# Average molecular weight?

1. Oct 28, 2010

### tag16

1. The problem statement, all variables and given/known data
I'm trying to figure out how to compute the average molecular weight of a gas mixture (nitrogen dioxide and dinitrogen tetroxide). I think I'm suppose to use this equation: M(mix)=X(NO2)M(NO2)+ X(N2O4)M(N2O4) , where I would just need to calculate the mole fractions and the molecular weight of the individual gases. Is this correct? If not assistance would be appreciated.

Also I need to determine the equilibrium constant, delta G standard and delta S standard. I think to find the equilibrium constant I need to use this equation: K= n(N2O4)/((2)n(NO2)) ? To find delta G I could then use this equation: detla G= -RTlnK ? and I have no idea how to find delta S standard.

Known: Pressure, Temp, Volume and three different weights of gas mixture.

2. Relevant equations
M(mix)=X(NO2)M(NO2)+ X(N2O4)M(N2O4)
K= n(N2O4)/((2)n(NO2))
detla G= -RTlnK
3. The attempt at a solution
Ok I'm trying to get the mole fraction and ran into a little bit of a problem:
nf= PV/RT= 0.95646 x105cm3x267.403 x10-6bar/ 8.31 x 303.15 K
= 0.010153 mol

ni+x=119.283g/(92 g/mol)
= 1.29655 ?

ni+x cannot equal nf when ni alone is larger. What am I doing wrong?

2. Oct 28, 2010

### Staff: Mentor

What is what of what? We are not a bunch of seers.

3. Oct 28, 2010

### tag16

I'm not sure what is unclear. I'm try to get the average molecular by using this equation:M(mix)=X(NO2)M(NO2)+ X(N2O4)M(N2O4) (unless this not the equation I should be using?)

To do that I need the mole fractions, which is what I was trying to do here: nf= PV/RT= 0.95646 x105cm3x267.403 x10-6bar/ 8.31 x 303.15 K
= 0.010153 mol

ni+x=119.283g/(92 g/mol)
= 1.29655 ?

ni+x cannot equal nf when ni alone is larger. What am I doing wrong?

If you need more information I would be happy to oblige, especially since your helping me, I'm just not sure what other information you need.

4. Oct 28, 2010

### Staff: Mentor

I have no idea what is 0.95646, 105cm3 is probably volume, but I have no idea of what, then comes pressure (of what?), R (that I know) and temperature (of what?), and finally 0.010153 mol is not a mole fraction, although you wrote that's what you are trying to calculate.

I have no idea what is ni and nf, I have no idea what is x (that is - it is probably molar fraction, but of what?)

What is 119.283g? Mass, but of what?

So - to quote Piglet once again - what is what of what?

5. Oct 28, 2010

### tag16

0.95646 cm^3 is the volume of the dinitrogen textroxide and nitrogen dioxide gas mixture, the pressure is the pressure in the lab at the time of the experiment, temperature of the gas mixture.

119.283g in the mass of the gas mixture

x is the mole fraction, ni is the number of moles of the components

To get the mole fractions of both gases I need to do this:

N2O4= 2(NO2)

n(N2O4)= ni -x
n(NO2)= 2x

but first I need to find ni and x, hence this:

n= PV/RT= 0.95646 x105cm3x267.403 x10-6bar/ 8.31 x 303.15 K
= 0.010153 mol

ni+x=119.283g/(92 g/mol) + x
119.283g/(92 g/mol)= 1.29655

6. Oct 28, 2010

### Staff: Mentor

Am I right guessing that the volume is in fact 0.65646x105cm3? If not, your gas has density over 120 g/mL, this is about five times more than the most dense metals.

I guess you are trying to use some information about reaction stoichiometry, but - sorry - it won't get you anywhere. I am almost sure this is wrong, but you have still not gave full information about the problem and known data, so I still have no idea what you are doing. Sorry to say that, but you are wasting my time. Please explain the question in full. Don't repeat for the fourth time the same calculation, describe PROBLEM and DATA.

7. Oct 28, 2010

### tag16

PROBLEM: Assume an ideal mixture of ideal gases. From your data, compute the average molecular weight
of the gas mixture at each temperature. Using these values, determine the degree of dissociation
and the equilibrium constant, Kp, ΔGo and ΔSo for the dissociation at each temperature.
http://homepages.wmich.edu/~dschreib/Courses/Chem436/I-4 N2O4 Dissociation.pdf

experimental data: bulb 1: 141.97 g
bulb 2: 99.9668g
bulb 3: 115.9037g

T=30Co
P= 717.41 Torr= 0.956468003 bar

8. Oct 29, 2010

### Staff: Mentor

Now we are getting somewhere.

115.9037g is not mass of the gas, it is mass of the bulb and gas inside. You should also know mass of the empty bulb (point 2, week two).

0.95646cm3 is not volume of the bulb - bulb volume is around 250 mL. No idea what 0.95646 is. Perhaps mass of the gass?

9. Oct 29, 2010

### tag16

Oh sorry, that is week 2 data, weighing the bulbs with the gas inside, this is week 1:

bulb 1: 141.82g
w/water:395.95g
bulb 2: 99.27g
w/water: 361.77g
bulb 3: 143.79g
w/water: 397.75g

To get the volume I did V= mass/ water density , don't know if that's what I should have done but that's what I did.

10. Oct 29, 2010

### Staff: Mentor

So, what is volume of the bulb, and what is mass of the gas inside? Can you use these two information to calculate density and average molar mass?

Let's stick to bulb number 1.

11. Oct 29, 2010

### tag16

Well the mass of the gas would just be the mass of the bulb filled with the gas minus the mass of the bulb itself, correct?

Do you mean just the volume of the bulb by itself? 250 cm^3 ?

12. Oct 29, 2010

### Staff: Mentor

Yes.

No, you have determined the volume experimentally on the first week of the lab. Have you actually read the document you have linked to?

13. Oct 29, 2010

### tag16

Yes, it was necessary in order to perform the experiment.

So the volume would be the mass w/ water minus w/o?

14. Oct 30, 2010

### Staff: Mentor

No, that will be mass of water, not the volume.

Have you read the document you have linked to? It contains plenty of hints.

15. Oct 30, 2010

### tag16

yes, that is what I did. Then I did V= mass/ water density, which must have been wrong.

16. Oct 30, 2010

### Staff: Mentor

What was the result and why do you think it was wrong?

17. Oct 30, 2010

### tag16

In a previous post I wrote this:
Since in a later post you asked me how to get the volume, I just assumed that what I had already posted must have been wrong.

When I calculated it I got V=267.403 cm^3. To get the mass to use in this equation would I want to take the average of the three bulbs w/water?

18. Oct 30, 2010

### Staff: Mentor

Do the calculations for each bulb separately.

Yes, now I see you have calculated masses of water earlier. Trick is, in some posts you write things that are half right and half wrong, it is hard to keep track of what was already correct and what was not. Besides, you have not listed the V value earlier, and when I asked what it was you suggested it was 250 mL, which was incorrect.

19. Oct 30, 2010

### tag16

I just assumed what I originally thought the volume was must be wrong or you wouldn't asked about the volume so I just came up with something random because I didn't know what else it could be, hence the 250 cm^3.

So I have to do the calculations for each of the three bulbs seperately as well as for each temperature, correct?

20. Oct 30, 2010

### Staff: Mentor

To quote the document you have read: