 Hello all you chemistry kids. Welcome to Tutor Terrific. Today we're gonna look at the following concept the mole and related concepts molar mass and mole conversions. So what is the mole? Well, the mole is a particularly sized group of objects, usually identical objects. We're looking at a way to categorize a giant group of objects. Now, why would this be useful? Well, for small particles like atoms and compounds, it's very useful, especially in chemistry. This particular group has a size and that size is 6.02 times 10 to the 23 objects. Okay? Compare this to something like a dozen or a gross. A dozen has 12 objects. So a group of 12 objects is called a dozen. A group of a hundred and forty four objects that all culinary students know has a hundred and forty four objects in it. So a gross of eggs is a hundred and forty four eggs. A dozen donuts is 12 donuts. Well, a mole of anything is 6.02 times 10 to the 23rd objects or examples of those things. This number is very specific and has been given the name Avogadro's number and it's an honorary name given to a medio Avogadro. He was an Italian chemist and he, there's a very famous law attributed to him that really doesn't have much to do with this. It has to do with gases and numbers of particles of gases. But based on his contributions to chemistry, he was given the honorary name to this number 6.02 times 10 to the 23. Now, it's defined specifically as the number of atoms that are present in exactly 12 grams of one isotope of carbon, carbon 12. That means carbon with 12 protons plus neutrons in its nucleus. So if I have 12 grams of carbon 12, I have exactly Avogadro's number of atoms or a mole of atoms of carbon 12 in that sample. So again, I have to reiterate. I can make a dozen of anything. I can make a gross of anything. I can make a mole of anything, but it has to have 6.02 times 10 to the 23 particles of that thing in that group to be called a mole. Now, I'm going to show you how this relates to chemistry and mass in the next slide. So here is the moles use in chemistry. It's really helpful to discuss a reasonable amount, a usable amount of a particular element or compound. Looking at individual molecules or atoms of a compound is virtually impossible. But if we have a large quantity of them that's handleable, we usually have a couple moles or a fraction of a mole of that compound. Remember, a mole is 6.02 times 10 to the 23 of anything, particles, for example. Now, where does mole come into play most valuably when it comes to mass, when it comes to mass? Molar mass is a very important definition I need to go over with you. It is the exact mass in grams of one mole of a single element or compound. OK, again, the exact mass in grams of one mole of anything. So an element or a compound, for example, the molar mass of individual elements is actually on the periodic table. Here are two particular elements, iron and silver. Their atomic numbers are in the top that tells you how many protons are in the nucleus. The number in the bottom, you might have first learned as the weighted average of the isotopes, basically saying the average by percent abundance. The mass number, which would be the number of neutrons plus protons in the nucleus. Well, it's actually also the molar mass of that element. That means the units of molar mass, by the way, are grams per mole based on the definition that makes sense, number of grams of something in a mole of that thing. These numbers down here at the bottom of each periodic table elements listing gives you that molar mass. So for iron, for example, if I have a mole of iron, I have 55.845 grams of iron. If I have a mole of silver, I have 107.87 grams of silver. So this is how we go from mole to gram or moles to mass. All right, I have some questions for you just to keep you on your toes. First one, how many atoms of silver are in one mole of silver? I'm asking how many atoms? OK, this isn't a trick question. Go back up here. If I have a mole of anything, I have 6.02 times 10 to the 23 of that thing. So the answer to this is 6.02 times 10 to the 23 atoms. OK, so you get the idea. If I'm asking for atoms, I am looking at Avogadro's number. Another example, how many atoms of iron are there in two moles of iron? OK, two moles. We know one mole of anything is 6.02 times 10 to the 23. Well, if I multiply that by 2, I would get 1.204 times 10 to the 24 atoms. OK, so I took that mole relationship there and I multiply it by 2 because I was asked how many atoms are in two moles of iron. Now, this last question asks you for something different. It says what is the mass of one mole silver? The mass of one mole silver is given by the molar mass itself, that number at the bottom. So no calculators needed for this. You could just write 107.87 grams or just G for short. 107.87 grams of silver are in one mole of silver. So now you see how the mole relates to mass. OK, now let's see how to find these numbers and use them on the periodic table when you're asked particular questions about molar mass. For example, what is the mass of 3.5 moles of aluminum? So when you're asked for the mass of a certain amount of moles of aluminum, you have to start by finding aluminum on the periodic table. Aluminum is in the post-transition metal section. So it's in the P block right after the transition metals. And the number you'll get for aluminum is 26.982. Now, that's the grams per mole. So I will start by writing my given information, 3.5 moles. And then I'm going to multiply that by the number of grams per mole of aluminum. I often write it like this, 26.982 grams per one mole. OK, so it's basically multiplying 3.5 times 26.982. Let's determine what that is. 3.5 times 26.982. I get 94.437. Now, my original number had 3.5, that's two significant figures. So that's all of the significant figures I'm allowed to have in my final answer, two. And so I have to round the number at the ones place, which rounds down to 94. So I will have what are my units after all? I know it's 94, but what are my units? Well, moles cancel. And so I have grams left. So I have 94 grams. All right. Next, what is the mass of one mole of NAF, which is known as sodium fluoride? In this example, I actually gave you the compound. Sometimes you just get the name of the compound. You have to figure out the exact balance of the elements in the compound. Here I have NA, one atom, and F, one atom. Remember, the molar mass works for compounds as well. It's the exact mass in grams of one mole of a compound as well. So a compound has multiple elements. And so what I have to do is I have to find sodium and fluorine on the periodic table and add one of each of their molar masses together. So I will have, since it's one mole, I don't have to do any multiplying. I just have to do some adding. So I'm going to have to add 22.990 grams per mole. That's sodium. I'm going to have to add that to 18.998 grams per mole for fluorine. I can add these two together because their units are the same. Now I'm not going to do that by hand. I'm going to do it on the calculator. 22.990 plus 18.998. Notice how I'm not rounding in the middle of a problem because I'm going to round at the end due to the significant figures. I get 41.988. Now my original given was not very accurate. It only had one sig fig. It was just the number one. I didn't say exactly one. So it doesn't have an infinite number of significant digits. It's just one sig fig. So I have to round this number to one significant figure, which will be that first four and I'll have to round down to 40 without a decimal place. So I will write 40 grams. OK, another example. What is the mass of exactly two moles of aluminum hydroxide? Ah, I need to know what aluminum hydroxide is. OK, know that aluminum is Al. And hydroxide, you should be memorizing these by now when you're in this section of chemistry is OH. It's a polyatomic ion. Now you need to know the charges. OK, for the polyatomic ions for a while, you're just going to have to memorize them. And a hydroxide ion has a charge of minus one. Aluminum is an element and it's a metal. It's in group 3A, which is right here. Right after the transition metals, we start counting our main group numbers again. Here is 1A, 2A, and then we skip all these. Then here's 3A. Aluminum has three valence electrons. If it's in group 3A and since it has less than four, it will give them away in an ion bonding situation. And so aluminum will have a net charge of three plus in ionic form. So here we've got a little issue. If we want to combine aluminum and a hydroxide ion, we're going to need the charges to be the same. So what we do is we find the least common multiple of the two charges. The least common multiple of three and one is three. And so it looks like we're going to have to have three hydroxide ions. That's how we notate three polyatomic ions. We put a three outside of parentheses. Then we will have a total negative charge of three minus three plus and three minus combined to get a neutral atom. So what we have in essence is AlOH3. I'm going to move that closer. This is aluminum hydroxide. I'm asked for the mass of exactly two moles of aluminum hydroxide. And it's definitely a compound. So we're going to have to add some numbers together. First, we're going to have to add one aluminum molar mass amount. So that's twenty six point nine eight two grams per mole. Then I've got oxygen. How many oxygens do I have since it's inside of parentheses with a three outside? I have to add three oxygens. So I'm going to have to add to that three times oxygen's molar mass. Fifteen point nine nine nine. Next, I'm going to have to add three hydrogen atoms. Hydrogen is one point zero zero eight grams per mole. So if I come over here, I have three times one point zero zero eight grams per mole. I just include that everywhere because I'm a stickler for units grams per mole. OK, so when we add these together. Add them like so. Twenty six point nine eight two plus in parentheses, I like to do three times fifteen point nine nine nine for oxygen plus parentheses. Three times one point zero zero eight for hydrogen. And I get seventy eight point zero zero three. I'm not going to round early. I haven't even done the two moles multiplication part. I just know that all of these together give me seventy eight. Point zero zero three grams per mole. That's per mole. Now I have two moles, so I'm going to multiply that by two. Now, since this is exactly, that means that the two has an infinite number of significant digits. And so when I multiply