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 IV
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General Chemistry: Concept Development and Application
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From the lesson
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.
Meet the Instructors
John Steven HutchinsonDean of Undergraduates and Professor of Chemistry
We're going to begin today with a so-called review of the concepts that we
developed in the previous lectures. Remember, our fundamental goal here is to
attempt to count numbers of particles, at least in relationship to one another,
because when we can count those particles, we could do things like determine the
formulas of individual molecules or do a chemical reaction calculations.
We figured out a rather systematic and clever way to count the numbers of
particles by counting the number of moles of particular particles.
Where a mole, you'll recall is a specific number, and it is the number of atoms in
12 grams of carbon. So, it's just some specific number.
And every time you hear us talk about moles, what you want to think is, that's a
specific number of particles just like a dozen, is a specific number of particles,
or a gross is a specific number of particles.
And if I know that I have one mole of carbon and I have one mole of helium, that
I know I have exactly the same number of atoms of carbon than I have of helium.
Furthermore, we can figure out the mass of a mole of any particles.
We know what the mass of a mole of carbon is.
By definition, it's 12 grams. But since the ratio of the mass of a
carbon atom to a helium atom is 12 to 4 , than the mass of a mole of carbon atoms to
a mole of helium atoms is 12 to 4. And therefore, a helium mole weighs 4
grams. So, it's straightforward to then figure
out what the mass of a single mole is and the mass of a single mole then is
something that we will call the molar mass and it is equal in grams to the mass of a
single atom or a single molecule weighed in the original atomic mass units.
Once we know the mass of one mole, then it's a relatively straightforward thing
for us to figure out how many moles we have from this ratio here.
We'll just take the ratio of the mass of the sample that we have weighed to the
mass of one mole and that ratio, is simply going to then be the number of moles.
And remember, the number of moles is a count of particles.
So, by weighing the mass of the particles, we've actually figured out how many
particles there are. This formula here, this simple ratio, is
an equation that we will use often. And why do we calculate the number of
moles? Because it's a way of counting the number
of particles. And counting particles is important to us,
as we saw in the previous lecture, to determine molecular formula.
As we will see in this lecture, to do calculations about chemical reactions.
And as we'll see in the next lecture, is a means of determining the concentrations of
solutions. Let's illustrate this with a particular
example. Let's say, we have substance, acetylene.
And we don't actually know what acetylene is at this point, other than it contains
hydrogen and carbon. We'll burn it in oxygen and when we burn
it, we're going to produce the products, carbon dioxide and water.
And we'll weigh the mass of those products that come out of burning, 10 grams of
acetylene. Well, how are we going to use these data?
Maybe the best way to do this is to attempt to count the atoms of carbon and
hydrogen. And we can do that because we can actually
count the number of particles of carbon dioxide by counting the number of moles of
carbon dioxide. And how do we do that?
We do it by measuring the mass of the carbon dioxide in our sample and dividing
by the molar mass of the carbon dioxide which is the mass of a single mole of CO2.
How do we work that out? Remember, from what we just said, we will
take the mass of a single particle of carbon dioxide, which is 12 for the carbon
plus 16 for each oxygen, so that totals up to 44.
And then, that is grams per mole because the mass of a single mole is the mass in
grams of a single particle in AMUs. Well, let's see then.
The mass of CO2 was determined to be 33.8 grams, the molar mass is 44.0 grams per
mole, and if we just plug that into the calculator to find out how many moles we
have, we'll discover that we have 0.76 moles of carbon dioxide.
Now notice, we weren't actually trying to determine the number of moles of carbon
dioxide. What we wanted to know was the number of
moles of carbon because we're trying to figure out how many carbon atoms there are
in acetylene. But, in each mole of carbon dioxide, since
each molecule of carbon dioxide contains a single carbon, then the number of carbon
atoms is simply equal to the number of moles of carbon dioxide.
The number of moles of carbon atoms that are in the reactants must be equal to the
number of carbon atoms in the products, and the number of carbon atoms in the
products is simply equal to the number of moles of carbon dioxide.
So, we have actually counted the number of atoms of carbon that we started off with
in the acetylene. We can do the same trick now with the
water. The number of moles of water is equal to
the mass of the water divided by the molar mass of the water.
The molar mass of the water is easy to find because the oxygen is 16 and each
hydrogen is 1, so it's 18.0 grams per mole.
And if we do the calculation, let's see the mass of the water according to the
problem here is 6.9 grams and the molar mass of the water is 18.0 grams per mole.
And if we just do the calculator math there and plug that in, we wind up with
0.38 moles of water. Again, that wasn't what we wanted.
We wanted to know the number of moles of hydrogen, but it's straightforward to
figure out the number of hydrogen, moles of hydrogen once we know the number of
moles of water, because every molecule of water contains 2 hydrogens.
So, we can then figure out that the number of moles of hydrogen must be equal to
double the number of moles of water because every water molecule has 2
hydrogen atoms in it. So, the number of moles of hydrogen is
then equal to 0.76 mole. And we notice that the number of moles of
carbon and the number of moles of hydrogen are exactly the same, and as a
consequence, we can say that the molar ratio of carbon to hydrogen is equal to 1
to 1, which means that the atomic ratio of carbon to hydrogen is 1 to 1.
As a consequence, we can answer the question, and say the empirical formula of
acetylene is, we can say it this way, C1H1 or just CH.
And it turns out acetylene ultimately is C2H2, so the ratio is, in fact, 1 to 1
just as we have calculated. We can actually push this a little bit
further and understand even more about chemical reactions by considering what's
called a balanced chemical equation. Remember that according to the atomic
molecular theory, the atoms in chemical reaction are always conserved.
We have exactly the same atoms that we started off with that we started off with
when we wind up the chemical reaction. All of the atoms that were present at the
outset, must be present in the products. And there can be no atoms in the products
that were not present in the reactants. So, what that tells us is we can just
count the number of atoms in the reactants and that number of atoms must also be
present in the products. Let's illustrate that with a simple
example. We're going to burn propane.
Using the kinds of calculations we've described here, propane can be determined
to be C3H8. When we burn something, we're reacting it
in oxygen and we wind up with our products being carbon dioxide and water, as shown
by the equation here on the screen. But there is something wrong with the
equation on the screen, as you can immediately tell.
Because if I count the number of carbon atoms, which are in the reactants, there
are 3. If I count the number of carbon atoms
which are in the products, there is only 1.
So, we can't have carbon atoms disappearing so we're going to have to do
something to adjust this reaction. One way that we could do that is to simply
say maybe, we should have 3 carbon dioxides.
That's supposed to be a 3. If that's the case, then I would have
simply 3 carbons on both sides of the equation.
And now the carbons are accounted for, but how about the hydrogens?
We started off with 8 hydrogens here and we wind up with only 2 hydrogens.
One way to simply solve that would be, if we were to put a 4 in front of the water,
and now we would have 8 hydrogens on both sides of the equation and 3 carbons on
both sides of the equation. The only thing left to look at is the
oxygens. Right now, we've got 6 oxgens from the
carbon dioxide and 4 oxygens from the water for a total of 10 oxygens in the
product, but there's only 2 oxygens in the reactants.
Well, we can fix that easily enough by simply putting a 5 in front of the O2.
The result, written rather more neatly over here on the page, is that a single
propane reacts with 5 oxygen atoms to produce 3 carbon dioxides plus 4 H20
molecules and if we look carefully at the, all of the 3 types of atoms, carbon,
hydrogen, and oxygen have the same number of atoms of each type on both sides of the
equations. We call this a balanced chemical equation.
Let's do another example here. What if we took copper metal and dissolved
it in nitric acid? The products of that reaction turn out to
be copper nitrate, CuNO32, nitrogen, nitrogen dioxide and water.
If we write that down symbolically, then what we've got here is, of course, the
copper with the nitric acid reacting to form the copper nitrate, nitrogen dioxide,
and water. So, that expresses a list of the reactants
in the products, but it's clearly not an equation.
I can look at the copper and immediately see that there's one copper on each side
of the equation but as soon as I look at, say, the nitrogens, there's only one
nitrogen listed there in the reactants and there are clearly 3 nitrogens listed in
the products. And if I look at the hydrogens, there's
only 1 hydrogen in the reactants, but 2 in the products, and the oxygens were all
over the place. We have 6 oxygens from the copper nitrate,
2 from the nitrogen dioxide, and 1 from the water, so that's 9.
But there are only 3 that we started off with in the original reactants.
In order to rectify this, we actually need to do as we did in the previous reaction
and what we'll actually see is that we can balance this equation by putting actually
a 4 in front the nitric acid. We don't need to put anything in front of
the cooper nitrate because the coppers are already balanced.
But now, let's see, I'm going to need a 2 in front of the nitrogen dioxide so that
there are 4 nitrogens on each side of the equation, and I will have a 2 in front of
the H2O so that there are 4 hydrogens on each side of the equation.
Slide that over. Let's see that equation on the next page
here. That's the balanced equation that we just
worked out and if you take a look at this, you'll easily see that there's 1 copper, 4
nitrogens, 12 oxygens, and 4 hydrogens on each side of the reaction.
We can actually work with this very systematically with some fairly
interesting , calculations which can be done.
Let's imagine we started off with a piece of copper metal which was 10 grams and
ask, what mass can be produce from the product?
So, we are trying to relate the amount of the copper we started off with to the
amount of the nitrogen dioxide we wind up with.
How should we proceed? Well, if I could determine the number of
moles of copper, then I would know the number of moles of carbon dioxide, because
note I'm sorry, of nitrogen dioxide. Because you'll notice that there's a 2 to
1 ratio for every 1 copper, there are 2 nitrogen dioxides produced.
As a consequence, one mole of copper will produce 2 moles of nitrogen dioxide.
