Understanding Moles and Molecular Mass

This is section 3.1 for our homework quiz number three slot notes. Chapter number three we're going to basically take a look at the composition of compounds and then we're going to take a look at solutions. So in section three point one we're going to take a look at formula mass and the mole concept. So again we're trying to connect the microscopic world of atoms and molecules to the macroscopic world. We can't see individual atoms so we need some way in order to understand what's going on. So we talked about atomic mass. in the last set of slide notes. And we talked about an AMU and that basically one atom of carbon-12 weighs 12 AMU. And so that on this scale, a hydrogen atom would have a mass of 1.008 AMU and an oxygen-16 atom would have a mass of 16.00 AMU. The problem that we run into is we don't have any... AMU scales in our labs. We measure mass in kilograms and grams not in AMU's so we need some way of connecting those two things together. So how we connect those together is we use this idea the mole. It's a unit to count the number of particles. So, I mean, you're used to, if you go to buy eggs, you buy them basically by the dozen or you buy 12 of them. You don't walk into the grocery store and go to the egg area and go, well, I'm going to buy one egg. No, normally you've got to buy either half a dozen, a dozen, a dozen and a half. You don't go into a shoe store and buy one shoe. You buy pairs of shoes. And the reason that we do that is for convenience. Well, in chemistry. We use the mole as the amount of a substance because it's convenient. It's convenient. So what we've done is the mass of a carbon-12 atom has exactly 12 grams, or a mole of carbon-12 atoms has a mass of 12 grams or 12 AMUs. So one mole, if we were to find out how many atoms fit into one mole, That would be 6.0221367 times 10 to the 23rd. You're going to find we're going to stick with 6.02 or 6.022 depending on whether we need two, whether we need, sorry, three or four significant figures. Okay, this number is called Avogadro's number. Why do we use this big, huge number? Because it's convenient for us to lump this number of atoms together to have an understanding of what's going on at the molecular level. So molar mass is the mass of one mole of anything in grams. Well, because this number is so huge, they're pretty much only a few things that it's really reasonable to measure 6.022 times 10 to the 23rd of. And that's typically atoms, ions, molecules. We wouldn't measure out the mass of tennis shoes. We wouldn't measure out a mole of eggs, but we do do things that are extremely small. And so if you'll notice the atomic mass units we were talking about on the periodic table in the last chapter, we can assume the exact same value for molar mass. And that's why we have this odd number of 6.022 times 10 to the 23rd. so that we can use the same set of numbers for amu and for molar mass so what we're going to do is we're going to create a couple conversion factors if you remember back in slot note set number one we did some metric conversions and we put a thousand milliliters in front of a liter and we put a thousand grams in front of a gram We put 100 centimeters in front of a meter so we could get the units to cancel out. Well, we're going to create similar conversion factors except now they're going to have the mole in it. And so if you'll notice our first one we're going to talk about is molar mass which is grams per mole. There's an asterisk because it's not always going to be a set number. So where are we going to get that number from? Well we're going to either get it from the problem or it's found on the periodic table. So if you remember on your periodic table underneath each of the symbols is the atomic mass for the element. Well, we now know that can be measured in AMUs if it's a one single atom. But if we want to have something that's in grams, if we have a mole's worth of it, the hydrogen would be 1.008 grams if we have a whole mole of it. And so our second conversion factor is Avogadro's number, which is the number in a mole. And so we're... pretty much going to use 6.02. Every now and then I may use 6.022 if I want four significant figures, but typically three will be enough for us in this class. Remember the clue is one, you're going to have this big ugly number. So if you got the big ugly number 10 to the minus 23, 10 to the 23, 22nd, 21st, 20th, that it's a pretty good clue that you're probably gonna have to use Avogadro's number in the problem. The other clue is you're going to be asked about the number of atoms, molecules, formula units, because pretty much these are the only things that are really small enough for us to actually use Avogadro's number with. So if we want to find the mass of 0.89 moles of calcium, if you remember back when we were doing problems in slot notes for quiz number one, I'm going to take... 0.89 moles of calcium. I'm going