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 V
Loading...
From the course by Rice University
General Chemistry: Concept Development and Application
219 оценки
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 wrap up our work now on atomic masses and molecular formula from
concept development study 2, by talking about one more example of chemical
stoichiometry. You might recall that what we have been
trying to do is count the number of particles which are participating in a
chemical reaction. That's important to be able to do because
we know that atoms and molecules react with each other in simple energy ratios.
So we have to be able to count the particles to know how many of them are
going to react with each other. Let's consider now, rather than the way we
have done things, which is to simply take a mass.
What do we do when we have a reaction taking place that's in a solution?
In that case, we're going to have to deal with something that we call the
concentration, or chemists will call the molarity.
And in this, we're going to need to distinguish what we mean by a solution.
So, what I have here, actually, is a solution of sodium chloride dissolved in
150 milliliters of water in this beaker. And the question that I might ask is, how
many moles of sodium chloride are there in this Erlenmeyer flask?
The answer is, we don't have a means to determine that yet.
I could, for example, just weigh the water, or the solution rather, which is in
this Erlenmeyer flask and try to determine the number of moles in a way that we have
but that won't do me any good. Because overwhelmingly what I am weighing
is the water, which is the solvent in this case, and not the sodium chloride, the
solute, which is dissolved in the water. In many instances I need to know how many
moles of solute are there in a particular solution.
Let's give an example here. Here's a chemical algebra, a chemical
stoichiometry problem that we need to work out.
We're going to try to dissolve some solid copper in some nitric acid.
So here's a nitric acid solution, copper dissolves in nitric acid to form copper
nitrate, so nitrogen dioxide. And some water as the products here.
Let's imagine that we're doing the same kind of chemical stoichiometry that we've
done before, where we weigh out the mass of one of the reactants.
In this particular case, we're going to take 10 grams of copper metal.
And the question is, how much nitric acid do I need to react with that 10 grams of
metal? And how would I find that number of moles
of nitric acid? If in fact I can't simply weight it.
So how do we know how many moles of nitric acid there are in a solution of nitric
acid? That fundamentally is the question we're
going to try to address and will address by the end of this particular lecture.
To do that, we're going to consider this idea of concentration that I discussed a
moment ago, by defining the concentration as being now the number of moles of solute
in a liter of solution. The formula that we're going to focus on
here is this one. The concentration is given by this bracket
so if S is the solute, might be sodium chloride for example dissolved in the
water, then. Putting S in brackets is the concentration
of the solute in the solution. And it's pretty clear when we look at this
formula that in this particular formula here, the number of moles of solute is a
number of moles, and the volume is going to be measured in liters.
So the units here are going to be moles per liter.
We'll describe this in units of moles of solute per liter of solution.
And sometimes that unit is simply given the capital M or molar.
We might refer to a 1 molar solution or a 0.5 molar solution.
For example. So let's imagine then that what we want to
be able to do is count the number of moles of solute using the concentration of the
solution. Here's a particular problem.
Let's say that I have a 0.1 molar solution, again here's that unit capital M
that means over here moles per liter. And let's say that I've measured out then
200 milliliters of the solution. What is n?
In other words, how many moles of solute are there?
If we look at the formula on the screen there the concentration of the solute is
equal to the number of moles of the solute which is what we're trying to find out.
Divided by the volume of the solute. I can easily rearrange that equation to
say that the number of moles is equal to the concentration multiplied by the
volume. And we've been given both of those
numbers. Let's look back at the problem here.
Here's the volume, there's the concentration.
So the number of moles is going to be equal to the concentration which is 0.10
molar or moles per liter. Multiplied by the volume.
And here we have to be just slightly careful because the problem has given us
the volume in terms of milliliters, whereas the units on molarity are moles
per liter. So we need to rewrite the volume as 0.200
liters. If we multiply those two together, we get
0.020 moles because moles per liter times liter gives us just units of moles.
So we've answered the question now. We've actually been able to determine how
many moles there are in a 200 milliliter sample of a solution, which has
concentration 0.1 molar. We could do this slightly differently as
well. What if we turned it around?
What if we needed to measure out 0.5 moles of the solute, what volume would be
required? That might be useful.
That's the same as trying to say what mass of material do I need.
To have a certain number of moles for the chemical reaction to take place.
How are we going to do this? Well, again, our formula back over here
says that the concentration is equal to the number of moles divided by the volume.
What have I been given in the problem here?
