Welcome, Guest. Please login or register.
Did you miss your activation email?

Login with username, password and session length
Welcome to our free

Chemistry Homework Help

forum.

Please read the forums rules before posting.

Find out how you can Help My Chemistry Tutor

   Home   Help Search Tags Login Register  

Live Tutoring 24/7 - Math, Statistics, Chemistry
Pages: [1] 2
  Print  
Author Topic: Hydrogenation of Methyl Oleate Post-Lab Questions  (Read 344 times)
chrisf
Global Moderator
Sr. Member
*****
Posts: 262



View Profile
« on: October 08, 2008, 07:11:42 PM »

A 0.150g sample of a pure compound subjected to catalytic hydrogenation takes up 25.0mL of H2 at 25.0C and 1atm pressure. Calculate the molecular weight of the compound assuming that it has only one double bond.

I wish I could say I've made a really good attempt at this... but I'm not sure where to start. I'd suppose you'd find the moles of hydrogen gas that corresponds to. Hydrogenation of only one double-bond would mean that you're replacing that pi bond with two hydrogen (to give 0 formal charge).

From there, I'm not sure where to go... and I think if I can figure out this one, I can figure out the other two questions.

Help??
Logged
SchoolBoyDJ
Full Member
***
Posts: 143



View Profile
« Reply #1 on: October 08, 2008, 11:39:36 PM »

So we know abosluting nothing about this molecule, other than it has 1 double bond?  We don't know if it's a hydrocarbon, or what functional groups it has?  I guess pure compound could be telling my something, but I'm not sure what you mean by it >_<

IF this is an alkene, with no other functional groups this problem wouldn't be so bad.

An alkane has the molecular formula CnH2n+2
Double bonds and rings each take away 2 hydrogens (if you can't visualize this, draw one)
A triple bond will take away 4 hydrogrens.

We have 1 double bond, out molecular formula will be CnH2n

Hydrogenation (H2 should break homolytically, both are pulling on electrons with the same force) will force that pi bond to break (Homolytically I think)..giving us no carbocation, but rather a hydrocarbon, with 2 C having a radical.  The 2 H radicals will form up with the 2 C radicals and we get an alkane.

Now here's the kicker.  We don't know how many carbons are in this molecule.

Lets start by figuring out how many moles of H2 we have...and I can't remember how the heck to convert mL of gas to grams >_<  Okay, I just finished looking it up.  I guess you can use the specific gravity of H2 to find the density, and then from there find grams.  The specific gravity of H2 is 0.070, evidently multiply this by the density of air at STP will give us the denisty of H2 at STP.  1.2m³ (air) x 0.07=0.084kg/m³.  Now I don't know why the used meters cubed as a unit...I have no idea how the heck we're supposed to use a density with meters as a unit when we have mL.  If you can figure out how to convert your mL H2 to grams, I may be able to help you with the rest of the problem.

0.150g (unknown) x  (1 mole (unknown) / x g (FW unknown) x ( 1 mol H2 / 1 mol unknown) x ( 2.01g H2 / 1 mol H2) = Grams H2 reacted.*

*We know how many mL H2 reacted, if we could figure out how to convert mL H2 to g H2, we'd only have 1 variable in the above work.
* I got the 1 mole Unknown to 1 mole H2 ratio from the simple fact that no matter the size of the hydrocarbon, it can only react with 2 H atoms (1 H2 molecule) per molecule of itself.  See:

CH3-CH2-CH=CH-CH2-CH2-CH2-CH3  +  H2  ---> CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3

No matter what you change the alkene to, it will only react with 1 molecule H2.

Here is a simplified version of the math above:
(.3015 / x) = z

Where
x is the formula weight of the unknown
z is the grams H2 (which we can solve for, but I don't know how to)

Okay so we know that .150g of this hydrocarbon can take up 25mL of H2 gas.  Now this is where is gets tricky.  Lets think:

1) If our hydrocarbon contains too many carbons, it will be too heavy and not be able to react with the 25mL H2 with only .150g of itself.
2) If our hydrocarbon contains too few carbons, it will be too light and there will be access of it left after reacting with H2.

So-  How can we get a perfect hydrocarbon.


Sorry everything is so messy.  I'm really scatterbrained when I'm trying to figure something like this out.  I hope I haven't confused you...

Good luck, and please let me know if you figure out how to convert that mL H2 to grams Smiley

Logged

Important***
I try to help as much as I can, but I am by no means a professional.  I am always open to corrections!
SchoolBoyDJ
Full Member
***
Posts: 143



View Profile
« Reply #2 on: October 09, 2008, 12:01:25 AM »

I thought this conversion was only for water, but I guess it's not.

m^3 = 1000 L

So from above, the density of H2 gas is 0.084kg/m³

I'm just gonna convert out mL, easier for me that way.

0.025 L x (1m3 / 1000 L) = 2.5x10^-5

0.084 = x kg / 2.5x10^-5   x=2.1x10^-6 kg or 2.1x10^-3g.

Now from above...
0.3015 /x = 2.1x10^-3

x=143.57=FW of our alkene.  Let's see if this is at least close to a possible alkene.  Remember is has to have the formula CnH2n  (I'm just gonna plug and punch numbers)

C10H20 does have a molecular weight of 140.268.  It's a bit off, but after those calculations it wouldn't surprise me if it would be.