to write down my numbers and my units, multiplication sign, division bar. Well, the only thing we can use is either molar mass or Avogadro's number. Well, I need to get rid of moles. That should look familiar. And then what I need to do is I'm looking for the mass of calcium. So I've got grams well I need a number well then what I'm going to do is I'm going to look on the periodic table because there's no number for calcium given and so if I look on the periodic table I'm going to look right underneath calcium and it says it's 40.08 grams per mole put an equal sign which means it's now put it in your calculator because we've got only thing we've got is grams are looking for grams so this says take 0.89 times 40 0.08. And when I multiply the two of them together, I end up with 36 grams or 36 rounded to two significant figures. Okay. So if we take a look, we're starting out with 1.65 times 10 to the 23rd atoms of zinc. I'm going to put a multiplication sign division bar. And so if you'll notice, I've got this big ugly number so I'm probably going to have to use Avogadro's number. So I'm going to put Avogadro's number in the bottom. Remember that can be used to cancel out atoms or molecules or formula units. But I have to put moles on the other side. And this says I've got moles but if you notice in my problem it's asking for grams. So that means I'm not done with the problem. So what I'm going to do is I'm going to put a multiplication sign division bar. I'm going to put moles in the bottom. Now it kind of looks like the previous problem, except the only difference is instead of the mass of calcium, I need the mass of zinc. Because notice I get rid of the moles, grams. And so I'm going to go to the periodic table. I'm going to find the molar mass of zinc, which is 65.38. And this says take 1.65 times 10 to the 23rd. divide by 6.02 times 10 to the 23rd, and then multiply by 65.39, and we end up with 17.9 grams of zinc. So what is the mass of one hydrogen atom? Let's find out what the numbers would look like if we just did the mass of single atoms on the periodic table. and find out why we use the mole. So I've got one hydrogen atom. Well, it's atoms and it's a number of something. So I'm going to have to divide by Avogadro's number because remember it goes with atoms. I'm going to put the mole on the top, but I don't want moles, I want mass. So this kind of looks like the previous problem again. We got moles and we want to go to mass. And so I'm going to put moles in the bottom and I'm going to put the mass of hydrogen in the top. And if we go to the periodic table, that's 1.008 or rounded to two decimal places is 1.01 because we're only going to take our masses out to two decimal places. um, a lot simpler for us to work with. If you took it out to 1.008, you would still get a very similar number, but I think it's just more convenient if we just take everything out to two decimal places, and then we won't run into issues in the second section of this chapter, or when we, or this set of slot notes, or chapter three, or the next set of slot notes. That's how I'm going to take one divided by 6.02 times 10 to the 23rd times 1.01. And what that means is I get 1.66 times 10 to the minus 24th grams for a single atom of hydrogen. So if I wanted to do the mass of single atoms on the periodic table for hydrogen, that 1.008 would be replaced with 1.66 times 10 to the minus 24th. I think it's a lot more convenient for us to work with moles of things. Just like it's a lot more convenient to buy a dozen eggs or a pair of shoes than to buy one egg or one shoe, it's a lot easier to take a look at the mass of a mole of atoms than it is to just take a look at the mass of just single ones. So now we've got grams and we want to go to atoms. So I've got 27.6 grams of argon. multiplication sign division bar well i need to get rid of the grams i'm going to put moles on top and i need the mass of argon so i'm going to go to the periodic table and see that it's 39.95 remember the asterisk was in front of the g so the 39.95 has to go in the bottom Multiplication sign, division bar, because if you take a look, we've got moles, but it's asking for atoms. What we're going to do is we're going to put moles on the bottom, and we're looking for atoms of argon. So we've got to use Avogadro's number. We cancel out the moles. So this is 27.6 divided by 39.95. times the 6.02 times 10 to the 23rd. And should be getting 4.16 times 10 to the 23rd atoms of argon. So if you notice, even going from grams to the number of atoms gives us really big, huge, cumbersome numbers to work with. The nice thing is, is once we get done with this section, pretty much Avogadro's number is going to disappear. We're not going to do a whole lot with it. So just make sure you can get this into your calculator now. We'll have to deal with scientific notation a little bit later on, but we won't be dealing with Avogadro's number, okay? Because