This course will cover the topics of a full year, two semester General Chemistry course. We will use a free on-line textbook, Concept Development Studies in Chemistry, available via Rice’s Connexions project. The fundamental concepts in the course will be introduced via the Concept Development Approach developed at Rice University. In this approach, we will develop the concepts you need to know from experimental observations and scientific reasoning rather than simply telling you the concepts and then asking you to simply memorize or apply them. So why use this approach? One reason is that most of us are inductive learners, meaning that we like to make specific observations and then generalize from there. Many of the most significant concepts in Chemistry are counter-intuitive. When we see where those concepts come from, we can more readily accept them, explain them, and apply them. A second reason is that scientific reasoning in general and Chemistry reasoning in particular are inductive processes. This Concept Development approach illustrates those reasoning processes. A third reason is that this is simply more interesting! The structure and reactions of matter are fascinating puzzles to be solved by observation and reasoning. It is more fun intellectually when we can solve those puzzles together, rather than simply have the answers to the riddles revealed at the outset. Recommended Background: The class can be taken by someone with no prior experience in chemistry. However, some prior familiarity with the basics of chemistry is desirable as we will cover some elements only briefly. For example, a prior high school chemistry class would be helpful. Suggested Readings: Readings will be assigned from the on-line textbook “Concept Development Studies in Chemistry”, available via Rice’s Connexions project. In addition, we will suggest readings from any of the standard textbooks in General Chemistry. A particularly good free on-line resource is Dickerson, Gray, and Haight, "Chemical Principles, 3rd Edition". Links to these two texts will be available in the Introduction module.
CDS 2 Atomic Masses and Molecular Formulas II
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En provenance du cours de Rice University
General Chemistry: Concept Development and Application
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À partir de la leçon
Atomic Molecular Theory and Atomic Masses
Chemistry can be understood fundamentally as the study of atoms and molecules. In this module, we will examine the experiments which reveal that all matter is composed of atoms which combine to form molecules. The clever analysis of these experiments illustrates scientific reasoning at its finest, allowing us to understand the existence and properties of particles which could not be directly observed. In addition, by measuring the relative masses of the different types of atoms, we can begin to predict the ratios of masses of reactants and products during a chemical reaction.
Rencontrer les enseignants
John Steven HutchinsonDean of Undergraduates and Professor of Chemistry
In the previous lecture, we began to develop a means by which we could
determine the relative masses of various atoms.
And from those to be able to determine what the formulas are, for various
molecules. This is material from the second concept
development study. And this is the second lecture in that
series. Let's take a quick review of what we found
in that lecture. We determined that using Avogadro's Law
for elements and compounds which were gases, we could measure the relative
volumes of gases and use those to determine the relative numbers of
particles. And corresponding then, we could
essentially count the numbers of atoms of each type that were in a particular
molecule. And once we knew the relative numbers of
atoms by using the mass proportions measured for that compound, we could
figure out what the relative mass is of the individual atoms were to one another.
And figure out a relative atomic mass scale.
And that worked really well for nitrogen and oxygen and hydrogen and chlorine,
which were the elements we were looking at before.
But notice that all of those elements are gaseous in their form and in addition all
the compounds we studied were gases as well.
What if instead we were looking at non-gaseous elements?
And most of the elements are non-gaseous, very few of them are gases.
How could we, for example, go about the business of counting carbon atoms when we
know that carbon is not a gas, it's a solid.
And therefore we cannot use the law of combining volumes and Avogadro's
hypothesis to count carbon atoms and to figure out what the relative number of
carbon atoms would be. Here is the approach we're actually going
to use. We're going to do is examine two compounds
which are carbon oxides. Combinations of carbon and oxygen
together. And here we have the relative mass
proportions for oxide A and for oxide B. And we don't know what the molecular
formulas of those are, so we're just going to call them A and B for now.
The first thing we'll note when we look at these data is, in fact, the masses of
oxygen when we fix the mass of carbon. That's in a simple 2 to 1 ratio.
