Hello everybody, in this video we're going to be covering section 3.1 in our textbook. This is going to be a big section for us. We're going to learn how to calculate the formula masses for covalent and ionic compounds. And we're going to define the amount unit called the mole, and the related quantity Avogadro's number.
And we're going to be able to explain the relation between mass, moles, and numbers of atoms or molecules and perform calculations deriving these quantities from one another. So the first thing we want to talk about, so far we've been talking about the masses of individual atoms, but we know that those atoms can bind together to form molecules, and these molecules have their own masses. Typically the general term, whether we're talking about a covalent or an ionic compound, is called the formula mass of a substance. And it's going to be the sum of the average atomic masses of all the atoms in the substance's formula. So if you give a formula for either an ionic compound or a covalent compound, we're going to add up all of the masses of those individual atoms, and that's going to be the mass of that molecule or formula unit.
Sometimes we do want to make a little bit of a distinction. here though because covalent substances as we know do actually exist as individual molecules and so then what we can talk about is that the formula mass could be correctly called the molecular mass it'll be the mass of a specific covalent molecule. And an example that we have here is going to be this guy here. This is chloroform.
Alright, it has one carbon, three chlorines, and a hydrogen. If we wanted to figure out the molecular mass or the formula mass for this molecule here, we're going to take that one carbon times its atomic mass, we're going to take that one hydrogen times its atomic mass, and we're going to take the three chlorines times their atomic mass. mass we're going to get all of those values and we're going to sum them together to give us the overall mass for this molecule here are a couple of other examples you can see that this can get to be quite intricate at times but in general all we're going to do is we're just going to come through here we're going to add up each individual type of atom and then we're going to add them all together. We're going to multiply by each of the atomic masses and add them together. These type of images that you're going to see here are going to become more and more familiar to you as you continue on in chemistry.
and organic chemistry and stuff, this is how they're going to start to represent molecules. Because it not only shows each and every different atom, but also how they're connected to one another, the connectivity that you see. But we could easily have done the exact same thing if we were just given a formula like C9H8O4, multiplied by those subscripts there.
So when we're dealing with ionic substances, they're composed of discrete cations and ions combined in ratios to yield electrically neutral bulk matter. So as we've talked about before, when you have these cations and anions, ions, okay, there's a smallest unit that then gets repeated through a bulk material, all right, but they don't exist as separate molecules, okay, smallest repeating unit is not an action. actually a separate unit or its own molecule. A lot of times they call that the formula unit and it's in place of using the term molecule.
So the formula mass for an ionic compound may not correctly be referred to as the molecular mass. The average atomic masses, so we just call it the formula mass for ionic compounds. The average atomic masses of ions, so now that we have ions, can be approximated to be the same as the average atomic masses of the neutral atoms and that's because to create those ions we just gained or lost electrons. Electrons have a very small mass therefore they're not going to have any appreciable difference from that of a neutral atom. If we take a look at some examples here we can see that when we're just looking at the bulk material determining the smallest repeating unit can be a little trickier than figuring out how to calculate the formula mass a little harder in the example of ionic compounds.
It's fairly simple if you have a simple cubic structure like this to identify the smallest repeating unit. In this case it would be a sodium atom and a chloride atom there that just repeats throughout the bulk. Look at this material.
Therefore, for simple table salt, we just have one sodium and one chloride ion, and then we multiply them by their average atomic masses. Again, these are just the same as the one for neutral sodium. one for neutral chlorine we add those together to get its formula mass but down here identifying the smallest repeating unit gets to be a little trickier but in general we're not really going to ask you these these kind of questions, I'm just wanting to point it out a little bit. Instead, you'll be given a formula.
For instance, this one is aluminum sulfate, and we're going to have two aluminums, three sulfurs, and 12 oxygens for each repeating unit. And then we just apply the exact same algorithm, add them together, and we can get its formula mass. So you would have been given the formula for aluminum sulfate in this case.
So this is a really big idea that's going to be very important to us. This term is going to come up over and over again. So we do want to make sure that we fully understand the idea of the mole.