hopefully what I've tried to show you in the past couple problems is it's really not a convenient number to work with in terms of mathematically, but it makes things convenient if we take a look at moles of atoms instead of individual atoms, okay? So molecular mass or molecular weight is the sum of the atomic masses of a molecule. So we've got SO2 and you could go to the periodic table and there's one sulfur and there's two oxygens. And if I added this together, I get 64.07. So for any molecule, the molecular mass AMU again, is also equal to the molar mass in grams. So the assumption is, is if I have one mole of SO2 molecules, it would be 64, they would have a mass of 64.07 grams. So one molecule of SO2 is 64.07 amu, and one mole will be 64.07 grams. So this is even, it's just as convenient with molecules as it was for atoms. Formula mass is the sum of the atomic masses in AMU in a formula unit. Basically, ionic compounds aren't considered molecules. They're considered formulas or formula units or ionic cubes. And even though we kind of still call them molecules and compounds anyways. But if you'll notice, mathematically finding the mass of sodium chloride was the exact same method as we use for the SO2. take one sodium, one chlorine, and add them together. So if we take a look at a couple problems, if we go to find the molar mass of KCl, I'm going to take potassium and chlorine, and then I'm going to go to the periodic table, and I'm going to find the mass of potassium. and I'm going to find the mass of chlorine. This is molar mass, so it's grams per mole, not amu. And then what I'm going to do is I'm just going to add the two of them together, and I end up with 74.55 grams per mole. If I've got sodium hydroxide, I'm just going to take three different atoms and add them together. Sodium is 22.99. Oxygen is 16.00. Hydrogen is 1.01. And so if we add them all together, we end up with 40.00. Notice how I'm taking the atomic masses out to two decimal places. I want to find the molar mass of calcium hydroxide. Well, if you notice in the problem... when we went to the SO2, it said to take the sulfur, multiply by one and take the oxygen and multiply it by two. Well, I find that can be cumbersome. So I find a better way of doing that is to split out the formula and then go find the masses. Let me show you what I mean. So I've got calcium and I've got two oxygens in my formula. And I've got two hydrogens. When I write this down this way, what this tells me is go find the mass of calcium, which is 40.08. Go find the mass of oxygen and multiply it by 2 before I write it down. Oxygen is 16.00. Multiply it by 2 and it's 32.00. The next one says take hydrogen, multiply it by 2, and then write it down. Hydrogen is 1.01, so I'm going to write this down as 2.02, add the three numbers together, and we end up with 74.10 grams per mole. Listing a formula vertically is something that we do a lot in chemistry for even other types of problems, okay? Breaking it out like what you've seen here by just finding all the different types of atoms. So now we can kind of connect this to what we were doing before using molar mass and Avogadro's number, but I'll be kind. I'm not going to mix Avogadro's number into this. We're just going to deal with molar mass, and we're going to be dealing with molar mass for a while until we get to the third section of this, and then we'll add our third conversion factor in. So it says calculate the number of moles of calcium carbonate, chalk, containing 14.8 grams of calcium carbonate. So I've given you a strategy that says find the molar mass of calcium carbonate and use it to convert 14.8 grams to moles. So I've got to first come up with a chemical formula. Oh that sounds like slot notes quiz number two. So we need calcium, carbon, and three oxygens because calcium carbonate is CaCO3. If we add all these together, I end up with 100.09 grams per mole. I was asked how many moles were in 14.8 grams of calcium carbonate. So the grams goes on the bottom. So the 100.09 goes on the bottom. The grams will cancel out. And we end up 0.148 moles of calcium carbonate. So in question number 16, they're asking us to find the number of moles of a compound, and version A asks for propylene, which is C3H6. So when I see a formula and it's talking about the number of moles, it kind of triggers in me that I'm going to need some sort of mass. And so what I'm going to do is I'm going to add three carbons and six hydrogens. and they get 42.09 again i'm taking things out to two decimal places and then i'm going to take 25 grams multiplication sign division bar i'm going to put the 42.09 grams in the bottom to get the grams to cancel out with the moles on top so this says to take 25 and divide by 42 0.09 because the number's in the bottom and we end up with 0.594 moles of C3H6. Problem number 17 is going in the, is starting out with the 0.0146 moles of KOH, kind of in the opposite direction. We