That's exactly the kind of thing we looked at before when we saw the law of multiple
proportions. So one of the things we can conclude
immediately from these data is oxygen is in fact an atom, that oxygen is atomic in
nature. The question is, can we use Avogadro
hypothesis or law to determine the molecular formula of compounds a and b?
And the answer of course is we are not going to be able to measure the volumes
and carbon and compare them because carbon is not a gas.
But what we can do alternatively, is to find out how much of oxide A can we make
from one liter of O2, and how much of oxide B can we make from one liter of O2.
And here are the data. What we see is that a single liter of
oxygen when we burn carbon with that single liter of oxygen we will produce 2
liters of oxide A. Now remember from Avogadro's hypothesis
since O2 and A are both gases then 2 liters of oxide A contain twice as many
particles as 1 liter of oxygen does, because each liter contains the same
number of particles. So if we compare these two, what we can
say is that 1 O2 molecule produces 2 molecules of oxide A, because since the
ratio of the volumes is double, the ratio of the number of particles must be double.
But if we know that that's true, and we have a single O2 making two molecules of
the product, then each molecule of the product must contain 1 oxygen atom because
the 2 oxygen atoms in the O2 must split up.
And form each separately in Oxide A molecule.
So now we do know something about Oxide A. We know that each Oxide A contains exactly
one oxygen atom. We ought to be able to do the same
analysis with Oxide B, and we can. So if we'll look carefully here, we'll see
1 liter produces 1 liter. That means a single molecule of O2 makes a
single molecule of B, therefore each B molecule must have both oxygen atoms in
it, and I can conclude that, that an oxide B molecule has two oxygen atoms in it.
So we've been able to determine now the relative numbers of oxygen atoms.
Notice that this is actually kind of reassuring, because the mass ratio of the
oxygen is also 2 to 1. But we have no idea how much carbon this
represent. Whether it's 1 atom, 2 atoms, 50 atoms.
We really don't have any way to know. So we need some other way to proceed.
And here's how we can do it. How could we determine the mass of a
single oxide A gas molecule? Well one way we could do that, is actually
to say take 1 liter of oxide A. So here is a container that is 1 liter and
its going to contain oxide A. And here is another container exactly the
same volume, 1 liter and it's going to contain O2.
And we can weigh them both and just find out how much is the mass of one relative
to the mass of the other. Turns out to be relatively easy to do.
Here's what the data say. There's a ratio of 0.875 between those
masses. What that might mean for example then, is
if the mass of the oxygen flask is 1,000 grams, then the mass of the flask
containing A contains 875 grams, for example.
But here's the key. There is the same number of molecules in
these two flasks. Same number of molecules, because they
have equal volumes. And since from Avogadro's hypothesis,
equal volumes contain equal numbers of particles.
Then this mass of oxygen molecules and this mass of A molecules contain exactly
the same number of molecules. What that mean is, each individual
molecule of A is in the same ratio as to oxygen as the overall ratio is.
So for example then, what we can figure out is that the single molecule of oxide A
is only 0.875 times as great of a single molecule of O2.
And a single molecule of O2 we've previously determined or set is 32.
So what does that mean? It means that the mass of 1 A molecule is
equal to 0.875 times the mass of 1 O2 molecule.
And if that's true, and the mass of a single O2 molecule is 32.
Then 32 times 0.875 is in fact 28. So, we've actually figured out what the
mass of a single, single molecule of A is. A mass of a single A molecule is 28.
But remember on the previous couple of slides, we figured out that each A
molecule contains a single oxygen atom. And that oxygen atom, of course, has mass
16. Well, if the total mass of the molecule is
28, and 16 of that is oxygen, the, the rest of it must be carbon.
So what we can conclude is that the mass of a single carbon atom, I'm sorry of the,
a mass of carbon in a single molecule oxide A is 12.
Because 28 minus 16 is 12. Now we don't know how many carbon atoms
that is right now. But we do know that the mass is 12.
