The next two sections will be about how we get chemical formulas from experimental data. This first section is on elemental analysis, which is an analytical process that chemists use all the time. Now typically what we tend to do as a synthetic chemist would be that you make something, you use other methods to convince yourself that you have what you think you do, and then you send it out to elemental analysis for confirmation. Other chemists will find a sample of something, say that they have an idea of what it is. So for example, I saw a really interesting random video from an archaeological chemist who found something that she believed was incense.
One of the things that could be done is that a piece of that incense could be sent out for elemental analysis, and that would help you determine some of the formulas of what's in there. Now we do kind of need a pure sample for what we're going to do because what we are going to do is take the data and turn it into a formula. Now there's a strategy I'm going to give you a strategy and then we're going to go through examples. So when you're doing these on your own I recommend having the strategy written down next to you and then gradually working to be able to do this without the strategy in front of you. So the steps in elemental analysis, the output is that you will get mass percent values.
So your first step is going to be to turn those percentages into grams, then turn the grams into moles, use the moles to write a preliminary formula, convert the preliminary formula to a regular formula with integers, which we will call the empirical formula. And then if you have enough information, you can find the molecular formula. The empirical formula, so this first thing that you get, shows the smallest number, the smallest whole number ratios of elements in a compound.
It's kind of a version of formula unit, but it works for both molecular and ionic compounds. So as an example, Something that has the molecular formula of C2H4 would have the empirical formula of CH2, right? The subscripts have been reduced to the smallest whole number ratios.
For every one carbon, I have two hydrogen atoms, which is true in C2H4, right? It's also true in C3H6. It's also true in C4H8. So the empirical formula...
doesn't necessarily give us exactly what we need, but it gives us a point at which we can use the information to get to a molecular formula. All right, we're going to use our strategy, and we are going to look at the elemental analysis of a compound that has a molar mass of 90.08 grams per mole, and it shows that it contains 40% carbon by mass. 6.71% hydrogen by mass and 53.3% oxygen by mass. We are asked to find its molecular formula. Now I've got the molar mass listed here. That is the piece of information you need to find the molecular formula.
If you don't have the molar mass, you can only find the empirical formula. So our first step in the strategy is that we want to turn the percentage into a value in grams. And the first step, the way that you do this and the way that we tend to teach it is we just say, assume a 100 gram sample. And if you're anything like me at this point, you are asking why?
Why am I assuming 100 grams? How do I know that there's 100 grams? Why do I know that I can assume anything if I don't know what I'm doing in chemistry thus far? Okay, it's okay. The reason that we assume 100 grams is because it makes the math easy.
We have a mass percent. So if my sample is 100 grams and 40% of it is carbon, then I have 40 grams of carbon. That's the only reason.
You can assume any mass that you want. and get this correct. The thing that we get out of this we're eventually moving to the amount of carbon in relation to the amount of hydrogen in relation to the amount of oxygen. That information isn't going to change based on the mass of the sample.
So you can assume literally any mass you want. So if you have the time and the interest, you can try this out. Assuming 100 grams just makes it easy for the very first math step.
So 40% of 100 grams is 40 grams of carbon. 6.71% of that 100 gram sample means I have 6.71 grams of hydrogen. 53.3% of 100 grams means I have 53.3 grams of oxygen. So in whatever size sample I have, in this case 100 grams, I have the specific gram amounts of each element, which I need to do step two, which is where I'm converting the grams to moles.
So grams of carbon, molar mass of carbon from the periodic table gives me moles of carbon. Same for hydrogen, same for oxygen. But again, the question here is, why are we now converting this to moles? Well, remember, from our C6H12O6 example, that those subscripts in the formula, you could relate that to moles. So those subscripts tell you how many moles of each type of atom are in one mole of the molecule.
We initially started by saying in the formula that it was the number of atoms, but now remember, number of atoms and number of molecules scales up. through Avogadro's number in the same way. So these relationships between numbers of atoms are also relationships between numbers of moles. So now I know how many moles of carbon related to how many moles of hydrogen, that's going to tell me, that's going to lead me directly into my whole number ratios that are in those subscripts.
Step three, I recommend this it's okay if you don't want to write a preliminary formula. I think it helps kind of hit home this point that I was just making, which is that those subscripts are mole relationships. Now, this is obviously preliminary because we have not whole numbers, 3.33 moles of carbon, but it's just preliminary.
It's just to remind you that those numbers are the subscripts. And this is our basic relationship between those atoms in the molecule. The next step is to convert to integers.
So divide by the smallest number, and then if you need to, multiply back up to get integers. So the smallest number is 3.33. If I divide everything by that, I get carbon with a subscript of one, hydrogen with a subscript of two, and oxygen with a subscript of one. You don't have to write the ones in, I just put it here to show that I didn't completely forget about those values. So my empirical formula is CH2O.
I have whole numbers, I don't need to do the multiply back up step, but we'll have examples where you need to. The next step is to convert it to the molecular formula, which I can only do because I have the molar mass of the molecule. The way to do this is...
is to calculate the molar mass of the empirical formula. So CH2O, the molar mass would be 30.025 grams per mole. Now my full molecule has a molar mass of 90.08. So it's quite a bit bigger.
So the comparison, if the empirical formula and the molecule have the same molecular mass or same molar mass, then... What you have is also the molecular formula. That's not the case here.
If you don't have that, you can divide to get the multiplication factor. Now in this case, it might have been a simple inspection to get the multiplication factor, which is also fine. You don't have to write the math out necessarily, except for where I say show all your work. And this, the way that you would set this up, if you do the actual molar mass divided by the empirical molar mass, The multiplication factor is 3. So what that means is that our empirical formula, we need three of them put together to be the whole molecule, which means that we multiply each subscript by 3. So 1 times 3, 2 times 3, and 1 times 3. So my molecular formula is C3H6O3.