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Chemistry: Boiling Point Elevation Problem

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  • Type: Video Tutorial
  • Length: 12:53
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 139 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Physical Properties of Solutions (14 lessons, $22.77)
Chemistry: Colligative Properties (5 lessons, $9.90)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

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Thinkwell
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When a solute is put in a solvent, the boiling point is raised and the freezing point is lowered. Now, remember this is an example of a colligative property. And again, we might want to ask ourselves, so what's the big deal. Well, there's a lot of practical applications of that simple idea. Remember that it's independent of the chemical nature of the thing you're dissolving. So, if you're worried out boiling point rising, for instance, one application we talked about is boiling in an engine. Let's suppose that you're in Denver and the atmospheric pressure is lower and so that means the water is going to boil at a lower temperature, as we have seen before. And you want to be able to somehow speed up the time it takes to cook spaghetti or bring yourself back up to a higher temperature. Well, now you know one way to do it, you can add salt, dissolve it in the water, that raises the boiling point of the water again. So, there's lots of practical things that you can do as far as manipulating the temperature at which something boils at. Another application that you're all familiar with, or certainly if you were in the East Coast. You're familiar with when the roads freeze, trucks go out and they sprinkle salt on the roads. They sprinkle the salt on the roads simply to dissolve into the liquid phase and that lowers the freezing point and that causes the roads to melt, as long as you don't get too low in temperature. Understand that if the temperature is too low, so that it is below the freezing point of the solution, well then you're not going to get any melting in that stage. So, again, there are lots of common applications for where you would want to lower a melting point or raise a boiling point of a substance.
Let me show you another one here. Let's suppose that one day, when I was upstairs, I found in my nine year-olds room, a mysterious white substance behind his socks in the drawer. And I was a little bit suspicious about the friends he had been playing with and just what this material might be. So, I very carefully removed it from the drawer and immediately took it to my lab. Well, I'm going to use the notion of boiling point elevation to determine what this substance is. Now I'm not going to just use boiling point elevation. I'm going to take some other information as well. I can use a different technique called combustion analysis and that will tell me simply the ratio of one atom type to another in this material. So, suppose I find that there is a certain percentage of carbon in this material and a certain percentage of hydrogen. And for the sake of argument, let's say that's all that there is in here, carbons and hydrogens. But I don't know what the substance is and in particular, I don't know what the molecular weight of this material is. How am I going to find that? Well, what I'm going to do is I'm going to take a solvent, in this case chloroform, that I know the boiling point for. And I can just read what the boiling point is, in fact. I'll weigh out a certain amount of material - an arbitrary amount - let's say 5 grams. And I'll put this stuff into the solvent. Now, that's going to cause a boiling point elevation when I do this. Let's suppose that I weighed that out exactly. Pretend I weighed out exactly five grams and I get a boiling point change as a result of that five grams being dissolved. Now, remember, this being a colligative property, it is independent of the chemical nature of this stuff. But it does depend on the amount of moles per kilogram solvent that I put in. So, remember this is the relationship that we're talking about. The change of temperature between the boiling point of the pure substance and then the solution. That is directly proportional to the concentration in terms of moles per kilogram solvent that I have. And what ties those ideas together, those two numbers, is this guy, the boiling point elevation constant. Which, for chloroform, I know and we will get to in just a moment. By knowing that number and by knowing the change in temperature, which I can read right off the thermometer, I can determine my concentration. Now, if I know how much mass I put in and I know how much moles per kilogram solvent I must have had to cause that temperature change, the only piece I'm missing is what my molecular weight. So I can determine my molecular weight by knowing those other things. Once I know the molecular weight, that gives me a good handle on what the substance actually is. So, again, what we're going to do is we've determined a boiling point change by putting 5 grams in. I am going to, first of all, figure out the ratio of carbon to hydrogen for you. Once we know that ratio, that will be an empirical formula but it won't tell us the molecular formula. To figure out the molecular formula, I need to know the number of moles that I put in there. The way I'm going to figure out the number of moles that I put in there in that 5 grams is by figuring out my concentration of moles per unit of volume of the solvent and that I'm going to be able to figure out from this colligative property, by knowing my change in temperature.
