to call the theoretical yield. And the theoretical yield is basically what we just calculated for ammonia. It is the quantity of product produced in principle.
What I mean by in principle is... Assuming everything went perfectly, that is the amount of product we'll get when the limiting reactant is finally consumed. When the limiting reactant is consumed.
So you can see why that's theoretical, right? Because that assumes that our lab practice is absolutely perfect and that the universe is truly smiling down on us and saying, in this reaction, you won't make any byproducts. Every atom that you start with here is going to go to your product. Yeah. And then we've got the real world which we're going to refer to as our actual yield Which is of course What you actually recover after you've completed the reaction.
So how much product did we actually end up with? And a very useful metric for how good our procedure was is something called the percent yield. So then we find the percent yield of our reaction to determine how successful it was. So with the actual yield when the reaction is complete we always compare always compare that to the theoretical yield and then find this percent yield and the percent yield is just the actual yield which will always be smaller than the theoretical yield and so then we compare that directly against our theoretical yield and multiply by 100%. And for those of you that have the absolute privilege to take organic chemistry, I'm actually not being sarcastic on that one.
It's a great class. In lab, you're going to do several reactions, and you're always going to calculate your percent yield. You'll measure the mass of your final product, and you'll always compare that against the amount.
that you calculated in a stoichiometry calculation like what we just did. So let's do an example. We're going to do one more example, and that's going to conclude, actually, chapter three. This is our last topic. So, all right, this example is a bit long-winded, but it's a relevant example, especially since we live in Iron County.
So let's suppose that we want to purify iron from iron ore. And iron ore binds iron in iron 3 oxide. And the way that we're going to do that is through a reaction with carbon monoxide and heat.
We'll generate CO2 and purified iron. Let's first... Write the balanced equation and then calculate the percent yield from 100 grams of iron oxide.
and 2 moles carbon monoxide. When we finally recover, at the end of the reaction, we actually recovered 50 grams of iron. Whew!
Sorry. Let me reread that to you to give you a second. We want to purify iron from iron ore. which is iron oxide. A reaction with carbon monoxide and heat will yield carbon dioxide in the thing we want which is the pure iron. So what we need to do is first always always in any one of these problems first start with writing the balanced equation and then we're gonna find our percent yield for this reaction when we're given the amounts of iron ore and the carbon monoxide that we start with and then we finally actually recover 50 grams okay that might feel a little bit confusing but let's start with the let's start with the chemical equation Alright, so what is iron 3?
I kept that in all words to pop quiz you. What is iron 3 oxide again? How do we represent that in a chemical formula?
Iron 3, remember, says that we've got iron in the 3 plus ion, the 3 plus charge state. And then oxide is... O2 minus, right? So our iron 3 oxide will be 2 irons and 3 oxides to balance the charge. This is an ionic compound.
Don't let that 3 fool you into putting a 3 as the subscript. Remember, it's the charge, not the subscript. Plus, so this is going to be a solid, plus carbon monoxide, which is going to give us iron plus CO2 gas.
So I've got two irons here on the reactant side. So I would start with that one and just say, hey, that's an easy one. That means I need two iron, two moles of iron on the product side.
And on the reactant side, I have 3, 4 oxygens. So let's see here. So we could put 2 here to make the oxygens work, but then we would not have the right amount of carbons.
But what if we do a 3 and a 3? That gives us 3 plus 3 oxygens. 6 oxygens.
3 times 2 is 6 oxygens. 3 carbons, 3 carbons. And then some heat up there. Okay. It's part A.
That's our balanced chemical equation. We always have to start there. Why do we have to start there? Why is that so important? Because the coefficients, alas, hold the key to enable us to calculate the theoretical yield.
So we need to calculate for part B, percent yield, and that's composed of two pieces. The actual yield of iron. relative to the theoretical yield times 100%. So this tells us the pieces we need for our final calculation.
