From what I read, buckminister fullerenes are composed of 60 carbon atoms, $\ce{C_{60}}$ and it is made up of 20 hexagons and 12 pentagons. So, I decided to do a back of the hand calculation with the knowledge that each of these were tri-atomic junctions. Since one hexagonal tri-atomic site would comprise a $\ce{C-C}$ bond angle of $\frac{2\pi}{3}$, thus just $\frac{1}{3}$rd of an atom and hence two atoms in a hexagon. Since there are 20 hexagons, this gives me 40 atoms. Now the $\ce{C-C}$ bond angle at the pentagon site is $\frac{3\pi}{5}$ which makes it $\frac{3}{10}$ atoms per site and hence $\frac{3}{2}$ atoms per pentagon. With 12 pentagons, it results in 18 atoms. So the total number of atoms according to this calculation amounts to 40+18=58. Where did I go wrong?

From what I read, buckminsterfullerene is composed of 60 carbon atoms, $\ce{C_{60}}$ and it is made up of 20 hexagons and 12 pentagons. So, I decided to do a back of the hand calculation with the knowledge that each of these were tri-atomic junctions. Since one hexagonal tri-atomic site would comprise a $\ce{C-C}$ bond angle of $\frac{2\pi}{3}$, thus just $\frac{1}{3}$rd of an atom and hence two atoms in a hexagon. Since there are 20 hexagons, this gives me 40 atoms. Now the $\ce{C-C}$ bond angle at the pentagon site is $\frac{3\pi}{5}$ which makes it $\frac{3}{10}$ atoms per site and hence $\frac{3}{2}$ atoms per pentagon. With 12 pentagons, it results in 18 atoms. So the total number of atoms according to this calculation amounts to 40+18=58. Where did I go wrong?

From what I read, buckminister fullerenes are composed of 60 carbon atoms, $C_{60}$ and it is made up of 20 hexagons and 12 pentagons. So, I decided to do a back of the hand calculation with the knowledge that each of these were tri-atomic junctions. Since one hexagonal tri-atomic site would comprise a $C-C$ bond angle of $\frac{2\pi}{3}$, thus just $\frac{1}{3}$rd of an atom and hence two atoms in a hexagon. Since there are 20 hexagons, this gives me 40 atoms. Now the $C-C$ bond angle at the pentagon site is $\frac{3\pi}{5}$ which makes it $\frac{3}{10}$ atoms per site and hence $\frac{3}{2}$ atoms per pentagon. With 12 pentagons, it results in 18 atoms. So the total number of atoms according to this calculation amounts to 40+18=58. Where did I go wrong?

From what I read, buckminister fullerenes are composed of 60 carbon atoms, $\ce{C_{60}}$ and it is made up of 20 hexagons and 12 pentagons. So, I decided to do a back of the hand calculation with the knowledge that each of these were tri-atomic junctions. Since one hexagonal tri-atomic site would comprise a $\ce{C-C}$ bond angle of $\frac{2\pi}{3}$, thus just $\frac{1}{3}$rd of an atom and hence two atoms in a hexagon. Since there are 20 hexagons, this gives me 40 atoms. Now the $\ce{C-C}$ bond angle at the pentagon site is $\frac{3\pi}{5}$ which makes it $\frac{3}{10}$ atoms per site and hence $\frac{3}{2}$ atoms per pentagon. With 12 pentagons, it results in 18 atoms. So the total number of atoms according to this calculation amounts to 40+18=58. Where did I go wrong?