2013 Spring
E344 Solution Set 5
1. 14.2 Compute repeat unit molecular weights for the following: (a) poly(vinyl chloride), (b) poly(ethylene
terephthalate), (c) polycarbonate, and (d) polydimethylsiloxane.
Solution
(c) For polycarbonate, from Table 14.3, each repeat unit has sixteen carbons, fourteen hydrogens, and three
oxygens. Thus,
m = 16(AC) + 14(AH) + 3(AO)
= (16)(12.01 g/mol) + (14)(1.008 g/mol) + (3)(16.00 g/mol)
= 254.27 g/mol
(d) For polydimethylsiloxane, from Table 14.5, each repeat unit has two carbons, six hydrogens, one silicon and one
oxygen. Thus,
m = 2(AC) + 6(AH) + (ASi) + (AO)
= (2)(12.01 g/mol) + (6)(1.008 g/mol) + (28.09 g/mol) + (16.00 g/mol) = 74.16 g/mol
For Nylon 6-6, from Table 14.3, each repeat unit has twelve carbons, twenty-two hydrogens, two nitrogens and two
oxygens. Thus, (C12H22N2O2)n
m = 12(AC) + 22(AH) + 2(AN) + 2(AO)
= (12)(12.01 g/mol) + (22)(1.008 g/mol) + 2(14.007 g/mol) + 2(16.00 g/mol) = 226.31 g/mol
2. 14.9 For a linear polymer molecule, the total chain length L depends on the bond length between chain atoms d,
the total number of bonds in the molecule N, and the angle between adjacent backbone chain atoms θ, as follows:
 
(14.11)
L = Nd sin  
2 
Furthermore, the average end-to-end distance for a series of polymer molecules r in Figure 14.6 is equal to

r = d N
(14.12)
A linear polytetrafluoroethylene has a number-average molecular weight of 500,000 g/mol; compute average values

of L and r for this material.
Solution
This problem first of all asks for us to calculate, using Equation 14.11, the average total chain length, L, for
a linear polytetrafluoroethylene polymer having a number-average molecular weight of 500,000 g/mol.
It is
necessary to calculate the degree of polymerization, DP, using Equation 14.6. For polytetrafluoroethylene, from
Table 14.3, each repeat unit has two carbons and four flourines. Thus,
m = 2(AC) + 4(AF)
= (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol
and
DP =
Mn
500,000 g/mol
=
= 5000
m
100.02 g/mol
which is the number of repeat units along an average chain. Since there are two carbon atoms per repeat unit, there

are two C—C chain bonds per repeat unit, which means that the total number of chain bonds in the molecule, N, is
just (2)(5000) = 10,000 bonds. Furthermore, assume that for single carbon-carbon bonds, d = 0.154 nm and  =
109 (Section 14.4); therefore, from Equation 14.11
 
L = Nd sin  
2 
 109 
= (10,000)(0.154 nm) sin 
 = 1254 nm

  2 
It is now possible to calculate the average chain end-to-end distance, r, using Equation 14.12 as

r = d N = (0.154 nm) 10,000 = 15.4 nm
The typical conformation of a PTFE molecule in the sample is random coil.

3. 14.26 The density and associated percent crystallinity for two nylon 6,6 materials are as follows:
ρ (g/cm3)
crystallinity (%)
1.188
67.3
1.152
43.7
(a) Compute the densities of totally crystalline and totally amorphous nylon 6,6.
(b) Determine the density of a specimen having 55.4% crystallinity.
Solution
(a) We are asked to compute the densities of totally crystalline and totally amorphous nylon 6,6 (c and a
from Equation 14.8). From Equation 14.8 let C =
% crystallinity
, such that
100
C =

c ( s  a )
 s (c  a )
Rearrangement of this expression leads to

c (C s  s ) + ca  C sa = 0
in which c and a are the variables for which solutions are to be found. Since two values of s and C are specified

in the problem, two equations may be constructed as follows:
c (C1 s1  s1)  ca  C1s1a  0

c (C2 s2  s2 )  ca  C2s2a  0
3
3
In which s1 = 1.188 g/cm
 , s2 = 1.152 g/cm , C1 = 0.673, and C2 = 0.437. Solving the above two equations for
a and c leads to
a 
=

(1.188
 s1  s2 (C1  C2 )
C1  s1  C2  s2
g/cm3)(1.152 g/cm3) (0.673 0.437)
3
(1.188 g/cm3)  (0.437) (1.152 g/cm3) = 1.091 g/cm
(0.673)
And
c =
=
 s1  s2 (C2  C1)
 s2 (C2  1)   s1(C1  1)
(1.188
g/cm3)(1.152 g/cm3) (0.437  0.673)
3
 g/cm3) (0.437 1)  (1.188 g/cm3) (0.6731) = 1.242 g/cm
(1.152
(b) Now we are asked to determine the density of a specimen having 55.4% crystallinity. Solving for s

from Equation 14.8 and substitution for a and c which were computed in part (a) yields
s =
=

c a
C (c  a )  c
 (1.242 g/cm3)( 1.091 g/cm3)

(0.554) (1.242 g/cm3  1.091 g/cm3)  1.242 g/cm3
= 1.170 g/cm3
4. The diagram below gives the monomer structure for a so-called Liquid-Crstal Polymer (LCP) referred to as
HNA. Note that the unlabeled vertices in this diagram correspond to C, and any carbon atom at a vertex without 4
bonds has the remain bonds made with H atoms (also not shown).
A) Show that the molecular weight of the HNA monomer is: 170 g/mole
B) What is the average degree of polymerization for a sample of polyHNA whose average molecular weight is
28,000 g/mole. How does this compare to an identical sample of polyethylene with the same molecular weight?
C) Briefly explain why the monomer for HNA is more rigid than that of polyethylene. What effect would this higher
rigidity have on the conformation of a polyHNA molecule relative to that of a PE molecule with the same molecular
weight?
Solution
A) There are 11 carbons, 2 oxygens and 6 hydrogens. Thus, the molecular weight of the HNA monomer is :
m = 11(AC) + 6(AH) + 2(AO)=170
B) DP =
M n 28,000 g/mol
=
= 165
m
170 g/mol
DPPE =
M n 28,000 g/mol
=
= 1000
m
28 g/mol
C) The backbone of the HNA monomer contains two aromatic rings, which are much less flexible than C-C bonds
in the monomer of polyethylene. This high rigidity will induce a much longer persistence length of a polyHNA
molecules
than
that
of
a
PE
molecule
with
same
molecular
weight.
5. 15.31 Name the following polymer(s) that would be suitable for the fabrication of cups to contain hot coffee:
polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why?
Solution
This question asks us to name which, of several polymers, would be suitable for the fabrication of cups to
contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum
temperature of hot coffee is probably slightly below 100C (212F). Of the polymers listed, only polystyrene and
polycarbonate have glass transition temperatures of 100C or above (Table 15.2), and would be suitable for this
application.
6. 15.32 Of those polymers listed in Table 15.2, which polymer(s) would be best suited for use as ice cube trays?
Why?
Solution
In order for a polymer to be suited for use as an ice cube tray it must have a glass-transition temperature
below 0C. Of those polymers listed in Table 15.2 only low-density and high-density polyethylene, PTFE, and
polypropylene satisfy this criterion.
7. Core 5.03 The boiling temperatures of methane (CH4), carbon tetrachloride (CCl4), and water (H2O) are
111 K, 350 K, and 373 K, respectively. For each of these three molecules:
a) make a 2-D sketch of the structure of the molecule; and
b) calculate the percent ionicity of the C-H; C-Cl, and O-H bonds; then
c) develop a brief argument justifying the magnitudes and general order of boiling temperatures for these three
molecules.
Solution
a)
b)
c) The key point here is to recognize the nature of the secondary bonding in each molecule. Methane has relatively
weak secondary bonding because the C-H dipole is not strong, and the molecule is symmetric. The boiling point of
carbon tetrachloride is much higher because the dipole strength is much higher. Water has the highest boiling
temperature of the three, because it is an asymmetric molecule and the positive and negative parts of the dipole can
align.
8. Core 5.10
Solution
a)
m=8(12.01g/mol)+8(1.01g/mol)=104.16g/mol
DP 
Mw 100,000 g / mole

 960
m
104.16 g / mole
b) One refers to an average molecular weight when describing a typical polymer sample, because
the molecules in the sample all have different lengths around this average. The reason they have
varying lengths is because of variations in the synthesis. Different molecules initiate at different
times. They terminate at different times and in different ways. And, the can grow at slightly
different rates depending on such things as the local temperature or pressure.
c) The specimen with the higher average degree of polymerization will have a much larger extent
of entanglements between different molecules. This is especially true for a polymer like high
molecular weight polystyrene where the conformation of the individual molecules is globular and
in the form of a random coil. For the specimen to significantly deform, the entangled molecules
must de-entangle, which is much easier in the case of short molecules than it is for long
molecules.