Lecture 24. Review Final Part I
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Transcript: English(auto-generated)
00:23
Dear, dearest friends, today we are going to start a review of the material, and for the next week we have a special meeting, called the finals week, and that week, of course, we are going to show all the educators how great the person he was. Time to get those A's, right?
00:50
But let me give you some information about our exams that may be helpful. So first, this class we evaluated Every class that this guy has evaluated through this system, I think you're talking about an invitation to fill out a questionnaire. It's
01:06
rather long, I mean, somewhat long. But it's actually very useful, and these evaluations are taken quite seriously. In fact, every time my performance is evaluated by my department chair, he has all these steps, all the comments that are made by students, and his scores, and his
01:28
doesn't look good, he's kind of smacking me. So it's a serious talk, it's not something that should be taken lightly, and so fill out your impression, if you like it or don't like it, you can just fill out without your experience lost.
01:45
Okay. In addition, next week Laura, one of the TAs, has kind of generously offered to do a review session. Next week, of course, there are no visual classes, so this is an extra review
02:03
session in this auditorium. So this place is at 7 o'clock on December 10th. And so she will be at her service to ask her questions, and just like we've done before, we'll do it again. Unfortunately, on Monday, I'm not going to be able to do it again.
02:21
I'm unable to be here. I have a very important engagement in North Carolina, and to the university, I have to be there. So I can't help but to myself, that's why Laura is stepping in, and she's highly qualified to do that. All right. So December 14th, it's on a Friday, it starts at 1.30. Like I said, the exam will be two hours in
02:44
total, so it's much longer than the mid-first, and that means there's a little bit more breathing space for a few questions, and you can look over your answers probably several times, and be sure you get the best possible performance on
03:03
this exam. So while I'll send you more information, again, you can allocate a seating, so keep an eye on the website, the class website, all the information will be there. It will be posted. The practice midterm has already been posted, by the way, so you can already start looking at that. Yes? The practice exam has been posted. I just said that, and it's a
03:26
you can just go there and look and get an idea of how long the exam will be, what kind of questions you may expect. As I said, it's going to be slightly different, but this is a very good impression of the level of difficulty, the length of the
03:41
exam. Things to bring. As usual, an index card, your calculator, cell phones, we don't like so much, so leave them in your bag, turn them off.
04:00
Oh! Very good question, people. How many index cards can you use? The answer is uno. And that is the same as before. So this is not because we have too many terms here, we have two cards. One card. You can use that wisely. Some people very beautifully find
04:24
liners and do a great job. A lot of people just print it out on a special printer and also do a great job. That's fine. You don't have multiple layers if you're flat open. That's not a lot. Okay. So this is
04:40
the list of topics. There's nothing new about this, I'm just going to do it for formality's sake. Everything we have talked about is going to be in final. Final is truly a cumulative kind of exam that basically considers the entire course in its entirety.
05:02
So basically, if you like the book as a main source, it runs from chapter one to chapter four. That includes the following topics. Dealing with numbers, okay? sig figs, units, prefixes, and so forth. Dealing with numbers, writing numbers in scientific notation
05:22
is part of this class, so we should do that on the exam. Even if it's not explicitly stated, we should do that because it's part of this exam. Unit conversions, ultimately important, converting one unit into another unit. Centimeters into meters, but also
05:42
density into mass, or, for instance, inches into kilometers. Those kind of conversions we have done many times over, and we should be able to do that on the exam as well. Atoms and isotopes, we talked about the basic architecture of an atom, isotopes, ions, and so forth,
06:04
and we should know something about the orbital and where the electrons are, remember? and how many electrons a certain atom has, how many protons it has, how many neutrons, because the basic architecture of an atom is something that's part of this exam, as is, naming these atoms, making molecules from them, and naming their ions probably.
06:26
Knowing the charges of the ions, and knowing the difference between a polyatomic anion and an elemental anion, for instance. Then we started to do some calculations with molecules, atoms, and ions, namely, we started to determine the molar mass of these guys
06:46
and mass percentages of elements in compounds. These calculations are part of the final. Then, we looked at chemical reactions. So we have molecules that are actually engaging into a chemical interaction, and there are reagents and products.
07:06
and we should be able to balance such a reaction and come up with the right stoichiometry. Why is the stoichiometry so important? Because we have to be able to determine mole ratios between things. Mole ratios
07:21
can be used to convert one type of molecule into another type of molecule. Another type of compound is to convert how many atoms. So we need to be able to do the mole ratio to follow the stoichiometric coefficients, given the mole ratios. Molarity, and again, stoichiometry in solution, we've done that, we've looked at many compounds that are dissolved in water, and then react in a solution.
07:51
Molarity is an expression of concentration, and along with it, we should be able to calculate, for instance, how many ions of a particular kind there are in a solution.
08:05
You should be able to calculate the number of chlorine ions in there. If you dilute it, put another liter in there, you should be able to calculate the number of chlorine ions again for calculating the concentration.
08:21
Those are very standard calculations that we've done. It's part of the problem. Now finally, three types of reactions that we've wrestled with precipitation reactions, acid-base reactions, and redox reactions as examples of very important chemical reactions. There are more types of chemical reactions, but these are three main ones you should be familiar with.
08:44
Dealing with the stoichiometry, being comfortable with the naming of the compounds would set you up for a successful M1A series. Okay. Now I'm going to say a few more words and details about exam taking, then I'm going to
09:01
say a few more words about redox, just to wrap that up. Here's kind of a snapshot of the exam two years ago Here's a question in terms of molarity We're not going to look exactly at how this is answered, I'm going to tell you that show your work, because we are grading your work.
09:21
In this particular question, you are asked to calculate the volume of a solution. And there are some details to it. But you have to do it the first thing you always have to do with stoichiometry is to determine how many moles of the stoichiometry you have. Always you want to know how many moles you have. And once you have the number of moles, you can easily calculate by the degree of grams, or molarity, and so forth. So that's the first step. You have to determine the number of
09:44
moles, check. That check means, if you're on the right way, that means you get points. Then, you have the number of moles, you have to do something correct with it, you have to convert that, in this case, to volume, the molarity is given, is the ratio, namely, the definition of molarity is the number of moles over volume. That means the volume is on number of moles.
10:05
Now, when one or two are given, you calculate the volume, that's exactly what you have to do here, and that's the second check. If you know how to do that, you get points. In other words, you get points for these steps, for instance, 5 and 5, total of 10 points. The final answer actually is not even the most important. Okay? So even if it's completely wrong, you still get
10:26
points for doing this correctly. The methodology is great methodology. It takes a lot of time for us to do that. But we think it's very important. So we go through all the calculations, all you guys, all 300 exams, we go through all the questions and see if there's something correct in there.
10:45
The better, the more correct we go like this. If you want to help us, and this will help you to get more points, be clear about what you calculate and what steps you get. If you convert it down like this, what brings in this example, where I say exactly what
11:01
I'm trying to do in each step, I'm going to calculate the number of moles in each step, Knead, and so forth, conversion to grandpa, I'm going to just explicitly write out what I did, then the gradients can see, oh, look, this is very nice. One, two, three, four steps, beautifully laid out, maybe a little mistake here and there, but at least the person knows what he or she is doing.
11:20
This greatly helps that's great, and helps you get more points. Because if this is all messy and jumbled up, we have a very hard time giving you partial credit. So, check, check, check, we can give you more points. That's not good. It's completely fine to do this first on the, you know, use a blank page, first write out what you think the answer
11:44
should be, and then reproduce the answer neatly and nicely with the correct steps in the blank that we are actually looking at. Just one tip. You don't have to do this, I'm just telling you it's an option that makes life a lot easier for us.
12:00
Okey-dokey. Keep that in mind. I'm going to wrap up the topic of redox very briefly by following, you know, we know what redox reactions are by now, redox reactions are reactions where there's an oxidation step and a reduction step. Something is losing electrons, something is gaining electrons. Here's an example, in this particular case, I can split this into half reactions, namely, the
12:26
iron is losing electrons. So that's an oxidation step. And the couple of things that I'm going to talk about are the copper here is gaining electrons. So that is the reduction step. So I can write this in that equation in the following
12:41
way. Forget about this sulfate, it's a spectator anion. It doesn't do anything. It doesn't partake in it whatsoever. Everything is aqueous, so I can just get rid of it, and just write this. It's very easy to see in this case that, look, this guy is changing to 2+, this 2+, copper is changing to solid copper, which has an oxidation state, 0.
