# Ch 8 - Iowa State University ```Leader: Grant DeRocher
Course: Chem 167
Supplemental Instruction
Instructor: Houk
Iowa State University
Date: 03/05/13
Do a problem from end of ch 7 first as review. Then 8.10, 8.13, 8.17, 8.19,
Next Exam is Wed March 13
2 hour Exam Review on Sunday March 10, Room TBD
Ch 8
4. Polonium is the only metal that forms a simple cubic crystal structure. Use the fact that the
density of polonium is 9.32g/cm3 to calculate its atomic radius.
In the simple cubic crystal structure there are four Po atoms at the corners of each unit cell. The
lengths of a unit cell, a, equals two times its radius, a=2r. The number of Po atoms resent in the
simple cubic unit cell is 8*(1/8)=1 Po atom. The mass of one Po atoms is
1 𝑚𝑜𝑙 𝑃𝑜
209 𝑔 𝑃𝑜
:1 𝑃𝑜 𝑎𝑡𝑜𝑚 𝑋 6.022𝑥1023 𝑎𝑡𝑜𝑚𝑠 𝑃𝑜 𝑋 1 𝑚𝑜𝑙 𝑃𝑜 = 3.47𝑥10−22 𝑔. The volume of a unit cell is
Vcell=(a)3
Using density we can solve for the volume in terms of the cell edge length, a, and then finally
find a.
𝑉=
d=m/v;
𝑚
𝑑
=
3.47𝑥10−22 𝑔
9.32 𝑔/𝑐𝑚3
= 3.72𝑥10−23 𝑐𝑚3 .
3
So 𝑎 = √3.72𝑥10−23 = 3.34𝑥10−8 cm.
Now the cell edge a=2r so r=a/2 = 3.34x10-8/2= 1.67x10-8 cm
5. Europium forms a body-centered cubic unit cell and has density of 4.68g/cm3. From this
information, determine the length of the edge of a cubic cell.
The body centered cubic unit cell contains 8*(1/8) +1= 2 Eu atoms.
1 𝑚𝑜𝑙 𝐸𝑢
152.97 𝑔 𝐸𝑢
The mass of these two atoms is: 2 𝐸𝑢 𝑎𝑡𝑜𝑚𝑠 𝑋 6.022𝑥1023 𝑎𝑡𝑜𝑚𝑠 𝐸𝑢 𝑋 1 𝑚𝑜𝑙 𝐸𝑢 = 5.080𝑥10−22 𝑔
The volume of this unit cell is Vcell= (a)3
Using density, we can solve for volume in terms of the cell edge length, a, and then finally find
a. d= m/V 𝑉 =
So 𝑎 =
𝑚
=
5.080𝑥10−22 𝑔
= 1.09𝑥10−22 𝑐𝑚3
4.68 𝑔/𝑐𝑚3
3
√1.09𝑥10−22 𝑐𝑚3 = 4.78𝑥10−8 𝑐𝑚
𝑑
```