Hello everybody. In this video we're going to cover section 3.2 in our textbook, where we're going to learn how to compute the percent composition of a compound, determine the empirical formula of a compound, and determine the molecular formula of a compound. So the first thing we're going to do is look at the percent composition.
This is the percentage by mass of each element in a compound. So we know that compounds are composed of multiple different kinds of elements a lot of the time and we can talk about what the percentage by mass of each one of the elements that make up that compound. An example, we might be given some data that looks like this.
A 10 gram sample of a compound is determined to contain 2.5 grams of hydrogen and 7.5 grams of carbon. The first thing I notice when I read this is that 7.5 grams plus 2.5 grams is equal to 10 grams. So I know that this sample is composed of just hydrogen and carbon.
To calculate the percent hydrogen, it's pretty simple. I just take the mass of hydrogen and I divide it by the total mass of the compound. I multiply that by 100 and I get 25%. To calculate the percent of... carbon.
I take the 7.5 grams, I divide it by the total mass, I multiply that by 100% and I get to that I find out that it's 75% carbon. If I add 25% plus 75% together I get 100% which is consistent when I know about this compound now that it is just composed of hydrogen and carbon. We don't actually have to be given that much information in order to find the percent composition of a compound.
We can actually do this just by being given the formula for that compound. And what we're going to do is we're going to go look up our values in the periodic table for all the elements in that compound. And then the first thing we're going to do is calculate its formula mass, which we learned how to do in the last section. for a little refresher. For instance, if I wanted to find the formula mass of this ammonia molecule here, I would look up the atomic mass of nitrogen.
I would look up the atomic mass of hydrogen. I would add the atomic mass of hydrogen to three times the atomic mass of hydrogen, and I would come up with 17.03 amu. After that, I can calculate the percent nitrogen by taking just the atomic mass of nitrogen, just the... portion of the formula mass that is due to nitrogen and dividing it by the overall formula mass for the compound. I multiply that by 100% and I can see that it's 82.27% by mass nitrogen.
I can do the same thing with hydrogen but there is a little caveat here. I have this 3 here so I need to recognize that I actually need to bring that down here and have 3 times the atomic mass of hydrogen in my numerator in order to get all of the bits in the formula mass that are due to the hydrogen. I divide that by the total formula mass, I multiply it by 100%, and I can see that it is 17.76% hydrogen.
Again, when I add these two together, I'm going to get roughly 100%. It does not always work out perfectly, like it doesn't in this case, for instance. But it will often be very close.
It should be very, very close to 100%. The reason it doesn't always work out perfectly just has to kind of do with our rounding and stuff sometimes. So let's talk about the difference first between an empirical and a molecular formula. An empirical formula shows the composition of a compound given as the simplest whole number ratio of atoms. And a molecular formula indicates the composition of a compound, giving the actual number of atoms in each element in a molecule of a compound.
So what does that really mean? What it means is that molecular formulas are the ones that we're really used to talking about. All right, for instance, I have this graphic here.
You can see that I have one, two, three, four, five, six carbons. Each of those has two hydrogens for 12 hydrogens. The molecular formula is going to be C6H12, but I can divide 6 by 6 and I can divide 12 by 12. If I were to do that, I would get CH2 over here, and that would be the empirical formula.
Now, why do we care about empirical formulas? Well, they really only enter in in this class with... a specific type of problem that's based on an idea called elemental analysis.
It's kind of an older technique, but it's still relevant in a lot of industries. And what it's going to allow us to do is to get the percent composition of a molecule, a compound, and from that percent composition, we're going to be able to figure out the empirical formula of that compound. With a little bit of additional information, we're going to be able to figure out the molecular formula from the empirical compound, or the empirical formula. So let's talk about how we're going to do that. The empirical formula of a compound can be derived from the masses of all the elements in the compound.
So the same kind of mass information we got in order to calculate the percent composition. We'll start with the mass of one particular atom, and we'll have the mass of another particular atom. Here I'm calling them A atoms and X atoms. We divide by the molar mass, which we know is going to give us the moles of atom A and the moles of atom X.
Next, we have to make a decision. We need to find the lowest of these two moles, so the smallest number here, and we're going to divide both of these values by that. What that's going to give us is the ratio of one of the atoms to the other. We then can convert those ratios to the lowest whole number values, and we'll arrive at the empirical formula.
This flow chart is a little confusing maybe, but we're going to look at an example and you can see that it's really, it's not that bad. So here is our question. A compound is determined to contain 1.71 grams of carbon and 0.383 grams of hydrogen.
So what we're going to do is we're going to start with the masses that we were given in each, in the problem of each of the elements. We're then going to divide by the molar mass. Again, what we're actually doing is multiplying by the reciprocal here because we like to see these units get cancelled.