13:04
In other words, this is the reduction step, copper, and that is the oxidation step. Both things take place at the same time. So you can nicely see in this way of writing that the electrons lost by the iron have been eaten up by the copper. Effectively, the only thing that happened here is that
13:26
two electrons lost by the iron have gone to the copper. So there's Homer Dew this week, Friday, as usual, and we'll have him talk about it later. It's about redox reactions in all the conditions that we have considered, namely just normal redox reactions, redox reactions in solutions, two flavors, acidic and
13:50
basic conditions. So here are a couple of settling questions that have come up during office hours and I'd like to
14:00
just quickly go over, hopefully it will clarify a few things. So I'm just going to see if we can lift this down.
14:31
So let's look at the first one, because the settling does stay slightly differently than we have talked about in class, and so we should know that. So the first one is O2, and then four copper, and then copper oxide
14:49
So you see, first of all, what settling does not do so nicely, which I really don't like, is there's no physical states. Okay? That's not nice. Don't do the same thing on the exam. We
15:03
have to write physical states, because actually it is somewhat confusing in some cases, but it really doesn't matter. So in this case, what they want you to do in the first equation is to write half of it. And the problem is, copper oxide, this thing actually is solid, and we have said in class that you should keep it as a solid.
15:26
But something doesn't stick to those rules. So what they say, fine. Consider this as an ionic compound, and in effect, I have two ions. Mainly this thing is oxygen, oxide, over 2-. And this copper, because this is 2-, is two of them,
15:45
must be copper plus. In other words, this is copper 1. Oxidation state 1, this is oxide. Standard oxide is always oxidation state 2-. Okay. So now I can write the last reaction, because what we have to do is, in this
16:05
case, oxygen, O2 turns into O2-. And copper, which was oxidation state 0, turns into O2- proper plus. And I want you to balance these two gases, relative to one another. So we can see
16:26
where from the oxidation to the reduction, this in fact is the reduction step, because this is 0, this is 2-, so it becomes more negative, that means it has taken up electron, so this is the reduction step, this is the oxidation step. Indeed, this one is losing electron. So some electrons appear here, and some electrons appear there.
16:48
So to that, I have two oxygens here, one here, so I need to put a two there to balance the oxygens. There's no need to put waters there. There's already oxygen there. There's already oxygen there. You just have to balance
17:01
out the oxygens. Then by doing so, I have four charges there. In total, four minus, that means I have to balance that out. So in this case, I have to have four electrons to that side. I have four minus on each side.
17:20
This is the reduction step. That's right, this is the reduction step, it should appear on this side. Copper, the elements are already balanced, the charges are not, I have to add an electron here. So that's it. And the only thing I have to do now is to make sure they're balanced relative to one another. Seth wants to do that. You can do that by putting a four there, a four
17:43
there, and a four there. Four electrons in each state, and this is the final answer. Those are the second ones. Funny one. Funny, funny, funny one, and not something you can expect on the midterm. Sorry, on the final.
18:01
So it's very much an exception, I just want to show so that I don't get stuck on the whole thing. The question is, if you balance this equation, okay, plus copper again, yeah, copper again, 2+, and it generates S4 of 6 to my
18:21
plus Cu plus. Okay, so what I'm going to do, I'm going to split this, again, with two steps, I'm going to write down copper, 2+, I'm going to go to copper plus oh, this one has taken up an electron. So an electron called this gives me that.
18:41
This is the reduction step. That means the other guy is the oxidation step. Let's just put it down. Now this is indeed an oxidation step, but there is something very funny going on with this particular
19:02
compound. If you try to find the oxidation state of sulfur, you are in for a big surprise. The oxidation state of the sulfur here is 2.5. That doesn't make any sense. And it doesn't make any sense because it's a very particular compound where the four different sulfurs actually have different roles and different oxidation states. So this goes way beyond
19:24
what we should know at this point, and it's also not relevant for our discussion. So I'm going to tell you what you should focus on with these kind of questions. You balance the elements. You can do that quite easily by putting the two there. So now I have four sulfurs, six oxygens, that's the same as here,
19:42
but in doing so, I haven't really balanced the charges, because two times minus two is minus four, I have minus two here, so I need to add two more. There. Okay? So now it's balanced. This is the oxidation reaction. It is indeed an oxidation-add reaction, because I'm generating electrons. It's products.
20:03
The balance is then relative to one another. I have to multiply this whole thing by two, two, two, two. And now I can add them up. And the final solution, I can put it in here, two, two, two, and that's it. That's the answer for that.
20:22
Let's do the final one, too. Why not? I'd like to do this also because I'm going to show you that even though these pressures may seem morborious, if you know your stuff, you can just simply write as if you write a letter to your mom, and in that
20:41
kind of case you can solve this problem. It is straightforward application of the rules that we've dealt with. So here this is again a permanganate. So I'm going to write down permanganate, it's changing into 2+, I can quickly look at the oxidation states here. This is 2+, I've identified.
21:10
This whole thing has to be minus, 4 times 2 is minus 8, this is going to be plus 7. Okay? If you don't know how to do that, teach yourself how to do it. We have done it a lot of times, it's essential you are able to determine how to do it.
21:24
So I'm going to write down oxidation states. Namely, oxidation states that all these guys added should add up to minus 1. This is minus 2, the score of them is minus 8, plus 7, equal to minus 1. This must be plus 7. It shouldn't take you a lot of time. Okay? The linear algebra should take you about 10 or 50 seconds to figure it out.
21:44
And I'll do sign on that. So I'm going to balance this equation, what do I have to do? This happens in the acidic environment, let's say, I have four corners, but I have to add up also 8, 8+, and I do that, so I put that on the other side.
22:05
And now I have balanced all the elements, but I have not balanced the charges. I can tell you what the charges should be, they aren't even moving at the charges. I'm moving at the charges on each side. I can tell you that since this is 2+, and this is 7+,
22:21
it needs five electrons. There has to be five electrons on this side. Because it's supposed to be 7+, it's 2+. So that's a check. But I can also do it like this. The whole total charge on this side is 2+, the total charge on this side is 8+, minus
22:40
1 is 7+, and I need to get to 2, so I need to subtract 5. That's exactly these five. So either way, it doesn't matter how you get there, but do this consistently. Okay. Now the second one, so this is the reduction because this one eats electrons,
23:02
this guy goes into NO3-, what's going on here? Let's look at the oxidation states very quick. The whole thing is minus 3 oxygens, minus 6, this one is plus 5. What is going on here? This thing here is a nitride anion, and it comes at it.
23:25
So this whole thing here is what you're going to say, charge minus 1, oxidation state minus 2 times 2, is minus 4, plus 3 is minus 1. So this is plus 3. So oxidation state of the nitride is changing here by 2.
23:44
by 2. So I know two electrons have to move here, you can only tell by that. You're not looking at anything else. So let's see if that actually works out. First of all, there's two oxygens here, three there, so I can add water here,
24:02
two values of oxygen, by doing so we produce two extra hydrogens, there's already one here, there's three in total, and there's plus 3 and there's plus. Let's look at the charges, if they add up correctly. No charge on this side, okay? No charge on this side. That means there has to be no charge on this side.
24:21
This is minus 1, 2 electrons, minus 3, 3+, correct. So the charges, values of atoms. Now I have two reactions, this thing has 5, this thing has 2, so this whole reaction is multiplied by 2, this whole reaction is multiplied by 5, and then you can
24:47
add them up, and that's it. So you can do that for yourself, and then cross out the H+, and cross out the water.
25:01
Okay. So that's just a little detour, but you see you can just write this add waters, add H+, balance with electrons, and this can be done relatively quick. I've been doing this several times, like doing the homework, it's really helpful in getting comfortable with these kinds of questions.
25:24
Typically people do pretty well on redox on the final. So I'm pretty confident you guys will step it up. Okay. Let's rewind and go back to the very beginning when we talked about dealing with numbers and dealing with units. One of these
25:47
things is writing numbers correctly, significant figures, is the way of writing numbers correctly. And we should do that. We should just do that. So if you do that, make sure you discriminate between multiplication and division and subtraction. They have different rules.
26:06
So, multiplication and division says you have to give the number that you are calculating the same number of significant figures as the least precise number in the calculation, but that rule
26:21
is different if you add and subtract, because in that case, the rule involves that. Your answer should have the same number of decimal places as the least precise number in the calculation, or the number with the least amount of decimal places, dictates how much you get the answer.
26:41
So, let me quickly give you an example of how that plays out in practice. Let's look at the following. Calculate the mass of the magnesium here, 0.643. Well, if you do that, you go into the figure of table and you look at the molar mass of the magnesium, and you find 24.305. This number comes from the figure of table, it has five significant figures.
27:11
This is not the limiting factor in the calculation. It has five, this one has three, so the final answer, since this is a multiplication, should have also three significant figures. Okay. I think we can all do that, and most of you have done that correctly in exams. Now
27:30
there's a situation where you may wonder what to do. And that happens when you have a calculation that involves both multiplication and an addition.
27:42
For instance, calculate the total mass of this much magnesium and this much chlorine. So what you have to do is, well I have an answer here already, 15.6, I do the same thing for chlorine, I take the number of moles, I multiply that by the molar mass of chlorine, that is 35.453, that gives me this number here, in grams,
28:08
okay, so that is the answer, but this should be given in three sig figs, because 123 is the least precise measurement in this particular question, and so the answer here is 1.61. I have two numbers, these two I have to add up. So what I do here is at first, finish the number that I'm getting to when I have
28:26
to add up. And I'm running this because this multiplication step is over now. And I have to make sure that when the number rolls out of the multiplication, it has to be the proper number that means something. It means something if it has the max right number, and so on. So what I'm doing, I'm taking this as the input in my addition. So I'm going to add this number to it, these
28:47
are two results that I added up, as you can see, this is what your calculator says, but you have to realize that this number here has one decimal place, and hence you should also have this number in one decimal place. That is the rule for subtraction.
29:03
So the answer is 70.3. It has three sig figs, not because there are three sig figs here and here, but because the addition says that the final number needs to have one decimal place. So first complete the multiplication and division, then add them up, and use the rules for each operation.
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Okay. Now let's do an example of the kind of calculation that we have done. Let's determine the mass percentage of each element in this particular formula. This particular molecule.
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So let's do that. Here's the chemical formula of this compound. From this formula, I can calculate the molar mass. 14 times the molar mass of carbon, 12 times the molar mass of hydrogen, and 3 times the molar mass of oxygen.
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It gives me the molar mass. What I do then is 14 times the molar mass of carbon divided by the total that I just calculated, times 100, and you get a percentage. What is the number of sig figs in this answer? Well, this is a division and a multiplication, hence the number, the least amount of sig figs in determining the answer
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and in this case, that is this number right here. Okay? This one has four. This number comes from my table. If this number would have five, then this would be in five. You would have three places after the decimal point. Okay? Three.
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But because this is in four, your answer is going to be in four sig figs total. Hydrogen, let's say, for instance, that the table you get the information from has three sig figs for this number the molar mass of hydrogen. Then your answer should also be in three.
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If this was four, the answer was in four. Then you get 5.30 something. There's one more digit in that case. In this case, there's four, that means the answer should be in four. But, maybe this is an example that's well chosen,
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if this is four, then you get three places after the decimal point, right? Three digits. Which is two here, two there. So it doesn't mean these calculations are independent of one number. This can have three digits after the decimal point, this can have two, this can have two at the same time. So you don't have to kind of line them up for some sort of reason.
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That's not what they take. Sig figs. You look at each calculation separately. No. This zero is after a decimal point. This is significant.
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Conversions! You have to be able to construct the right unit factors for converting one unit into something else. and one of the most, you know, I would say, involved calculations are calculations of volume. Because you have to multiply them.
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by the length, to the power 3. So for instance, here you have a box, you have to calculate the volume, the box is given in inches, as you can see, but I have to give my answer in liters.
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So liters is decimeter cubed, right? That's the definition. One liter equals one decimeter cubed. So I have to go from this in case into something that is metric. So you want to go probably from inches to centimeters, and centimeters to decimeters, and then decimeter cubed. So
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what you can do this in a variety of ways, this is one way to do it. These are the two definitions that you need. The relation between inches and centimeters, and the relation between liter and decimeter cubed. So, for instance, I can convert each length separately. That's a possibility.
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This is 1.5 inch, I can convert it to centimeters by multiplying by this unit factor. 1 inch, 2.45 centimeters, this one goes on top, this one at the bottom, inches is crossing out, I get my answer in centimeters. Everything between parentheses here now is in centimeters. And I can do it
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for each of the lengths that I multiply. For instance, width, length, and depth. I do that for each one. I do each of these guys, I can do the conversion separately. That's completely fine. I could also have done it in one step, of course. I could have modified these numbers and then have this conversion step raised to the power of three. Which is exactly the same calculation.