Now the mole is an amount. All right, and there are other ones that you are familiar with. Okay, for instance pair, dozen, or gross. So we all know that a pair is two things, a dozen is twelve things, a gross is 144 things. What makes this is an amount unit is that it could be a pair of anything you can have a pair of shoes okay you could have a pair of hats you could have a pair of cars it doesn't define what it is just how what amount of those items we have and the mole works the exact same way okay i could have a mole of cars i could have a mole of shoes um what is a mole well it's defined as the amount of substance containing the same number of discrete entities as the number of atoms in a sample of pure carbon-12 weighing exactly 12 grams.
Alright, now that seems like a really arbitrary kind of weird thing to define it as, okay, but what this is actually going to allow us to do is to have a really easy conversion between the atomic mass or formula mass of a substance and a mole of that substance in grams and we're going to see how that plays out in just a second here. How many things is a mole? Okay, for instance a pair is two things, a dozen is twelve things. Well, a mole is an Avogadro's numbers worth of things. Okay, where Avogadro's number is this really huge number.
right here 6.02214179 times 10 to the 23rd power okay so it's a really huge number of things for instance if you had a mole of shoes it'd probably be like the size of the earth. Alright, so you can only have a mole of really really small things like atoms and molecules and stuff and for that to be any reasonable amount of stuff. It's named after a scientist named Amadeo Avogadro. It's often abbreviated with a capital N with a subscript A. Okay, and usually you don't need to use this whole long number.
We typically are just going to use 6.022 times 10 to the 23rd. All right, so one mole of an element is going to contain 6.022 times 10 to the 23rd atoms. The mass of this quantity, though, will depend on the element's identity. Okay, just like if I had a pair of shoes, that would be lighter than if I had a pair of bowling balls.
Okay, so because these atoms weigh different amounts, that means that a mole of those atoms is going to weigh different amounts. And we can see that in this image. here where we have a mole of various different molecules and atoms and stuff and how those are varying different amounts depending on the identity of the molecule So now that we've talked about formula mass, we can now talk about molar and moles.
We can talk about molar mass. The molar mass of an element or compound is the mass in grams. of one mole of that substance. This property gets expressed in units of grams per mole. So we're going to have grams per mole here.
And this is what's clever about the way that we defined the mole. The molar mass of any Any substance is numerically equivalent to its atomic or formula mass in AMU. So what we've set up here by cleverly defining what a mole is, is the fact that a gram per mole is the same unit as AMU.
Okay, so a single carbon-12 atom has a mass of 12 AMU, but a mole of those carbon-12 atoms is going to have a mass of 12 grams per mole. Alright, so the Big lesson to take away is that you can just replace AMU with grams per mole for a specific substance or molecule, or you could replace grams per mole with AMU. They're basically interchangeable units. And we can see that here and how it relates back to their identities.
Okay, one mole of carbon weighs 12 grams, right? If you went to the periodic table you'd see that the atomic mass of carbon is roughly 12 amu and if you went to the periodic table and you looked up zinc you'd see that it has an atomic mass of roughly 65.4 therefore one mole of it would be 65.4 grams and you can see that this is going to be different for each and every one and as they get to be heavier and heavier like lead or 10 you're going to have larger and larger masses Here's just a little table with some common values for the average atomic mass and the molar mass. And this is just, again, to highlight that these are numerically equal values with these different units.
AMU is going to describe the atomic mass of a single atom of these elements. And the molar mass is going to describe the amount of grams in a mole. of those atomic, of those atoms.
So let's talk a little bit about how we're going to go about doing these calculations and converting between these three different quantities. Okay, mass, moles, and number of atoms. First let's talk about how we're going to go from mass to moles if we were given a value in grams.
So according to nutritional guidelines from the U.S. Department of Agriculture, the estimated average requirement of dietary potassium is 4.7 grams. grams.
What is the average requirement of potassium in moles? So one thing that's going to be really helpful and it's a step that a lot of people want to skip when they're doing these kind of problems, but I really recommend you don't, is to kind of diagram it out. Alright, start off on the left hand side by saying what is it that I know, right? What was told to me in the problem? I know that I have 4.7 grams of what?