still need the mass of KOH. That's like how we did the sodium hydroxide except I'm going to use 39.10 for K instead of the 22.99 that we use for the sodium. And that 0.0146 moles of KOH. Put moles on the bottom so the moles will cancel out this time. Notice the 56.11 grams ends up in the top so it's going to be multiplication. And I'm going to end up with 0.819 grams of KOH. Folks, please be careful. Notice this is 0.0146. Students can sometimes accidentally put 0.146 and then if it's a multiple choice question you look and you don't think you have an answer that matches any of them then you don't think you know what you're doing with the problem. When in reality, mathematically, you know what you're doing with the problem you just accidentally put it wrong in the calculator. Number 18, we're starting out with grams. And so what we're going to do is we're going to find the mass of KBr, potassium and bromine. And then we're going to take that 2.21 grams. If you'll notice, the grams goes in the bottom. So the number goes in the bottom. So this is division. And so we end up with 0.0179 moles of KBr. Number 19 says determine the mass. So this is going in the opposite direction. And so we'd add up the FeS, O11, and 14 hydrogens. If you'll notice on some of these problems, I've already been giving you some of the masses. Okay. And so we got 278.06. So we got 0.5758 moles. Put the moles on the bottom, which means the mass goes on the top. So we're going to multiply these two numbers. And we end up with 111.8 grams. So let's take a look at a couple of practical problems. Question number two, we've got a 55 kilogram woman. Well, they're about 2.2 pounds in a kilogram, so we're talking about 120 pound female. Has 7.5 times 10 to the minus third moles of hemoglobin. It's got a whopping molar mass of 64,456 grams per mole in her blood. How many hemoglobin molecules is this, and what is the quantity in grams? That's going to be an interesting one to get. the quantity in grams. So first we've got 7.5 times 10 to the minus third moles and we're going to find the number of molecules. So we multiply that by Avogadro's number and we end up with 4.5 times 10 to the 21st. And if you wanted to know the number of iron atoms you needed, there are five iron atoms in each hemoglobin. molecules so you would multiply that by four and you'd have an idea of the number of molecules of iron or atoms of iron you would need but we're taking a look at hemoglobin and so we've got 7.5 times 10 to the minus third moles and this time we want mass so we're going to put moles on the bottom and we're going to put 64,456 grams on the top and we're going to end up with you 4.8 times 10 to the second grams okay which is 480 grams which is about a half a kilo a kilogram is 2.2 pounds so basically 120 pound female is going to have about a pound of um of hemoglobin molecules floating around in their blood kind of makes a little bit more interesting when you start thinking about it in that direction than just putting some numbers together And we'll be doing a fair number of practical problems at the end of our problem sets in each of these sections on your in your slide notes. So let's take a look at one more practical problem. We're going to take a look at the Cullinan diamond. It's the largest natural diamond ever found. It weighs 3,104 carats and a carat is 200 milligrams. How many carbon atoms were present in Well, let's first of all, we've got 3,104 carats. And so it tells us there are 200 milligrams per carat. Notice the carat goes in the bottom, so the carats will cancel out. And if you remember from the slot notes for quiz number one, there are 1,000 milligrams in a gram. And so that gives us a 620. 0.8 grams of carbon or basically two-thirds of a kilo and so when we go to take a look we're talking about this is about the diamond itself weighs about a pound and a half. Boy that would make a nice big mega ring for somebody to wear on their finger or to put as a diamond pendant around their neck. If you take a look at the picture it's not a real pretty stone but it is pretty good size if you see it in his hand. But if we want to know how many carbon atoms, we take the 6.20 times 10 to the 8th grams. Diamonds are pure carbon. And so if we divide by the 12.01 grams per mole, that would give us the moles. And then if we want to get rid of the moles and plug in Avogadro's number, I did this to four significant figures just to show you what one with the four significant figures would look like. And we end up with 3.11 times 10 to the 25th carbon atoms. I really don't think the number of carbon atoms really mean, but when we talk about 620.8 grams or two thirds of a kilogram, so we're talking about basically about one and a half, a little bit more than one and a half pounds. I think that's something that you kind of relate to. And this ends our 3.1 section of figuring out and working with molar mass and Avogadro's number.

