Chemistry: Converting density to molar mass given other values.

AI Thread Summary
The discussion revolves around calculating the molar mass of a gaseous compound using its density, temperature, and pressure. The ideal gas law (PV=nRT) is applied, with an initial assumption of a volume of 1 m^3 leading to a calculated molar mass of 168.72 g/mol. Participants clarify the use of the formula d = MP/RT, noting it can be applied outside of STP conditions. There is some confusion regarding calculations, with one participant arriving at approximately 153 g/mol, prompting a review of the temperature used in the calculations. Ultimately, the correct molar mass is confirmed to be around 168.72 g/mol based on the initial calculations.
LakeMountD
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Homework Statement



The density of a gaseous compound was found to be 1.23 kg/m^3 at 330K and 20,000 Pa. What is the molar mass of this compound.


Homework Equations



PV=nRT

The Attempt at a Solution



I don't think this is right but I assumed a volume of 1 m^3. So I used ideal gas law as follows:

n = (20,000Pa * 1m^3)/(8.314 (Pa*m^3/mol.K) * 330K) = 7.29 mol.

Then what I did was took density and multiplied it by the assumed volume of 1m^3 and got 1.23 kg/m^2. Then I took 1230g/m^3 and divided it by 7.290 mol. go get grams/mol*m^2. The answer was 168.72 but I can almost guarantee I am being retarded and overlooking something.
 
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you can also use this equation

d = MP/RT

let me know what you get
 
I take it that is the molar volume formula?

The reason why I didn't use that is because it says in my book that the formula for molar volume (above) is used under STP conditions. Did I read that wrong?
 
Rocophysics is right. \rho =\frac{\mu p}{RT} for an ideal gas.
 
Okay so d = MP/RT where M is the molar mass (or molar volume?)?

Sorry I just want to make sure. And thanks for the help guys. Like I said I saw that formula in the book but I thought it said it only applied at STP.
 
I ended up getting ~ 153g/mol
 
that's not what i ended up getting, is this that the correct answer in the book? i'll re-do my calc.

i'm off by ~ 15
 
Last edited:
LakeMountD said:
I ended up getting ~ 153g/mol

I didn't get this answer either. I got the 168.72 g/mol that was given as the answer in the first post.
 
hage567 said:
I didn't get this answer either. I got the 168.72 g/mol that was given as the answer in the first post.
lol i didn't even notice the answer in the first post, it's correct tho. different method but valid and good.

gj!
 
  • #10
Wait so I did it right the first time? I don't have the answer in the book, unfortunately it is an even numbered problem.

Why didn't the formula you told me work?
 
  • #11
LakeMountD said:
Wait so I did it right the first time? I don't have the answer in the book, unfortunately it is an even numbered problem.

Why didn't the formula you told me work?

It should have. Double check your calculation.
 
  • #12
hage567 said:
It should have. Double check your calculation.

M = d*R*T/P = 1230g/m^3 * 8.314 * 300k / 20,000 = 153.39 g/mol?
 
  • #13
It's 330 K, not 300 K. :wink:
 
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