Chemistry: Converting density to molar mass given other values.

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Discussion Overview

The discussion revolves around calculating the molar mass of a gaseous compound given its density, temperature, and pressure. Participants explore different approaches and equations related to the ideal gas law and molar volume, while addressing potential discrepancies in their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant uses the ideal gas law to calculate the number of moles and subsequently derives the molar mass, expressing uncertainty about their method.
  • Another participant suggests using the equation d = MP/RT to find the molar mass, prompting questions about its applicability under non-STP conditions.
  • Some participants confirm that the equation for density in terms of molar mass is valid for ideal gases, regardless of STP conditions.
  • Multiple participants report different results for the molar mass, with values around 153 g/mol and 168.72 g/mol being mentioned, indicating varying calculations and methods used.
  • There is confusion regarding the temperature used in calculations, with one participant mistakenly using 300 K instead of the provided 330 K.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct molar mass, as different values are reported based on varying methods and calculations. There is also uncertainty regarding the applicability of certain equations under the given conditions.

Contextual Notes

Some participants express uncertainty about the assumptions underlying their calculations and the definitions of terms used in the equations. There is also mention of a lack of reference answers for verification, as the problem is from an even-numbered set.

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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