TBR Stoichiometry Passage IV - Question 26 - (WRONG ANSWER!?)

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saoj

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I am probably the one wrong here what makes me mad. But there is a tiny chance TBR is wrong, probably not. :(. But why ???

I believe it is wrong simply because air REMAINED in the flask so the mass must have been greater. But somehow TBR says the mass will be smaller. I think it is answer A but TBR things it is answer D.

Any hope for me? :xf:

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If air remained in the flask you would measure the weight of less than one liter of the unknown (in its gas form). Since you'll assume that you have collected a full liter, your calculated MW will be too low. Answer D is correct.
 
I am probably the one wrong here what makes me mad. But there is a tiny chance TBR is wrong, probably not. :(. But why ???

Here is the passage:
http://notes.soliveirajr.com/docs/TBR-IV.png

Here is question:
http://notes.soliveirajr.com/docs/question-26.png

I believe it is wrong simply because air REMAINED in the flask so the mass must have been greater. But somehow TBR says the mass will be smaller. I think it is answer A but TBR things it is answer D.

Any hope for me? :xf:


D is right. If you don't fill the flask with vapor (because there is air taking up some of the space) you won't get as much liquid. Air does not condense at room temperature, so instead of having 1L of unknown vapor condensing to the liquid that you weigh, you have <1L of unkown vapor and therefore less unknown liquid.
 
D is right. If you don't fill the flask with vapor (because there is air taking up some of the space) you won't get as much liquid. Air does not condense at room temperature, so instead of having 1L of unknown vapor condensing to the liquid that you weigh, you have <1L of unkown vapor and therefore less unknown liquid.

I see I have a tendency to complicate the simple. That's not good. The mass of air does not matter. I am after the mass for ONE litter. So the liquid most evaporate completely and fill the flask completely to get 1L.

If air remains inside you will get LESS mass for 1L, in other words, the real mass would be greater than the actual mass measured (for 1 L). So the mass in your calculation would be less than the real one, making the molecular weight less than the real one.

It is worth mentioning that less molecular weight it is due to the fact that:

no of moles will = 1 L / 24.96 L

no of moles = mass / MW

MW = mass / no of moles

Same number of moles and LESS mass, makes MW smaller.

Had he asked: "Will the number of moles change?" The answer would be NO.

THANK YOU VERY MUCH TO MILSKI AND MEDPR.

I guess to put it in a way that silly me can understand:

"You are after the mass for ONE LITER. If you have less than one liter in there you will get a SMALLER mass for one liter. Number of moles will be the same and when you go calculate the MW, you get a smaller numerator and consequently a smaller MW."

I guess that's the silly-proof explanation. :-|
 
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Keep in mind that it is only in your calculations that the number of moles does not change. In the real experiment, the number obviously changed. That explains the discrepancy between your results and the real value of MW.

That seems to be popular type of questions: If we do experiment X and something unaccount happens, what will happen to your results. You need to think carefully about what you're calculating and which of your assumptions are different from what really happened.
 
Yes, you're correct. It's not a silly way to explain it, that's the same type of thing I do when I'm confused about a concept or answer. I explain it in lay terms, then go back and try to apply formulas/concepts to make sure I really get it.

You can think of it this way.

I have 500mL of water. I give it to you and say "this is 1000mL of an unknown liquid. Weigh it and figure out its density." Simple, right? All you have to do is get the mass, then divide to find the density.

However, I lied to you and told you it was 1000mL when it was really only 500mL. When you do your calculation, you won't get the right answer because you will be calculating the density based on the assumption that I gave you 1L instead of 0.5L. So, instead of getting a density of 1g/cc, you will get 0.5g/cc.

But yes, your reasoning is correct.
 
Keep in mind that it is only in your calculations that the number of moles does not change. In the real experiment, the number obviously changed. That explains the discrepancy between your results and the real value of MW.

We have two scenarios:

1) Wrong volume (1L when it was less)

The number of moles measured was 1 L / 24.96 L

2) Right volume (1L)

The number of moles measure was again 1 L / 24.96 L

What you are saying is that although we measured the same number of moles in both scenarios, the number of moles in the first scenario was calculated incorrectly. The real volume is less than 1L, so the REAL number of moles is smaller than what was incorrectly calculated.

So perhaps, and again I have a feeling I am complicating the simple, the correct option D could have also said:

"The mass of unknown liquid collected would be too small, so the calculated molecular mass would be to low, even though the *calculated* number of moles would be the same."

Do you agree or disagree?

One thing is the CALCULATED number of moles in the experiment, either the correct experiment or the wrong experiment. Another thing is the REAL number of moles. Obviously less mass OR less volume leads to a smaller number of moles.
 
We have two scenarios:

1) Wrong volume (1L when it was less)

The number of moles measured was 1 L / 24.96 L

2) Right volume (1L)

The number of moles measure was again 1 L / 24.96 L

What you are saying is that although we measured the same number of moles in both scenarios, the number of moles in the first scenario was calculated incorrectly. The real volume is less than 1L, so the REAL number of moles is smaller than what was incorrectly calculated.

So perhaps, and again I have a feeling I am complicating the simple, the correct option D could have also said:

"The mass of unknown liquid collected would be too small, so the calculated molecular mass would be to low, even though the *calculated* number of moles would be the same."

Do you agree or disagree?

One thing is the CALCULATED number of moles in the experiment, either the correct experiment or the wrong experiment. Another thing is the REAL number of moles. Obviously less mass OR less volume leads to a smaller number of moles.


Yes, if you put 1L into your calculation, it doesn't matter what the real value was. I don't know why you're trying to argue that point. If you have 50 billion liters, but you only put 1L into your calculation, you will get the same answer as if you had 10^-50billion liters and put 1L into your calculation.
 
So perhaps, and again I have a feeling I am complicating the simple, the correct option D could have also said:

"The mass of unknown liquid collected would be too small, so the calculated molecular mass would be to low, even though the *calculated* number of moles would be the same."

Do you agree or disagree?

That would also be a correct answer, I agree. There are no reasons to talk about any other number of moles, so don't expect to ready something like that in a real question.
 
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