What I've been given is a, a number of moles needed.
And a concentration. So let's see.
I need the concentration to be equal to 0.10 moles.
I need the number of moles to be 0.50 moles.
The unknown here is the volume. I can rewrite this equation fairly
straightforwardly, just simply say that the volume, then, is equal to the number
of moles, 0.50 moles, divided by the concentration, 0.10 molar.
And that simply then turns out to be 5.0 liters.
So that tells us that if we need a certain amount of moles, we can find that number
of moles by measuring a particular volume, or if we have a particular volume of a
known concentration solution, we know the number of moles which are in that
solution. Let's apply this now in a particular
chemical contub, context. Here's a chemical reaction.
We're going to react two solutions now, each of which has a dissolved solute in
it. The first solution's going to have sodium
chloride in it. The second solution's going to have silver
nitrate in it. When sodium chloride and silver nitrate
are both in a solution together, they react.
And the product of the reaction is a solution of sodium nitrate and solid
silver chloride. Notice what we've produced here is a solid
which is actually no longer in the solution.
The sodium nitrate remains behind in the solution.
Let's imagine now that what we're going to do is mix together two solutions.
We are going to mix together a solution of silver nitrate and a mixture of sodium
chloride, but they don't have the same concentration of what we have here, don't
have the same concentration, and we're not taking the same volumes either.
How are we going to work this particular problem?
Well, it turns out to be a limited reactant problem, of the type we have
solved before. What we need to do is to calculate how
many moles, of the sodium, of the silver nitrate do I have.
And then I should know how many moles of the sodium chloride do I have so that I
can know whether or not I have equal numbers of moles or if one of the moles is
in greater proportion than the other. Because we can go back and notice that
they react in a one to one ratio, so I need to know which one of these I have
more moles of. Well, according to the formula we had a
little while ago the number, the number of moles, remember we had the concentration
of a solute is equal to the number of moles divided by the volume.
So in each one of these cases I can say that the number of moles of silver nitrate
is equal to the concentration of the silver nitrate, which was given, times the
volume of the silver nitrate. And let's see, what are those numbers?
The concentration of the silver nitrate, back from before, recall, is actually 0.15
molar. And the volume of the silver nitrate is
given here as well, is 0.5 liters. And if we multiply those together we get
0.075 moles of silver nitrate, which are in the solution.
How about the sodium chloride? Well again, we need the concentration of
the sodium chloride, which was given, we can go back over and look at that.
It is 0.20 moles, and the volume of the sodium chloride, which was given as 0.25
liters. Plug those numbers in.
The concentration of the sodium chloride is 0.20 molar; the volume of the sodium
chloride is 0.25 liters. And if we multiply those two numbers
together, we get 0.050 moles of sodium chloride.
So, do we have enough sodium chloride to reaction with the sil, silver nitrate?
Yes, we do. Uh,er, actually, no, sorry, we don't.
We need 0.075 moles of silver nitrate to react, I'm sorry, of sodium chloride to
react with 0.075 moles of the silver nitrate.
We don't have it. Therefore, we can conclude that sodium
chloride is the limiting reactant in this particular reaction.
Because we'll run out of it first. So if I now ask the question let's say how
much silver chloride am I going to produce?
The answer than turns out to be, let's see.
From back in the original formula here there's clearly a one to one ratio between
the number of moles of sodium chloride which react and the number of moles of
silver chloride which are produced. So, if in fact I have 0.50 moles of sodium
chloride. The number of moles of silver chloride,
which will be produced will be equal to 0.050 moles of solid which will be
produced. So if we ask the question now, how much is
produced, we've got the answer. What remains in the solution behind?
Well, we precipitated out quite a bit of silver chloride.
But notice, there was a lot more silver in that solution than there was chlorine.
Buy 0.25, by 0.025 moles. So left in the solution is some sodium
ions, some nitrate ions. But also some silver ions.
There are silver ions left behind in the solution.
And we can actually calculate the numbers of silver ions left over in solution.
But also, importantly, we could calculate the concentrations of the silver ions
which remain in the solution. So, to do that, we have to take into
account a new phenomena as well. Something we're going to call dilution.
Dilution's a pretty familiar kind of thing in everyday life.
We take some sort of a solution and we add some solvent to it.
This might be, for example, taking a soap solution and watering it down.
Taking an alcohol solution and watering it down.
