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CHAPTER 4

POLYMER STRUCTURES

PROBLEM SOLUTIONS

Hydrocarbon Molecules
Polymer Molecules
The Chemistry of Polymer Molecules

4.1 The mer structures called for are sketched below.

(a) Polychlorotrifluoroethylene

(b) Poly(vinyl alcohol)

Molecular Weight

4.2 Mer weights for several polymers are asked for in this problem.
 (a) For polytetrafluoroethylene, each mer unit consists of two carbons and four fluorines (Table 4.3). If
AC and AF represent the atomic weights of carbon and fluorine, respectively, then

m = 2(AC) + 4(AF)

= (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol

(b) For poly(methyl methacrylate), from Table 4.3, each mer unit has five carbons, eight hydrogens, and
two oxygens. Thus,

m = 5(AC) + 8(AH) + 2(AO)

= (5)(12.01 g/mol) + (8)(1.008 g/mol) + (2)(16.00 g/mol) = 100.11 g/mol

(c) For nylon 6,6, from Table 4.3, each mer unit has twelve carbons, twenty-two hydrogens, two nitrogens,
and two oxygens. Thus,

m = 12(AC) + 22(AH) + 2(AN) + 2(AO)

= (12)(12.01 g/mol) + (22)(1.008 g/mol) + (2)(14.01 g/mol) + (2)(16.00 g/mol)

= 226.32 g/mol

(d) For poly(ethylene terephthalate), from Table 4.3, each mer unit has ten carbons, eight hydrogens, and
four oxygens. Thus,

m = 10(AC) + 8(AH) + 4(AO)

= (10)(12.01 g/mol) + (8)(1.008 g/mol) + (4)(16.00 g/mol) = 192.16 g/mol

4.3 We are asked to compute the number-average degree of polymerization for polypropylene, given that
the number-average molecular weight is 1,000,000 g/mol. The mer molecular weight of polypropylene is just

m = 3(AC) + 6(AH)

= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol

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If we let nn represent the number-average degree of polymerization, then from Equation 4.4a

Mn 106 g/mol
nn = = = 23, 700
m 42.08 g/mol

4.4 (a) The mer molecular weight of polystyrene is called for in this portion of the problem. For
polystyrene, from Table 4.3, each mer unit has eight carbons and eight hydrogens. Thus,

m = 8(AC) + 8(AH)

= (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, nw, is 25,000, using Equation 4.4b

M w = nw m = (25, 000)(104.14 g/mol) = 2.60 x 106 g/mol

4.5 (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

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
____________________________
Mn = ∑
xi M i = 33, 040 g/mol

(b) From the tabulated data, we are asked to compute M w , the weight-average molecular weight.

Molecular wt.

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 Range Mean Mi wi wiMi

8,000-16,000 12,000 0.02 240

16,000-24,000 20,000 0.10 2000
24,000-32,000 28,000 0.20 5600
32,000-40,000 36,000 0.30 10,800
40,000-48,000 44,000 0.27 11,880
48,000-56,000 52,000 0.11 5720
___________________________
Mw = ∑ wi M i = 36, 240 g/mol

(c) Now we are asked to compute nn (the number-average degree of polymerization), using Equation 4.4a.

For polypropylene,

m = 3(AC) + 6(AH)

= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol

And
Mn 33, 040 g/mol
nn = = = 785
m 42.08 g/mol

(d) And, finally, we are asked to compute nw, the weight-average degree of polymerization, as

Mw 36, 240 g/mol

nw = = = 861
m 42.08 g/mol

4.6 (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 Mi xi xi M i

8,000-20,000 14,000 0.05 700

20,000-32,000 26,000 0.15 3900
32,000-44,000 38,000 0.21 7980
44,000-56,000 50,000 0.28 14,000
56,000-68,000 62,000 0.18 11,160

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 68,000-80,000 74,000 0.10 7400
80,000-92,000 86,000 0.03 2580
_________________________
M n = ∑ x i M i = 47, 720 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

8,000-20,000 14,000 0.02 280

20,000-32,000 26,000 0.08 2080
32,000-44,000 38,000 0.17 6460
44,000-56,000 50,000 0.29 14,500
56,000-68,000 62,000 0.23 14,260
68,000-80,000 74,000 0.16 11,840
80,000-92,000 86,000 0.05 4300
_________________________
M w = ∑ w i M i = 53, 720 g/mol

(c) We are now asked if the number-average degree of polymerization is 477, which of the polymers in
Table 4.3 is this material? It is necessary to compute m in Equation 4.4a as

Mn 47, 720 g/mol

m = = = 100.04 g/mol
nn 477

The mer 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

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 PET--192.16 g/mol
Polycarbonate--254.27 g/mol

Therefore, polytetrafluoroethylene is the material since its mer molecular weight is closest to that calculated above.

(d) The weight-average degree of polymerization may be calculated using Equation 4.4b, since M w and

m were computed in portions (b) and (c) of this problem. Thus

Mw 53, 720 g/mol

nw = = = 537
m 100.04 g/mol

4.7 This problem asks if it is possible to have a poly(vinyl chloride) homopolymer with the given
molecular weight data and a number-average degree of polymerization of 1120. The appropriate data are given
below along with a computation of the number-average molecular weight.

Molecular wt.
Range Mean Mi xi xi M i

8,000-20,000 14,000 0.05 700

20,000-32,000 26,000 0.15 3900
32,000-44,000 38,000 0.21 7980
44,000-56,000 50,000 0.28 14,000
56,000-68,000 62,000 0.18 11,160
68,000-80,000 74,000 0.10 7440
80,000-92,000 86,000 0.03 2580
_________________________
Mw = ∑ xi M i = 47, 720 g/mol

For PVC, from Table 4.3, each mer unit has two carbons, three hydrogens, and one chlorine. Thus,

m = 2(AC) + 3(AH) + (ACl)

= (2)(12.01 g/mol) + (3)(1.008 g/mol) + (35.45 g/mol) = 62.49 g/mol

Now, we will compute nn using Equation 4.4a as

Mn 47, 720 g/mol

nn = = = 764
m 62.49 g/mol

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Thus, such a homopolymer is not possible since the calculated nn is 764 not 1120.

4.8 (a) For chlorinated polyethylene, we are asked to determine the weight percent of chlorine added for
5% Cl substitution of all original hydrogen atoms. Consider 50 carbon atoms; there are 100 possible side-bonding
sites. Ninety-five are occupied by hydrogen and five are occupied by Cl. Thus, the mass of these 50 carbon atoms,
mC, is just

mC = 50(AC) = (50)(12.01 g/mol) = 600.5 g

Likewise, for hydrogen and chlorine,

mH = 95(AH) = (95)(1.008 g/mol) = 95.76 g

mCl = 5(ACl) = (5)(35.45 g/mol) = 177.25 g

Thus, the concentration of chlorine, CCl, is just

177.25 g
CCl = x 100 = 20.3 wt%
600.5 g + 95.76 g + 177.25 g

(b) Chlorinated polyethylene differs from poly(vinyl chloride), in that, for PVC, (1) 25% of the side-
bonding sites are substituted with Cl, and (2) the substitution is probably much less random.

Molecular Shape

4.9 This problem first of all asks for us to calculate, using Equation 4.7, the average total chain length, L,
for a linear polyethylene polymer having a number-average molecular weight of 300,000 g/mol. It is necessary to
calculate the number-average degree of polymerization, nn, using Equation 4.4a. For polyethylene, from Table 4.3,

each mer unit has two carbons and four hydrogens. Thus,

m = 2(AC) + 4(AH)

= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol

and

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 Mn 300, 000 g/mol
nn = = = 10, 695
m 28.05 g/mol

which is the number of mer units along an average chain. Since there are two carbon atoms per mer unit, there are
two C--C chain bonds per mer, which means that the total number of chain bonds in the molecule, N, is just
(2)(10,695) = 21,390 bonds. Furthermore, assume that for single carbon-carbon bonds, d = 0.154 nm and θ = 109°
(Section 4.4); therefore, from Equation 4.7

⎛θ ⎞
L = Nd sin ⎜ ⎟
⎝ 2⎠

⎡ ⎛ 109° ⎞ ⎤
= (21, 390)(0.154 nm) ⎢sin ⎜ ⎟ ⎥ = 2682 nm
⎣ ⎝ 2 ⎠⎦

It is now possible to calculate the average chain end-to-end distance, r, using Equation 4.8 as

r = d N = (0.154 nm) 21, 390 = 22.5 nm

4.10 (a) This portion of the problem asks for us to calculate the number-average molecular weight for a
linear polytetrafluoroethylene for which L in Equation 4.7 is 2000 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 θ = 109°. Thus

