question in chemistry berkeley review

[batista_123]

I dont understand this even after reading the answer
my question is on page 36 of volume 1 of general chemistry from the berkeley review
since we may have different editions, i will give a summary of the passage:
a researcher places 5mL of an unknown liquid in a capped 1 L flask. the cap has a tiny hole. the compound is heated until it reaches a gentle boil. the vapor escapes thru the tiny hole. The liquid continues boiling at 304K, until none of it remains visible in the flask. The heat source is removed, and the contents are allowed to cool back to room temperature. As the flask cools, the vapor in it becomes liquid. The liquid inside the flask weighs 2.32 grams. It is assumed that at the moment when the heat source was removed, the flask was completely filled with vapor from the liquid and that all of the air originally in the flask was displaced. At 273K, 1 mol of any gas occupies 22.4 L at 273K and 24.96L at 304K.

Now the question:
if the organic vapor had not fully displaced all of the air from the flask by the time the heat was removed from the flask, how would the results have been different?
a) The mass of unknown liquid collected would be too high, so the calculated molecular mass would be too high
b) high, low
c) low, high
d) low, low
I cannot relate this to any concepts in gchem, but i suspect that it has something to do with if you have impurities in a liquid, the boiling point goes up?? but i am not sure. can you help me understand why the answer is d?

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[Fort]

I thought about this one for a while, and to be honest I would've said "A" as well. I think I know what they're getting at, however.

We need to think about how condensation of the unknown liquid will be affected by the air particles lingering in the flask. As you thought, the air particles are impurities, and will affect the extent to which the liquid condenses. What I'm thinking is the impurities will prevent a small fraction of the unknown liquid to condense due to intermolecular interactions (collisions with other particles may prevent the molecules from slowing down enough to condense). This may be similar to the "freezing point depression, boiling point elevation" idea with impurities dissolved in liquids. Not sure if this can be applied to gasses, though.

Anyway, that's the best I have to explain "D". If not as much liquid condenses, then the mass will be lower, as well as the calculated molecular mass.

Anyone else want to chime in?

 

[BerkReviewTeach]

I always hated error analysis questions. If you consider the way the Dumas experiement is supposed to work, then you know that ideally the flask has 100% vapor (no liquid left and no air present) at its boiling point.

If there is any air present in the flask (an open system), then that means that there is less vapor present than if it were pure. When you condense the vapor into liquid, you get less collected than expected because there's less vapor than there should be. With less liquid condensed, there will be less mass of liquid, which eliminates choices A and B.

The experiment requires that you measure the mass of the vapor (which equals the mass of the collected liquid) and get the moles of vapor using PV = nRT to determine n. The molecuar mass has mass in the numerator, so the error results in a numerator that is too small. The final calculated number for molecular mass is lower than expected. This makes choice D the best answer.

Again, I hate error analysis questions, but unfortunately the MCAT has them.

 

[batista_123]

a liquid at its boiling point is all vapor? i thought liquid and vapor are at equilibrium, meaning half vapor but half remains liquid, no?

also another question that is bugging me. i thought about this for a while, and i thought i would just forget about it, but i know if i forget about it, the mcat writers will put like 5 of them on my test.
somewhere at the beginning of the BR chemistry it says if you have 2 molecules that have C, H, and O only, the one that has fewer Os always has the higher carbon mass percentage.
you know what i mean?
suppose the carbon mass percentage in one molecule is 55%.
in another molecule its 45%
then in the first molecule, you have fewer oxygens
is this true?

 

[batista_123]

another question
you say "If there is any air present in the flask (an open system), then that means that there is less vapor present than if it were pure. When you condense the vapor into liquid, you get less collected than expected because there's less vapor than there should be."
i dont understand why. i think if there is air in there, its going to prevent the liquid from boiling, right? so you would have some liquid in there that shouldnt be there. I understand that the presence of air (or any impurity) prevents the liquid from boiling, but that's where i get lost.

 

[loveoforganic]

What's your rationale for the liquid not boiling if there's air in the container?

