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Molar Mass of a Volatile Substance

  • Thread starter absci2010
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  • #1
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Homework Statement


We were given a volatile substance and a Dumas tube and told to find the moles of the unknown as well as its molar mass. We also had to measure the volume of the tube, air pressure, and the temperature at which all of the substance was a gas.
Pressure=750mmHg=0.987atm
Volume=mass of water/density at 24 Celsius=17.8641g/.997296g/ml=17.9125ml
T=373K (given)
mass of the substance=.59g

Homework Equations


PV=nRT (ideal gas equation)


The Attempt at a Solution


n=PV/RT=(.987atm*.0179125L)/(0.0821*373K)=.000577mol
molar mass=mass/number of moles=.59g/.000577mol=1000g/mol which seems ridiculously high. Please help me! What did I do wrong?
 

Answers and Replies

  • #2
Borek
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Can you please elaborate on the experimental procedure?

While it is not the first time I hear about Dumas tube, so far I have never seen an experiment using it.
 
  • #3
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Sure. We weighed the empty dumas tube to determine the mass of the tube plus the air in it. Then, a small amount of an unknown volatile substance was put in the tube. We heated the tube in a beaker of boiling water until all of the substance became gas. Then we let it cool to room temperature to determine the mass of the tube plus air plus the unknown. We subtracted the first mass from this second one to determine the mass of the unknown. This is the value we were to use to find the molar mass of the unknown.
Hope that clarifies this experiment. Thanks!
 
  • #4
Borek
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OK. So the first stage of the heating is to vaporize the substance so that it fills whole tube and replaces air - at the same time part of the substance leaves the tube as well. Is 0.59g amount of substance calculated taking into account fact that part was lost, or is it just the amount that you put into the tube?
 
  • #5
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No, a dumas tube has a very narrow neck, so we are assuming that none of the substance was lost.
 
  • #6
Borek
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So what is the neck for, if not to allow exchange of mass?

Think about it: volume is always the same, isn't it? So if you take a sample of 0.1g molar mass calculated your way will be different from the molar mass calculated for 1g sample? But we are talking about the same substance, same molar mass, just different sample sizes. So either whole procedure doesn't make sense, or your understanding of what is going on and how to calculate is wrong.

Or another way: if you put 0.59g of the substance into the tube (and you know its mass), why the heck heat it up and cool it down, weight twice and do all that operations, if you can just use known mass of 0.59g?

But you have not answered my question: where did the 0.59g come from?

--
 
  • #7
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The .59g is the weight of the unknown. We DON'T know the weight of the unknown before we put it in the tube, so we have to measure it.
We use the ideal gas law equation (PV=nRT) to find the number of moles. However, we can't use the ideal gas law for a liquid (obviously). So we heat it up so that all the liquid vaporizes and becomes a gas. At the point when all of the unknown is vapor, we can use PV=nRT. We also use Avogadro's hypothesis that equal volumes of gases contain equal numbers of molecules at the same temperature and pressure.
 
  • #8
Borek
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How is 0.59g calculated? Are you sure it is not 0.059g? I agree 1000g/mol is ridiculous.

Note: your description of the procedure still suggest that some of the substance is lost from the tube prior to weighting.
 

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