seventy eight point zero zero three times two, I can use all of the digits I get as a result. One hundred and fifty six point zero zero six grams per mole. Grams and it's grams, actually, because I'm looking at two moles. OK, guys, this is how we go from a certain number of moles to a certain mass using molar mass. OK, let's continue by looking now at reactions. We are going to learn how to convert between moles of one element of the reaction to moles of another element of the reaction using what's called the molar ratio. The molar ratio is a ratio of the coefficients of one element or compound in a reaction to that of another element or compound in the reaction. So here I have a generic reaction of four objects, A, B, C and D. And according to the balanced reaction, I have one mole of A plus two moles of B that yields three moles of C and four moles of D. That's what those coefficients really mean. Now you can look at this as an individual reaction. We have one particle of A plus two particles of B equal three particles of C plus four particles of D. You're willing to do that. You're welcome to do that. But you can also think of it in moles as well. So a ratio, of course, is a fraction or stated as a column where you do a direct comparison, but usually in chemistry, we like to think of it as a fraction of equal quantities based on the reaction. So I'm going to ask you what the molar ratio of A to C is in this reaction. Well, we will put the moles of A on top and we'll put the moles of C on bottom in a fraction based on this reaction. So what I'll write, if this is my fraction line here, I will write on top one mole A and I use MOL for short, if you haven't seen that yet, over three mole of C. OK, now if I'm asked for the molar ratio of D to B, I will put the four moles of D on top. See, I'm looking at the coefficients here. Four mole of D divided by now it's D to B, so two moles of B. Now you're welcome to simplify fractions that can be, but I would steer clear of that so that you really get in your head the connection between these numbers in the ratio and these coefficients in the reaction. So let's look at a real reaction. Now this is called the Haber process. It's the production of ammonia and we have a balanced reaction here to nitrogen atoms, six hydrogen atoms on each side. Now the reason this yield sign goes both ways is because this is actually a reversible reaction, but we're just going to think about it going forward. I have an example for you. If six moles of N2 are used in this Haber process, how many moles of NH3 are produced? Okay, we are going to use the molar ratio in order to do this. See how mass is not involved in this setup at all. So what we often do in chemistry is we set up what's called the factor label method, which allows us to convert between units or different quantities or different elements. Often you see it like this, a bunch of fractions multiplied together like so, or you could see it like this. A bunch of lines and vertical slashes through them. We are able to use either of these because these represent fractions multiplied together. I'm going to use the latter because it's a little cleaner and easier to understand for beginning students, in my opinion. So what I do here is I start by writing a first part of this fraction because I know I have to put the given first in the top left. And I have two elements and two compounds involved, but one given piece of information. That's six mole and two. Normally what I do is I put a one on the bottom. Nothing needs to go there. I like to have something there, so I don't think has zeros there, which would cause problems. Now I need to use the molar ratio of the reaction to convert that to moles of NH3. So this is a one step process. We'll see later we can have multiple steps in our processes for more complicated calculations. This is a pretty simple one. We want to convert to moles NH3 because that's what we're asked for. So that unit has to be at the top of the final section of this factor label, this set of fractions that I'm multiplying together has to be in the top. And on bottom needs to go the previous unit that was on top. So down here needs to go moles N2. The reason I have to do that is because they have to cancel so that the only unit I have left is the one that I want, moles of NH3. So now we look at the coefficients and we place the coefficients next to their particular elements or compounds in my factor label fraction method here. So I have NH3. It looks like two moles of NH3 per one. Why do I put a one down here? Because one is implied by an empty space like it was in this generic reaction up here. OK, now I cancel the mole N2s like so. Now, since these are really fractions, we are multiplying straight across. Six times two divided by one times one. Whenever one is involved in multiplying or dividing, it can be ignored because dividing or multiplying by one does nothing for you. So we're going to multiply six by two. We're going to get 12. The only unit I have left is mole NH3. The reason I'm not rounding this down to 10 like I technically should is I'm going to say exact. Sometimes you can assume you are working with exact numbers. So we can turn this into two sig figs here because we assume this as an infinite number