 

[sleepy425]

another question
you say "If there is any air present in the flask (an open system), then that means that there is less vapor present than if it were pure. When you condense the vapor into liquid, you get less collected than expected because there's less vapor than there should be."
i dont understand why. i think if there is air in there, its going to prevent the liquid from boiling, right? so you would have some liquid in there that shouldnt be there. I understand that the presence of air (or any impurity) prevents the liquid from boiling, but that's where i get lost.

The question has nothing to do with the air preventing boiling. Nothing. Nothing nothing nothing.

Think about the experiment. What happens? You take this liquid, and you start to boil it. As the liquid goes into the vapor phase, the vapors start to displace the air that was above the liquid, and the air gets pushed out of the pinhole. Also, some of the unknown sample in vapor form will also get pushed out of the pinhole. You take it off of the heat when all of the liquid is gone. At that point, there is supposed to be only unknown vapor in the flask. Since we're at 304 degrees, there are 24.96 L in one mole of our vapor. Since the flask is 1 L, that means we have 1/24.96 moles. Ok, now we let it cool and then weigh the liquid that condenses. So now we have a mass in grams, and the number of moles (1/24.96). So you take the mass in grams and divide by the number of moles and you have your molecular weight in g/mol.

Ok, so that's how the experiment is supposed to work. Now we do it again but with air left in it. So we take the liquid and start to boil it. Vapors start to displace the air, but for some reason, we don't displace all of the air by the time all the liquid is gone (the experimenter doesn't realize that he hasn't gotten rid of all of the air though. he is still operating under the assumption that all of the air is gone, even though that assumption is wrong). So now we have a flask that has, idk, say 90% unknown vapor, and 10% air. Let's do our calculation again. We're still at 304 degrees, there are 24.96 L in one mole of our vapor, and since our flask is a 1 L flask, that means the number of moles we calculate in the flask is 1/24.96. So that's still the same. Ok, we cool it down, the unknown vapor condenses. But last time, the entire flask was full of this unknown vapor. This time, we have 90% as much in the flask, so when it condenses, our mass will be 90% what we measured when we got all of the air out. So that means that the mass of the liquid will be less when we don't get all of the air out (that's the first part of the question). Now we do the calculation for molecular weight. It was mass/moles last time, so this time, our mass is lower, but number of moles is the same, so our calculated molecular mass is going to be lower than it was when we got all of the air out. So the answer is D.

I have a few things to point out/emphasize. First, remember that the experimenter doesn't know that he didn't get all of the air out. There's really no way for him to know. So the question is asking, what happens to the calculations when the assumptions you make are wrong? The problem arises because when the experimenter goes to calculate the number of moles, he is assuming that all of the vapor in there is the unknown, so no matter how much air is in there, that calculation will always be the same. Then when the experimenter lets the stuff condense, the mass is going to be less if there were air there, because the air takes up some of the space that would otherwise be occupied by the unknown vapor. So there's less of the unknown vapor in the flask.

Remember, this has nothing to do with any sort of interactions between the air and the unknown vapor. It is simply asking you to think about the experiment and work out what happens if an assumption is wrong.

 

[batista_123]

sleepy425, you said, "But last time, the entire flask was full of this unknown vapor. This time, we have 90% as much in the flask, so when it condenses, our mass will be 90% what we measured when we got all of the air out."

ok, very nice explanation, but i have a problem. if only 90% of the contents is the unknown liquid vapor, that doesn't necessary mean it weighs less than if all of it was the unknown vapor. because what if the molecules are now much closer to each other when you have 90%?

i will explain it again if that didn't make sense:

in the first case when there was no air, 100% vapor.

in the second case, 90% of the container is vapor, 10% air. but maybe now those vapor molecules are much closer to each other so although there is only 90% vapor, there is more of them so they weigh more.

 

[sleepy425]

sleepy425, you said, "But last time, the entire flask was full of this unknown vapor. This time, we have 90% as much in the flask, so when it condenses, our mass will be 90% what we measured when we got all of the air out."

ok, very nice explanation, but i have a problem. if only 90% of the contents is the unknown liquid vapor, that doesn't necessary mean it weighs less than if all of it was the unknown vapor. because what if the molecules are now much closer to each other when you have 90%?

i will explain it again if that didn't make sense:

in the first case when there was no air, 100% vapor.

in the second case, 90% of the container is vapor, 10% air. but maybe now those vapor molecules are much closer to each other so although there is only 90% vapor, there is more of them so they weigh more.

that would be true if the container were completely sealed so that pressure could build up. since there is a pinhole, the pressure doesn't build up. thus, the pressure and temperature will be the same in both cases, so the amount of vapor molecules is directly proportional to the volume, which is 90% of the total in this case.

