Freezing-Point Depression to Determine an Unknown Compound | Protocol
Freezing-point depression is the decrease of the freezing point of a solvent on addition of a The molar mass of a solute is determined by comparing mB with the amount of Consider the problem in which the solvent freezes to a very nearly pure a simple linear relationship with the cryoscopic constant ("Blagden' s Law"). Problem # What is the molar mass of g of an unknown substance that depresses the freezing point of kg of water °C? Kf for water is °C/ m. Problem: g of naphthalene (C10H8) is dissolved in g of benzene. The freezing point of pure Step 1: Calculate the freezing point depression of benzene. delta Tf = (Freezing point molality = moles of solute / kg of solvent moles of.
The water molecules are going to have a regular structure where the hydrogen bonds dominate any kind of kinetic movement they want to do, and all the kinetic movement, they're just vibrating in place. So you have to get a little bit orderly right there, right? And then, obviously, this lattice structure goes on and on with a gazillion water molecules.
- Freezing-Point Depression to Determine an Unknown Compound
- How does molality affect the freezing point?
- Boiling point elevation and freezing point depression
But the interesting thing is that this somehow has to get organized. And what happens if we start introducing molecules into this water? Let's say the example of sodium-- actually, I won't do any example.
Let's just say some arbitrary molecule, if I were to introduce it there, if I were to put something-- let me draw it again. So now I'll just use that same-- I'll introduce some molecules, and let's say they're pretty large, so they push all of these water molecules out of the way.What's the Point of Molality?!?
So the water molecules are now on the outside of that, and let's have another one that's over here, some relatively large molecules of solute relative to water, and this is because a water molecule really isn't that big. Now, do you think it's going to be easier or harder to freeze this? Are you going to have to remove more or less energy to get to a frozen state? Well, because these molecules, they're not going to be part of this lattice structure because frankly, they wouldn't even fit into it.
They're actually going to make it harder for these water molecules to get organized because to get organized, they have to get at the right distance for the hydrogen bonds to form. But in this case, even as you start removing heat from the system, maybe the ones that aren't near the solute particles, they'll start to organize with each other.
But then when you introduce a solute particle, let's say a solute particle is sitting right here. It's going to be very hard for someone to organize with this guy, to get near enough for the hydrogen bond to start taking hold. This distance would make it very difficult. And so the way I think about it is that these solute particles make the structure irregular, or they add more disorder, and we'll eventually talk about entropy and all of that. But they make it more irregular, and it's making it harder to get into a regular form.
And so the intuition is is that this should lower the boiling point or make it-- oh, sorry, lower the melting point. So solute particles make you have a lower boiling point. Let's say if we're talking about water at standard temperature and pressure or at one atmosphere then instead of going to 0 degrees, you might have to go to negative 1 or negative 2 degrees, and we're going to talk a little bit about what that is.
Now, what's the intuition of what this will do when you want to go into a gaseous state, when you want to boil it? So my initial gut was, hey, I'm already in a disordered state, which is closer to what a gas is, so wouldn't that make it easier to boil?
But it turns out it also makes it harder to boil, and this is how I think about it. Remember, everything with boiling deals with what's happening at the surface, and we talked about that in our vapor pressure.
So at the surface, we said if I have a bunch of water molecules in the liquid state, we knew that although the average temperature might not be high enough for the water molecules to evaporate, that there's a distribution of kinetic energies.
And some of these water molecules on the surface because the surface ones might be going fast enough to escape. And when they escape into vapor, then they create a vapor pressure above here. And if that vapor pressure is high enough, you can almost view them as linemen blocking the way for more molecules to kind of run behind them as they block all of the other ambient air pressure above them.
So if there's enough of them and they have enough energy, they can start to push back or to push outward is the way I think about it, so that more guys can come in behind them. So I hope that lineman analogy doesn't completely lose you.
Now, what happens if you were to introduce solute into it? Some of the solute particle might be down here. It probably doesn't have much of an effect down here, but some of it's going to be bouncing on the surface, so they're going to be taking up some of the surface area. And because, and this is at least how I think of it, since they're going to be taking up some of the surface area, you're going to have less surface area exposed to the solvent particle or to the solution or the stuff that'll actually vaporize.
You're going to have a lower vapor pressure. And remember, your boiling point is when the vapor pressure, when you have enough particles with enough kinetic energy out here to start pushing against the atmospheric pressure, when the vapor pressure is equal to the atmospheric pressure, you start boiling.
But because of these guys, I have a lower vapor pressure. So I'm going to have to add even more kinetic energy, more heat to the system in order to get enough vapor pressure up here to start pushing back the atmospheric pressure. Such creatures have evolved means through which they can produce high concentration of various compounds such as sorbitol and glycerol.
This elevated concentration of solute decreases the freezing point of the water inside them, preventing the organism from freezing solid even as the water around them freezes, or as the air around them becomes very cold. Examples of organisms that produce antifreeze compounds include some species of arctic -living fish such as the rainbow smeltwhich produces glycerol and other molecules to survive in frozen-over estuaries during the winter months.
In the case of the peeper frog, freezing temperatures trigger a large-scale breakdown of glycogen in the frog's liver and subsequent release of massive amounts of glucose into the blood.
The degree of dissociation is measured by determining the van 't Hoff factor i by first determining mB and then comparing it to msolute. In this case, the molar mass of the solute must be known. The molar mass of a solute is determined by comparing mB with the amount of solute dissolved.
In this case, i must be known, and the procedure is primarily useful for organic compounds using a nonpolar solvent. Cryoscopy is no longer as common a measurement method as it once was, but it was included in textbooks at the turn of the 20th century.
As an example, it was still taught as a useful analytic procedure in Cohen's Practical Organic Chemistry of in which the molar mass of naphthalene is determined using a Beckmann freezing apparatus.
Freezing-point depression can also be used as a purity analysis tool when analysed by differential scanning calorimetry. This is also the same principle acting in the melting-point depression observed when the melting point of an impure solid mixture is measured with a melting-point apparatussince melting and freezing points both refer to the liquid—solid phase transition albeit in different directions.
Chemical potential is the molar Gibb's energy that one mole of solvent is able to contribute to a mixture.
ChemTeam: Freezing Point Depression Problems #
The higher the chemical potential of a solvent is, the more it is able to drive the reaction forward. Consequently, solvents with higher chemical potentials will also have higher vapor pressures. Boiling point is reached when the chemical potential of the pure solvent, a liquid, reaches that of the chemical potential of pure vapor. Because of the decrease of the chemical potential of mixed solvents and solutes, we observe this intersection at higher temperatures.
In other words, the boiling point of the impure solvent will be at a higher temperature than that of the pure liquid solvent. Freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent.