4.6 Molecular weight data for some polymer are tabulated here. Compute (a) the number-average molecular
weight and (b) the weight-average molecular weight. (c) If it is known that this material's degree of polymerization
is 710, which one of the polymers listed in Table 4.3 is this polymer? Why?
Molecular Weight
Range g/mol
15,000–30,000
xi
0.04
wi
0.01
30,000–45,000
0.07
0.04
45,000–60,000
0.16
0.11
60,000–75,000
0.26
0.24
75,000–90,000
0.24
0.27
90,000–105,000
0.12
0.16
105,000–120,000
0.08
0.12
120,000–135,000
0.03
0.05
Solution
(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 Mi
xi
xiMi
15,000-30,000
22,500
0.04
900
30,000-45,000
37,500
0.07
2625
45,000-60,000
52,500
0.16
8400
60,000-75,000
67,500
0.26
17,550
75,000-90,000
82,500
0.24
19,800
90,000-105,000
97,500
0.12
11,700
105,000-120,000
112,500
0.08
9000
120,000-135,000
127,500
0.03
3825
_________________________
Mn =

 xi M i = 73,800 g/mol
(b) From the tabulated data, we are asked to compute M w , the weight-average molecular weight. This
determination is performed as follows:

Molecular wt.
Range
Mean Mi
wi
wiMi
15,000-30,000
22,500
0.01
225
30,000-45,000
37,500
0.04
1500
45,000-60,000
52,500
0.11
5775
60,000-75,000
67,500
0.24
16,200
75,000-90,000
82,500
0.27
22,275
90,000-105,000
97,500
0.16
15,600
105,000-120,000
112,500
0.12
13,500
120,000-135,000
127,500
0.05
6375
_________________________
Mw =
wi M i = 81,450 g/mol
(c) We are now asked if the degree of polymerization is 710, which of the polymers in Table 4.3 is this

material? It is necessary to compute m in Equation 4.6 as
m =
Mn
73,800 g/mol
=
= 103.94 g/mol
DP
710
The repeat unit molecular weights of the polymers listed in Table 4.3 are as follows:

Polyethylene--28.05 g/mol
Poly(vinyl chloride)--62.49 g/mol
Polytetrafluoroethylene--100.02 g/mol
Polypropylene--42.08 g/mol
Polystyrene--104.14 g/mol
Poly(methyl methacrylate)--100.11 g/mol
Phenol-formaldehyde--133.16 g/mol
Nylon 6,6--226.32 g/mol
PET--192.16 g/mol
Polycarbonate--254.27 g/mol
Therefore, polystyrene is the material since its repeat unit molecular weight is closest to that calculated above.
4.10 Using the definitions for total chain molecule length, L (Equation 4.9) and average chain end-to-end distance r
(Equation 4.10), determine the following for a linear polyethylene:
(a) the number-average molecular weight for L = 2500 nm;
(b) the number-average molecular weight for r = 20 nm.
Solution
(a) This portion of the problem asks for us to calculate the number-average molecular weight for a linear
polyethylene for which L in Equation 4.9 is 2500 nm. It is first necessary to compute the value of N using this
equation, where, for the C—C chain bond, d = 0.154 nm, and  = 109. Thus
N =
=

L
 
d sin  
2 
2500 nm
= 19,940
109 
(0.154 nm) sin 

 2 
Since there are two C—C bonds per polyethylene repeat unit, there is an average of N/2 or 19,940/2 = 9970 repeat

units per chain, which is also the degree of polymerization, DP. In order to compute the value of M n using
Equation 4.6, we must first determine m for polyethylene. Each polyethylene repeat unit consists of two carbon and
four hydrogen atoms, thus

m = 2(AC) + 4(AH)
= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol
Therefore
M n = (DP)m = (9970)(28.05 g/mol) = 280,000 g/mol
(b) Next, we are to determine the number-average molecular weight for r = 20 nm. Solving for N from

Equation 4.10 leads to
N =

r2
(20 nm)2
=
= 16,900
d2
(0.154 nm)2
which is the total number of bonds per average molecule. Since there are two C—C bonds per repeat unit, then DP
= N/2 = 16,900/2 = 8450. Now, from Equation 4.6
M n = (DP)m = (8450)(28.05 g/mol) = 237,000 g/mol
4.18 An alternating copolymer is known to have a number-average molecular weight of 250,000 g/mol and a

degree of polymerization of 3420. If one of the repeat units is styrene, which of ethylene, propylene,
tetrafluoroethylene, and vinyl chloride is the other repeat unit? Why?
Solution
For an alternating copolymer which has a number-average molecular weight of 250,000 g/mol and a degree
of polymerization of 3420, we are to determine one of the repeat unit types if the other is styrene. It is first
necessary to calculate m using Equation 4.6 as
m =

Mn
250,000 g/mol
=
= 73.10 g/mol
DP
3420
Since this is an alternating copolymer we know that chain fraction of each repeat unit type is 0.5; that is fs = fx =

0.5, fs and fx being, respectively,
the chain fractions of the styrene and unknown repeat units. Also, the repeat unit
molecular weight for styrene is
ms = 8(AC) + 8(AH)
= 8(12.01 g/mol) + 8(1.008 g/mol) = 104.14 g/mol
Now, using Equation 4.7, it is possible to calculate the repeat unit weight of the unknown repeat unit type, mx. Thus
mx =
=
m  f s ms
fx
73.10 g/mol - (0.5)(104.14 g/mol)
= 42.06 g/mol

0.5
Finally, it is necessary to calculate the repeat unit molecular weights for each of the possible other repeat

unit types. These are calculated below:
methylene = 2(AC) + 4(AH) = 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol
mpropylene = 3(AC) + 6(AH) = 3(12.01 g/mol) + 6(1.008 g/mol) = 42.08 g/mol
mTFE = 2(AC) + 4(AF) = 2(12.01 g/mol) + 4(19.00 g/mol) = 100.02 g/mol
mVC = 2(AC) + 3(AH) + (ACl) = 2(12.01 g/mol) + 3(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol
Therefore, propylene is the other repeat unit type since its m value is almost the same as the calculated mx.
4.24
The density of totally crystalline polypropylene at room temperature is 0.946 g/cm 3.
Also, at room
temperature the unit cell for this material is monoclinic with lattice parameters
a = 0.666 nm
α = 90
b = 2.078 nm
β = 99.62
c = 0.650 nm
γ = 90
If the volume of a monoclinic unit cell, Vmono, is a function of these lattice parameters as
Vmono = abc sin 
determine the number of repeat units per unit cell.
Solution
For this problem we are given the density of polypropylene (0.946 g/cm3), an expression for the volume of
its unit cell, and the lattice parameters, and are asked to determine the number of repeat units per unit cell. This
computation necessitates the use of Equation 3.7, in which we solve for n. Before this can be carried out we must
first calculate VC, the unit cell volume, and A the repeat unit molecular weight. For VC
VC = abc sin 
= (0.666 nm)(2.078 nm)(0.650 nm) sin (99.62)
= 0.8869 nm3 = 8.869  10-22 cm3
The repeat unit for polypropylene is shown in Table 4.3, from which the value of A may be determined as follows:
A = 3(AC) + 6(AH)
= 3(12.01 g/mol) + 6(1.008 g/mol)
= 42.08 g/mol
Finally, solving for n from Equation 3.7 leads to
n =
=

(0.946 g/cm3)( 8.869 

VC N A
A
10 -22 cm3/unit cell)( 6.022  1023 repeat units/mol)
42.08 g/mol
= 12.0 repeat unit/unit cell