Calculations using Avogadro's number (part 2) | Chemistry | Khan Academy

Let's solve a few numerical on Avogadro number and moles. Here's the first one: how many glucose molecules are in 2.37 moles of glucose?

Let's quickly remind ourselves what moles are. Moles are like dozens. Just like how one dozen equals 12, a mole represents an Avogadro number of things. A dozen eggs equals 12 eggs; a dozen carrots means 12 carrots. Similarly, a mole of eggs would represent these many eggs, a mole of carrots would represent these many carrots, and a mole of glucose would represent these many glucose molecules.

Okay, so one mole of glucose equals these many molecules of glucose. What if you have two moles of glucose? It'll be twice the amount. If you have three moles, it'll be thrice the amount, and so on and so forth.

Now, what are we given? We are given there are 2.37 moles of glucose, and we are asked how many glucose molecules are there. Now, since one mole equals these many molecules, 2.37 moles of glucose would just be 2.37 times these many molecules. Intuitively, that makes sense, right? But guess what? This is a slightly simple numerical, and that's why it's easier to do it in our heads, and we can do it intuitively.

But in general, there might be more conversions required, more calculations needed. So, a better way to do this is to think in terms of conversion factors. Okay, so we can say that, hey, we are given moles of glucose, and we need to convert from moles to molecules. So how do we do that?

Okay, so here's how we do it. We start with what's given to us. We are given 2.37 moles of glucose. Now, from moles of glucose, how do we go to molecules of glucose? Well, for that, we'll build a conversion factor. Here's how I think about it: since I want glucose molecules, my conversion factor should have molecules of glucose in the numerator, and since I want to get rid of moles, my moles should be in the denominator. So this is what my conversion factor looks like: molecules per mole.

But how many molecules per mole? Well, I know that there are 6.022 * 10^23 molecules per mole, so I can fill this in. Now, I have built my conversion factor. So look, the moles of glucose cancels out, and what I now end up with is molecules of glucose.

So I just need to plug this in my calculator: 2.37 * 6.022 * 10^23. The way you input that is you use this "e" okay, not this one, this one: 10 to the 23. So you put "23," and that's it! So equals 1.1427, which I can round off to three significant figures there because there are three significant figures here. So 1.43 * 10^24. Don't forget this, okay? So 1.43 * 10^24.

And look, as we saw earlier, it makes intuitive sense as well. One mole has these many molecules, so 2.37 mol will have 2.37 * these many molecules. But now, let's see the true power of thinking it this way.

Okay, let's say the question was how many atoms of hydrogen are in 2.37 moles of glucose? How would we solve this? Well, again, let's think about this intuitively. We already know that 2.37 moles of glucose has these many molecules. So I have these many molecules of glucose. But now, I'm asking how many atoms of hydrogen are there in these many molecules of glucose? How do I do that?

Well, for that, I know that if I have one molecule of glucose, then there are 12 hydrogen atoms. If I take two molecules of glucose, I will have 12 + 12, that is 2*12 hydrogen atoms. If I take three molecules of glucose, 12 * three hydrogen atoms. So if I take these many molecules of glucose, it'll be 12 * these many atoms of hydrogen. Makes sense, right?

But here, we first calculated the molecules of glucose, and then we are now calculating the atoms of hydrogen. So multiple times we'll have to crunch in the numbers in the calculator. But what if I had to do this from scratch, not worrying about the earlier calculations?

Then again, if you think in terms of conversions and building conversion factors, we can calculate very efficiently. So again, we start with what's given to us. We are given 2.37 moles of glucose, and then we try to go from moles to molecules of glucose. I know how to do that: I need molecules in the numerator, moles in the denominator.

So I have Avogadro's number, so I know there are these many molecules of glucose per mole, and so then this cancels out. Now comes the question: how do I go from molecules of glucose to atoms of hydrogen? I need another conversion factor. I need atoms of hydrogen in the numerator because that's what I want to end up with, and I need molecules of glucose in the denominator so that I can cancel it with this one.

So my new conversion factor would be atoms of hydrogen per molecule of glucose, and I know how much that is: 12 atoms of hydrogen per molecule of glucose. And so now I can cancel the molecules of glucose out. Look, I'm done! I end up with atoms of hydrogen, and now in one shot, I can crunch all of these numbers in my calculator.

So 2.37 * 6.022 * 10^23, and the way you do that, remember, you use this one to the 23 time 12 divided by 1. So that's it! And again, we now have to round this off to how many significant figures? Well, there are three here, there are four here, so the minimum is three significant figures. But wait a second, what about this one? Aren't there two significant figures here? No, because this is not a measurement. Every glucose molecule has exactly 12 hydrogen atoms, so this is not a measurement, and therefore our significant figures are not limited by this number; it's limited by this number.

So there are three significant figures, so we have to round it off to three significant figures. So it would be 1.71 * 10^25.

All right, let's kick things up a little bit. Let's go to the next question: how many formula units of calcium carbonate are in 4 g of calcium carbonate?

All right, first things first! What exactly are formula units? I used to get always confused with them, so let's clarify that. When we're dealing with non-metals, which are usually covalently bonded, we have individual molecules. So for example, if you look inside a glucose, then we can say that, hey, there is this glucose molecule and there's that glucose molecule, so you know they're made of molecules, okay? You can identify individual molecules and we can talk about them.

But when it comes to ionic compounds like calcium carbonate, or let's take a simpler one, say sodium chloride, okay? Because it's much simpler, then you don't find individual molecules over there. If you were to look inside them, you'll probably see a crystal structure that looks like this.

But if there are no molecules over here, then what is this? What does this represent? Or over here, if you look at it, what does it mean to say that there are three oxygen atoms over here? What does it represent? Well, think about it this way.

Think about it this way, okay? Let's start with the sodium chloride. If I would take a chunk of sodium chloride, there are lots and lots of sodium cations and chlorine anions, but they will be in the ratio 1:1. That's what it means to write it this way. Similarly, if I were to take a chunk of calcium carbonate, its crystal structure is much more complicated—that's why I didn't draw that over here—but if I were to take a chunk of calcium carbonate, then I'll find that the ratio of the amount of calcium to the amount of carbon to the amount of oxygen that I'll find in that chunk is 1:1:3. There will be twice as many oxygen atoms as carbon or as calcium. That's what it means!

So a formula unit is a representation of the relative ratios of the different atoms or elements that make up your ionic compound. All right, but more importantly, how does it change things when it comes to the mole? The answer is: it doesn't.

See, over here when you're dealing with the mole, we would say that a mole of glucose would be Avogadro's number of glucose molecules. Over here, we would say a mole would be an Avogadro's number of formula units, and that's it. So it doesn't change anything.

Okay, all right. So we need to find how many formula units of calcium carbonate are in 4 g of calcium carbonate.

Okay, so this time we are given the mass of the calcium carbonate. So look, this time you need to convert from mass to formula units. How do we do that? Well, here's what I'm thinking: if you can convert from mass (m) to moles, we're done. If I can figure out how many moles of calcium carbonate this represents, we are done, because then just like what we did in our previous numerical, we went from moles to molecules.

In the same way, using the Avogadro number, in the same way over here, we can go from moles to formula units. So the big question is: how do I go from grams to moles? We can figure that out using the periodic table. See, the periodic table will give me the molar mass of individual elements.

So, for example, here, calcium has an atomic mass of 40.8 u. This means that 1 mole of calcium will have a mass of 40.8 g. Remember that a mole is a conversion from u to g, okay?

So I know the molar mass of calcium is 40.8 g per mole. Similarly, carbon is 12.01 g per mole, and oxygen is 16 g per mole. So what will be the molar mass of calcium carbonate? What will be the mass of 1 mole of calcium carbonate? Well, 1 mole of calcium carbonate will have 1 mole of calcium, 1 mole of carbon, and 3 moles of oxygen.

So it'll be a great idea to pause the video and see if you can use these numbers to calculate first the molar mass of calcium carbonate.

All right, here's how we can do it. The molar mass of calcium carbonate, meaning one mole of calcium carbonate, its mass will be the mass of 1 mole of calcium plus the mass of 1 mole of carbon plus 3 times the mass of 1 mole of oxygen because there are three oxygen atoms here.

So you plug this into our calculator. Let me just do this first because there's a multiplication over here. So it's 3 * 16 + 12.01 + 40.8 = 100.9. We'll not round it off right now because we're still in the middle of the calculation. So its molar mass is 100.9 g per mole.

So I can use that as a conversion factor to go from grams to moles, and then I know the Avogadro number; from there I can go from moles to formula units. Again, it'll be a great idea to pause the video and see if you can try this yourself first.

All right, let's do it! So we'll start with what's given to us. We are given 4 g of calcium carbonate. Now, to convert this into moles, I need to have moles in the numerator and my conversion factor and grams in the denominator to cancel this out.

So I need moles per gram—the moles of calcium carbonate per gram of calcium carbonate—and I have that conversion factor over here. I just have to do a reciprocal because I want the mole on the numerator. So I will write this as 1 mole of calcium carbonate per 100.9 g of calcium carbonate. You see why I wrote it that way?

So that now I can cancel this. So now this gives me the moles. Now, how do I go from moles to formula units? Build a new conversion factor: I need a formula unit in the numerator and mole in the denominator.

So how many formula units do I have per mole? Well, the Avogadro number. I have Avogadro's number of formula units of calcium carbonate per mole of calcium carbonate. And boom! This cancels out.

Let me cancel out slightly differently. Okay, this cancels out, and there you have it. I am now left with formula units of calcium carbonate. I can plug this into the calculator now.

So it's 4 * 1 * 6.022 * 10^23—which again, we'll use the exponent over here—23. Divide by 100.9. Divide by 100.9. This gives us 2.46, and again, we'll round it off to three significant figures. That's the minimum one over here.

So 2.41 * 10^22, don’t forget this! 2.41 * 10^22. There you have it— that many formula units of calcium carbonate.

Okay, here's the last question: how many oxygen atoms are in 4 g of calcium carbonate? Again, pause the video and see if you can do this as the last one.

Okay, this is just like the previous numerical. We already know 4 g of calcium carbonate has this many formula units, and we know a single formula unit has three oxygen atoms.

So these many formula units will have three times this much—that's it. That’s our answer: three times this much. But again, if this was asked to us to do it from scratch, how would we do it?

Well, we'll start from here. So we'll do all of this. Let me just copy and paste that over here, and then we'll have one more conversion factor that helps me cancel out formula units and have oxygen atoms.

So over here, I would need oxygen atoms in the numerator and formula units of calcium carbonate in the denominator. How many oxygen atoms do I have per formula unit of calcium carbonate? Three.

So my last conversion factor would be three atoms of oxygen per formula unit of calcium carbonate. This cancels out, and I'm done. And again, we can crunch it in one last time: 4 * 6.022 * 10^23.

So, use the exponent 23 * 3 divided by 100.9. That's it! So that gives us 7.219; again rounding it off to three significant figures: 7.22. Don't forget this: * 10^22. Atoms of oxygen!

And this is the same number as multiplying this number by three, but if we were asked from scratch and we didn't have this number, this is how we could do it in just one shot!