We're going to have grams cancelling in the top and bottom there, and we're going to wind up with the moles of carbon and the moles of hydrogen. Now a nice little mnemonic here is to actually just take these decimal values and stick them as the subscripts in the formula. right away.
Even though this does look silly, that's what these really ultimately mean. And it helps you to realize that what you really want to do is find a way of making those whole values. So what we're going to do is we're going to divide by the smaller of the two, in this case, 0.0974. All right. And when you do that, what's nice is you're always going to wind up with one of these that is going to wind up being one.
And the other one will still maybe wind up, it'll either be a whole number or it'll be something pretty close to a whole number most of the time. All right. And, but you will wind up having to do a little bit of rounding.
For instance, 3.9, we're going to round that to four. We're not going to let that sway us too much. All right.
That being said, sometimes in more complicated examples, it doesn't work out quite as nicely. So in this example, We're given some information where a compound contains 5.31 grams of chlorine and 8.4 grams of oxygen. I divide by the molar masses, which I looked up again in the periodic table. I made sure that my units canceled, that I'm in moles. I place them in here.
I divide by the smallest of those subscripts. And what I wind up with is something that's... got like a 0.5 on it. So this would have been 3.5.
I've written it here as 7 halves. All right. We can't really just round 3.5.
That's just, it's too in the middle to just say it rounds up or it rounds down, for instance. In a case like this, what we can do is we can multiply the subscripts by a value that will give us a whole number. Okay.
That's why I wrote it at 7.5, because it's pretty clear when I write it at 7.5 that if I multiply this subscript by 2 and I multiply this subscript by 2, I'm going to clear that 2 in the denominator, and I wind up with an empirical formula that looks like CL2O7. Now, I know what you're wondering. When do I know when I can round and when I can't?
In this class, most of these problems are going to work out really, really nicely for you. And you're not even going to have to bother with multiplying by some value to get to this. In general, you should only really consider having to do that if you could rewrite one. subscripts with either a 2 or a 3 in the denominator.
Okay, if you're getting to 4, 5, other larger fractions there, you really don't need to consider those. Just go ahead and round anything like that. So we've now seen two examples where we can calculate the empirical formula given mass data, but as I alluded to before, we can actually also calculate the empirical formula from the percent composition. And all that's really going to happen in this case is we're going to have a little additional step at the beginning, all right, where we need to obtain the mass of each compound.
And then we're going to start with data that looks very similar to those other examples. And the most convenient way to do this is to assume that you started with 100 grams of sample. And I'll show you how that works out. So here, a bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% carbon and 72.71% oxygen. And this is a typical thing where we can actually analyze a gas pretty easily for their elemental composition.
We might see data like this, okay, and we're being asked what is the empirical formula for this gas. Now if I assume that I started with a hundred grams of gas and 27.29% of it was carbon, that means that there's 27.29 grams of carbon in that sample. And if I had 100 grams of the gas and 72.71% of it is oxygen, then I had 72.71 grams of oxygen. And now I have mass data that is exactly the same as what I started with in those previous examples.
And all we need to do is follow through with the rest of the steps. where we divide by the molar masses. We see those units cancel. We get the moles of carbon, we get the moles of oxygen. We place those values as the subscripts in our formula.
We divide by the smallest one. One of them goes to one and the other one very conveniently winds up being almost exactly two. And we see that that gas was carbon dioxide.
Now, once we're given the empirical formula, we need an extra little bit of information in order to get the molecular formula. What we need is we need that empirical formula and we need its molecular or molar mass of the compound. And there are a lot of clever ways to experimentally get the molar mass of a compound as well. And basically what you're going to do is you're going to take that molecular molar mass or formula mass, as we've also been calling it, We're going to divide it by the empirical formula mass and that then we're going to get a whole number value from that.
And typically this doesn't really require any rounding it really will just work out to be a very very close or exactly a whole number. And then what we're going to do is we're just going to multiply all of the subscripts by that n value. So let's see how this works out. A compound has an empirical formula of CH2O. The empirical formula mass is 30 AMU.
So if I added up the atomic mass of one carbon, two hydrogens, and one oxygen, I'd get 30 AMU. And it has a molecular mass of 180 AMU. So I take that 180 AMU, I divide it by 30 AMU, I get 6. If I multiply all of the subscripts by 6, which I can actually use this parentheses to indicate doing that.
It works with the six will distribute across the subscripts, just like if you were doing algebra. Then I can see that the molecular formula is C6H12O6. You could be asked to do a problem that combines the two types of problems together.
OK, for instance, I could give you percent composition information. Ask you to find the empirical formula, all right? Ask you to go and look up the atomic masses and tell me what the formula mass of that empirical formula is, and then give you the molecular mass of the compound and ask you to find the molecular formula. But just remember that even if you're given that, it's still just those same steps one right after another.
They're just not being broken up into different problems.