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Meadock

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  1. Alright, first, thanks for answering. Next, why multiply by 1000cm cubed? by not 1cm cubed? It is 2.23 grams per cubic centimeter. So lets pretend we have only one cubic centimeter. Also, I do believe the grams cancel each other in the end of your calculations. So it should be a quantity of moles, not a mass. I plugged your calculations into my own and it decreased the size of the radius by a factor of 10:-\ Thanks. Andrew
  2. Out of pure curiosity I decided to try to determine the radius of a Carbon atom plus the average distance between atoms in Graphite to give me the total distance between the center of two Carbon Atoms in Graphite. Here's my data: Graphite Sample Density: about 2.23 g/cm cubed. Carbon Atomic Weight: 12.0107 g/mol. Avogadro constant(L): 6.02214279(30)×10^23 Start by determining how many moles are in our sample: 2.23 g / 12.0107 g/mol = 0.187 mol Then multiply it by Avogadro constant to find how many molecules of carbon are in our sample. 0.187 mol * (6.02*10^3) = 1.13*10^23 molecules in 1cm cubed Find the cube root of this: = 4.817*10^7 molecules in 1cm Then convert it to pm = 0.04817 molecules per 1pm The inverse of this should be the radius of a Carbon Atom = 10.38pm but... Carbon Atomic Radius is actually about 70 pm. What did I do wrong?
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