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 π‘šπ‘œπ‘™ π‘ƒπ‘œ
209 𝑔 π‘ƒπ‘œ
:1 π‘ƒπ‘œ π‘Žπ‘‘π‘œπ‘š 𝑋 6.022π‘₯1023 π‘Žπ‘‘π‘œπ‘šπ‘  π‘ƒπ‘œ 𝑋 1 π‘šπ‘œπ‘™ π‘ƒπ‘œ = 3.47π‘₯10−22 𝑔. The volume of a unit cell is
Using density we can solve for the volume in terms of the cell edge length, a, and then finally
find a.
3.47π‘₯10−22 𝑔
9.32 𝑔/π‘π‘š3
= 3.72π‘₯10−23 π‘π‘š3 .
So π‘Ž = √3.72π‘₯10−23 = 3.34π‘₯10−8 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 π‘šπ‘œπ‘™ 𝐸𝑒
152.97 𝑔 𝐸𝑒
The mass of these two atoms is: 2 𝐸𝑒 π‘Žπ‘‘π‘œπ‘šπ‘  𝑋 6.022π‘₯1023 π‘Žπ‘‘π‘œπ‘šπ‘  𝐸𝑒 𝑋 1 π‘šπ‘œπ‘™ 𝐸𝑒 = 5.080π‘₯10−22 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 π‘Ž =
5.080π‘₯10−22 𝑔
= 1.09π‘₯10−22 π‘π‘š3
4.68 𝑔/π‘π‘š3
√1.09π‘₯10−22 π‘π‘š3 = 4.78π‘₯10−8 π‘π‘š