an unknown element crystallizes in a face-centered cubic unit cell. with an edge length of 392.4 pm. The solid has a density of 21.09 g/cm^3. what is atomic weight of solid?
so what i did was :
FC cubic uit --> 4 atoms
edge length--> l = 2*root2*r = 392.4 pm = 392.4 * 10^-8 cm.
d= 21.09 g/cm^3
m= l^3 * d
m= 8.2757*10^-5 g
1amu = 1.66054 * 10^-24 g
atomic weight = 4.98 * 10^19 amu.
and that is not the correct answer at all. where am i going wrong? please help :/
the options are :
a) 6.9 amu
b) 241.7 amu
d) 191.8 amu
chemistry - Tom, Thursday, November 8, 2012 at 1:53pm
ryeng i suppose?
this is a very easy question
392.4*10-10= x cm
ycm^3* 21.09= mass in grams
since face centered unit cell has 4 atoms
mass/4, then use the product of that to divide by 1.66*10^-24 = amu
your result should be 191.84 amu
chemistry - DrBob222, Thursday, November 8, 2012 at 2:49pm
One place you erred is in the volume. You have a given to you (the edge length) so volume = a^3.
I agree with Tom's answer although I don't follow all of the work. My answer is 191.77 which rounds to 191.8 and that is one of the answers.