Quick melting/boiling point question?

[timtamboy63]

6. The molar masses of C2H6, CH3OH and CH3F are very similar. How do their boiling points compare?

A. C2H6 < CH3OH < CH3F

B. CH3F < CH3OH < C2H6

C. CH3OH < CH3F < C2H6

D. C2H6 < CH3F < CH3OH

The answer is D, anyone know why? I would have said A, as the difference in electronegativities between H and F are greater than the difference in electronegativities between H and O resulting in a 'more' polar molecule, hence stronger Hydrogen bonds

Thanks

[Sandwich]

If the bonds between H and F are stronger, surely it WOULD be D, though? Ahem, scrap that, my brain's not working properly today. I remember understanding this once, but I can't recall how it worked (hurray time). Check out the course companion?

Edited November 18, 2010 by Sandwich

[Drake Glau]

6. The molar masses of C2H6, CH3OH and CH3F are very similar. How do their boiling points compare?

A. C2H6 < CH3OH < CH3F

B. CH3F < CH3OH < C2H6

C. CH3OH < CH3F < C2H6

D. C2H6 < CH3F < CH3OH

The answer is D, anyone know why? I would have said A, as the difference in electronegativities between H and F are greater than the difference in electronegativities between H and O resulting in a 'more' polar molecule, hence stronger Hydrogen bonds

Thanks

The F and OH groups on the molecule allow it to form hydrogen bonds which are the hardest to break This cuts it to A and D.

I'm not sure on how to pick between the two honestly, all I can think of is maybe the structure? I THINK the VSEPR structures might be different and could cause something, but I'm currently kind of too lazy to do the structures right now so you might want to check that out

Edited November 20, 2010 by Drake

[larry]

The answer is D, anyone know why? I would have said A, as the difference in electronegativities between H and F are greater than the difference in electronegativities between H and O resulting in a 'more' polar molecule, hence stronger Hydrogen bonds

I think it's because in CH3F the Hs aren't directly bonded to the F (all 3 Hs and F are bonded to the carbon) so there is no H-bonding between the molecules (they can form H-bonds with water). In D, however, there's an OH group (oxygen directly bonded to hydrogen); hence there is inter-molecular H-bonding.

CH3F has a higher boiling point than C2H6 because it is polar.

(so in your question, the intermolecular bonding goes: dispersion forces in C2H6 (Van der Waals / instantaneous dipole-dipole), polar in CH3F, H-bonding in CH3OH)

[Hedron123]

The answer is D, anyone know why? I would have said A, as the difference in electronegativities between H and F are greater than the difference in electronegativities between H and O resulting in a 'more' polar molecule, hence stronger Hydrogen bonds

I think it's because in CH3F the Hs aren't directly bonded to the F (all 3 Hs and F are bonded to the carbon) so there is no H-bonding between the molecules (they can form H-bonds with water). In D, however, there's an OH group (oxygen directly bonded to hydrogen); hence there is inter-molecular H-bonding.

CH3F has a higher boiling point than C2H6 because it is polar.

(so in your question, the intermolecular bonding goes: dispersion forces in C2H6 (Van der Waals / instantaneous dipole-dipole), polar in CH3F, H-bonding in CH3OH)

Exactly, in CH3F fluorine is not attached to hydrogen so no hydrogen bonding is formed. Its boiling point is higher than the C2H6 because, although it has no hydrogen bonds, the molecule is polar due to the difference in electronegativity (dipoles-dipoles). C2H6 will only have Van der Waals forces and, as the molar masses of all the compounds is very similar, the strength of the Van der Waals will also be very similar. The compound with oxygen, on the contrary, will be forming hydrogen bonds so its boiling point is the highest of all.

[CristinaV]

It all pretty much has to do with the intermolecular forces of the molecules. When you're melting a compound, you're not actually breaking the bonds between the atoms of the molecule and forming new compounds; you're only changing its state. So you have to look at the type of intermolecular bonding of each and check its relative strength. For this, I really recommend you memorize this:

Van der Waals' < Dipole:dipole < hydrogen bonding < covalent bonding < ionic bonding < metallic bonding < giant covalent

Note that not all of these are intermolecular, they're placed in order of strength in general. So we can see that Van der Waal's has the lowest melting point, followed by dipole to dipole, and topped by hydrogen bonding.

C2H6 is completely saturated, there are no "gaps" or differences in electronegativities (between the C and the surrounding atoms), so it's Van der Waal's, which is the easiest to "break apart" because it is really only attraction depending on size.

CH3F, on the other hand, doesn't have equal electronegativities between the bonds. Fluorine has a much higher electronegativity than Carbon and Hydrogen, so it will attract more electrons and have a negative charge. Because of this charge difference, there will be a permanent dipole to dipole moment and the negative Fluorine part will attract positives, and the other C and H parts will attract negatives. So the melting and boiling points are higher.

And then there's CH3OH. Like they others mentioned before, it has hydrogen bonding. When you said that "resulting in a 'more' polar molecule, hence stronger Hydrogen bonds" I don't know if you're saying that all the molecules have hydrogen bonding. But just in case, I'll tell you that they don't. For hydrogen bonding to occur, you need a hydrogen covalently attached to Nitrogen, Oxygen, or Fluorine, and that the atom it's attached to (N, O, or F) has a lone pair of electrons. Only CH3OH meets this requirement, so it has hydrogen bonding and, therefore, the highest melting point.

Remember, if they ask you about the melting and boiling points, it's the intermolecular forces! In a practice test I had to do a short answer on something similar and wrote on intramolecular forces instead (like ionic and covalent), and got no points for it. Just learn that little equation with the strengths and you'll be fine.

I hope this was helpful

[Mahuta ♥]

Yup, you covered everything.

Timtam always remember these two points:

Van der Waals' < Dipole:dipole < hydrogen bonding < covalent bonding < ionic bonding < metallic bonding < giant covalent

and

if they ask you about the melting and boiling points, it's the intermolecular forces!

I remember this question on a quiz, and my chem teacher told me those exact two points.