• Molybdenium has BCC structure Calculate atomic radius and mass density if lattice
constant is 3.15 A and atomic weight is 95.94.
Answers
Answer:
B
C
C
c
r
y
s
t
a
l
s
t
r
u
c
t
u
r
e
r
M
o
a
t
o
m
=
0.1363
n
m
a
t
o
m
i
c
m
a
s
s
M
o
=
95.94
g
m
o
l
The density is the quotient of the mass and the volume. First we need the cell geometry to find the volume of the unit cell of molybdenum using the atomic radius. The closest approach of atoms is along the cell diagonal, the distance from A to C'.
Find the volume of the unit cell
We can relate the lattice parameter a, the length of the side of the cube to the atomic radius using the Pythagorean Theorem. The cell diagonal makes a triangle with the side length, C to C' and the face diagonal, which is AC in the figure. The relation of atomic radius to lattice parameter is
a
2
+
(
√
2
a
)
2
=
3
a
2
3
a
2
=
(
4
r
)
2
a
=
4
√
3
r
3
The volume of the unit cell is the cube of the lattice parameter. It will be useful in the calculation to convert nanometers to picometers.
0.1363
n
m
×
1000
p
m
n
m
=
136.3
p
m
Inserting this value into the equation for volume of the cell gives
a
3
=
(
4
√
3
×
136.3
p
m
3
)
3
=
3.119
×
10
7
p
m
3
Find the mass of the Mo atoms
There are two atoms per unit cell for the BCC structure because the central atom contained is contained within the cell and each corner atom is shared with seven other atoms.
1
+
1
8
×
8
=
2
The mass of the atoms is 95.94 atomic mass units because a mole of the atoms has a mass of 95.94 grams per mole. The mass of atoms in the cell is
95.94
a
m
u
×
2
=
191.88
a
m
u
Convert the mass to grams using the conversion from amu to grams,
1.66
×
10
−
24
g
a
m
u
191.88
a
m
u
×
1.66
×
10
−
24
g
a
m
u
=
3.18
×
10
−
22
g
Theoretical density
The relationship among density, volume, and mass is
d
e
n
s
i
t
y
=
m
a
s
s
v
o
l
u
m
e
Place the calculated mass and volume into this relationship after converting cubic picometers to cubic centimeters.
3.12
×
10
7
p
m
3
×
(
1
×
10
−
10
c
m
p
m
)
3
=
3.12
×
10
−
22
c
m
3
d
e
n
s
i
t
y
=
3.18
×
10
−
22
g
3.12
×
10
−
23
=
10.19
g
c
m
3
Comparison to accepted value
The accepted value for the density of molybdenum is 10.22. The percent error between the two values is
(
10.22
−
10.19
10.22
)
×
100
%
=
0.29
%
The agreement between calculated and accepted value is excellent with a 0.29 percent error.