of sig figs like 6.0 repeating. Now, we never use or never have to use the molar relationship coefficients to figure out how many sig figs we have because those are assumed to be exact numbers. So the answer is 12 moles NH3 are produced when we use six moles of N2. OK, I have another example for you. A little bit different structure to this question. Now I'm going to ask you if I have 16.7 moles of NH3 being produced, how many moles of H2 were required? OK, I'm thinking this way. So the yield is the NH3. And I'm asking how many moles of H2 were required to produce 16.7 moles of NH3? So again, we're going to set this up. Simple factor label method in the top left corner. We're going to start with what we're given, which is 16.7 mole NH3 over one. OK, we're going to keep that over one process going there for that first fraction. Next, we're going to use the molar ratio to convert to the unit we want, which is mole of H2. And on bottom, we'll go the previous unit, what was on top? Mole NH3. OK, so we use the molar ratio between NH3 and H2 given the coefficients in the reaction. Next to H2, I see a three. So three moles of H2 for every two moles NH3, simple ratio problems, guys. So in this situation, I'm going to have to multiply 16.7 by three and then divide that result by two. So I have 16.7 times three and then divided by two. I can go straight across with the order of operations on these scientific calculators or graphing calculators for that. And I get 25.05. I had three sig figs to begin with, so I have to round that up to three sig figs. 25.05 will become 25.1. Now, units, again, cancel all units that appear twice, one on top and on bottom, and I will have left only moles of H2. 25.1 mole H2 was required to produce that many moles of NH3. So this is molar ratios in a nutshell. All right, guys, we have come to the crux of the entire matter. Now we are going to convert between mass and moles and moles and mass, often called mass to mole conversions, multi-mass conversions or mass to mass conversions. We're going to use both molar mass and molar ratios to do so. And here's our roadmap. We could start and stop at any spot on this map. And I show you how to get from one place to the other. Sometimes you're asked to find the moles of a substance given the mass of another substance. The first substance that represents the given information is called Substance A in this model. And the required or requested information about another substance, we call that other substance B. So if we start with the mass of Substance 1 given, we will use molar mass of that substance to convert to moles of that substance. Then we will look at the reaction and use the molar ratio of B, second substance to A, first substance, to get from moles of A to moles of B. Then when we finish with getting to moles of B, then we will use the molar mass of B to convert from moles to B to mass of B. This is the extent of the roadmap. And like I said, you could start and stop anywhere on the map. Now, where you start and stop is based on the question you are asked. For example, if 6.7 moles of K potassium are used, how many grams of AL aluminum are produced? We will look at this particular reaction for this and all following example questions. So I start with my given information, 6.7 moles of K potassium. So it looks like I'm starting at the mole spot. And then I'm asked how many grams of another substance aluminum are produced. So I end all the way over here at the mass of my second substance, aluminum. So as you can see, this is a two step process. Each arrow in this map represents a step. So I'm going to have two particular spots in this, two particular lines in this factor label method. OK. The top left corner, I put the given information, 6.7 mole potassium K for potassium over one. Now I'm going to move from mole A to mole B using a molar ratio. So again, just like in the previous examples, I'm going to put the mole of what I had in the previous step on top, the unit on bottom in the second step. And on top, I'm going to put the unit of the desired amount as I move along the map. So the mole of aluminum is what I want as I move forward to get to eventually grams of aluminum. Now let's look at the molar ratios. Mole aluminum, I don't see any number by it. So that's assumed one mole of aluminum and then three in front of the potassium. That means for every one mole of aluminum produced, that was three moles of potassium required. So a one to three ratio. Now, the last step is to convert the moles of aluminum to grams of aluminum. And we will use molar mass of aluminum to do that. OK. So on top, we want grams of aluminum. And on bottom, we will put the previous unit that was on top, mole aluminum. It's decided to get all the ones involved in all the elements, all the molar masses and write them down at the top there for reference. For aluminum, we have 26.982.9. OK. 26.92 grams of aluminum per mole of aluminum. Now I cancel all units that show on top and bottom called cross canceling. And it looks like I'm going to take 6.7, multiply it by 26.92 and then divide by three. OK. I prefer you do the multiplying before the dividing. As you'll see later, it's very important. 