What is the relationship between atomic density and mass density in aluminum?

[mvr01]

Can anyone help.

How do I calculate mass density of aluminium, given the atomic density of 6.04 X 10^22.

I understand that Density = Mass/Volume. but the question confuses me.

Thanks

 

[Dickfore]

the mass density is equal to:

[tex]
\rho = \frac{\Delta m}{\Delta V}
[/tex]

where [itex]\Delta m[/itex] is the mass contained in the volume [itex]\Delta V[/itex]. But, at the same time, the total mass is equal to:

[tex]
\Delta m = m_{0} \, \Delta N
[/tex]

where [itex]m_{0}[/itex] is the mass of one particle (atom, molecule) and [itex]\Delta N[/itex] is the total number of particles in that volume. So, we can write:

[tex]
\rho = \frac{m_{0} \, \Delta N}{\Delta V} = m_{0} \, \frac{\Delta N}{\Delta V}
[/tex]

But, by definition, the number density is:

[tex]
n_{0} \equiv \frac{\Delta N}{\Delta V}
[/tex]

So, we have:

[tex]
\rho = m_{0} \, n_{0}
[/tex]

So, all you need is to find the mass of one atom of aluminum. You will need to review the concept of atomic mass unit and relative atomic (molecular) mass.

 

[mvr01]

Thanks dickfore...

so is this right?

p = (6.04 X 10^22 / 6.023 X 10 ^23) x 26.98 = 2.7g

 

[Dickfore]

Thanks dickfore...

so is this right?

p = (6.04 X 10^22 / 6.023 X 10 ^23) x 26.98 = 2.7g

in what volume are there [itex]6.04 \times 10^{22}[/itex] particles? You will need to divide with that volume to get a density. As it is now, your result has units of mass.

 

[mvr01]

Sorry,, I hope I am not annoying you. I am getting mixed answers. Can you please work it out for me...

Im gettng volume of 6.04 X 10^22 = 4.470 X10^-23, therefore the mass density equaling 3.645 X 10^46..

or

(6.04 X 10^22 / 9.98 ) X 26.98 = 1.6g/cm^3

Im lost

 

[Dickfore]

no, i won't. In your original post, you hadn't specified a correct unit for the number density of Al atoms. This is why your mass density is in incorrect units. You should look back in your problem formulation again.

 

[mvr01]

Yes you're right,,, The question is the atomic density of 6.04 X 10^22 atoms per cm^3.
I missed out the units.

so would this mean, (6.04 X 10^22 cm^3 / 1.023 X 10^23) x 26.98 amu = 2.7 g/cm^3

 

[Dickfore]

yes. look up density of aluminum and you compare with what you have found. However, I urge you to think about which one is easier to measure, the density of a bulk material or the number density of the atoms in the material?