We're given the actual yield in the problem. The problem stated, hey, we actually recovered 50 grams from this process. We want to know, how good is that? How well did we do?
So we need to find the theoretical yield first. That's the missing piece. And so we're going to start with what we know and in order to calculate the theoretical yield we would need to first find the limiting reactant here. So we're going to calculate our iron yield if we used all 100 grams of the iron oxide and then we'll compare that number to if we consumed All two moles of our carbon monoxide. So we can start with either one.
I'm just going to start with 100 grams of iron 3 oxide. What do we do first? Step one, we need to get ourselves out of grams and into moles, right?
And we're going to do that with the molar mass. I'll leave you to check, but I believe the molar mass is 160 grams per mole. Now we're in moles of iron oxide.
Is that what we want? Heck no! We want moles of iron!
How are we going to get from moles of iron oxide into moles of iron? You got it! through the balanced chemical equation.
So this says that for every one mole of the iron oxide we can we're going to yield two moles of iron. Right, we're in moles of iron. We want grams.
So only one more step and that is the molar mass of iron is 55.8 grams and that will give us 69.8 grams of iron. Now remember, does that mean that that's actually how much we're going to make? Well, we're not sure yet.
What we've just calculated is the amount we'd make theoretically if the iron oxide is in fact the limiting reactant. And we don't know that yet. So we'll know after one more of these calculations.
And that's from asking the theoretical yield from consuming two whole moles of carbon monoxide. And so let's start with the two moles of carbon monoxide. We need to get into moles of iron from there.
How are we going to do that? Balance chemical equation. This states that for every 3 moles of carbon monoxide, we're going to yield 2 moles of iron.
And the last step. I'll give you, oh wait just a second there. I feel like I'm going a little bit fast today.
Right, so then what's the last step? It is to go from moles of iron to grams through the molar mass. And when we finally do that calculation, We get 74.4 grams of iron. Okay, which one of these is the correct answer for the theoretical yield?
What do you think? Always the smaller one. So 69.8 grams, which was the value that we calculated by consuming all of the iron 3 oxide, is less than 74.4 grams, which was the amount if we would consume all the carbon monoxide. Therefore, The iron oxide is the limiting reactant and the theoretical yield is therefore 69.8 grams. Okay, so now we're finally at a place where we're ready to calculate the percent yield.
We've got our theoretical yield, 69.8 grams of iron. So now to calculate the percent yield. What do we do? We take our actual yield and divide it by that number. Our actual yield was 50 grams.
We're going to divide that by the theoretical yield, which should always be larger. Excuse me. And multiply by 100. We get... We get 71.6%. 71.6% was our percent yield.
How good is that? What do you think? That's actually not too bad. What you're going to find is in the chemical synthesis literature, especially when people are reporting a new methodology for doing a reaction and generating a particular product that you care about, this percent yield is going to be really important because we might say, gosh, we've been using the same procedure since 1967 and that gives us a typical percent. yield of about 50%, this new method which doesn't use any more reactants gives us a percent yield of 70%.
Hey that's great! If I'm making this thing on a big scale that's a whole lot of money I just put into my pocket because I'm making a lot more product for the same amount of starting materials. So this is a highly useful number to have in the chemical synthesis industry.
research tells us something about that and typical yields that you find tend to be in the seventies to low nineties for sure if you can get into the nineties as a percent yield you're doing really well you're probably a pretty talented chemist Okay, that concludes chapter three, folks. We're going a little bit slower compared to the schedule, you might have noticed, and that's okay at the end of the day. I think this is probably a good stopping place before we get into the next topic. And we're going to buy a little bit of change back from this lecture.
We might go over on the next one a little bit. Either way, we'll conserve the time one way or another. You'll notice these lectures aren't all the same time duration. Anyway, work through these homework problems. And this is a really good launching point for the next chapter in particular, which is going to be getting into chemical reactions and aqueous solution.
All right. Thanks, folks.