To put it more simply, given a certain temperature and pressure of gases, you will have a fixed number of molecules. if you add a different gas under these conditions, it will simply displace some of the other gas. if you seal the container so that none of the gas can escape, and then add a different gas, then you will increase the number of molecules and pressure will build up

 

[altitude]

The question has nothing to do with the air preventing boiling. Nothing. Nothing nothing nothing.

Think about the experiment. What happens? You take this liquid, and you start to boil it. As the liquid goes into the vapor phase, the vapors start to displace the air that was above the liquid, and the air gets pushed out of the pinhole. Also, some of the unknown sample in vapor form will also get pushed out of the pinhole. You take it off of the heat when all of the liquid is gone. At that point, there is supposed to be only unknown vapor in the flask. Since we're at 304 degrees, there are 24.96 L in one mole of our vapor. Since the flask is 1 L, that means we have 1/24.96 moles. Ok, now we let it cool and then weigh the liquid that condenses. So now we have a mass in grams, and the number of moles (1/24.96). So you take the mass in grams and divide by the number of moles and you have your molecular weight in g/mol.

Ok, so that's how the experiment is supposed to work. Now we do it again but with air left in it. So we take the liquid and start to boil it. Vapors start to displace the air, but for some reason, we don't displace all of the air by the time all the liquid is gone (the experimenter doesn't realize that he hasn't gotten rid of all of the air though. he is still operating under the assumption that all of the air is gone, even though that assumption is wrong). So now we have a flask that has, idk, say 90% unknown vapor, and 10% air. Let's do our calculation again. We're still at 304 degrees, there are 24.96 L in one mole of our vapor, and since our flask is a 1 L flask, that means the number of moles we calculate in the flask is 1/24.96. So that's still the same. Ok, we cool it down, the unknown vapor condenses. But last time, the entire flask was full of this unknown vapor. This time, we have 90% as much in the flask, so when it condenses, our mass will be 90% what we measured when we got all of the air out. So that means that the mass of the liquid will be less when we don't get all of the air out (that's the first part of the question). Now we do the calculation for molecular weight. It was mass/moles last time, so this time, our mass is lower, but number of moles is the same, so our calculated molecular mass is going to be lower than it was when we got all of the air out. So the answer is D.

Sorry to bump a really old post, but I cannot understand (and it's really bothering me) how the mass would change based on whether or not the organic vapor completely displaces all the air or not. If the organic vapor doesn't displace the air, how is mass lost? If in each of the two scenarios the same mass of liquid is used and is completely evaporated into a liquid, what would account for a decrease in mass (where does this lost mass come from)?

 

[altitude]

This is related to the last question of passage IV in section I of TBR general chem. Please help!

 

[altitude]

Sorry to bump a really old post, but I cannot understand (and it's really bothering me) how the mass would change based on whether or not the organic vapor completely displaces all the air or not. If the organic vapor doesn't displace the air, how is mass lost? If in each of the two scenarios the same mass of liquid is used and is completely evaporated into a liquid, what would account for a decrease in mass (where does this lost mass come from)?

Never mind- I re-read a previous post a couple more times and figured it out.

 

[anomedic]

Why wouldnt the air condense to form water and cause an apparent increase in mass of the unknown substance?
never mind it says the substance boils at 31 C in the passage

 

[BerkReviewTeach]

Why wouldnt the air condense to form water and cause an apparent increase in mass of the unknown substance?
never mind it says the substance boils at 31 C in the passage

Because air is mostly N2 gas and O2 gas, which won't condense at room temperature.

 

[IcemanMD]

I have a question about this one...If all of the liquid is evaporated as stated in the passage but there is less vapor present due to the air, is the vapor just more compressed due to the air? In this case wouldn't the mass of liquid collected be the same as the original experiment?