ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Martoman on April 22, 2010, 04:08:15 pm
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Does increasing weight mean it will travel slower and hence have a smaller Rf value? Nelson chem doesn't think so, but my intuition does. :-\
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Yeah it should
Higher molar mass = bigger molecule = more area for molecule to adsorb onto stationary phase = longer retention time
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"Heavier molecules, particularly those with fewer polar sites, tend to travel further
up the paper, despite the downward pull of gravity. This implies that dispersion forces
play a less important role than other kinds of bonding." page 104.
Your logic is the exact same as mine.
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Oh? Thats strange
When comparing substances with simliar properties (e.g Alkanes) the heavier will come out with the longest retention time. i.e. Pentane would have a longer retention time compared to Methane.
Im not sure, in my head, that paragraph just defies the laws of mankind, if a big molecule travels "despite the downward pull of gravity", wouldnt a smaller molecule take even less gravity into account?
Maybe its trying to say weight makes hardly any difference compared to other forces or maybe theres another rule that governs bigger molecules. I dunno
Lets wait for someone smarter to explain ;P
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It is my understanding that those hydrocarbons can only be used in GC. Then, this holds true. Nelson agrees with me. Higher MM then longer retention time, but the retardation factor is different. hmmm. It seems that my retardation factor here is high. :uglystupid2:
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Both paper and glass have a polar surface (paper is essentially cellulose). Even if you have a large Mr, if you have very few polar sites, you won't adsorb onto the stationary phase all that much (cf a much smaller but polar molecule).
Nelson tried to demonstrate that disperson forces don't play a significant role when the main intermolecular forces are dipole-dipole attraction. However, the trend you proposed (larger Mr --> smaller Rf) is valid for molecules with the same dipole moments (a measurement of how 'polar' a molecule is).
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Both paper and glass have a polar surface (paper is essentially cellulose). Even if you have a large Mr, if you have very few polar sites, you won't adsorb onto the stationary phase all that much (cf a much smaller but polar molecule).
Nelson tried to demonstrate that disperson forces don't play a significant role when the main intermolecular forces are dipole-dipole attraction. However, the trend you proposed (larger Mr --> smaller Rf) is valid for molecules with the same dipole moments (a measurement of how 'polar' a molecule is).
I thought cellulose was non-polar!
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Both paper and glass have a polar surface (paper is essentially cellulose). Even if you have a large Mr, if you have very few polar sites, you won't adsorb onto the stationary phase all that much (cf a much smaller but polar molecule).
Nelson tried to demonstrate that disperson forces don't play a significant role when the main intermolecular forces are dipole-dipole attraction. However, the trend you proposed (larger Mr --> smaller Rf) is valid for molecules with the same dipole moments (a measurement of how 'polar' a molecule is).
I thought cellulose was non-polar!
cellulose has polar -OH groups.
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Both paper and glass have a polar surface (paper is essentially cellulose). Even if you have a large Mr, if you have very few polar sites, you won't adsorb onto the stationary phase all that much (cf a much smaller but polar molecule).
Nelson tried to demonstrate that disperson forces don't play a significant role when the main intermolecular forces are dipole-dipole attraction. However, the trend you proposed (larger Mr --> smaller Rf) is valid for molecules with the same dipole moments (a measurement of how 'polar' a molecule is).
I thought cellulose was non-polar!
cellulose has polar -OH groups.
Yes, but why is it insoluble in water?
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Ahh, very interesting point. Here, we must distinguish between 'soluble in water' and 'polar'
Cellulose is polar, [look at all these OH groups]. Each molecule of cellulose is linear, and they hydrogen-bond to each other to form a highly crystalline structure. The intermolecular attraction between cellulose molecules are so strong, that water at 320 degrees and pressure 250 times greater than standard atmosphere is required to 'dissolve' cellulose in water [actual dissolution may occur before this, but you get my point]. At room temperature [or any normal conditions], cellulose is not soluble in water.
However, on the surface of a cellulose particle, there are quite a few exposed OH groups. Other polar molecules can adsorb onto the surface. Especially water, cellulose is hydrophillic.