So again we're trying to connect the microscopic world of atoms and molecules to the macroscopic world. We can't see individual atoms so we need some way in order to understand what's going on. So we talked about atomic mass. in the last set of slide notes.

And we talked about an AMU and that basically one atom of carbon-12 weighs 12 AMU. And so that on this scale, a hydrogen atom would have a mass of 1.008 AMU and an oxygen-16 atom would have a mass of 16.00 AMU. The problem that we run into is we don't have any... AMU scales in our labs. We measure mass in kilograms and grams not in AMU's so we need some way of connecting those two things together.

So how we connect those together is we use this idea the mole. It's a unit to count the number of particles. So, I mean, you're used to, if you go to buy eggs, you buy them basically by the dozen or you buy 12 of them. You don't walk into the grocery store and go to the egg area and go, well, I'm going to buy one egg. No, normally you've got to buy either half a dozen, a dozen, a dozen and a half.

You don't go into a shoe store and buy one shoe. You buy pairs of shoes. And the reason that we do that is for convenience. Well, in chemistry. We use the mole as the amount of a substance because it's convenient.

It's convenient. So what we've done is the mass of a carbon-12 atom has exactly 12 grams, or a mole of carbon-12 atoms has a mass of 12 grams or 12 AMUs. So one mole, if we were to find out how many atoms fit into one mole, That would be 6.0221367 times 10 to the 23rd. You're going to find we're going to stick with 6.02 or 6.022 depending on whether we need two, whether we need, sorry, three or four significant figures. Okay, this number is called Avogadro's number.

Why do we use this big, huge number? Because it's convenient for us to lump this number of atoms together to have an understanding of what's going on at the molecular level. So molar mass is the mass of one mole of anything in grams.

Well, because this number is so huge, they're pretty much only a few things that it's really reasonable to measure 6.022 times 10 to the 23rd of. And that's typically atoms, ions, molecules. We wouldn't measure out the mass of tennis shoes.

We wouldn't measure out a mole of eggs, but we do do things that are extremely small. And so if you'll notice the atomic mass units we were talking about on the periodic table in the last chapter, we can assume the exact same value for molar mass. And that's why we have this odd number of 6.022 times 10 to the 23rd. so that we can use the same set of numbers for amu and for molar mass so what we're going to do is we're going to create a couple conversion factors if you remember back in slot note set number one we did some metric conversions and we put a thousand milliliters in front of a liter and we put a thousand grams in front of a gram We put 100 centimeters in front of a meter so we could get the units to cancel out. Well, we're going to create similar conversion factors except now they're going to have the mole in it.

And so if you'll notice our first one we're going to talk about is molar mass which is grams per mole. There's an asterisk because it's not always going to be a set number. So where are we going to get that number from?

Well we're going to either get it from the problem or it's found on the periodic table. So if you remember on your periodic table underneath each of the symbols is the atomic mass for the element. Well, we now know that can be measured in AMUs if it's a one single atom. But if we want to have something that's in grams, if we have a mole's worth of it, the hydrogen would be 1.008 grams if we have a whole mole of it. And so our second conversion factor is Avogadro's number, which is the number in a mole.

And so we're... pretty much going to use 6.02. Every now and then I may use 6.022 if I want four significant figures, but typically three will be enough for us in this class.

Remember the clue is one, you're going to have this big ugly number. So if you got the big ugly number 10 to the minus 23, 10 to the 23, 22nd, 21st, 20th, that it's a pretty good clue that you're probably gonna have to use Avogadro's number in the problem. The other clue is you're going to be asked about the number of atoms, molecules, formula units, because pretty much these are the only things that are really small enough for us to actually use Avogadro's number with.

So if we want to find the mass of 0.89 moles of calcium, if you remember back when we were doing problems in slot notes for quiz number one, I'm going to take... 0.89 moles of calcium. I'm going to write down my numbers and my units, multiplication sign, division bar. Well, the only thing we can use is either molar mass or Avogadro's number. Well, I need to get rid of moles.

