Finding the gas based off of mass

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In summary, the conversation was about determining the gas based on the given kinetic energy and impulse of a single molecule. Through calculations, it was determined that the mass of the molecule is approximately 8.29 times the mass of a proton or neutron. This suggests that the molecule is either krypton or hydrogen bromide, with krypton being the more likely option. The conversation also included some clarification on basic arithmetic and the concept of molecular weight.
  • #1
mk9898
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Homework Statement


Die average kinetic energy and impuls squared of a single molecule are:
$$<E_{kin}>= 6.5*10^{-21} J$$ and $$<p^2> = 1.8*10^{-45} kg^2m^2s^{-2}$$

Which gas ist it?

3. The Attempt at a Solution

##\frac{m<v^2>}{2}=\frac{<p^2>}{2m}## ##\Leftrightarrow## ##\frac{<p^2>}{<v^2>} = m^2##

Therefore is ##m^2 = 2.77*10^{-25}kg ## and ##m = 5.26*10^{-13}kg##

I searched and attempted to find with gas it is based off of the mass and also multiplying it by a mole but to no avail.
 
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  • #2
The equation for kinetic energy reads ##<E>=\frac{m <v>^2}{2} =\frac{<p^2>}{2m} ## and not ## \frac{m}{2 <v>^2} ##. Try again=this one should be fairly straightforward.
 
  • #3
Yea that's just a typo in the latex of course. You can tell from the next step that the p and v variables are where they need to be. The value for the mass above is correct.
 
  • #4
mk9898 said:
Yea that's just a typo in the latex of course. The value for the mass above is correct.
There's no ## m^2 ## in this calculation. You should be able to get ## m=\frac{<p^2>}{2 <E_{kin}>} ##.
 
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  • #5
Ah yea of course. The value I get is 1.3846*10^-25. Even then I'm not sure how I can tell which gas that is. If I multiply a mole with it, it's still smaller than hydrogen.
 
  • #6
mk9898 said:
Ah yea of course. The value I get is 1.3846*10^-25. Even then I'm not sure how I can tell which gas that is. If I multiply a mole with it, it's still smaller than hydrogen.
The proton has mass ## m_p=1.67 \cdot 10^{-27} \, kg ##. The molecule is much heavier than that.
 
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  • #7
How can that be the mass of a proton when it states in the exercise, that the molecule has the kinetic energy and the impulse? Meaning, m would be the entire molecule? Or why is it the mass of a proton?
 
  • #8
mk9898 said:
How can that be the mass of a proton when it states in the exercise, that the molecule has the kinetic energy and the impulse? Meaning, m would be the entire molecule? Or why is it the mass of a proton?
How many protons and/or neutrons (both have approximately the same mass) do you need to get the mass ## m ## of the molecule?
 
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  • #9
Hmm. I have to be honest here...I have no idea. This wasn't covered in the lecture and from the readings I've done, it doesn't touch on this.
 
  • #10
If you take the mass ## m ## you computed, and divide by 1.67 E-27, this will tell you the molecular weight of the molecule. I computed this already, but I so far haven't identified what the molecule most likely is.
 
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  • #11
I would guess it's oxygen but where does the 1.67e-27 come from? My apologies if I'm asking rudimentary questions. I really want to learn this stuff. Thanks a lot for the effort.
 
  • #12
The proton and neutron mass have ## m_p \approx m_n \approx 1.67 \cdot 10^{-27} ## kg. If you divide the molecular mass by the mass of the atomic building block, it will tell you how many neutrons and protons you have in the molecule. The problem isn't complete there though. I'm looking over the periodic table and trying to get a match. The closest thing I have paired up is Krypton, which is simply an atom. ##SO_2 ## and ## Cl_2 ## are both quite a ways off.
 
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  • #13
Thanks for the clarification. The number I get is 8.29 which I would assume would give us atomic number 8.
 
  • #14
mk9898 said:
Thanks for the clarification. The number I get is 8.29 which I would assume would give us atomic number 8.
Try your arithmetic again. I get 83. ## \\ ## When they tell you it is a molecule, it really should be a molecule, but the only gas I could find with atomic weight near 83 that is at all common is Krypton, at 83.8. They only give 2 sig figs in the statement of the problem. Another possibility though is ## HBr ##=hydrogen bromide=molecular weight=81. ## \\ ## As a homework helper, I'm not supposed to supply answers, but, assuming our arithmetic is correct, this question is somewhat sticky for trying to teach concepts. They could have made it something simple like ## O_2 ## or ## SO_2 ## or ## CO_2 ##, but that doesn't appear to be the case.
 
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  • #15
Hm. 1.38e-25/1.67e-27 is roughly 8.29. What did you use to get 83?
 
  • #16
mk9898 said:
Hm. 1.38e-25/1.67e-27 is roughly 8.29. What did you use to get 83?
(.83)100=83
 
  • #17
But from the division of the two variables it gives us 8.29. I don't understand where .83*100 comes from.
 
  • #18
mk9898 said:
But from the division of the two variables it gives us 8.29. I don't understand where .83*100 comes from.
## \frac{10^{-25}}{10^{-27}}=10^2=100 ##, and 1.38/1.67=.83.
 
  • #19
Yes but 1.38 is smaller than 1.67. Therefore it's only one decimal place higher.
 
  • #20
mk9898 said:
Yes but 1.38 is smaller than 1.67. Therefore it's only one decimal place higher.
Yes. It's less than 1.0. 1.38/1.67=0.83 .
 