L
N =
⎛ θ⎞
d sin ⎜ ⎟
⎝ 2⎠

2000 nm
= = 15, 900
⎛ 109°⎞
(0.154 nm) sin ⎜ ⎟
⎝ 2 ⎠

Since there are two C--C bonds per PTFE mer unit, there is an average of N/2 or 15,900/2 = 7950 mer units per
chain, which is also the number-average degree of polymerization, nn. In order to compute the value of M n using

Equation 4.4a, we must first determine m for PTFE. Each PTFE mer unit consists of two carbon and four fluorine
atoms, thus

m = 2(AC) + 4(AF)

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 = (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol

Therefore
M n = nn m = (7950)(100.02 g/mol) = 795, 000 g/mol

(b) Next, we are to determine the number-average molecular weight for r = 15 nm. Solving for N from
Equation 4.8 leads to

r2 (15 nm) 2
N = = = 9490
d2 (0.154 nm) 2

which is the total number of bonds per average molecule. Since there are two C--C bonds per mer unit, then nn =

N/2 = 9490/2 = 4745. Now, from Equation 4.4a

M n = nn m = (4745)(100.02 g/mol) = 474, 600 g/mol

Molecular Configurations

4.11 We are asked to sketch portions of a linear polypropylene molecule for different configurations
(using two-dimensional schematic sketches).
(a) Syndiotactic polypropylene

(b) Atactic polypropylene

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 (c) Isotactic polypropylene

4.12 This problem asks for us to sketch cis and trans structures for butadiene and chloroprene.
(a) The structure for cis polybutadiene is

The structure of trans butadiene is

(b) The structure of cis chloroprene is

The structure of trans chloroprene is

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 Thermoplastic and Thermosetting Polymers

4.13 This question asks for comparisons of thermoplastic and thermosetting polymers.
(a) Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers,
harden upon heating, while further heating will not lead to softening.
(b) Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the
structures will normally be network or crosslinked.

4.14 (a) It is not possible to grind up and reuse phenol-formaldehyde because it is a network thermoset
polymer and, therefore, is not amenable to remolding.
(b) Yes, it is possible to grind up and reuse polypropylene since it is a thermoplastic polymer, will soften
when reheated, and, thus, may be remolded.

Copolymers

4.15 This problem asks for sketches of the mer structures for several alternating copolymers.
(a) For poly(ethylene-propylene)

(b) For poly(butadiene-styrene)

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 (c) For poly(isobutylene-isoprene)

4.16 For a poly(styrene-butadiene) alternating copolymer with a number-average molecular weight of

1,350,000 g/mol, we are asked to determine the average number of styrene and butadiene mer units per molecule.
Since it is an alternating copolymer, the number of both types of mer units will be the same. Therefore,
consider them as a single mer unit, and determine the number-average degree of polymerization. For the styrene
mer, there are eight carbon atoms and eight hydrogen atoms, while the butadiene mer consists of four carbon atoms
and six hydrogen atoms. Therefore, the styrene-butadiene combined mer weight is just

m = 12(AC) + 14(AH)

= (12)(12.01 g/mol) + (14)(1.008 g/mol) = 158.23 g/mol

From Equation 4.4a, the number-average degree of polymerization is just

Mn 135, 000 g/mol

nn = = = 8530
m 158.23 g/mol

Thus, there is an average of 8530 of both mer types per molecule.

4.17 This problem asks for us to calculate the number-average molecular weight of a random nitrile
[poly(acrylonitrile-butadiene) copolymer]. For the acrylonitrile mer there are three carbon, one nitrogen, and three
hydrogen atoms. Thus, its mer molecular weight is

mAc = 3(AC) + (AN) + 3(AH)

= (3)(12.01 g/mol) + 14.01 g/mol + (3)(1.008 g/mol) = 53.06 g/mol

The butadiene mer is composed of four carbon and six hydrogen atoms. Thus, its mer molecular weight is

mBu = 4(AC) + 6(AH)

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 = (4)(12.01 g/mol) + (6)(1.008 g/mol) = 54.09 g/mol

From Equation 4.5, the average mer molecular weight is just

m = fAc mAc + fBu mBu

= (0.70)(53.06 g/mol) + (0.30)(54.09 g/mol) = 53.37 g/mol

Since nn = 2000 (as stated in the problem), M n may be computed using Equation 4.4a as

M n = m nn = (53.37 g/mol)(2000) = 106, 740 g/mol

4.18 For an alternating copolymer that has a number-average molecular weight of 100,000 g/mol and a
number-average degree of polymerization of 2210, we are to determine one of the mer types if the other type is
ethylene. It is first necessary to calculate m using Equation 4.4a as

Mn 100, 000 g/mol

m = = = 42.25 g/mol
nn 2210

Since this is an alternating copolymer we know that chain fraction of each mer type is 0.5; that is fe = fx = 0.5, fe
and fx being, respectively, the chain fractions of the ethylene and unknown mers. Also, the mer molecular weight

for ethylene is

ms = 2(AC) + 4(AH)

= 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol

Now, using Equation 4.5, it is possible to calculate the mer weight of the unknown mer type, mx. Thus

m − feme
mx =
fx

45.25 g/mol - (0.5)(28.05 g/mol)

= = 62.45 g/mol
0.5

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 Finally, it is necessary to calculate the mer molecular weights for each of the possible other mer types.
These are calculated below:

mstyrene = 8(AC) + 8(AH) = 8(12.01 g/mol) + 8(1.008 g/mol) = 104.16 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, vinyl chloride is the other mer type since its m value is almost the same as the calculated mx.

4.19 (a) This portion of the problem asks us to determine the ratio of butadiene to acrylonitrile mers in a
copolymer having a weight-average molecular weight of 250,000 g/mol and a weight-average degree of
polymerization of 4640. It first becomes necessary to calculate the average mer molecular weight of the copolymer,
m , using Equation 4.4b as

Mw 250, 000 g/mol

m = = = 53.88 g/mol
nw 4640

If we designate fb as the chain fraction of butadiene mers, since the copolymer consists of only two mer types, the
chain fraction of acrylontrile mers fa is just 1 – fb. Now, Equation 4.5 for this copolymer may be written in the form

m = f b mb + fa ma = f b mb + (1 − f b ) ma

in which mb and ma are the mer molecular weights for butadiene and acrylontrile, respectively. These values are

calculated as follows:

mb = 4(AC) + 6(AH) = 4(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol

ma = 3(AC) + 3(AH) + (AN) = 3(12.01 g/mol) + 3(1.008 g/mol) + (14.01 g/mol)

= 53.06 g/mol.

Solving for fb in the above expression yields

m − ma 53.88 g/mol − 53.06 g/mol

fb = = = 0.80
m − ma 54.09 g/mol − 53.06 g/mol
b

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Furthermore, fa = 1 – fb = 1 – 0.80 = 0.20; or the ratio is just

fb 0.80
= = 4.0
fa 0.20

(b) Of the possible copolymers, the only one for which there is a restriction on the ratio of mer types is
alternating; the ratio must be 1:1. Therefore, on the basis of the result in part (a), the possibilities for this
copolymer are random, graft, and block.

4.20 For a copolymer consisting of 60 wt% ethylene and 40 wt% propylene, we are asked to determine the
fraction of both mer types.
In 100 g of this material, there are 60 g of ethylene and 40 g of propylene. The ethylene (C2H4) molecular

weight is

m(ethylene) = 2(AC) + 4(AH)

= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol

The propylene (C3H6) molecular weight is

m(propylene) = 3(AC) + 6(AH)

= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol

Therefore, in 100 g of this material, there are

60 g
= 2.14 mol of ethylene
28.05 g / mol

and
40 g
= 0.95 mol of propylene
42.08 g / mol

Thus, the fraction of the ethylene mer, f(ethylene), is just

2.14 mol
f (ethylene) = = 0.69
2.14 mol + 0.95 mol

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Likewise,

0.95 mol
f (propylene) = = 0.31
2.14 mol + 0.95 mol

4.21 For a random poly(isobutylene-isoprene) copolymer in which M w = 200,000 g/mol and nw = 3000,

we are asked to compute the fractions of isobutylene and isoprene mers.