That means, I should figure out how many moles of copper did we actually start off
with. And let's see, we've got 10 grams of
copper. And if we divide that by the molar mass of
copper, which is just equal to the mass in grams of the mass of a single atom of
copper in atomic mass units, which is, from the periodic table, 63.55 grams per
mole. And if I just plug that into the
calculator, I'll discover that is 0.76 moles, I'm sorry, that's the wrong number,
it is equal to 0.157 moles of copper that we started off with.
But if there's one mole of copper, we would produce 2 moles of nitrogen dioxide
so the number of moles of nitrogen dioxide produced is always num 2 times the number
of moles of copper that we started off with, and therefore, it must be the case
that we have produced 0.314 moles of nitrogen dioxide.
And if I wanted to know what mass of the nitrogen dioxide that was equal to, then I
could go back to the formula that we reviewed earlier in this lecture, which is
that the mass of the sample divided by the molar mass of the substance is equal to
the number of moles. Notice we know this number.
It's 0.314. We've determined that.
We know what the molar mass of nitrogen dioxide is, because each nitrogen is 14
and each oxygen is 16, so that's 44 grams per mole.
Notice that we know this number and we know this number, then it's fairly
straightforward to calculate that the mass of the nitrogen dioxide is the number of
moles of the nitrogen dioxide multiplied by the molar mass of the nitrogen dioxide.
And if I multiply those numbers together, I determine that the mass of the nitrogen
dioxide I've produced is 14.5 grams. So, this is an illustration of what is
called chemist, chemical stoichiometry. Stoichiometry is kind of a complicated
word here, but I defined it for you to simply mean chemical algebra.
It's basically just doing algebra involving the numbers of atoms of each
type and their associated masses used in a balanced chemical equation.
Now, there's something else interesting that can show up here.
Notice that all we calculated was the mass of the copper that we started off with
assuming essentially that we have plenty of nitric acid around to dissolve that
amount of copper. But we don't always have that.
We might actually have limited amounts of different materials.
So, if I take a look at that particular reaction and say, what if we don't have an
unlimited amount of nitric acid, would we actually produce as much nitrogen dioxide
as we have just predicted from the previous calculation?
Let's do an example. Let's imagine that we have 0.20 moles of
nitric acid and let's ask the question, how much copper can I dissolve in 0.2
moles of nitric acid? Notice that the ratio of the number of
moles of copper to the number of moles of nitric acid, in the chemical equation, is
1 to 4. Here they are, 1 to 4.
So, what that means is I need to have 4 times as many molecules of nitric acid as
I have atoms of copper or, stated differently, I need to have 4 times the
number of moles of nitric acid that I have of the number of moles of copper.
Well, let's see, what we had at the outset you'll recall, is that the number of moles
of copper is 0.157 so the number of moles of nitric acid, which are needed in order
to react with 0 I'm sorry, 10 grams of copper, must be equal to 4 times 0.157
moles which is equal to 0.628 moles. That's how many moles of, of nitric acid
are needed in order for me to dissolve that mass of copper.
Do I have that? The answer is no.
Because if we go back over here and take a look, we only gave 0.2 moles of nitric
acid. Consequently, there's not enough nitric
acid to dissolve all of the copper. The result is that the production of the
product nitrogen dioxide is now limited by the nitric acid.
We now call the nitric acid the limiting reactant because it limits the amount of
product that we can produce. It's the thing that we will run out of
before we run out of anything else. So, if I now ask the question, how many
moles of nitrogen dioxide can I produce, it's determined by the number of moles of
nitric acid, not the number of moles of copper because I've got plenty of copper,
but I'm going to run out of nitric acid first.
Notice that the ratio between the number of molels between nitric acid that we
start off with, and the number of moles of nitrogen dioxide we produced is in ratio 4
to 2, which is also the ratio 2 to 1. Therefore, for every 2 moles of nitric
acid I start off with, I produce 1 mole of the nitrogen dioxide.
As a consequence then, since the number of moles of nitric oxide sorry, of nitrogen,
nitric acid that we started off with was given to be 0.20 moles, then the number of
moles of nitrogen dioxide that we can produce is simply equal to 0.10 moles,
much smaller than what we were able to produce before.
And now, if we ask what is the mass of that number of moles of nitrogen dioxide,
it's simply equal to the number of moles of the nitrogen dioxide multiplied by the
mass of a single mole of nitrogen dioxide. And again, since the mass of a single mole
of nitrogen dioxide is 46 grams per mole, I have an error up here, 46 grams per
mole. Then, if I multiply those two numbers
together, I wind up with the mass being 4.6 grams of nitrogen dioxide.
So, we can actually do predictions of the masses of materials produce from the
masses of the materials that we started off with.
But only, if we do calculations of the numbers of moles of reactants and use that
to determine the number of moles of products, and take into account this idea
that we might run out of one of the reactants before we run out of the other
type of reactant. We're going to continue these kinds of
calculations in the next lecture.
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