Need help on CHEMISTRY

[Xylitol]

Apparently he never learned it, and I havent either (I have a diff teacher). The AP chem students dont know the answer to this either, so I am posting it here. I'm grateful for any amount of help I can receive.

Question: Place the following substances in order of increasing volatility.

Answers: CHCl3 CH3Cl CCl4 CH3Br CH2Br2 CH2Cl2

I think it would help the most (without telling me the answer) to tell me what volatility is

 

[Ika]

Volatility is the tendency of a substance to vaporize. The weaker the bonds of a molecule are, the more it wants to vaporize.

Pretty basic concept in AP Chem.

 

[Brainonska511]

CHCl3 CH3Cl CCl4 CH3Br CH2Br2 CH2Cl2

Think about their polarity. Then think about their size. The most polar will be the least volatile, while the smallest/least polar will be the most due to the weakest intermolecular forces.

H-bonding (which you don't have in this case) > Dipole-Dipole > London Dispersion (aka Van der Waals).

By the way: Volatility: more volatile substances evaporate easier due to weaker intermolecular forces.

 

[Xylitol]

Guess not

 

[Vertimus]

Tell your friend to join AT.

 

[Xylitol]

Originally posted by: Brainonska511
CHCl3 CH3Cl CCl4 CH3Br CH2Br2 CH2Cl2

Think about their polarity. Then think about their size. The most polar will be the least volatile, while the smallest/least polar will be the most due to the weakest intermolecular forces.

H-bonding (which you don't have in this case) > Dipole-Dipole > London Dispersion (aka Van der Waals).

By the way: Volatility: more volatile substances evaporate easier due to weaker intermolecular forces.

Oh I thought the problem was talking about the bonding inside of the molecule not intermolecular bonding which bonds different molecules together.

Still though, I dont remember, do you tell the polarity by the difference in electronegativity?

 

[Brainonska511]

Originally posted by: Xylitol

Originally posted by: Brainonska511
CHCl3 CH3Cl CCl4 CH3Br CH2Br2 CH2Cl2

Think about their polarity. Then think about their size. The most polar will be the least volatile, while the smallest/least polar will be the most due to the weakest intermolecular forces.

H-bonding (which you don't have in this case) > Dipole-Dipole > London Dispersion (aka Van der Waals).

By the way: Volatility: more volatile substances evaporate easier due to weaker intermolecular forces.

Oh I thought the problem was talking about the bonding inside of the molecule not intermolecular bonding which bonds different molecules together.

Still though, I dont remember, do you tell the polarity by the difference in electronegativity?

Think about the polarity of each of the bonds in each molecule and then think about how structure plays a part. Certain structures cause polar bonds to negate each other, making the molecule have a neutral polarity. Electronegativity is used to determine bond polarity.

 

[Ika]

Originally posted by: Xylitol
Oh I thought the problem was talking about the bonding inside of the molecule not intermolecular bonding which bonds different molecules together.

It is. A molecule such as CH2Cl2 is held together by intramolecular forces. The forces that Brainonska mentioned are intramolecular forces. The intermolecular forces are ionic and covalent (along with the three intramolecular forces, I believe).

 

[Xylitol]

CCl4 since its the most non-polar will be the weakest I assume and that CH3Cl will be the strongest?

 

[Brainonska511]

Originally posted by: Xylitol
CCl4 since its the most non-polar will be the weakest I assume and that CH3Cl will be the strongest?

Yes.

 

[Xylitol]

CH3Cl, CH3Br, CH2Cl2, CH2Br2, CHCl3, CCl4?

 

[Brainonska511]

Originally posted by: Xylitol
CH3Cl, CH3Br, CH2Cl2, CH2Br2, CHCl3, CCl4?

Looks about right.

I would go with this way though:

CH3CL, CH3Br, CH2Br2, CH2Cl2, CHCl3, CCl4

I put the CH2Br2 ahead of the CH2Cl2 because it is a larger molecule. It will have higher London Dispersion forces holding the molecules together as compared to the CH2CL2.

 

[Xylitol]

Originally posted by: Brainonska511

Originally posted by: Xylitol
CH3Cl, CH3Br, CH2Cl2, CH2Br2, CHCl3, CCl4?

Looks about right.

I would go with this way though:

CH3CL, CH3Br, CH2Br2, CH2Cl2, CHCl3, CCl4

I put the CH2Br2 ahead of the CH2Cl2 because it is a larger molecule. It will have higher London Dispersion forces holding the molecules together as compared to the CH2CL2.

And that wont happen between CH3Cl adn CH3Br because of Di-Di? Or am I off

 

[Brainonska511]

Originally posted by: Xylitol

Originally posted by: Brainonska511

Originally posted by: Xylitol
CH3Cl, CH3Br, CH2Cl2, CH2Br2, CHCl3, CCl4?

Looks about right.

I would go with this way though:

CH3CL, CH3Br, CH2Br2, CH2Cl2, CHCl3, CCl4

I put the CH2Br2 ahead of the CH2Cl2 because it is a larger molecule. It will have higher London Dispersion forces holding the molecules together as compared to the CH2CL2.

And that wont happen between CH3Cl adn CH3Br because of Di-Di? Or am I off

The dipole-dipole bonding is the more dominant force in the CH3Cl and the CH3Br because there is no other Cl or Br (respectively speaking) in the molecule to negate the overall polarity of the molecule.

A good site to double check your work is Chemfinder.com. You can check out the boiling points of the molecules to see if you were correct (higher boiling point generally = less volatile).

Edit: Looking at that site, monochloromethane has a lower boiling point than monobromomethane. I would have thought it would have been the other way around due to the polarity of the bonds. I was in AP chem last year, but some of the stuff is blanking my mind right now (and I don't have any of my notes from AP or from the O-Chem class I took this quarter). I could be wrong, but I still hold to the monochloromethane being less volatile than the monobromomethane.