I have no idea if I did this problem correctly-  I've never seen anything like it before and I simply thought through it and did what I though was logical.  (Don't hate me if I'm wrong xD)

P.S.  Let me know if you find out the answer, I'd like to know if I did it right or not Tongue
« Last Edit: October 09, 2008, 12:03:48 AM by SchoolBoyDJ » Logged

Important***
I try to help as much as I can, but I am by no means a professional.  I am always open to corrections!
chrisf
Global Moderator
Sr. Member
*****
Posts: 262



View Profile
« Reply #3 on: October 09, 2008, 08:02:06 AM »

Man, that's a heck of a reply! Seems pretty logical.. I come up with an answer of 146.793g

PV=nRT

First, what volume does one mole of hydrogen occupy at 25C, 1atm?

(1atm)(V) = (1mol)(0.08205)(298.15K)
V=24.465L = 24465mL H2

25mL H2 x (1 mol H2 / 24465mL H2) = 0.00102185mol H2

1mol H2 : 1mol product so 0.00102185mol H2 : 0.00102185mol product

0.150g product / 0.00102185mol product = x g product / 1mol product

x g product = 146.793g product

The closest thing I can come up with is C10H20 at 140.268g as well. Hmm....

« Last Edit: October 09, 2008, 08:07:49 AM by chrisf » Logged
spock
Global Moderator
Hero Member
*****
Posts: 1457



View Profile
« Reply #4 on: October 09, 2008, 10:04:17 PM »

The question isn't asking for the formula, only the molecular weight.  I didn't check your math, but your method looks good and if your math is correct than the molecular mass should be 147 g/mol (since your data only contains 3 sig. digits, your answer can only be reported to 3 sig. digits).
Logged
chrisf
Global Moderator
Sr. Member
*****
Posts: 262



View Profile
« Reply #5 on: October 09, 2008, 10:06:06 PM »

Oh, I know it only asks for the molecular mass.. we were just trying to do a sanity-check.
Logged
SchoolBoyDJ
Full Member
***
Posts: 143



View Profile
« Reply #6 on: October 09, 2008, 10:37:43 PM »

Yes, what he said.  Gotta make sure your head's on straight every now and then xD
Logged

Important***
I try to help as much as I can, but I am by no means a professional.  I am always open to corrections!
SchoolBoyDJ
Full Member
***
Posts: 143



View Profile
« Reply #7 on: October 09, 2008, 11:04:24 PM »

The first one would be hexane.  Once again, with H2 gas I'd expect homolytic clevage, making 2 H radicals, per mole.  The electrons from the double bonds would split homolytically as well giving 2 C radicals per double bond...Giving us a total of 4C radicals and 4H radicals.  Covalent bonds will form =D

What's wrong with the last one before I attempt it? xD
Logged

Important***
I try to help as much as I can, but I am by no means a professional.  I am always open to corrections!
chrisf
Global Moderator
Sr. Member
*****
Posts: 262



View Profile
« Reply #8 on: October 09, 2008, 11:17:22 PM »

Actually, I removed it because I discovered both answers were wrong when I really thought about it. Smiley

The first should be cyclopentane with two double-bonds (cyclopentadiene) and the second should be a cyclohexane with one double-bond.

Logged
SchoolBoyDJ
Full Member
***
Posts: 143



View Profile
« Reply #9 on: October 09, 2008, 11:19:27 PM »

You mean you put the wrong structures up, or that di alkene formed a cyclohexane?

Logged

Important***
I try to help as much as I can, but I am by no means a professional.  I am always open to corrections!
chrisf
Global Moderator
Sr. Member
*****
Posts: 262



View Profile
« Reply #10 on: October 09, 2008, 11:19:46 PM »

I put the wrong structures up. I thought about the problem a bit more. Smiley
Logged
SchoolBoyDJ
Full Member
***
Posts: 143



View Profile
« Reply #11 on: October 09, 2008, 11:20:32 PM »

Ah alright Smiley  Well have a good night, I'm off to bed.  Class in the morning >_<
Logged

Important***
I try to help as much as I can, but I am by no means a professional.  I am always open to corrections!
chrisf
Global Moderator
Sr. Member
*****
Posts: 262



View Profile
« Reply #12 on: October 09, 2008, 11:21:52 PM »



These are my proposed solutions.
Logged
cregstephen
Labrat
*
Posts: 6



View Profile WWW
« Reply #13 on: January 05, 2009, 04:02:22 PM »

Can any one answer this question if not i will post answer to this after two days.
A 0.150g sample of a pure compound subjected to catalytic hydrogenation takes up 25.0mL of H2 at 25.0C and 1atm pressure. Calculate the molecular weight of the compound assuming that it has only one double bond.
« Last Edit: January 05, 2009, 08:20:20 PM by chrisf » Logged

chrisf
Global Moderator
Sr. Member
*****
Posts: 262



View Profile
« Reply #14 on: January 05, 2009, 08:21:47 PM »

Solve for moles of hydrogen at the temperature and pressure. Since it has only one double bond, you know that only one mole of hydrogen reacts. Stoichiometrically relate the mass of hydrogen to the mass of the compound and that should be the answer.
Logged
Pages: [1] 2
  Print  
 
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.7 | SMF © 2006-2007, Simple Machines LLC | Sitemap Valid XHTML 1.0! Valid CSS!
Green Web Hosting by HostGator