#### The Mole

# The Mole

I have a handout that deals with conversions. You can download it here. There are some conversions we will be using right away, and others we will use in the future.

But I want you to think in terms of the concepts. The key thing here are:

- Reactions occur at the level of atoms and molecules interacting.
- These atoms and molecules have to come together in the right proportions to assure most efficient reactions.
- We need a way to determine that we have the right proportions.

Let's take hydrogen combining with Fluorine to make HF. Both hydrogen and fluorine come as diatomic molecules, so the reaction is:

H

_{2}+ F

_{2}→ 2HF

You can see that there are two H atoms on the left and two on the right. There are two F atoms on the left and right as well. That satisfies the requirement of conservation of matter. To make this "recipe," because that's really all it is, I will always need the same ratio: 1H

_{2}and 1F

_{2}to make 2 HF molecules.

The proportions by mass of this are 1g Hydrogen plus 19g F give you 20g HF. Based on Dalton's Atomic Theory, we know that a Fluorine atom must weigh 19-times as much as a Hydrogen. Here, since there are two atoms of each in each molecule, 2 Fluorine atoms weigh 19 times as much as 2 Hydrogen atoms. Or, an F

_{2}molecule weighs 19 times as much as an H

_{2}molecule. Let's make up a new unit of mass, we'll call the Atomic Mass Unit (AMU). We'll say that it's equal to the mass of a hydrogen atom (actually, there is a more refined definition…but, this is close to correct still).

We would say the H

_{2}has a mass of 2 AMU and F

_{2}has a mass of 38 AMU. Still, that's a 1:19 ratio.

10 F atoms would weigh 19 times as much as 10 H atoms, and 1,000,000,000 F atoms would weigh 19 times as much as 1000,000,000 H atoms. And, 6.022x1023 atoms of F will weigh 19-times as much as 6.022x1023 atoms of H. I think you get the picture.

So, for any number, let's say "X" of H atoms, the same number of F atoms would have a mass 19 times as much, right? That's a necessary consequence of Dalton's law of definite proportions.

Now, 1g of H atoms would have some number of atoms in it…Let's make that "X" atoms. How many grams of F atoms would have X atoms? If "X" hydrogen atoms have a mass of 1g, then X Fluorine atoms would have a mass of…19g.

Well, those X hydrogen atoms wouldn't really be separate atoms. They would form H

_{2}molecules. Each molecule requires 2 H atoms, X H atoms would form 1/2X H

_{2}molecules. To make X H

_{2}molecules, I would need 2g of hydrogen. To make X F2 molecules, I would need 38g Fluorine.

This idea used to be called the "gram molecular weight." It's an old term, but a useful one. Because all you need to do is measure stuff out in grams using the values of their atomic masses, which are fundamentally the ratio of the masses of the atoms.

I don't actually need to know exactly how many molecules are in 2g of hydrogen gas (H

_{2}). But I do know that 38g of Fluorine gas will have the same number of molecules. I could figure it out if I knew how to convert grams to AMU. But, even if I didn't know the conversion, it would be a useful idea. Think about it, if you have a recipe for a cake in which the volumes are measured cups, say, a cup of milk and two cups of flour, it actually doesn't matter what cup you use, does it? You'll have the right ratio whether you use a teacup or a bucket.

It turns out that 1.00AMU= 1.6606x10

^{-24}grams (this can be determined in a machine called a mass spectrometer). To do the same calculation for any other element, all you do is plug in the atomic mass in AMU and take the same number of grams ("X" in equation below). You will again get 6.022x10

^{23}So, Figure out how many atoms there are in 24.31g of magnesium.

This number is called Avagadro's Number and is the number of particles in one "mole" (abbreviated mol). For any atom or molecule, if you determine the molecular mass of the molecule in AMU and take that number of grams of the substance, you will have 1 mole. A mole is no different from one "dozen." If you set up any of the calculations as though you are dealing with "dozens," it will be fine. Then, at the last step, change the word dozen to "mole" and change "12" to 6.022x10

^{23}, you will get the right answer. Remember the conversions are just like you would do for "dozens" If you know the number of moles you have and want to know molecules, multiply by Avagadro's number (2.5 dozen is 2.5x12 eggs/dozen or 30. 2.5 mol is 2.5x 6.022 x 1023) If you know number of molecules and want moles, you divide by Avagadro's number (or multiply by 1 mol/6.o22 x1023 molecules). If you know the number of grams, and want the number of moles, divide by the molar mass (grams/mol). If you know mol and want to know grams, multiply by grams/mol.

## Calculations with compounds

If we know we can find the molar mass (the mass, in grams, of one mole of any element) simply by taking the average atomic mass off the periodic table, what do we do about compounds? Well, that’s really simple. If we were talking about a potato head again (sooner or later I have to let that one go), you could determine the mass of the finished head easily enough, simply by adding up the mass of the skull, 2x the mass of each ear, likewise for arms, then add the mass of each of the other components. It works the same for compounds. So for water, you have one atom of oxygen (16AMU, actually 15.9994AMU) and 2 atoms of hydrogen (1AMU each for a total of two). Because of conservation of mass, you would put them together to make a molecule of water with a

*molecular*mass of 18AMU (same concept as the atomic mass). Therefore, the molar mass is just 18g (the same number of the molecular mass but in grams). If those numbers, 16g of oxygen combines with 2 grams of hydrogen, sound familiar, they should. These are just exactly the same as the numbers we used when we talked about definite proportions. So, let’s take CO2 as another example. It has one atom of Carbon (12.01AMU) and two of Oxygen (2x16AMU). The molecular mass must therefore be 12+2x15.9994=44.009. Once again, these are the same numbers we used when we were talking about definite proportions. So, here is a practice problem: How many moles of oxygen (atoms) are there in 15.0g of C2H4O2? Hint: you know there are two moles of oxygen atoms in every mole of the compound (because of the formula). So, all you need to know is how many moles of C2H4O2 you have. To do that, you need to determine the molar mass of the compound. Just add them up. Here’s a followup: oxygen usually comes as O2 molecules, not as oxygen atoms. So, how many moles of O2 are there in the same 15.0g?