From Table 4.5, the isobutylene mer has four carbon and eight hydrogen atoms. Thus,

mib = (4)(12.01 g/mol) + (8)(1.008 g/mol) = 56.10 g/mol

Also, from Table 4.5, the isoprene mer has five carbon and eight hydrogen atoms, and

mip = (5)(12.01 g/mol) + (8)(1.008 g/mol) = 68.11 g/mol

From Equation 4.5

m = fibmib + fipmip

Now, let x = fib, such that

m = 56.10x + (68.11)(1 − x)

since fib + fip = 1. Also, from Equation 4.4b

Mw
nw =
m
Or

200, 000 g / mol

3000 =
[56.10x + 68.11(1 − x)] g / mol

Solving for x leads to x = fib = f(isobutylene) = 0.12. Also,

f(isoprene) = 1 – x = 1 – 0.12 = 0.88

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 Polymer Crystallinity

4.22 The tendency of a polymer to crystallize decreases with increasing molecular weight because as the
chains become longer it is more difficult for all regions along adjacent chains to align so as to produce the ordered
atomic array.

4.23 For each of four pairs of polymers, we are asked to (1) state whether it is possible to decide which is
more likely to crystallize; (2) if so, which is more likely and why; and (3) it is not possible to decide then why.
(a) No, it is not possible to decide for these two polymers. On the basis of tacticity, the isotactic PP is more
likely to crystallize than the atactic PVC. On the other hand, with regard to side-group bulkiness, the PVC is more
likely to crystallize.
(b) Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is
more likely to crystallize than crosslinked cis-isoprene since linear polymers are more likely to crystallize than
crosslinked ones.
(c) Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely
to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones
crystallize relatively easily.
(d) Yes, it is possible to decide for these two copolymers. The block poly(acrylonitrile-isoprene)
copolymer is more likely to crystallize than the graft poly(chloroprene-isobutylene) copolymer. Block copolymers
crystallize more easily than graft ones.

4.24 Given that polyethylene has an orthorhombic unit cell with two equivalent mer units, we are asked to
compute the density of totally crystalline polyethylene. In order to solve this problem it is necessary to employ
Equation 3.5, in which n represents the number of mer units within the unit cell (n = 2), and A is the mer molecular
weight, which for polyethylene is

A = 2(AC) + 4(AH)

= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol

Also, VC is the unit cell volume, which is just the product of the three unit cell edge lengths in Figure 4.10. Thus,

nA
ρ =
V N
C A

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 (2 mers/uc)(28.05 g/mol)
=
( 7.41 x 10 -8 cm)( 4.94 x 10 -8 cm)( 2.55 x 10-8 cm)/(unit cell)( 6.023 x 10 23 mers/mol)

= 0.998 g/cm3

4.25 For this problem we are given the density of nylon 6,6 (1.213 g/cm3), an expression for the volume
of its unit cell, and the lattice parameters, and are asked to determine the number of mer units per unit cell. This
computation necessitates the use of Equation 3.5, in which we solve for n. Before this can be carried out we must
first calculate VC, the unit cell volume, and A the mer molecular weight. For VC

VC = abc 1 − cos2 α − cos2 β − cos 2 γ + 2 cosα cosβ cosγ

= (0.497)(0.547)(1.729) 1 − 0.441 − 0.054 − 0.213 + 2 (0.664)(0.232)(0.462)

= 0.3098 nm3 = 3.098 x 10-22 cm3

The mer unit for nylon 6,6 is shown in Table 4.3, from which the value of A may be determined as follows:

A = 12(AC) + 22(AH) + 2(AO) + 2(AN)

= 12(12.01 g/mol) + 22(1.008 g/mol) + 2(16.00 g/mol) + 2(14.01 g/mol)

= 226.32 g/mol

Finally, solving for n from Equation 3.5 leads to

ρVC N A
n =
A

(1.213 g/cm 3 )( 3.098 x 10 -22 cm 3 /unit cell)( 6.023 x 10 23 mers/mol)

=
226.32 g/mol

= 1 mer/unit cell

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 4.26 (a) We are asked to compute the densities of totally crystalline and totally amorphous polyethylene
% crystallinity
(ρc and ρa from Equation 4.6). From Equation 4.6 let C = , such that
100

ρ c (ρ s − ρ a )
C =
ρ s (ρ c − ρ a )

