m = 3(A
C
) + 6(A
H
)
= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol
If we let
n
n
represent the number-average degree of polymerization, then from Equation (14.4a)
n
n
=
M
n
m
=
10
6
g/mol
42.08 g/mol
= 23, 700
14.5
(a)
The mer molecular weight of polystyrene is called for in this portion of the problem.
For
polystyrene, from Table 14.3, each mer unit has eight carbons and eight hydrogens.
Thus,
m = 8(A
C
) + 8(A
H
)
= (8)(12.01 g/mol) + (8)(1.008 g/mol) = 104.14 g/mol
(b)
We are now asked to compute the weight-average molecular weight. Since the weight-average
degree of polymerization,
n
w
, is 25,000, using Equation (14.4b)
M
w
= n
w
m = (25,000)(104.14 g/mol) = 2.60 x 10
6
g/mol
14.6
(a)
From the tabulated data, we are asked to compute
M
n
, the number-average molecular weight.
This is carried out below.
Molecular wt
Range
Mean M
i
x
i
x
i
M
i
8,000-16,000
12,000
0.05
600
16,000-24,000
20,000
0.16
3200
24,000-32,000
28,000
0.24
6720
32,000-40,000
36,000
0.28
10,080
40,000-48,000
44,000
0.20
8800
48,000-56,000
52,000
0.07
3640
____________________________
M
n
=
x
i
M
i
= 33, 040γ/μολ
109