Calculating Carbon Atomic Radius

[Meadock]

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?

[thedarkshade]

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:

as you surely know the formula is: [math]n=\frac{m}{M}[/math], and as you have only the density, then you find it through [math]m=\rho \times V[/math], when V=1[math]1dm^3[/math]. So (I think) it goes like this:

 

[math]m=\rho \times V= 2.23\frac{g}{cm^3} \times 1000cm^3=[/math]

[math]m=2230g[/math]

 

Then you find the moles by:

 

[math]m=\frac{m}{M}=\frac{2230g}{12 \frac{g}{mol}}=185.8 g[/math]

 

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

 

What you did here is you put density instead of mass. According to your datas, it's density that is 2.23 not the mass!

[Meadock]

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

[thedarkshade]

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

In my chemistry classes (which I hardy take now:-() when we had to find the mass when only density was given, we had to multiply the density with the volume of 1[math]dm^3[/math] , and as 1[math]dm^3[/math] is equal to 1000[math]cm^3[/math] that is why I multiplied with 1000!

 

And the grams cancel each other only when finding moles (if I haven't done that then I must have probably missed it) but they do not cancel each other by this formula: [math]m=\rho \times V[/math]!

 

as you surely know the formula is: [math]n=\frac{m}{M}[/math], and as you have only the density, then you find it through [math]m=\rho \times V[/math], when V=1[math]1dm^3[/math]. So (I think) it goes like this:

 

[math]m=\rho \times V= 2.23\frac{g}{cm^3} \times 1000cm^3=[/math]

[math]m=2230g[/math]

 

Then you find the moles by:

 

[math]n=\frac{m}{M}=\frac{2230g}{12 \frac{g}{mol}}=185.8 mol[/math]

 

 

What you did here is you put density instead of mass. According to your datas, it's density that is 2.23 not the mass!

edit: you're right , I put m instead of n.

Sorry I didn't do that on purpose, I must have just not paid attention to it! Sorry again!