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Because effectively, that's what you do here. One conversion, two conversion, three, all multiplication signs, you basically raise this conversion to the power of three. Right? That's exactly the same operation. Wherever you prefer is fine. These things are equivalent.
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The answer is 446 centimeter cubed. Why is it cubed? Well, we have 1, 2, 3 centimeters multiplied, it becomes centimeter cubed. Right? The next thing I have to do is relate this to decimeters. I convert that by using the conversion between centimeters and decimeters.
34:42
10 centimeters to one decimeter. One decimeter equals 10 centimeters. Centimeter goes at the bottom, this one goes on top, because I go from centimeters to decimeters, and raised to the power of three, so the unit factor also goes to the power of three. Okay? So that means you can convert this number now to
35:03
decimeter cubed and one decimeter cubed equals a liter, so you can just replace decimeter cubed by liter, and this is your final answer. I'm going to put that in two sig figs, because this one has two sig figs, this one too. Okay? So be familiar with this kind of calculation. If you are on the exam and you see this volume conversion, and you
35:26
have no idea what to do, then you haven't prepared it. We'll probably do something like this on the final if you don't want to see if you can do this. This is how it works. This is one way you should solve it.
35:43
Math. Throughout this class, we have used different formulas, the main formulas, the formulas from which you can derive everything else are the following. Density equals mass over volume, molar mass equals mass, total
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mass, over number of moles, okay? That's how you calculate the molar mass. Or, you can also say you can circle this through, right? You can basically put this mole right here, number of moles, equals total mass over molar mass. That's the same. And the definition of molarity, which is the number of moles in the number of moles of the
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solute in liters of the solution. Almost all calculations make use of either this, this, or this in one way or another form.
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So, typically, something is given, two variables are given, you calculate the third. Be it mole, be it liter, or molarity, it doesn't matter. These relations can be used and rotate through to calculate any of the missing variables. This is really all the math there is to it. It's either division or multiplication. That's it. And your relations will be the same.
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This will tell you what it is. So don't flip things. Again, there are lots of situations on the last midterm where people simply guessed that it was a multiplication or division. If you do that, you have 50% chance of doing it. Two ways to avoid it is to probably think about
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these relations and you are able to tell yourself what you have to do, if that is not enough. Look at the units. The units should cross the line. They should cross out properly. If they do not, you probably would make a mistake. So look at your approach carefully. If you
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know what to do, of course, no issues, but if you are stuck, you feel like, I know what to do. Multiplying and dividing, you have to use logic. Never ever guess, because the chances are pretty high that you make the wrong guess. This is
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not how we do science. In science we can't guess. So you are in the hospital, and everybody says, give this to me. This person is this many molecules of this drug, and you have to guess when you have to multiply and divide, and you are in big trouble. If you are guessing wrong, the patient may die. You don't want that on your resume. You will never find the job again.
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So these things are things you can avoid by just staying calm, and there is a all these things are logic, there are no blurry lines. They are not there. The blurry lines are only in your head. Get rid of them.
38:42
Okay. An example of a quick calculation could be anything I have 1.2 liters of calcium sulfate solution and a certain concentration now the question could be, for instance, calculate how many moles you have or how many grams of this stuff do you have
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how many grams were dissolved in this solution? How do you do that? Well, I want to know how many moles, I know liters times molarity equals moles. Boom, boom, boom. Okay. This is something I have done so many times that I trust in this, okay. Liters times molarity equals moles. If you don't believe that, suddenly, it may happen. Suddenly you say, oh, hold on a second. I don't believe
39:24
that. And you can always say, look, molarity equals moles over volume. That's the unit of molarity. Molarity is moles over volume. M over L. L is here, L is at the bottom, L crosses out, your unit is going to be moles.
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That's called dimensional analysis. You can use that unit trick to figure out if you are doing the right thing. Always check out if you have time. If you know the number of moles, you can do a simple multiplication with the molar mass to find the amount of grams of calcium sulfate that were dissolved in this particular solution to get to this concentration.
40:02
Number of moles times the molar mass equals your answer in grams. If you don't forget about this, look at the units. mole here, the unit of molar mass is grams to mole, mole crosses out, your answer is going to be in grams.