That should look familiar. And then what I need to do is I'm looking for the mass of calcium. So I've got grams well I need a number well then what I'm going to do is I'm going to look on the periodic table because there's no number for calcium given and so if I look on the periodic table I'm going to look right underneath calcium and it says it's 40.08 grams per mole put an equal sign which means it's now put it in your calculator because we've got only thing we've got is grams are looking for grams so this says take 0.89 times 40 0.08. And when I multiply the two of them together, I end up with 36 grams or 36 rounded to two significant figures.

Okay. So if we take a look, we're starting out with 1.65 times 10 to the 23rd atoms of zinc. I'm going to put a multiplication sign division bar.

And so if you'll notice, I've got this big ugly number so I'm probably going to have to use Avogadro's number. So I'm going to put Avogadro's number in the bottom. Remember that can be used to cancel out atoms or molecules or formula units.

But I have to put moles on the other side. And this says I've got moles but if you notice in my problem it's asking for grams. So that means I'm not done with the problem.

So what I'm going to do is I'm going to put a multiplication sign division bar. I'm going to put moles in the bottom. Now it kind of looks like the previous problem, except the only difference is instead of the mass of calcium, I need the mass of zinc.

Because notice I get rid of the moles, grams. And so I'm going to go to the periodic table. I'm going to find the molar mass of zinc, which is 65.38. And this says take 1.65 times 10 to the 23rd. divide by 6.02 times 10 to the 23rd, and then multiply by 65.39, and we end up with 17.9 grams of zinc.

So what is the mass of one hydrogen atom? Let's find out what the numbers would look like if we just did the mass of single atoms on the periodic table. and find out why we use the mole.

So I've got one hydrogen atom. Well, it's atoms and it's a number of something. So I'm going to have to divide by Avogadro's number because remember it goes with atoms. I'm going to put the mole on the top, but I don't want moles, I want mass. So this kind of looks like the previous problem again.

We got moles and we want to go to mass. And so I'm going to put moles in the bottom and I'm going to put the mass of hydrogen in the top. And if we go to the periodic table, that's 1.008 or rounded to two decimal places is 1.01 because we're only going to take our masses out to two decimal places. um, a lot simpler for us to work with.

If you took it out to 1.008, you would still get a very similar number, but I think it's just more convenient if we just take everything out to two decimal places, and then we won't run into issues in the second section of this chapter, or when we, or this set of slot notes, or chapter three, or the next set of slot notes. That's how I'm going to take one divided by 6.02 times 10 to the 23rd times 1.01. And what that means is I get 1.66 times 10 to the minus 24th grams for a single atom of hydrogen. So if I wanted to do the mass of single atoms on the periodic table for hydrogen, that 1.008 would be replaced with 1.66 times 10 to the minus 24th. I think it's a lot more convenient for us to work with moles of things.

Just like it's a lot more convenient to buy a dozen eggs or a pair of shoes than to buy one egg or one shoe, it's a lot easier to take a look at the mass of a mole of atoms than it is to just take a look at the mass of just single ones. So now we've got grams and we want to go to atoms. So I've got 27.6 grams of argon. multiplication sign division bar well i need to get rid of the grams i'm going to put moles on top and i need the mass of argon so i'm going to go to the periodic table and see that it's 39.95 remember the asterisk was in front of the g so the 39.95 has to go in the bottom Multiplication sign, division bar, because if you take a look, we've got moles, but it's asking for atoms. What we're going to do is we're going to put moles on the bottom, and we're looking for atoms of argon.

So we've got to use Avogadro's number. We cancel out the moles. So this is 27.6 divided by 39.95.

times the 6.02 times 10 to the 23rd. And should be getting 4.16 times 10 to the 23rd atoms of argon. So if you notice, even going from grams to the number of atoms gives us really big, huge, cumbersome numbers to work with. The nice thing is, is once we get done with this section, pretty much Avogadro's number is going to disappear. We're not going to do a whole lot with it.

So just make sure you can get this into your calculator now. We'll have to deal with scientific notation a little bit later on, but we won't be dealing with Avogadro's number, okay? Because hopefully what I've tried to show you in the past couple problems is it's really not a convenient number to work with in terms of mathematically, but it makes things convenient if we take a look at moles of atoms instead of individual atoms, okay?

