We're going to pick up where we left off in the last lecture.

You'll recall what we were doing was developing the atomic molecular theory by

examining mass measurements, that will lead us to a conclusion that all material

is made of atoms and molecules. This is the material from the first

concept development study, which we did not complete in the first lecture, but

which we will complete in this one. If you haven't read it yet, it will be

worth your while to familiarize yourself with the material from that reading before

watching the rest of this lecture. In the first lecture, we actually

developed something based upon mass measurements called the Law of Definite

Proportions. What we did was to decompose individual

compounds into their constituent elements, weighed amounts of the, relevant, relevant

amounts of each of the elements by mass. Take the ratios of those, and we

discovered that those ratios are fixed from sample to sample for any given

compound. We can't vary those ratios by mixing

together different amounts of those compounds.

Instead we have to, take them just in the proportions which we're allowed by the Law

of Definite Proportions essentially. We're going to examine that a little but

further now, because we haven't really developed the atomic molecular theory out

of that, by looking at a few other examples of the Law of Definite

Proportions. Here we have a few of them.

Here are some hydrocarbons, meaning that these are compounds contained entirely or,

composed entirely of the elements carbon and hydrogen.

The simplest of those hydrogen carbons is called Methane, and just like with the Law

of, Definite Proportions, we can decompose each of these compouds into their Carbon

and Hydrogen components, take the masses of each and determine the mass

composition. So, for example notice here that Methane

is 74.9% Carbon and it is 25.1% oxygen. And what's also interesting here is that

we noticed that are couple of other compounds we have listed here, both Ethane

and Benzene. Turns out there were enormous number of

compounds composed just of carbon and hydrogen.

Carbon and hydrogen combined in an enormous number in different ways, each

one of which will have its own mass percentages.

That does not violate the Law of Different Proportions though, even though the Law of

Definite Proportion said, in a particular compound, the elements combine in a

different ratio. These are different compounds, they've

distinctively different chemical and physical properties, so we can

differentiate them. There is nothing in the Law of Definite

Proportions that says we couldn't have more than one set of proportions, and that

those different proportions could lead to different compounds.

So, in fact, we're going to look at multiple proportions as we analyze these

data. And here we've got, as it illustrates for

example, different ratios for Ethane than we had for, for Benzine, and likewise, we

can look at different ratios for Benzine and for Methane.

Staring at those numbers, there's no obvious pattern in those numbers at all,

nothing that jumps out at us that says okay, I can understand why I have those

particular ratios. So what we need to do is actually figure

out is there some way that we can analyze these data, make it easier to understand,

rather than just looking at percent compositions by mass.

One way to do that is to say instead of taking 100%, let's take 100 grams.

So instead you'll notice here we have taken a column of data, in which every

element has 100, or I'm sorry, every compound has 100 grams of that compound.

And then by taking the percentages and multiplying by 100 grams we can wind up

with the masses of carbon and hydrogen. Well the numbers don't really change

right? All we've done is to look at different

masses rather than to look at different compositions.

Again, there's really not anything obvious here that we could get out of these data,

that jump at us and say there's some kind of a pattern.

But one way we could do this, is to try to now look more carefully at the ratios,

somewhat more simply in each of these compounds.

So for example, what if we were to take n Methane, not 100 grams of Methane, but

rather, imagine instead what we did was to take 1 gram of hydrogen and ask how much

carbon will be associated with that 1 gram of hydrogen.

We're going to actually do that calculation essentially all we need to do

is divide the mass of hydrogen by 25.1, to get 1.

An so to keep the ratio the same we would have to take 74.9 grams carbon divided by

25.1, and we would wind up with a number that turns out actually to be equal to

2.98 grams for one carbon. For the carbon, when there's a mass of

hydrogen which is 1. So let's make a new table now.

In the new table you notice, this time, we're going to take the masses of the

hydrogens to all be 1, rather than the total mass to be equal to 100.

And we can do calculations just like the ones that we said a little while ago,

except now what we're going to do with these calculations, where we'll take the

previous masses and divide by the mass of the hydrogen.

And in this case we get 2.98 grams of carbon, just like we showed before.

How about for the, Ethane? Let's go back and look at the Ethane then.

And in Ethane we will then take for 1 gram of the Hydrogen, we need to take 79.9

grams divided by 20.1 and that turns out to then be equal to, if we get our

calculators out, 3.97 grams. So let's put that into our table back here

as well, 3.97 grams of each. Notice that the masses of the compounds

are varying all over the place now, right? This 3.98 grams of, of, for the total mass

of Methane, and this now 4.97 grams for the total mass of the Ethane, so we're no

longer holding the total masses of the compounds comparable.

