Avogadro’s Number

 

Topics discussed:

·        An analogy between the “mole” and straws of hay

·        The definition of a mole and Avogadro’s number

·        How to convert grams of an element into moles or atoms

·        Definition of atomic and molecular weight

·        How Avogadro’s number was discovered.

 

Student:    Professor, is there any chance we could talk about Avogadro’s number?  I know a few things about it, but I don’t know what it’s good for, or how to work any of the problems that go along with it.

Professor:  Okay, sure.  Tell me what you know so far.

Student:    Well, I know that Avogadro’s number is 6.02 x 10²³, and I know that Avogadro once said that equal volumes of gas at the same pressure and temperature have the same number of molecules in them, and then after that it gets a bit fuzzy, especially when my chemistry teacher starts using the word “mole”.  For all I can tell, a “mole” is some kind of furry animal.

Professor:  I agree with the first two things that you said – and no, “mole” is not an animal – we’ll get to the definition of “mole” in a few minutes.  You also mentioned that you aren’t sure what we do with Avogadro’s number.  Consider this: when we ship freight, we don’t always count the exact number of boxes.  Sometimes we measure things in truckloads, because it’s more convenient to group things that way.

Student:    Oh, yeah.  That’s just like when I was growing up on the farm.  We didn’t count actual straws of hay when we measured hay, we just sold it by the bail.

Professor:  Good example!  So why didn’t you count actual straws of hay?

Student:    Hah!  No idiot would be dumb enough to count individual straws of hay, and it wouldn’t make sense to tell anyone that you had 3,745,339 straws of hay in the back of the truck.  It was easier to visualize bails of hay, which come in big rolls (or sometimes in large brick shaped things).

Professor:  I agree that it wouldn’t be very practical to count individual straws, but couldn’t you just weigh the hay to determine how much hay you had, and how much hay you were buying.

Student:    Well, I suppose that would work, but we generally didn’t have large scales around on my farm – it just wasn’t very practical.

Professor:  So it was more practical to group the straws of hay into really large chunks or groups or bails or whatever.  Did you ever have to convert from bails of hay to number of actual straws of hay?

Student:    No, I don’t recall ever having needed to do that.  We just worried about the bails.  One bail will feed so many pigs or cows, and a truck can carry so many bails.

Professor:  But what if you needed to for some reason.  Could you figure out the number of straws of hay in your truck if you had 25 bails and knew that there were, oh -- let’s say 3.28 x 10⁶ straws of hay in each bail?

Student:    Yeah, I guess, if you really needed to do that.

Professor:  Well, you see, the deal is with Avogadro’s number – chemists are in the same position as the farmers.  Instead of dealing with hay, they’re dealing with atoms and molecules.  But the chemists don’t want to count individual atoms any more than the farmer wants to count straws of hay.  But at the same time, both the farmer and the chemist need to be able to have some way of counting things up so that we know how many straws of hay (or in the chemists case, how many atoms or molecules) we’re dealing with.

Student:    What are some of the reasons why a chemist would need to know how many atoms are actually in something.  I mean, you can’t even see them.

Professor:  I agree that you can’t see them, but some reactions only take place if you mix “A” and “B” so that “A” starts out with the same number of atoms as “B”.  And besides, sometimes we don’t just want to know what the outcome of a chemical reaction is going to be, we want to know how much stuff we’re going to get when we know how much stuff we put in it.  The only way to know how much stuff (in grams, I suppose) is to know how many atoms went into the reaction in the first place.

Student:    Oh okay.  I guess even though the atoms are changing, I think my professor said that in a chemical reaction the atoms don’t disappear or reappear, they just rearrange and hook up to each other differently.

Professor:  Exactly!  Earlier I said we’d talk about what a “mole” is.  A mole is to a chemist what a “Bail” of hay is to the farmer.  It’s just a way of grouping things so to make the counting a little easier.  The chemist doesn’t want to deal with huge, unwieldy numbers of atoms just like the farmer doesn’t want to deal with large, unwieldy numbers of straws of hay.  If the farmer knows how many bails of hay he has, and he knows that there are 3.28 x 10⁶ straws in each bail, then if he has 25 bails in his truck, he could figure out how many straws of hay were in his truck – if for some unusual reason he needed to know that.

Student:    Oh, okay.  So if a bail of hay is 3.28 x 10⁶ straws, then how many atoms are in this “mole” thing?

Professor:  That’s where Avogadro’s number comes in.  There are 6.02 x 10²³ atoms in a “mole”.  If the atoms are hooked up together to make molecules, then there are 6.02 x 10²³ molecules in a mole.

Student:    So I guess on the four 6’s ranch, the entire ranch might produce a mole of straws of hay – that would be 6.02 x 10²³ straws of hay.