How can we make some more progress? Let's repeat this whole analysis with B
instead. In this particular case, again, we're
going to take a liter of oxide B and a liter of oxygen and then we're going to
get the mass ratio between those two, 1.375.
And since each liter of O 2 and each liter of B containing exactly the same number of
particles, then that ratio of the masses of the liters is the same as the ratio of
the mass of the particles. Correspondingly then we can also say that
the mass of 1 B molecule is equal to 1.375 times the mass of 1 O 2.
And that mass of O 2 is of course 32, if we multiply those together we'd get 44.
So we can conclude then that the mass of one molecule of oxide B is 44.
Now remember, we have already figured out that each oxide B molecule has 2 oxygen,
and each oxygen is 16, so the total mass of oxygen in the oxide B molecule is 32.
That leaves 12 overall. So the mass of carbon in a single oxide B
is 12, which the same as the mass of carbon in a oxide A.
That certainly seems to suggest that a carbon atom weights 12, but we don't know
that. It could well be that both A and B have 2
carbon atoms. In which case the mass of the carbon is 6.
Or maybe they have 3. In which case, the car, mass of a carbon
atom is 4. Maybe there's 5.
In which case, the mass of a carbon atom is 2.4.
There's no way to know. So what could we do?
One of the things that we could do is simply do lots and lots of experiments
measuring the massive carbon in molecules that contain carbon.
And here is what we discovered. In every case we get 12 or multiple of 12,
24, 36, 48 but we never get a fraction of 12.
We never get 6 or 4 or 3 or 2.4 or 1 we never see any of those kinds of numbers.
What that means is that the minimum mass of carbon that can be in a single molecule
is 12. And the only reasonable conclusion to draw
from that is that the mass of carbon is 12 in any molecule and therefore the mass of
the carbon atom is 12. Using the same kind of procedure then, we
can actually figure out what all atomic masses are at least relative to each other
on a scale where carbon is 12 Oxygen is 16, nitrogen is 14, hydrogen is 1 chlorine
is 35.45. And we can actually figure out the
relative masses of all of the elements or the atoms of all of the elements.
Those relative masses are measured in a particular unit of mass, we call it
naturally enough atomic mass units because the these are the units in which we
measure atomic masses. Now where do we use that, well we're going
to develop that more extensively in the next lecture but as a preview what we're
going to look at is something called the empirical formula of a compound.
The empirical formula simply gives me the ratio off the number of atoms of one type
to the number of atoms of another type, or to all other types.
It's simply the ratio of the atoms that combine to form a molecule.
And how can we figure that out? Well here's an example.
It's actually an example that we studied in an earlier lecture, which is lead
sulfide. And if we take a certain amount of lead
sulfide, we knew from the law of definite proportions that out of a 100 grams, 86.6
is lead and 13.4 grams is sulfur. From the work we just concluded little
while ago we know that we can measure the mass of every atom.
And so in fact we could also measure that the mass of the lead atom is 207.2 And the
mass of a sulfur atom is 32.06. But what can we do with that information?
One of the things we can do with that information is to say that if I compare in
lead sulfide, the mass of lead which is 86.6 divided by the mass of sulfur, which
is 13.4, and take that ratio, that's easy to do.
It turns out that's exactly equal to the mass of 1 lead atom in ratio to the mass
of 1 sulfur atom. You could actually work that out for
yourself. You could simply use your calculator, take
that ratio, and take that ratio, and you'll discover they're exactly the same
ratio. What does that mean?
It means that we must have exactly the same number of atoms and 86.6 grams of
lead that we have in 13.4 grams of sulfur. And therefore the ratio of the number of
atoms of lead to the number of atoms of sulfur in a compound of lead sulfide is 1
to 1. And therefore we've actually figured out
what we will call the empirical formula of lead sulfide.
We're going to develop a more systematic means of determining an empirical formula
based upon atomic masses in the next lecture.
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