Let's do the Benzene as well. Go back and look at the numbers here, for

Benzene. Let's see, again we want 1 gram of the

Hydrogen. So for the benz-, or for the carbon, we

need 92.3 grams divided by 7.7, and if we do that math we wind up with 11.98 grams.

So let's fill that into our table here as well.

We have 11.98 grams, and back over in the first column then this going to be 12.98.

And those numbers don't particularly add anything to our intuition from the total

mass, but if we look at the masses of the carbon we discover something rather

interesting here, right. Those numbers are not quite so random any

more, in fact, you can see a pattern. Basically, that first number there looks

pretty much looks like 3, right? And the second number looks like 4, and

the third number looks like 12. We have, in fact, discovered a simple

integer ratio of the masses of the carbons.

That's a particularly interesting outcome, right?

Because ordinarily, we wouldn't expect to see some kind of a simple ratio.

We didn't see simple ratios when we were examining things for the masses of the,

when we had 100 grams. Let's try again than, and consider, what

if, instead of fixing the mass of the hydrogen, we fixed the mass of the carbon

at say, 1 gram, we could certainly do exactly the same kind of work.

To do that, we need to go back and ask, okay, what were the mass ratios?

So if instead for the methane, I'm going to take the mass of the carbon to be 1

gram, then for the hydrogen, I need to take 25.1 grams divided by 74.9, and if i

do that math, it turns out to be 0.335. So we'll go back now and look in our new

table, and we will enter a 0.335 grams for the mass of the hydrogen, which will

combine with the mass of carbon to make methane.

And the total mass over here clears 1.335, not a particularly noteworthy number.

If we do the exact same calculations now for ethane and benzene, since we're

starting to get the pattern here, we have 0.251 and 1.251 as the masses.

And for benzine then we have 0.0, I'm sorry, 0.083 and 1.083.

And again now, if we look at these numbers, these numbers are not quite

conspicuous as the numbers we're for our previous analysis, right.

However, if we were to divide each of these masses of hydrogens, by the mass of

the lowest hydrogen, we discover a really simple ratio here, in fact.

It turns out that the ratio here is, if this is 1, then this one is in fact 3, and

this one is in fact 4. So again, we have simple integer ratio, of

the masses of the hydrogens. What we have in fact discovered is a

rather interesting pattern here. This interesting pattern is something that

we actually refer to now as the Law of Multiple Proportions.

The Law of Multiple Proportions actually is a somewhat complicated thing, because

you notice we had to do a lot of analysis here to figure out, you know, how we would

see these simple integer ratios. But here we go.

Notice what we did, was we took two different compounds, or in this case even

three compounds, of the same two elements carbon and hydrogens, and we mixed the,

fixed the mass of one of the elements. Either carbon or hydrogen at 1 gram and

then we calculated the mass of the second element using the Law of Definite

Proportions. Once we had that mass of the second

element, we compared those masses of the second elements, between the two

compounds. And it's really important that we look at

the mass from one compound next to the next compound.

And when we do that, we discover that we get these simple integer ratios.

Now that's an interesting conclusion, but before we really try to do too much from

that, we should attempt to verify whether or not this happens in other compounds.

So let's actually look at a set of nitrogen oxides for example.

So here we've got three different compounds formed from nitrogen and oxygen.

If we take a hundred grams of each one of them, then we can use in the Law of

Definite Proportions figure out, do the mass analysis.

How much nitrogen and how much oxygen are in each of these compounds.

And here are the numbers, and if we look at these numbers just sort of raw, eh,

there's not very much interesting here, right?

If I look at this set of numbers, I don't see a pattern.

If I look at this set of numbers, I don't see a pattern there either.

But remember in the Law of Multiple Proportions, what we really need to do, is

we need to fix the mass of one of the elements.

So, in fact, what we're going to do is we're going to fix the mass of the

nitrogen, and then compare it to the masses of the oxygen.

It's the same analysis we did a little while ago, so we won't go through the

details of it. But here's what we wind up with, in each

case now, I have the mass of nitrogen is 1, and the mass of oxygen are these set of

numbers which are back here, right. And if I look at those numbers, then in

fact I can do a simple calculation, and I can compare the masses of the oxygen, of

the oxygens, or the mass of the oxygen component, in the different compounds.

And notice they are in the ratio 2.28 to 1.14 to 0.57.

And if your eyes are good, then you'll actually notice that 1.14 is double 1.-,

0.57, and 2.28 is double 1.14. So this is, in fact, a simple integer

ratio of 4 to 2 to 1. Once again, we see simple integer ratios.

Well what is the significance of these simple integer ratios, why do we care?

The answer has to do with asking the question, what are the integers.

What is so particularly significant when we observe integers, rather than just any

other kind of number? So for example, if I measure my own

height, my own height can't really be measured in terms of integers.

I might say I'm 5 foot 10 inches, or that, that would be a total of then, what, 70

inches tall, but I'm not precisely 70 inches tall, you know, there's some

fraction in there. Likewise, whatever my mass is, it's not

some integer. And in fact, the integers that we, the

masses that we are observing in the compounds themselves are not integers.

And again, if I go back and look at these particular numbers, the number themselves

are not integers, right. It's the ratios that are integers.

So, what does that tell me? It tells me in fact that when I take a

mass of oxygen to combine with the specific amount of nitrogen, I can only do

so in a multiple of a fixed mass unit. So the integers then are used for

counting, we always count things with integers.

And what we must be doing in this particular case, these integers, the ones

that we are observing, here the 4 to 2 to 1 ratio that we observe.

Those integers themselves must be counting, something.

What are they counting? What they are counting, is in fact a

multiple of some fixed unit of mass. These are multiples, integer multiples, of

a fixed unit of mass. And apparently, oxygen can only be given

to us when we react it with a specific amount of nitrogen, in one of those fixed

units of mass. Well quite frankly, as soon as I've used

the word fixed unit of mass, I've discovered that mass is particulate.

That it only comes in little particles of mass and that the amount of mass of oxygen

it can have, can only be some multiple of that amount.

We'll call that fixed unit of mass an atom, and in this particular case, an

oxygen atom. What that means is that the Law of

Multiple Proportions, which we just stated a little while ago, is fundamentally the

proof that we need to know that there are, there, is absolutely the case that mass

comes in integers, or in integral little particle units.

And we can only take integer amounts of those units when we combine them together.

Now we could repeat this whole analysis, fix the mass of oxygen at one, calculate

the mass of nitrogen, and we will discover that nitrogen must also come in those

fixed integer units. What that means is that we've actually now

developed, the Atomic Molecular Theory. And here are those principles, and we had

stated them in a previous lecture. That each element is composed of these

little identical particles of atoms. And that these atoms, themselves have a

same characteristic mass such that no matter what I do, when I react these

materials together, there have to be a simple integer ratio of those masses,

because I can only take integer amounts of particles.

I can't take parts of particles, I can't divide these particles up.

Notice, that of course, according to the Law of Conservation of Mass, the, the

numbers and masses of these cannot change during a chemical reaction.

And furthermore, when I combine these atoms, I can only combine them in very

specific ratios. That means in fact, that I am only going

to be able to do combinations in simple whole number ratios and that that's what's

going to give rise to our molecules. So these data now establish clearly the

existence of atoms and molecules, but they leave behind some really important

questions. Particularly, if atoms combine in simple

integer ratios to form molecules, what are those integer ratios?

What, in fact, are these integers? We don't know, the data don't tell us.

Because we don't have the means to count them.

Let's go back and look at these data here, and say okay, it looks like the oxygen's

are in ratio 4 to 2 to 1. Does that mean, that for example, compound

A has twice as many oxygen's as compound B?

And the answer is no. It may or may not tell us that, because

what we have done here is to fix a mass of nitrogen.

We have fixed a mass of nitrogen, but we have not in fact fixed how many Nitrogen

atoms there are in any particular molecule, we don't know the answer to

that. We only know in ratio 2, a fixed number of

nitrogens. There is a ratio of the oxygens, which is

4 to 2 to 1. What that tells us then, when we go back

here, is we do not know those integer ratios.

We've determined that there are molecules, but we don't have a means to analyze those

molecules and figure out how many of each type there are.

What we fundamentally then need, is some way to count atoms.

I need to be able to count the numbers of nitrogens, and count the numbers of

oxygens and then take the ratio between those two.

Because if then I could do that, I could know what the molecular formula were, the

ratios of atoms in each of compound A, compound B and compound C of those

nitrogen oxides. So in the next lecture, which will begin

concept development study 2, we will figure out how we actually can count

atoms, and we'll use a couple of different techniques to do that.