Timeline for Total number of carbon atoms in buckminsterfullerene
Current License: CC BY-SA 3.0
14 events
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| Jun 4, 2017 at 13:15 | history | edited | orthocresol | CC BY-SA 3.0 |
deleted 2 characters in body; edited title
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| Jun 4, 2017 at 12:59 | history | edited | Mithoron |
edited tags
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| Aug 11, 2016 at 0:10 | comment | added | f'' | Fun fact, Descartes' theorem of angular defect states that if you do this on any polyhedron without holes, you'll always be off by 2. | |
| Aug 10, 2016 at 23:06 | comment | added | Ghosal_C | @f'' Yes, I realized a second folly in my approach. There will be parts of an atom not within any closed structure and I will miss out on them too. The approach only works if all my atoms are enclosed in a circuital structure and if all the circuital structures are planar. Thank you very much for your answer. It was a great help. | |
| Aug 10, 2016 at 23:01 | comment | added | Ghosal_C | @IvanNeretin Yes the problem is that I was treating these atoms as if they were planar and was hence landing up with a situation where I was defining the fraction as $\frac{\alpha}{2\pi}$ and at every pyramidized junction $\frac{\alpha}{2\pi}+\frac{\beta}{2\pi}+\frac{\gamma}{2\pi} < 1$ and hence I would never have a full atom counted. The sum of errors I guess added up to 2 atoms. Still, thank you very much for being patient with me and for clearing my doubt about the subject. | |
| Aug 10, 2016 at 14:01 | comment | added | f'' | Let's try counting the vertices of a cube this way. Each face has 4 corners with an angle of $\frac\pi2$, which is $\frac14$ of a full angle, so 1 vertex per face. There are 6 faces so a total of 6 vertices, but obviously a cube has 8 vertices... clearly this method doesn't work correctly. | |
| Aug 10, 2016 at 9:38 | comment | added | Ivan Neretin | Surely the angles need not add up to $2\pi$. The fractions, however, do need to add up to 1. This is a certain place in the molecule; it is occupied by one atom. You can't split atoms. | |
| Aug 10, 2016 at 9:34 | comment | added | Ghosal_C | Yes, but because the fullerenes are spherical balls and not graphene sheets, hence there needs to be some pyramidization, which is why the sums of the fractions need not necessarily be equal to one. Think of it as the apex angles of every triangle of each of the faces of the pyramid of Giza, the sum of the apex angles need not add up to $2\pi$ but if I flatten the pyramid out into a mesh of 4 2D triangles, then the apex angles would need to add up to $2\pi$, similarly the fractions of the out of plane pyramidized figure need not add up to 1. | |
| S Aug 10, 2016 at 9:24 | history | suggested | mhchem | CC BY-SA 3.0 |
usage of \ce
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| Aug 10, 2016 at 9:23 | comment | added | Ivan Neretin | Because if you count them as $3\over10$, then each atom (which, BTW, is a junction of two hexagons and a pentagon) is counted as ${1\over3}+{1\over3}+{3\over10}={29\over30}$, rather than a full solid $\mathbf1$. | |
| Aug 10, 2016 at 9:19 | comment | added | Ghosal_C | @IvanNeretin but, the atoms for the corners of the pentagons cannot be counted as $\frac{1}{3}$rd since those atoms are just $\frac{3}{10}$ths. If we do that then we get $20*6*\frac{1}{3} + 12*5*\frac{3}{10}$ equalling 58. What you're telling me to do is $20*6*\frac{1}{3} + 12*5*\frac{1}{3}$ equalling 60, but I can't assume a $\frac{1}{3}$rd for the atoms within the pentagon or maybe I think I can't and I should where in the problem becomes, why should I have to consider a fraction of $\frac{1}{3}$ for the atoms in the pentagons as well and not $\frac{3}{10}$ ? | |
| Aug 10, 2016 at 9:14 | review | Suggested edits | |||
| S Aug 10, 2016 at 9:24 | |||||
| Aug 10, 2016 at 8:56 | comment | added | Ivan Neretin | Now that's a nice trick indeed. Think of it this way: forget the angles, just count the atoms, as if there were six in each hexagon and five in each pentagon. That would make 180. But wait, we've just counted each atom more than once. Like you said, each atom is on the junction of three polygons, hence it must have been counted three times, so ${180\over3}=60$. | |
| Aug 10, 2016 at 8:37 | history | asked | Ghosal_C | CC BY-SA 3.0 |