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Another one, a 1.2 liter solution is a density. It's not given in molarity, but in density. Is that a reason for me to panic? Because I'm used to working with molarity. Suddenly, the concentration is given in density. Kilograms over liters. No? It's not a reason to panic whatsoever.
40:41
I can do exactly the same calculation because I can calculate how many kilograms I have in the solution another relation of density, liter, and mass. So that means that if you multiply volume with density, you should get an answer in kilograms. The unit of density is kilograms per liter.
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So if you multiply it with liters, L will cross out, your answer is going to be in grams. So here I have a measure of mass, thank you very much, and the measure of mass can be related to how many moles of that stuff you have, for instance.
41:21
All right. So this is again molarity, it was an important quantity, and we should be able to do quick calculations with it. This is a typical question, this is a question that was on the midterm, on the last midterm, I would say probably 50% of people did this incorrectly, you shouldn't do that again. This is a very standard and very practical situation. You have a certain volume of the
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solution, 1.2 liters, and you have to add a certain amount of sodium phosphate, actually sodium phosphate is the main ingredient of the brain.
42:01
This is a very important buffer system that people use in the biology lab all the time. It's also a very important buffer in medical sciences. So this is not a ridiculous question. Five grams of that stuff dissolved in 1.5 liter, your advisor wants to know what is the molarity of that. The molarity of sodium ions he wants to know.
42:21
Okay. So you have to realize that, look, sodium phosphate is this. Phosphate is 3-, you need 3 Na+, to create a neutral compound. So Na3PO4. That is sodium phosphate. If you dissolve that in water, what you get is
42:41
three sodium ions, one phosphate, and another. In other words, with each one mole of this, you generate three moles of sodium. So, 5 grams divided by the molar mass of sodium phosphate okay, gives you the number of moles of sodium phosphate. This is the amount of sodium phosphate in moles that has been dissolved in 1.2 liters. But that does not tell you what
43:08
the concentration of sodium ions are. Yes. So you can take this and divide it by liters and give you the molarity of sodium phosphate, but you have three moles of sodium ions for each one mole of the compound. So you have to multiply that by three.
43:25
So you have to multiply that by three. So you have to multiply that by three. So you have to multiply that by three. That three should not come as a total mystery to you. This is not a mystery. This is- there's no other way this can be. Because there's three sodium ions for one mole of the compound. You have to multiply this by three. You get three times as much sodium as you have this. As you have sodium phosphate.
43:46
So, the answer is in this case, 7.2 times 10 to the minus 2. Give your answer in the correct number of sig figs, preferably in scientific notation, and with the unit. Please add a unit so we know what you are talking about.
44:05
Moles per liter is basically the unit for molarity. So you have to put a big hand there. Okay. Two more minutes, naming compounds is something we have to be able to do. Here's one. What is that?
44:25
Beryllium sulfide. Yep. Beryllium, by the way, is 2+. Sulfide is always 2-. Good to know. What is this? Phosphorus
44:41
tribromide. Tribromide, because this is a forbidden compound, a molecular compound, this is not an ionic compound. So you have to tell the reader exactly how many anions you have in this case. How about this?
45:02
Ammonium hydroxide. Yes, ammonium hydroxide. This is the ammonium cation if the H plus splits off, you get NH3 which is a molecule that's the ammonium molecule. The ammonium molecule we have seen many times. Ammonium is a molecule if you add a proton to it, it becomes ammonium.
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NO2. What is that? Nitrogen dioxide. Nitrogen dioxide. It's not nitrite. Why not? Because there is no minus there. Polyatomic anions always have their charges indicated.
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Always. Some people ask me, do I have to write down the charges of ions? The answer is yes! Is that clear? Yes! Always add the charge. The charge of the ions. If you don't do that, you just write down something else. For instance, if you forget the
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if this was nitrite and you forget the charge, it becomes nitrogen dioxide. This is completely different. What is this? Sodium nitride. Sodium nitride. Because this is an anion that has three negative charges, right?
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Let's do one more. What's that? Aluminum oxide. Yes. It's aluminum oxide, but do I have to say, do I have to stipulate an oxidation state of aluminum? No. You don't. Because aluminum is a type 1 metal, not a type 2. It only has one
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oxidation state, which is 3 plus ammonium oxide. All right. See you on Friday for the second session of the review.