So, let's go ahead and do that. So here's our problem. We have an unknown sample, elemental analysis tells us that it's 93.7% carbon and 6.3% hydrogen. Now, remember, this is just percent by mass. That says the total mass of carbon divided by the total is 93.7%, and likewise for hydrogen. When I take 5 grams of this material, and again, pretend that I weighed out exactly a certain amount, I dissolve it, let's say, 100 grams of chloroform so I know my mass of solvent that I had so it's going to be easy for me to figure out the molal concentration. The boiling point is raised from 61.2 degrees to 62.6 degrees. So that's going to give me my information about what my temperature change is for the boiling point. My goal is to find the molecular formula for this stuff. And of course what I need is that tie between concentration and boiling point change and that is the boiling point elevation constant given for chloroform down here.
So, first of all, let's go ahead and figure out what our moles of carbon to hydrogen would be. That's going to be the first important clue here. I know I've got 93.7% carbon. So let's take a convenient mass that is going to make this work out easy, because I could take any arbitrary amount. Let's suppose that I had 100 grams of this stuff. Well, 93.7 grams of that 100 grams would be due to carbon. So, I'm going to take that 93.7 grams, divide it by the atomic mass of carbon and I'm going to end up with 7.81 moles of carbon, if I had a 100 grams of this stuff. Now, I'm going to take 6.3 grams, which is the hydrogen, remember 6.3% hydrogen. So 6.3 grams of that stuff would be hydrogen. Divide that by its atomic mass and that gives me 6.3 moles of hydrogen. Now, in that 100 grams then, that hypothetical 100 grams, I've got a ratio of carbon to hydrogen of 7.81 to 6.3. That turns out to be a ratio of 5:4. If I just simply take that ratio again. So, what I know so far is that I have 5 carbon atoms for every 4 hydrogen atoms. But what I don't know is, does the molecule actually have as its formula , or does it have a formula, or maybe its . See I don't know the molecular weight of the thing, only the ratio of one atom type to another. So, I'm going to use the colligative property to give a good estimate for what the actual molecular mass of this material is.
So, once again, we're going to start with the idea that I can measure directly the change in temperature caused by me putting that 5 grams in here. It turns out to be 1.4 degrees and that's 1.4 degrees Celsius, which is the same as 1.4 degrees Kelvin, and that remember is equal to or constant for chloroform times the concentration. That's what I want to know, my concentration. Adding 5 grams gave me a certain concentration causing that temperature change and I need to find out what that is. So, jut rearranging then, 1.4 3.63 gives me my moles per kilogram solvent, remember concentration in molal is equal to 0.386 molal. So that's my concentration of the substance, 0.386 molal and I want to know specifically how many moles I put in that 5 grams. So I got my concentration but I want to know how much stuff I actually put in this beaker. So, to do that, I'm going to take my molal concentration and multiply it by the actual mass of the solvent. So, if I have 0.386 moles per kilogram of chloroform, then I would have 0.386 times point one kilograms. That's 100 grams and that's going to give me 0.038 moles. So, again, all I did was I know I have 0.386 moles per kilogram. I'm going to have a tenth of that if I only have a tenth as much of my mass of solvent. So, now I know my moles of this white stuff that I put in, in that 5 grams that I put in to cause that boiling point change.
So, I'm almost there. 0.0386 moles is in that 5 grams. So, 5 grams divided by the number of moles that it is equivalent to. And that is going to give me my molar mass. Remember, after all, molar mass is just grams per mole. So, I'm taking five grams of stuff and how many moles I have in that 5 grams. That gives me my molar mass. So my molar mass is 128 grams per mole and if I indeed had ; that would only be a molar mass of 64 grams per mole. So that's telling me that I have something twice as big as that simple empirical formula. In other words, in order to have that molar mass, I need to have a molecular formula of . Well, that turns out to be the molecular weight or molar mass of naphthaline. Naphthaline is mothballs. And so my son had some mothballs stashed away in his drawer, probably not put there by son but by his mother and so I guess he's out of the doghouse.
Anyway, the connection that we made here is using a colligative property, the change in this case in temperature of boiling and how that directly connects to concentration of the solute that we put in to cause that change. By knowing concentration, we can go from there to moles because we knew how much stuff we put in, the mass of it, we could figure out the number of moles by knowing the moles and how much stuff we had. That gives us then the molecular weight of the material, or the molar mass of the material.
In fact we can use this strategy for a number of different colligative properties. We could have used it for vapor pressure lowering, we could use it for freezing point depression, but in fact the one colligative property we have not spoken about yet that is coming next is the most sensitive to small changes in concentration of solute and that's the one you use most actually in laboratories to determine molar masses of materials. That last colligative property is osmosis.
Physical Properties of Solutions
Colligative Properties
Boiling Point Elevation Problem Page [1 of 2]

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