Using data (Pressure vs. density) to find molecular mass .

  • Chemistry
  • Thread starter terp.asessed
  • Start date
  • #1
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3

Homework Statement


Use the data:
P/bar
a: 0.1000
b: 0.5000
c: 1.000
d: 1.01325
e: 2.000
vs.
density/(g/L)
a: 0.1771
b: 0.8909
c: 1.796
d: 1.820
e: 3.652


...for an unknown gas at 300K to determine the molecular mass of the gas...

Homework Equations


PV = nRT (b/c of very low pressure)

The Attempt at a Solution


..so, first I plotted the data on a graph, and it seemed like a linear line....considering Pressure is very low, I assumed ideal gas law, as in PV = nRT where n = mass/(molecular mass):

PV = (mass/mm)RT --> mm = (mass/V)(RT/P) = (density)*RT/P...so, I chose one middle data point (c), where:

density = 1.796 g/L
P = 1bar= 0.9869 atm
in addition to given values:
R = 0.082058 L*atm/(mol*K)
T= 300K

and I got mm = 44.80g/mol, which is not the same as 44.10 on the back of the textbook....could someone hint where I made the mistake?
 

Answers and Replies

  • #2
Quantum Defect
Homework Helper
Gold Member
495
116

Homework Statement


Use the data:
P/bar
a: 0.1000
b: 0.5000
c: 1.000
d: 1.01325
e: 2.000
vs.
density/(g/L)
a: 0.1771
b: 0.8909
c: 1.796
d: 1.820
e: 3.652


...for an unknown gas at 300K to determine the molecular mass of the gas...

Homework Equations


PV = nRT (b/c of very low pressure)

The Attempt at a Solution


..so, first I plotted the data on a graph, and it seemed like a linear line....considering Pressure is very low, I assumed ideal gas law, as in PV = nRT where n = mass/(molecular mass):

PV = (mass/mm)RT --> mm = (mass/V)(RT/P) = (density)*RT/P...so, I chose one middle data point (c), where:

density = 1.796 g/L
P = 1bar= 0.9869 atm
in addition to given values:
R = 0.082058 L*atm/(mol*K)
T= 300K

and I got mm = 44.80g/mol, which is not the same as 44.10 on the back of the textbook....could someone hint where I made the mistake?
What you did looks very good. BUT... You made an important assumption when you constructed the plot. For a real gas, where is this approximation likely to be best? Can you extrapolate the data back to this point?
 
  • #3
127
3
Wait, I thought this problem was related to ideal gas? I am sorry, but how am I to apply real gas in the question? Does it mean I have to find a 'general' equation as in y = mx +b?
 
  • #4
Quantum Defect
Homework Helper
Gold Member
495
116
Wait, I thought this problem was related to ideal gas? I am sorry, but how am I to apply real gas in the question? Does it mean I have to find a 'general' equation as in y = mx +b?
I think that I assumed that you plotted something that you did not.

What happens when you plot MM (just like you calculated) versus Pressure for all of the data? Do that first, and share the graph with us, and we will think about what to do next.
 
  • #5
20,844
4,543
This is supposed to be real experimental data, so there is supposed to be some uncertainty involved. So no single data point should deliver the exact value of the molar mass. But, if you plot the density as a function of the pressure, the slope of the line you get should be your best estimate of mm/RT. The best straight line you plot should not pass through each and every data point. From the slope of this line, what value do you get for the molar mass?

Incidentally, include the additional data point 0,0.

Chet
 
Last edited:
  • #6
127
3
Hello I did what you all have suggested, and got a slope of value 1.8145 = mm/RT from the graph of density as a function of pressure, but only got mm of 44.668, NOT 44.10.......

1.8145 = mm/(RT) = mm/[(0.082058)(300K)]
mm = 44.668

So, I tried for another slope W/OUT an additional data point (0,0) and got slope of 1.8243, which gave mm = 44.9095.....

Where am I making mistake?
 
  • #7
20,844
4,543
Hello I did what you all have suggested, and got a slope of value 1.8145 = mm/RT from the graph of density as a function of pressure, but only got mm of 44.668, NOT 44.10.......

1.8145 = mm/(RT) = mm/[(0.082058)(300K)]
mm = 44.668

So, I tried for another slope W/OUT an additional data point (0,0) and got slope of 1.8243, which gave mm = 44.9095.....

Where am I making mistake?
I had Kaleidograph do a linear fit to determine the slope with (0,0) included, and I got 1.8243. It looks like you did everything correctly.

But, when I used this value in the formula, I got mm=45.5, rather than 44.9. I think you may have neglected the difference between bars and atm.

I wouldn't worry about this too much. Your methodology is correct. That's all that really matters. We have no idea how they got their result in the back of your textbook. I have confidence in what you did.

Chet
 
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  • #8
127
3
Thank you very much! Knowing that not only me but also you got the value makes me feel better :)
 

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