Professor:  Hah!  I like how you think.  And yes, that might just add up that way.  But don’t forget that there are only 10²⁴ some-odd stars in the observable universe – all the straws of hay ever grown and bailed by humans probably doesn’t add up to anywhere near that amount – but other than that, yes you’re right.

Student:    Well, okay then.  I think I’m cool with the mole.  So maybe I can work this problem now:

                        “How many moles of atoms are in 32 grams of Iron?  How many atoms is this?  How many grams does one Iron atom weigh?”

                I’m not even really sure of where to start on this one.

Professor:  Okay, well this type of problem is where the periodic table comes in.  Find Iron on the table and tell me that large number beneath it.

Student:    I found it.  The number beneath it is 55.845.  I think my teacher said that number is the molecular weight.

Professor:  I agree that you have found the correct number.  But actually, that number is the atomic weight – we’ll get to molecular weight later.  The 55.845 means that a mole of Iron atoms weighs 55.845 grams – or that 6.02 x 10²³ atoms of Iron weigh 55.845 grams.

Student:    Oh, okay, so I guess the answer to the first part of the question would go like this:

Professor:  Excellent.  Let’s go to the next part.  How many atoms are in 32 grams of Iron?

Student:    Well, if I know how many moles there are in 32 grams of Iron, and I know that there are 6.02x10²³ atoms in a mole, then I guess it goes like this:

Professor:  You nailed it!

Student:    Thanks, but the last one is a little harder.  How many grams does one Iron atom weigh?  Hmm, let me think . . . we need our answer to have units of grams per atom, so I think it might go like this:

Professor:  Rock on!  Here’s another bit of trivia that may not have been discussed in class.  The 55.845 number actually serves two purposes.  Its primary purpose is the fact that there are 55.845 grams in a mole of Iron atoms.  The other way that number works is to say that one Iron atom weighs 55.845 amu, or “atomic mass units.”  Chemists defined the units of amu in such a way that the numbers would work out that way.  The atomic weight ( in grams per mole) of an element is the same number as the weight of one atom of that element in amu.  You won’t see amu much in this course, but you should at least be aware of the existence of this obscure unit.

Student:    So what about molecular weight?  What is that part about?

Professor:  Oh okay.  Well, in the Iron example, Iron atoms are just atoms.  But what if you have oxygen, which is diatomic (N₂), or ozone, which is (O₃).  Or even worse, what if you have ammonia, which is NH₃ or methane, which is CH₄.  In something like CH₄, it wouldn’t make sense to talk about atomic weight.  In a mixture of pure methane, there are no free atoms floating around – there are only molecules.  So we talk about molecular weight instead of atomic weight.  Now the hydrogen by itself has a certain atomic weight of its own, and the carbon if it were by itself would have a certain atomic weight of its own, but in pure methane the C’s and four H’s always occur together. 

So if you have 843 grams of CH₄, how many moles of molecules do you have?

Student:    I’m guessing that to get the molecular weight of methane, you just add up the atomic weights of each of the elements in it.

Professor:  Absolutely!  C = 12.00 g/mol and H = 1.00 g/mol, so CH₄ = 16.00 g/mol.

Student:    Then for the problem you just stated about how many moles of molecules I have, I would do:

Professor:  Awesome!

Student:    One last question.  How did Avogadro figure out that there are 6.02 x 10²³ atoms in 55.845 grams of Iron?

Professor:  Well, actually, he didn’t.  The only thing that Avogadro contributed to this picture is that he believed that equal volumes of gas at the same temperature and pressure contained equal numbers of molecules.  We didn’t actually get a number on how many molecules that was until long after he was a goner.

The first time anyone ever actually got an actual number on Avogadro’s number was when scientists started studying aerodynamics.  There’s an equation for the viscosity of a gas flowing through a nozzle that has Avogadro’s number in it.  Fortunately, the scientists knew or were able to measure all the other variables in the equation and could solve for Avogadro’s number.  Later, scientist figured out other formulas that had Avogadro’s number in it in a way that we could solve for.  Go ask your chemistry teacher if he knows how chemists first determined Avogadro’s number.  I’ll bet you 10 bucks they won’t know.  It’s seldom discussed in chemistry books.  If he says anything about the “gas law constant” or “Boltzmann’s constant” (the first of which we’ll get to later) he or she is wrong, because the gas law constant and Boltzmann’s constant were figured out once we knew Avogadro’s number.

Student:    Well, I think I’m all set to go to tackle some more problems.  Thanks a bunch!

Professor:  That’s why I’m here!

 

 

 Links of interest:

www.chem1.com/acad/webtext/virtualtextbook.html -- This website is so good, I almost considered not putting a link to it on here because I'm giving away my audience.  A very comprehensive site on chemistry, far superior to any textbook I've seen on general chemistry.

 

tannerm.com -- rather good website about inorganic chemistry, with quite a bit of detail on a number of difficult topics.