Mass percents and empirical molecular formulas. That'll be the topic in this second lesson in a chapter on stoichiometry in my high school chemistry playlist. Now if this is your first time joining me, my name is Chad.
Welcome to Chad's Prep, where my goal here is simply to make science both understandable and learning it even enjoyable. Now I'll be releasing the lessons in my high school chemistry playlist here weekly throughout the 2020-21 school year, so if you don't want to miss one, subscribe to the channel, click the bell notifications, you'll be notified every time I post a new lesson. All right, so we're going to start off here with a 3.2 gram sample of calcium carbonate here, CaCO3, and I chose that on purpose because that's a really super convenient sample for the calculations we're about to do, as we'll see. So, but in this case, we want to find the percent carbon and the percent oxygen in this compound. We could also find the percent calcium and stuff like this.
Now, in this case, to find... percent anything, it's usually parts over total times 100. So in this case, we'd find if I want the percent carbon, I'd find the number of grams of carbon in the sample divided by the total weight of the sample times 100. Except we're not going to do that. So it turns out these mass percents are what we call an intensive property. And back in, you know, one of the early chapters, you learn that an intensive property is one that the sample size does not matter. It'll be the same regardless of what size sample you use.
And so in this case, I've got a 3.2 gram sample size, doesn't matter. And it turns out I can use any sample size and it would work out exactly the same for these mass percents. So if I want to find the percent carbon here yet again though, so percent carbon here is going to be grams of carbon over your total grams or your total mass and then times 100. The only question that is what sample size total here do I want to actually use? Well, if you have the formula it turns out a really convenient sample size to use is the actual one mole sample. So that's what a molar mass corresponds to is one mole.
And so we'll look up on the periodic table and we'll find out that calcium has a molar mass close to 40 grams, really close to a whole number. Carbon's close to 12 grams, and each of the three auctions are close to 16 grams. Three times 16 is 48, plus 12 is 60, plus 40 is 100 grams. And that's why I used it.
It's gonna make the math really easy here. In this case, there's one carbon in a mole, and one carbon weighs 12 grams out of a total of 100 grams. times 100 and we can see that that is exactly 12 percent carbon and again that's why i chose it with a with having a molar mass of 100 grams it just made the math easy now if you decided to figure out you know and use the 3.2 gram sample it would have been a little more challenging to figure out well how many grams of carbon are in a 3.2 gram sample whereas In one mole, I know that the one mole weighs 100 grams, and it's really easy to figure out very quickly that it has exactly 12 grams of carbon in that one mole sample size. And so this is not so bad. We can do the same thing for percent oxygen here.
In this case, you just got to take into account that you've got three oxygen atoms in the formula. So instead of just taking oxygen's molar mass of 16 grams, in this case, I'm just going to continue all the way down here. So 16 grams per 100 grams, that would not be correct. Because it's not 16 grams, that's for one oxygen, but there are three in the formula, so it's really three times 16, which is 48 grams of oxygen per 100 gram sample size, times 100, and again, the math is nice, and that's 48% oxygen in the sample.
And again, that would be true for any size sample of pure calcium carbonate. Alright, so now I want to introduce what's called the empirical formula. Now, molecular formula is just the exact formula.
It gives all the number of atoms of every single type in a molecule. So, whereas an empirical formula, though, is the most reduced whole number ratio of those same atoms. So, it turns out benzene here has a formula C6H6.
And so, it turns out in every single molecule of benzene, there are six carbon atoms and six hydrogen atoms. But the corresponding empirical formula here would just be simply CH. The most reduced whole number ratio is 1 to 1. That 6 to 6 ratio can get reduced down to 1 to 1. And so that's the difference between molecular and empirical formula.
So same thing here with P4O10, a 4 to 10 ratio can be reduced down to a 2 to 5 ratio. And so once again, so our empirical formula is going to be somewhat different than the molecular. But what you'll find out is that for the vast majority of compounds, your molecular empirical formula end up being one in the same. And that's true for most compounds. I had to cherry pick a couple examples here so you could see the difference.
So if you take a look at like ethanol here, C2H6O, well, it's a two to six, and don't forget that means to a one ratio. Well, a two to six to one ratio can't be reduced down any further. It's simply a two to six to one ratio.
And so in the case of ethanol, its molecular formula and its empirical formula are exactly the same. So the example with calcium carbonate we did a little bit ago, we started with an actual formula and then went and found the mass percents. And that's actually the easier way to do it. So this time around we're going to start with the mass percents and then try and get back to a formula.
And we'll see that we'll start by getting back to an empirical formula and we'll need a little more data or a little more info to figure out how to get that to a molecular formula as well. But this is definitely the harder of the two ways to do this. So we'll start with mass percents. So we've got a compound that is 80.0% mass or percent carbon by mass and 20.0% hydrogen by mass. And the question is, what is the empirical formula?
And a lot of students see this and they're like, oh, sweet, Chad. You gave me an easy one. I can do this one in my head. I don't even need my calculator.
And they're like, yep, that's C4H. And the problem is that, again, these are the mass percents. They're the percent by mass of carbon and hydrogen, which is not what the numbers here represent.
So this four to one ratio is a ratio of either atoms or moles. So I like to think of it as a mole to mole ratio for convenience, not a gram to gram ratio, which is what this represents. And that's why this is not going to work.
And I picked the numbers, I picked a compound whose numbers would be really nice that students would think this. That's a common way to ask a multiple choice question, is pick one that you think students can get a common answer wrong. very quickly. So, but you're not going to be fooled by this anymore.
So the question is then what do you do? Well, once again, so mass per cents are an intensive property. Doesn't matter what sample size you start with. So you can start with any for the approaching this calculation. So, because one was not provided now in this case, you can't choose one mole though, because choose one mole, you'd have to know its formula to be able to know its molar mass, but that's Kind of the point in question, we want to find the formula, we don't know the formula.
So you can't actually use one mole because you have no idea what the formula is. So in this case, the convenient sample size is 100 grams. And again, you could use any sample size you want. The reason I'm choosing 100 grams is because it just makes the math easy.
For a 100 gram sample size, well, if 80% of those grams are carbon, well, then that's going to correspond to 80.0 grams of carbon. And if 20% of those grams are hydrogen, well, 20% of 100 grams would be 20.0 grams. of hydrogen.
And so I just saved myself some math. And again, I could have made this, you know, 4.376 grams and then figured out what 80% of that was and what 20% of that was. But by choosing a 100 gram sample size, I just saved myself a little bit of math here.
All right. But I still have only a gram to gram ratio here. And what I really need again is a mole to mole ratio and a proper formula.
And so what are you going to do? You're going to convert your grams of carbon, your grams of hydrogen to moles. And to convert grams to moles, you need your molar mass. And so in this case, I'll take my grams of carbon and convert it to moles of carbon, and one mole of carbon weighs 12 grams. And I'll convert my grams of hydrogen to moles of hydrogen, and one mole of hydrogen weighs one gram.
And so in this case, the math is a little bit easier down here. This is going to be 20.0 moles of not oxygen. Let's get that right, moles of hydrogen.
So but here, 80 divided by 12, you might plug that into your calculator. So and you'll see that this comes out to 6.67 moles of carbon. And so now we want to take a look at getting our empirical formula.
And we're going to get C6.67H20.0. Life is good. Wait a minute. Life is not good.
Why is this wrong? What's wrong with this? Why isn't this a good empirical formula?
So for a proper empirical formula, you can only have whole numbers. And obviously you can make 20.0 just 20, but that's not a whole number. No decimals allowed here.
And you can't just round that. That's not very close to a whole number here. And so in this case, you're going to have to figure out, well, how do I turn this into whole numbers? Well, what you want to do is...
By the time you get to this part in the calculation, you just want to divide all the numbers by a common factor. And the common factor you're going to use is whatever one is your smallest number. And so in this case, because I'm dividing them all by the same factor, the fundamental ratio is still the same.
But I guarantee that I'm going to... for sure get at least one whole number because 6.67 divided by 6.67 is one. That's a whole number. So, and in this case, if you carry this out, you get C1, obviously, and then 20 divided by six and two thirds.
If you actually plug this into your calculator. is like almost exactly three. And because it's close to a whole number, we will round it. Now, had that not been, you know, a whole number, we would have taken a different approach here, but it was a whole number.
And we don't usually write C1H3, we just write CH3. And that is our empirical formula. So in whatever compound this is, the carbon and hydrogen atoms are in a one to three ratio. That again is a mole to mole ratio.
Now let's just say for a second, so that by the time we'd actually carried out this process, instead of coming out to a one to three ratio, let's just say it had come out to C1H2.5. It didn't, but had it done so, we wouldn't have been done. You shouldn't just round this up to three or round it down to two or anything like that.
If you're not close to a whole number, you're gonna have to carry this out one step further and multiply by some common factor till you get a whole number ratio. And because 2.5 is- 2 and a half, well, if you've got halves for fractions, you can avoid having fractions by doubling everything. If you had 2 and 1 third or 2 and 2 thirds, like this was 2.33 or 2.66, then you could multiply by 3. You get rid of thirds by multiplying by 3 to get whole numbers. Well, in this case, I'd just double everything and I'd get C2H5 as my formula, as it turns out. But again, we didn't actually have to worry about this problem.
This was not our issue in this one as well. Ours came out to whole numbers right from the get-go at that point. Life was good. Cool, now this is an empirical formula.
My next question for you is, is this the molecular formula? Well, maybe. Could be. I mean, it could also be any multiple of this, right? As long as it's fundamentally a 1 to 3 ratio, so C2H6, C3H9, C100H300, and as long as it's a 1 to 3 ratio of carbon to hydronutrients, it's a 1 to 3 ratio.
It could be anything. There's an infinite number of possible combinations all the way up to infinity. So how do we figure out the right one? Well, we figure out the right one by being supplied with a molar mass.
So if you know the molecular weight or molar mass, well then you can figure it out. Now one thing you should realize here is that CH3, carbon weighs 12, hydrogen weighs one each, so for a total of three and 12 plus three is 15. And so this is going to have a weight of 15. And since all of these are multiples of that formula, their corresponding weights would all be multiples of 15. Double the formula, doubles the weight. Triple the formula, triples the weight.
Hundred times the formula... 100 times the weight. So I know whatever this compound is, its molar mass and molecular weight is going to have to be a multiple of 15. And once I supply it with you, you just figure out what multiple of 15 is it. And so in this case, second part of this question, if the compound above has a molecular weight of 30 AMUs, what is its molecular formula?
And you just be like, well, 15 goes into 30 twice, so I just need to double the formula. And so here is our molecular formula. One of the examples we went with right from the the get-go. Cool. But unless I had supplied you with that molecular weight or molar mass, you could not have actually figured out the molecular formula.
You would have only been stuck with the empirical. If you have benefited from this lesson, consider giving me a like and a share with some of the best things you can do to support the channel. And if you've got questions, feel free to leave them in the comment section below.
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