What we're doing, is we're adding solvent. But we are not changing the number of
moles of the solute, because we're only adding solvent.
The cons, consequence of that, is that we've actually changed the concentration
of the solution because the concentration of the solution is the number of moles of
the solute, divided by the volume of the solution.
I don't change the number of moles in the solute and I do change the volume of the
solution, then I have changed the concentration.
But note the number of moles of the solute is fixed so doing a dilution, when I in
fact dilute something before or after dilution.
The number of moles of the solute is fixed.
But recall the concentration, let's call it the concentration before a dilution is
equal to the number of moles of the solute divided by the volume before dilution.
And likewise the concentration after the dilution is the number of moles of the
solute divided by the volume after the dilution.
If I set the number of moles equal to each other then I can actually write down a
formula and here's the rather simple formula.
That the concentration before dilution. Multiplied by the volume before dilution,
equals the concentration after the dilution, multiplied by the volume after
the dilution. So for example, if I start off with a
solution which is let's say 0.1 molar, and I have 1 liter of that solution.
And I dilute it to double it's original volume and I want to know what the final
concentration it's equal to after I double the volume.
Well all I gotta do is divide both sides of this equation by 2 and the final
concentration is half of the original concentration.
Which makes sense. And it comes out to, to be 0.05 molar
after a dilution by a factor of 2. So if we go back and think about the
previous problem here. And we know that we have left behind from
our previous calculations, we'll go back and recover those.
We know that we still have the number of moles of silver, which are in the
solution, is still equal to now, comparing the numbers up above I had 0.075 moles of
silver nitrate and 0.05 moles of silver, sodium chloride.
And all of the sodium chloride all of the chlorine has precipitated out with the
chloride. There still remains 0.025 moles of silver
back in the solution. The concentration of the silver ions in
the solution is then equal to the number of moles of the silver, divided by the
volume of the solution. But notice that the volume of the solution
didn't remain fixed. Because we mixed together a half a liter
of a solution. Go back here and take a look.
Here's a half a liter with a quarter of a liter.
So now we've got 0.75 liters of solution. So we could write this now as being equal
to 0.025 molar divided by 0.75 liters. And if we actually divide those two things
out, we discover that the concentration of the solution in the end, checking my notes
here, is actually equally to 0.033 molar, moles per liter, of silver ions in
solution. In other words not only have we
precipitated out some of the silver, with the chlorine.
We've also diluted the solution, which remains, because there's not nearly the
concentration that we had started off with.
Now let's finally go back and look at the problem we've described at the very
beginning here. At the outset, one of the things we were
interested in is. How much, nitric acid is required to
dissolve 10 grams of copper? Well, to work that problem out, what we
would need to know is what is the number of moles of copper that we started off
with. Well let's see, because it's a solid, I
can find its number of moles by taking its mass divided by its molar mass.
Its mass is equal to 10 grams. Its molar mass, I can look that up in the
periodic table, is 63.55 grams per mole, the number of moles turns out to be 0.167
moles. From the stoichiometry, the balance to
chemical equation here. I know I need 4 moles of nitric acid for
every mole of copper that I started off with.
So, the number of moles of nitric acid which are needed.
Is going to be equal to 4 times the number of moles of copper, I started off with,
and that number turns out to be 0.628 moles of, sorry 0.668.
Moles of, of nitric acid. And let's see now what we've given us is
that there is that our solution is 1 molar.
So there's 1 mole per liter now. If we go back and remember the
concentration of the nitric acid. Must be equal to the number of moles of
the nitric acid, divided by the volume of the solution, and let's see, what do we
know? We know this number is 1.
We know this number is 0.668. So what we conclude is that the volume
required is 0.668 liters of the nitric acid solution.
The point here is that it is easy to do chemical stoichiometry, chemical algebra,
in the same way that we have done it before, except when we're dealing with
solutions we're more likely to know the concentration of a solution and use volume
measurements to count the numbers of moles.
That actually gives us a much more robust way to do calculations of chemical
reactions, and figure out how many moles of one material are required react with a
number of moles of another material. And to not just determine that by weighing
pure materials like say copper, but also by taking volumes of solutions.
Ознакомьтесь с нашим каталогом
Присоединяйтесь бесплатно и получайте персонализированные рекомендации, обновления и предложения.
Coursera делает лучшее в мире образование доступным каждому, предлагая онлайн-курсы от ведущих университетов и организаций.
© Coursera Inc., 2018 Все права защищены.