Potassium. Okay, then draw an arrow and this arrow is going to be something that you're typically going to multiply or divide by. It's some mathematical manipulation that you're going to do.
All right, and then name it out in words. Divide by the molar mass. Okay, grams per mole. Then you're going to go over here and then you're going to say this is my output from that.
In this case, the output is what we wanted, what we were being asked for, moles. But if it wasn't, we could do another manipulation to get to the output. that we wanted. Alright, so diagramming these things out is really helpful.
In this, generally what we're going to do is we would have gone to the periodic table, we looked up potassium, we would have said, okay potassium as an element has a atomic mass of of 39.10 amu. I know that that's the same as 39.10 grams per mole. Now here it does say divide by the molar mass and some people will look at this and say what are you talking about you're not dividing you're taking this and you're multiplying it by this fraction you're getting to this. Okay remember that that number was 39.10 grams divided by moles that was the unit grams divided by moles.
Okay you So flipping that fraction, all right, is the same as doing a division. And what we're doing here is that we're just always writing it as if it's a multiplication. And that has one really big benefit for us, okay? We know that one mole is going to be one of these either numerator or denominator.
And we know 39.10 grams is going to either be the numerator or denominator. denominator. I can, when I always write it as a multiplication, I can visually see very clearly that I should put the grams in the denominator because that's going to cancel these grams.
And I can very visually see that the unit that I picked up now is going to be moles. And that's the unit that my final answer should have. So when you always write it as multiplication, you can follow this simple rule that I'm going to cancel the unit that I want to get rid of by putting it in the denominator.
And I'm going to pick up whatever unit is in the numerator. So let's talk now about how to go the other way, moles to grams. A liter of air contains 9.2 times 10 to the 4 moles of argon. What is the mass of argon in a liter of air?
Again, we're going to diagram this out. We have the moles here. All right.
This is essentially the reverse of the problem that we just did. And we're going to see that there's a nice symmetry here in our methodology. And when we were going from mass to moles, we divided by the molar mass.
Here we're going to have to multiply by the molar mass. And that's going to give us the mass of argons that we're being asked for in the question. Let's go back. So I start off by writing out my moles of argon. Okay, I can clearly see that I need to put that one mole of argon in the denominator so I can cancel that unit.
And I'm going to pick up the unit of grams up here. Again, 39.95, that is the atomic mass of argon from the periodic table, and I can see that in the end I wind up with grams of argon. Now, here it said what what is the mass, and so we were kind of left open to put this value in grams because that's what easily comes out of this, right? But if they had specified a different unit, we could have always taken these grams and then converted that to it. different mass unit that the question was asking for and when we were back here had we not been given this value in grams we could have easily converted to grams before doing this manipulation here One other important thing to keep in mind when you're putting this stuff into your calculator is that it's always a good practice to put things like these scientific numbers in their own little parentheses.
I would have put parentheses around this number before going and multiplying by that fraction there. I would have put this fraction in parentheses too. It's just good practice.
In this case, it wouldn't have mattered if you had done that. still would have gotten the right value. But as these problems get more complicated, you'll get into a situation where you're going to have a problem with your order of operations.
So it's always good practice to just take these scientific numbers and put them into parentheses. Okay, we've done mass to moles and moles to atom. Let's see how we're going to go from mass to atoms. And our question here is going to be how many atoms are there in a 24 gram sample of copper wire? Now this when we look at this diagram, it brings up a very important point.
I've had a lot of students who try and come up with some clever way of going directly from mass to number of atoms or from the number of atoms to the mass without going through the moles as an intermediate and it just doesn't work. It won't ever work out correctly. So it's best to just not fight it and just recognize that moles are going to be our gatekeepers whenever we're converting from mass to number of atoms or from number of atoms over to mass.
All right. So the first step is going to be the exact same one that we did from mass to moles where we're just going to divide by the molar mass. And the second one is that now that we have the number of moles we're going to multiply by Avogadro's number to give us the number of atoms. This would be no different than me telling you you have three pairs of shoes, how many shoes do you have in total, and you multiply by two in that case.
It's just this time we have Avogadro's number which is much much larger because a mole is a much much larger collection of things. So we start off with the information we were given in the problem 24.0 grams of copper. We go to our periodic table.
