10 replies to this topic

[Spliknot13]

The density of a saturated hydrocarbon containing 80% wt carbon was measured against pressure at 20 degrees Celcius according to the following data:

density(g/L)   | 0.298 | 0.692 | 1.188 | 1.784 | 2.491 |
P(atm)           |   0.2   |    0.4  |   0.6   |   0.8   |  1.0    |

what is the exact molecular weight of the gaseous hydrocarbon? Also what is the molecular formula of the gas?
How do you necessarily solve these type of problems where there are many given data? and by the way the answer in the book is 29.675 g/mol and thanks in advance

Edited by Spliknot13, 25 April 2013 - 01:24 AM.

[ankur2061]

Spliknot13,

 

For any ideal gas if you know the relation between density, pressure and temperature you can calculate the molecular weight. In your case the pressure and temperature are given so you can calculate your molecular weight. Have a look at the following link:

 

http://www.cheresour...ard-conditions/

 

Regards,

Ankur.

[Spliknot13]

thanks for the help Ankur but what im trying to ask for is how do i solve the molecular weight given multiple variations of density with pressure? i mean when I calculate it individually the values of molecular weight are different. (by the way they are 35.82 g/mol,41.62 g/mol, 47.63 g/mol 53.644 g/mol and 59.92 g/mol respectively.) which doesnt even come close to the answer in the book which is 29.675 g/mol.

[Pingue2008]

Spliknot13,

 

The only way I think we can solve the problem is graphically. This is what I mean:

1- assume Ideal gas

2- PV = nRT= (Mass/MW)*RT

3- Divide by the volume P =(Mass/V)*(RT/MW)

                                          = Density *(RT/MW)

 

So, you can plot Pressure Vs Density (given data) and the slope will be RT/MW

you know R and T solve for MW. Try this and let us know.

 

Thank you,

[thorium90]

Perhaps you might want to share which book the question is from?
 
My workings:
 
80wt% / 12 = 6.67mol
20wt% / 1 = 20mol
6.67 / 20 = 3
 
Molecule is of type CH3
15 g/mol CH3
 
0.298 g/l * 0.0821 L.atm/mol.K * 293.15K / 1 atm = 35.86 g/mol
 
Repeat for the rest
This is the part which you get stuck on right?
Probably as the question states, they are measured values, ie: experimental values, subject to a degree of error.
 
Therefore, as one know the general formula to be CH3, one gets about between 2.4 and 4.0 units of CH3 in the molecule.
 
If there are 2 units of CH3, the molecular weight would be 15 g/mol * 2 = 30 g/mol

Edited by thorium90, 26 April 2013 - 09:45 PM.

[Spliknot13]

@Pingue2008, I tried solving it graphically but what  I got was around 66.72 g/mol...

 

@Thorium90, its a reviewer text that I bought from a local bookstore. Anyways the problem is in the physical chemistry section. I didnt quite get the part about the CH3 molecule that you said. Mind explaining it again please?

[thorium90]

Hi Kerrigan,

 

Type CH3 as in, C_nH_3n. Therefore, possible molecules are C2H6, C3H9 etc

[Pingue2008]

Spliknot13,

 

Please explore the possibility that the answer from the book is wrong. Could you also double check the problem statement? (suggestion)

 

Thank you,

[Bobby Strain]

Thorium90,

      Your general formula for a paraffin is a bit off. You should recognize there is no C3H9. Try again.

Bobby

[thorium90]

Thanks bobby for pointing it out. There isnt a CH3 either. Therefore the most plausible answer is C2H6 which has a MW of 30g/mol. And as the ratio is between 2.4 and 4.0 with the average closer to 2, the most likely answer would be C2H6. Although your book gives it as 29.675? Well i used 12 for C and 1 for H so i got 15...

If one assumes the hydrocarbon to be just hydrogen and carbon, one would be hard pressed to find one whose MW match each of the 5 values calculated by Kerrigan in post #3 using the appropriate equation. Im sure when you tried it out you got the same 5 values for MW too right?

This question appears to be one that expects the student to make an intelligent judgement rather than blindly applying equations. It seems to test the student on ideas like experimental error and fundamental concepts in chemistry like the fact that the chemical formula will only contain whole numbers of atoms.

 

Well, of course, I could be wrong, but as one needs only one set of density and pressure values to get the MW, there would thus appear to another reason why 5 were given instead.

Edited by thorium90, 26 April 2013 - 11:23 PM.

[Spliknot13]

@Thorium90 yeah. it says here in the book that the right answer is 29.675. the other choices are so near each other that is 30.187, 30.891 and 31.042 g/mol that I could not just rule out the other choices when I was trying to figure this out, but the queen of blades is indeed very thankful for the help.