So molecular mass or molecular weight is the sum of the atomic masses of a molecule. So we've got SO2 and you could go to the periodic table and there's one sulfur and there's two oxygens. And if I added this together, I get 64.07. So for any molecule, the molecular mass AMU again, is also equal to the molar mass in grams. So the assumption is, is if I have one mole of SO2 molecules, it would be 64, they would have a mass of 64.07 grams.

So one molecule of SO2 is 64.07 amu, and one mole will be 64.07 grams. So this is even, it's just as convenient with molecules as it was for atoms. Formula mass is the sum of the atomic masses in AMU in a formula unit. Basically, ionic compounds aren't considered molecules.

They're considered formulas or formula units or ionic cubes. And even though we kind of still call them molecules and compounds anyways. But if you'll notice, mathematically finding the mass of sodium chloride was the exact same method as we use for the SO2.

take one sodium, one chlorine, and add them together. So if we take a look at a couple problems, if we go to find the molar mass of KCl, I'm going to take potassium and chlorine, and then I'm going to go to the periodic table, and I'm going to find the mass of potassium. and I'm going to find the mass of chlorine.

This is molar mass, so it's grams per mole, not amu. And then what I'm going to do is I'm just going to add the two of them together, and I end up with 74.55 grams per mole. If I've got sodium hydroxide, I'm just going to take three different atoms and add them together. Sodium is 22.99. Oxygen is 16.00.

Hydrogen is 1.01. And so if we add them all together, we end up with 40.00. Notice how I'm taking the atomic masses out to two decimal places. I want to find the molar mass of calcium hydroxide.

Well, if you notice in the problem... when we went to the SO2, it said to take the sulfur, multiply by one and take the oxygen and multiply it by two. Well, I find that can be cumbersome.

So I find a better way of doing that is to split out the formula and then go find the masses. Let me show you what I mean. So I've got calcium and I've got two oxygens in my formula.

And I've got two hydrogens. When I write this down this way, what this tells me is go find the mass of calcium, which is 40.08. Go find the mass of oxygen and multiply it by 2 before I write it down. Oxygen is 16.00.

Multiply it by 2 and it's 32.00. The next one says take hydrogen, multiply it by 2, and then write it down. Hydrogen is 1.01, so I'm going to write this down as 2.02, add the three numbers together, and we end up with 74.10 grams per mole.

Listing a formula vertically is something that we do a lot in chemistry for even other types of problems, okay? Breaking it out like what you've seen here by just finding all the different types of atoms. So now we can kind of connect this to what we were doing before using molar mass and Avogadro's number, but I'll be kind. I'm not going to mix Avogadro's number into this. We're just going to deal with molar mass, and we're going to be dealing with molar mass for a while until we get to the third section of this, and then we'll add our third conversion factor in.

So it says calculate the number of moles of calcium carbonate, chalk, containing 14.8 grams of calcium carbonate. So I've given you a strategy that says find the molar mass of calcium carbonate and use it to convert 14.8 grams to moles. So I've got to first come up with a chemical formula. Oh that sounds like slot notes quiz number two. So we need calcium, carbon, and three oxygens because calcium carbonate is CaCO3.

If we add all these together, I end up with 100.09 grams per mole. I was asked how many moles were in 14.8 grams of calcium carbonate. So the grams goes on the bottom. So the 100.09 goes on the bottom.

The grams will cancel out. And we end up 0.148 moles of calcium carbonate. So in question number 16, they're asking us to find the number of moles of a compound, and version A asks for propylene, which is C3H6. So when I see a formula and it's talking about the number of moles, it kind of triggers in me that I'm going to need some sort of mass.

And so what I'm going to do is I'm going to add three carbons and six hydrogens. and they get 42.09 again i'm taking things out to two decimal places and then i'm going to take 25 grams multiplication sign division bar i'm going to put the 42.09 grams in the bottom to get the grams to cancel out with the moles on top so this says to take 25 and divide by 42 0.09 because the number's in the bottom and we end up with 0.594 moles of C3H6. Problem number 17 is going in the, is starting out with the 0.0146 moles of KOH, kind of in the opposite direction.

We still need the mass of KOH. That's like how we did the sodium hydroxide except I'm going to use 39.10 for K instead of the 22.99 that we use for the sodium. And that 0.0146 moles of KOH.

Put moles on the bottom so the moles will cancel out this time. Notice the 56.11 grams ends up in the top so it's going to be multiplication. And I'm going to end up with 0.819 grams of KOH.

Folks, please be careful. Notice this is 0.0146. Students can sometimes accidentally put 0.146 and then if it's a multiple choice question you look and you don't think you have an answer that matches any of them then you don't think you know what you're doing with the problem. When in reality, mathematically, you know what you're doing with the problem you just accidentally put it wrong in the calculator.

Number 18, we're starting out with grams. And so what we're going to do is we're going to find the mass of KBr, potassium and bromine. And then we're going to take that 2.21 grams.

If you'll notice, the grams goes in the bottom. So the number goes in the bottom. So this is division. And so we end up with 0.0179 moles of KBr.

Number 19 says determine the mass. So this is going in the opposite direction. And so we'd add up the FeS, O11, and 14 hydrogens. If you'll notice on some of these problems, I've already been giving you some of the masses.

Okay. And so we got 278.06. So we got 0.5758 moles. Put the moles on the bottom, which means the mass goes on the top.

So we're going to multiply these two numbers. And we end up with 111.8 grams. So let's take a look at a couple of practical problems. Question number two, we've got a 55 kilogram woman. Well, they're about 2.2 pounds in a kilogram, so we're talking about 120 pound female.

Has 7.5 times 10 to the minus third moles of hemoglobin. It's got a whopping molar mass of 64,456 grams per mole in her blood. How many hemoglobin molecules is this, and what is the quantity in grams? That's going to be an interesting one to get. the quantity in grams.

So first we've got 7.5 times 10 to the minus third moles and we're going to find the number of molecules. So we multiply that by Avogadro's number and we end up with 4.5 times 10 to the 21st. And if you wanted to know the number of iron atoms you needed, there are five iron atoms in each hemoglobin. molecules so you would multiply that by four and you'd have an idea of the number of molecules of iron or atoms of iron you would need but we're taking a look at hemoglobin and so we've got 7.5 times 10 to the minus third moles and this time we want mass so we're going to put moles on the bottom and we're going to put 64,456 grams on the top and we're going to end up with you 4.8 times 10 to the second grams okay which is 480 grams which is about a half a kilo a kilogram is 2.2 pounds so basically 120 pound female is going to have about a pound of um of hemoglobin molecules floating around in their blood kind of makes a little bit more interesting when you start thinking about it in that direction than just putting some numbers together And we'll be doing a fair number of practical problems at the end of our problem sets in each of these sections on your in your slide notes. So let's take a look at one more practical problem.

We're going to take a look at the Cullinan diamond. It's the largest natural diamond ever found. It weighs 3,104 carats and a carat is 200 milligrams.

How many carbon atoms were present in Well, let's first of all, we've got 3,104 carats. And so it tells us there are 200 milligrams per carat. Notice the carat goes in the bottom, so the carats will cancel out. And if you remember from the slot notes for quiz number one, there are 1,000 milligrams in a gram.

And so that gives us a 620. 0.8 grams of carbon or basically two-thirds of a kilo and so when we go to take a look we're talking about this is about the diamond itself weighs about a pound and a half. Boy that would make a nice big mega ring for somebody to wear on their finger or to put as a diamond pendant around their neck. If you take a look at the picture it's not a real pretty stone but it is pretty good size if you see it in his hand. But if we want to know how many carbon atoms, we take the 6.20 times 10 to the 8th grams.

Diamonds are pure carbon. And so if we divide by the 12.01 grams per mole, that would give us the moles. And then if we want to get rid of the moles and plug in Avogadro's number, I did this to four significant figures just to show you what one with the four significant figures would look like. And we end up with 3.11 times 10 to the 25th carbon atoms. I really don't think the number of carbon atoms really mean, but when we talk about 620.8 grams or two thirds of a kilogram, so we're talking about basically about one and a half, a little bit more than one and a half pounds.

I think that's something that you kind of relate to. And this ends our 3.1 section of figuring out and working with molar mass and Avogadro's number.

Transcript for:Understanding Moles and Molecular Mass

Transcript for:
Understanding Moles and Molecular Mass