We say copper has an atomic mass of 63.546 amu and I know that that's the same as grams per mole. Okay, I arrange my conversion factor here so that I'm going to cancel my grams of copper and retain the unit of mole. Okay, next I'm going to multiply by Avogadro's number which is basically saying that we have 6.022 times 10 to the 23rd atoms or things, it doesn't really matter, but in this case it's atoms of copper for every one mole of copper. And when I arrange it like this I can see that I'm canceling that mole of copper and I'm getting the unit of atoms of copper out of that.
And in the end I get this quite large number here of atoms. And again it's going to be parentheses are going to be your friend when you're putting this into your calculator. I would have put this scientific number in parentheses and then I would have put parentheses on the outside of this whole thing when multiplying it together just to make sure that my order of operations is the way that I want it to be.
So let's think a little bit here about this one. This is going to be a little bit more challenging one. And on the face of it, it's not always obvious to people why that's true.
Okay. What we're going to be asked for here is how many moles are there in a 28.35 gram sample. of glycine, alright?
And then we're given a chemical formula for glycine. So what we need to do here is still just go from mass to moles and divide by the molar mass, okay? But first we're going to need to know what the molar mass of glycine is. Now we're talking about a molecule that is made out of multiple different atoms and not just a single atom. So we can't just go to the periodic table and just look up glycine like we did with those elements before.
Instead, we need to start off by going through that chemical formula, multiplying each one of those subscripts by the molar mass of those different elements, and then summing those all together to get the molecular mass of glycine. And we find that that is. 75.07 grams per mole. Now we can go and arrange that conversion factor here. Again, we're going to put grams in the denominator to cancel the grams of lysine that we were told about in the problem.
We're going to pick up that unit of moles of glycine, and then we're going to arrive at our answer with the proper units. Here we can get just a little bit trickier with it here. A packet of artificial sweetener contains 40 milligrams of saccharin. It has this formula here and has this structural formula here. Given that saccharin has a molar mass of 183.18 grams per mole, how many saccharin molecules are in a 40 milligram?
.0400 gram sample of saccharin. How many carbon atoms are in the same sample? Okay, so a lot of people when they see a larger problem like this start to shut down. Okay, what we need to do though is just really think about what we know now. The first thing is you're going to look at the structural formula and maybe say, hey I don't really understand that.
The thing is is that this doesn't even really play into the problem at all. Okay, we've been given the chemical formula here. We know how many of each atom there is already.
Okay, and that's the only information that we need in order to figure out the second part of this problem, which is going to be how many carbon atoms are in the sample. All right, if I can figure out how many of these molecules I have, I know I'm going to have seven carbon atoms for each molecule. The second part is that we don't even have to calculate the... molar mass from this chemical formula, we've actually been given that in the problem. So them giving us this extra bit of information actually saved us a step that we had to have in the previous problem.
So let's go through this and see how we're going to diagram it out. So we've been told the mass of saccharin that we have, all right, we know that we need to get to the number of moles of saccharin, all right. Because we, moles are the only thing that are going, moles and atoms are the only thing that are going to allow us to use these subscripts as a conversion factor for figuring out the number of, the number of atoms that we have. Okay.
And then to figure out using these subscripts to figure out the number of atoms of carbon that we have. So first here, we're going to take. the mass that we were given in the problem, 0.04 grams of saccharin, all right?
And we're going to divide that by the value that we were given in the problem, all right? One mole of saccharin for every 183.18 grams. We're going to cancel those grams.
To figure out how many molecules we have, because now we're in units of moles, we're going to multiply by Avogadro's number, and we get this large number of molecules of saccharin, okay? Next, we need to go to our formula here, and we need to recognize that for every one of these molecules of saccharin, we have seven atoms of carbon. So if I want to know how many carbon atoms I have, I need to use this as my conversion factor. So for all these molecules that I have here, I take the number of molecules that I have, and I multiply that by seven atoms of carbon for every one of those molecules. And I arrive at the answer of how many...
carbon atoms I have in that one 40 milligram sample of saccharin.