CHEM...Calculation ..structure

Question

Q: The edge of the copper unit cell is found by X-ray diffraction to be
3.58x10^-8 cm.Calculate the density of the solid copper.
R.M.A of Cu :63.5
Avogadro constant=6.02x10^23

ANSWER: Vol= 6.02x10^23 (3.58x10^-8)^3
Density of Cu = 4(63.5)/[6.02x10^23 (3.58x10^-8)^3 ]

我唔明點解計 vol. 要 乘Avogadro constant...??
可否解釋下...........萬分感激
有興趣可睇睇呢題~
http://hk.knowledge.yahoo.com/question/?qid=7007122501199

Answer 1

Cu has abcabc.... close-packed structure.
The unit cell of Cu is face-centred cubic.

In each unit cell :
No. of Cu atoms at the corners = 8 x (1/8) = 1
No. of Cu atoms on the surfaces = 6 x (1/2) = 3
Total no. of Cu atom in the unit cell = 1 + 3 = 4
Volume of the unit cell = (3.58 x 10-8)3 cm3

==========
Method I for the calculation of the density of Cu :
(The method used in the question)

Consider 1 mol of unit cells. There are 4 mol of Cu atoms.
(This is because there are 4 Cu atom in each unit cell.)


No. of unit cells = 6.02 x 1023
Total volume = (6.02 x 1023) x (3.58 x 10-8)3 cm3
Mass of 4 mol of Cu atoms = 4 x (63.5) g

Density
= Mass / Volume
= 4 x (63.5) / [(6.02 x 1023) x (3.58 x 10-8)3] g cm-3

==========
Method II for the calculation of the density of Cu :

Consider 1 unit cell. There are 4 Cu atoms.

Volume = (3.58 x 10-8)3 cm3
Mass of 6.02 x 1023 Cu atoms = 63.5 g
Mass of 4 Cu atoms = 4 x (63.5) / (6.02 x 1023) g

Density
= Mass / Volume
= 4 x (63.5) / [(6.02 x 1023) x (3.58 x 10-8)3] g cm-3

==========
Both the methods give the same answer.