Typically density is defined in units of mass per volume. In the case of graphene it is mass per area, i.e. surface density. What would be the correct way to calculate the surface density (mass per unit area) of graphene, for example?

Would it be correct to multiply the density of graphite by the van der Waals gap of graphite?

  • $\begingroup$ Let's fix the terminology and go from there. Density is grams per cubic centimeter. Surface area is centimeters squared per gram. Now what do you want? $\endgroup$ – MaxW Feb 18 '17 at 16:01
  • $\begingroup$ @MaxW what is the "density" of graphene in grams per centimeter squared? $\endgroup$ – Sparkler Feb 18 '17 at 16:19
  • $\begingroup$ $\dfrac{1}{\text{Surface Area}} = \dfrac{\text{grams}}{\text{cm}^2} $ $\endgroup$ – MaxW Feb 18 '17 at 16:22
  • $\begingroup$ @MaxW ok so now how to actually calculate this "surface area" given carbon-carbon bond length etc.? $\endgroup$ – Sparkler Feb 18 '17 at 17:16
  • $\begingroup$ I'm still not sure what you are trying to do... // I think you're looking for what the unit cell is in an infinite plane of graphene. $\endgroup$ – MaxW Feb 18 '17 at 17:26

The C-C length in graphene is l = 0.142nm and the area of a hexagon can be calculated with the formula:

$ A = \frac{3\sqrt{3}}{2} l^2 = 0.0523nm^2$

In each hexagon, there are 2 full carbon atoms(1/3*6) so the surface density of one single layer is:

$ S_d = \frac{2*massCarbon}{A} = \frac{2*1.994 × 10^{-26}Kg}{0.0523× 10^{-18}m^2} = 76.26 × 10^{-8} Kg/m^2 = 7.63× 10^{-8} g/cm^2$

If you are considering 2, 3, etc. layers than the surface density it's twice, three times etc. the surface density of the single layer.

Additional note: The distance between layers is h = 0.335 nm and therefore its density can be calculated as:

$ d = \frac{S_d}{h} = \frac{7.63 × 10^{-8} g/cm^2 }{0.335× 10^{-7}cm} = 2.28 g/cm^3 $

This is very close to the experimental value I have found online says that the density of graphene is $2.267 g/cm^3$

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  • $\begingroup$ Which is the value for graphite as well... $\endgroup$ – Stian Yttervik Sep 19 '19 at 15:59

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