6.7 times 26.982 divided by three. It gives us 60.2598. How many sig figs can I use? Well, my initial measurement had only two. And remember, these guys are exact numbers, so they don't factor into it. Only the number of sig figs of my original measurement, two. So I'm going to have to round to the ones place. And it rounds down to exactly 60. I can use a decimal if I want to show that zero is significant. 60 grams aluminum. OK. Now the next example over here. If 84 grams of potassium chloride KCl is produced, how many moles of aluminum chloride was used? OK. So we're looking at a mass of potassium chloride. So we're going to start with mass A, which is potassium chloride. And then it asks us to go to how many moles of another substance was produced, was used. Ah, mole B. Now don't think that B has to be a product and A has to be reacted. It can be reversed. It does not have to be from right to left all the time. All right. So as you can see, we have two steps again. So we are going to do a two-stepper, as they say. Start by writing what you're given, the top left corner. 84 grams KCl over one. And now we have to use the molar mass of KCl to convert to moles KCl. So we want mole KCl on top. Get the new unit and the old unit from the previous step on top goes down below. OK. KCl is a compound. And so we're going to need to add one potassium and one chlorine together to get the number of grams we're dealing with here. Let's make some more space here. All right. So that just requires us to add 35.45239.098. Oops, excuse me. I did that wrong. So that requires us to just add 35.45239.098. So that's one chlorine and one potassium. 74.548. No rounding in the middle. 74.548. OK. So now we're in moles potassium chloride. And now we need to convert using the molar ratio to moles of aluminum chloride. OK. I'm going to make a little more space here. So we put the old unit that was on top on bottom, mole KCl. And we put the moles of the one we're looking for, moles AlCl3, aluminum chloride. AlCl3. And now we look at the reaction itself. It looks like there's a 3 next to the KCl. So that means 3 moles KCl. And there's no number in front of AlCl. So we just put a 1 there. All right. Now it looks like we have to divide twice. This is where you need to be careful. In the calculator, on most of these Texas Instruments Scientific Calculators, if you're dividing by multiple things, you can do that in one of two ways. You can either do 84 divided by 74.548. And then press divide again, divide by 3. Or you can do the following. I'm dividing 84 divided by, and in parentheses, put 74.548 times 3. And then close the parentheses. These will give you the same result. Large decimal, that starts with 0.375596. How many significant figures can I have? Two, according to my original measurement. So that would round to 0.38, round up. What are the units? If we cancel all these units that appear on top and bottom, I get moles. AlCl3. All right. That's my answer. All right. My last example for this lesson is the big doozy. You'll see in the previous examples that we did mole to mass and mass to mole. This one is a mass to mass. Let me show you. If 255 grams of aluminum chloride, stop right there. You were given a mass to start. And so we're going to start at the mass on our roadmap. And then it asks how many grams of KCl stop again. Grams of another substance. So we're going all the way to mass B is produced. That means we've got three steps. So our factor label method here is going to have three steps in it. So three vertical lines. Okay. As before, put our given in the first slot, top left. 255 grams of aluminum chloride with a 1 on bottom. Now I need to convert using molar mass to moles aluminum chloride. Mole aluminum chloride. One mole of aluminum chloride. Now we have to add, it looks like one aluminum to three chlorides. Okay. So one aluminum, 26.982 plus. Now we've got to multiply three chlorides together. Three times 35.45. 133.332. On bottom we'll go the previous top unit. Grams of aluminum chloride. 133.332. Again, I'm not rounding in the middle of a problem. Okay. Now that I have this, it's time to go from mole A to mole B using a molar ratio. The first element is mole is ALCL3. We're going from moles ALCL3 to moles KCL. Again, the molar ratio, three moles KCL per one mole aluminum chloride. Okay. Now we've converted to moles of KCL. Now we need to go to mass of KCL using the molar mass of KCL. Alright, so we want grams KCL as our final. And we will have mole KCL on the bottom since it was the previous steps top. One mole of KCL is how many grams of KCL in the previous problem. We got it. It was 74.548. So we could put that right there. Okay. As you can see, we're going to multiply twice and divide once. Remember the ones don't need to be included. So carefully make sure, take your time here, 225 grams times three from the molar ratio times 74.548 for the molar mass of KCL. Then we divide by the molar mass of aluminum chloride, 133.332. And we get 377.403, et cetera, grams KCL. How many significant figures am I allowed? Three based on the original measurement, which means I have to round to 377. It's a round down, 377 grams KCL. You guys made it. You made it through this video. Thanks so much for watching. I hope this helps you. Good luck on all your chemistry. This is Falconator, signing out.