Rearrangement of this expression leads to

ρ c (C ρ s − ρ s ) + ρc ρ a − C ρ s ρ a = 0

in which ρc and ρa are the variables for which solutions are to be found. Since two values of ρs and C are specified

in the problem statement, two equations may be constructed as follows:

ρ c (C1 ρ s1 − ρ s1 ) + ρ c ρa − C1 ρ s1 ρ a = 0

ρ c (C 2 ρ s2 − ρ s2 ) + ρ c ρa − C 2 ρ s2 ρ a = 0

In which ρs1 = 0.965 g/cm3, ρs2 = 0.925 g/cm3, C1 = 0.768, and C2 = 0.464. Solving the above two equations for
ρa and ρc leads to

ρ s1 ρs2 (C1 − C 2 )
ρa =
C ρ − C ρ
1 s1 2 s2

(0.965 g/cm 3 )( 0.925 g/cm 3 )(0.768 − 0.464)

= = 0.870 g/cm 3
(0.768) ( 0.965 g/cm 3 ) − (0.464) ( 0.925 g/cm 3 )

And

ρ s1 ρ s2 (C 2 − C1 )
ρc =
ρ s2 (C2 − 1 ) − ρ s1 (C1 − 1)

( 0.965 g/cm 3 )( 0.925 g/cm 3 )( 0.464 − 0.768)

= = 0.998 g/cm 3
( 0.925 g/cm 3 ) (0.464 − 1.0) − (0.965 g/cm 3 )(0.768 − 1.0)

(b) Now we are to determine the % crystallinity for ρs = 0.950 g/cm3. Again, using Equation 4.6

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 ρ c (ρ s − ρ a )
% crystallinity = x 100
ρ s (ρ c − ρ a )

( 0.998 g/cm 3 )( 0.950 g/cm 3 − 0.870 g/cm 3 )

= x 100
( 0.950 g/cm 3 )( 0.998 g/cm 3 − 0.870 g/cm 3 )

= 65.7%

4.27 (a) We are asked to compute the densities of totally crystalline and totally amorphous polypropylene
% crystallinity
(ρc and ρa from Equation 4.6). From Equation 4.6 let C = , such that
100

ρ c (ρ s − ρ a )
C =
ρ s (ρ c − ρ a )

Rearrangement of this expression leads to

ρ c (C ρ s − ρ s ) + ρ cρ a − C ρ sρ a = 0

in which ρc and ρa are the variables for which solutions are to be found. Since two values of ρs and C are specified

in the problem, two equations may be constructed as follows:

ρ c (C1 ρ s1 − ρ s1 ) + ρ cρ a − C1 ρ s1 ρ a = 0

ρ c (C 2 ρ s 2 − ρ s2 ) + ρ cρ a − C 2 ρ s2 ρ a = 0

In which ρs1 = 0.904 g/cm3, ρs2 = 0.895 g/cm3, C1 = 0.628, and C2 = 0.544. Solving the above two equations for
ρa and ρc leads to

ρ s1 ρ s2 (C1 − C 2 )
ρa =
C1 ρ s1 − C 2 ρ s2

( 0.904 g / cm 3 )( 0.895 g / cm 3 ) (0.628 − 0.544)

= = 0.841 g/cm 3
(0.628) (0.904 g / cm 3 ) − (0.544) (0.895 g / cm 3 )

And

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 ρ s1 ρ s2 (C 2 − C1 )
ρc =
ρ (C2 − 1) − ρ (C1 − 1)
s2 s1

(0.904 g / cm 3 )( 0.895 g / cm 3 ) (0.544 − 0.628)

= = 0.946 g/cm 3
( 0.895 g / cm 3 ) ( 0.544 − 1.0) − ( 0.904 g / cm 3 ) (0.628 − 1.0)

(b) Now we are asked to determine the density of a specimen having 74.6% crystallinity. Solving for ρs
from Equation 4.6 and substitution for ρa and ρc which were computed in part (a) yields

− ρc ρ a
ρs =
C (ρ c − ρ a ) − ρ c

− ( 0.946 g / cm 3 )( 0.841 g / cm 3 )
=
( 0.746) (0.946 g / cm 3 − 0.841 g / cm 3 ) − 0.946 g / cm 3

= 0.917 g/cm3

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