Chemistry: Converting density to molar mass given other values.

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SUMMARY

The discussion centers on calculating the molar mass of a gaseous compound with a density of 1.23 kg/m³ at 330K and 20,000 Pa using the ideal gas law. Participants utilized the equation PV=nRT to derive the number of moles and subsequently calculated the molar mass, arriving at values around 168.72 g/mol and 153.39 g/mol. The formula d = MP/RT was debated, with clarification that it applies to ideal gases regardless of STP conditions. Ultimately, the correct molar mass calculation confirmed the initial value of 168.72 g/mol as accurate.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Knowledge of gas density and its relation to molar mass
  • Familiarity with the concept of molar volume
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation and application of the Ideal Gas Law
  • Learn about the conditions under which the formula d = MP/RT is applicable
  • Explore the differences between STP and non-STP conditions in gas calculations
  • Practice additional problems involving gas density and molar mass calculations
USEFUL FOR

Chemistry students, educators, and professionals involved in gas law applications and molar mass calculations will benefit from this discussion.

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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