  • #22
Ah crap sorry. Haha oh boy. I had my calculator on scientific and thought that the 8.3e1 just meant 8.3. Silly. Sorry about that. I understand the problem now. I could have just taken a second and thought about it but just let the calculator do the simple work.
 
  • #23
mk9898 said:
Ah crap sorry. Haha oh boy. I had my calculator on scientific and thought that the 8.3e1 just meant 8.3. Silly. Sorry about that. I understand the problem now. I could have just taken a second and thought about it but just let me calculator do the simple work.
Very good. Now that we agree the "molecular weight" is 83 or something close to that, this one needs some guesswork, because none of the common molecules have M.W.## \approx ## 83. The krypton atom could also be considered a molecule, so perhaps that is the correct answer. And yes, krypton is a gas.
 
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  • #24
I think the Krypton answer is a good one and I think they wanted the answer to deviate a bit from an actual element. They have an additional short question asking, "What additional information could help better determine the gas?" My answer would be knowing the temperature and the degrees of freedom so then we could use the E = fkT/2 equation. And that would perhaps help in determining the element.
 
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  • #25
Charles Link said:
Try your arithmetic again. I get 83. ## \\ ## When they tell you it is a molecule, it really should be a molecule, but the only gas I could find with atomic weight near 83 that is at all common is Krypton, at 83.8. They only give 2 sig figs in the statement of the problem. Another possibility though is ## HBr ##=hydrogen bromide=molecular weight=81. ## \\ ## As a homework helper, I'm not supposed to supply answers, but, assuming our arithmetic is correct, this question is somewhat sticky for trying to teach concepts. They could have made it something simple like ## O_2 ## or ## SO_2 ## or ## CO_2 ##, but that doesn't appear to be the case.

Hm. Yes I understand your point and I'm going to have to look into this. Why isn't it the case that ##O_2## or ##SO_2## are valid options?
 
  • #26
mk9898 said:
Hm. Yes I understand your point and I'm going to have to look into this. Why isn't it the case that ##O_2## or ##SO_2## are valid options?
##O_2 ## has M.W.=32 . ## SO_2 ## has M.W.=64 .
 
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  • #27
Ok thanks I thought that's what you meant.
 
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  • #28
Or I could just begin to add various elements together and see if it is "the" molecule. But then that doesn't say anything of the validity of such a molecule existing.
 
  • #29
mk9898 said:
Or I could just begin to add various elements together and see if it is "the" molecule. But then that doesn't say anything of the validity of such a molecule existing.
I've got a reasonably good chemistry background, and looked it over, (the periodic table, as well as googling for molecular weights of common substances), pretty carefully. You need to make sure what you assemble is a gas. HBr =hydrogen bromide with M.W.=81 is also a possibility. I couldn't come up with any molecule that has M.W.= 83. It's possible I missed something...
 
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  • #30
Thanks for your expertise. Just to remove any doubt I have two questions: 1) how is the division from the mass/proton mass = atomic weight? That would just be a scalar. 2) How does one find the total amount of the protons of an element?
 
  • #31
mk9898 said:
Thanks for your expertise. Just to remove any doubt I have two questions: 1) how is the division from the mass/proton mass = atomic weight? That would just be a scalar. 2) How does one find the total amount of the protons of an element?
The only information we had was the molecular weight. You asked a good question, but other than by looking at the periodic table, I don't know of a simple experiment that will tell you what the atomic number (number of protons) is given the atomic weight.
 
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  • #32
Yea apologies I meant molecular weight. But molecular weight/proton weight = atomic weight? How can that be?
 
  • #33
mk9898 said:
Yea apologies I meant molecular weight. But molecular weight/proton weight = atomic weight? How can that be?
We computed the number of neutrons plus protons. (molecular weight). Since they both have very nearly the same mass, the only thing we knew is that the sum of the neutrons and protons was 83(1.67 E-27) kg. There was no info as to whether we had 40 protons or 41 or 42, etc. Usually the number of protons and neutrons in an atom are about equal, plus or minus one or two or so. ## \\ ## The quickest way to determine the number of protons is to identify the chemical properties of the material, so that you know what the molecule or atom is, and then simply look it up in the periodic table. Experiments with molecular dynamics determine the mass (molecular weight), but these experiments do not give any info on the number of protons.
 
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  • #34
Ok I understand now thank you. I just watched a video that gave some more background on the subject. Thank you a lot for your effort!
 
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Related to Finding the gas based off of mass

What is the process of finding the gas based off of mass?

The process of finding the gas based off of mass involves using a mass spectrometer, which separates gas particles based on their mass-to-charge ratio. The results are then compared to a database of known gas masses to identify the gas present.

Why is it important to find the gas based off of mass?

Finding the gas based off of mass is important because it allows scientists to identify and quantify the different gases present in a sample. This can provide valuable information about the composition of a substance or environment.

What is the difference between mass spectrometry and other methods of gas detection?

Mass spectrometry is a highly sensitive and accurate method of gas detection, as it can identify and measure individual gas particles. Other methods, such as gas chromatography, may only provide a general idea of the types of gases present.

Can mass spectrometry be used to detect all types of gases?

While mass spectrometry can detect a wide range of gases, it may not be able to detect very small or highly reactive gases. In these cases, other methods may be more suitable for gas detection.

How is the data from mass spectrometry analyzed and interpreted?

The data from mass spectrometry is analyzed using specialized software and compared to a database of known gas masses. The results are then interpreted to identify the specific gases present and their quantities in the sample.

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