CHAPTER 14
POLYMER STRUCTURES
PROBLEM SOLUTIONS
Hydrocarbon Molecules
Polymer Molecules
The Chemistry of Polymer Molecules
14.1 On the basis of the structures presented in this chapter, sketch repeat unit structures for the following polymers: (a) polychlorotrifluoroethylene, and (b) poly(vinyl alcohol).
Solution
The repeat unit structures called for are sketched below.
(a) Polychlorotrifluoroethylene
(b) Poly(vinyl alcohol)
Molecular Weight
14.2 Compute repeat unit molecular weights for the following: (a) poly(vinyl chloride), (b) poly(ethylene terephthalate), (c) polycarbonate, and (d) polydimethylsiloxane.
Solution
(a) For poly(vinyl chloride), each repeat unit consists of two carbons, three hydrogens, and one chlorine (Table 14.3). If AC, AH and ACl represent the atomic weights of carbon, hydrogen, and chlorine, respectively, then
m = 2(AC) + 3(AH) + (ACl)
= (2)(12.01 g/mol) + (3)(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol
(b) For poly(ethylene terephthalate), from Table 14.3, each repeat 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
(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
14.3 The number-average molecular weight of a polypropylene is 1,000,000 g/mol. Compute the degree of polymerization.
Solution
We are asked to compute the degree of polymerization for polypropylene, given that the number-average molecular weight is 1,000,000 g/mol. The repeat unit 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
Now it is possible to compute the degree of polymerization using Equation 14.6 as
14.4 (a) Compute the repeat unit molecular weight of polystyrene.
(b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000.
Solution
(a) The repeat unit molecular weight of polystyrene is called for in this portion of the problem. For polystyrene, from Table 14.3, each repeat 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 number-average molecular weight. Since the degree of polymerization is 25,000, using Equation 14.6
14.5 Below, molecular weight data for a polypropylene material are tabulated. Compute (a) the number-average molecular weight, (b) the weight-average molecular weight, and (c) the degree of polymerization.
Range (g/mol) / xi / wi
8,000–16,000 / 0.05 / 0.02
16,000–24,000 / 0.16 / 0.10
24,000–32,000 / 0.24 / 0.20
32,000–40,000 / 0.28 / 0.30
40,000–48,000 / 0.20 / 0.27
48,000–56,000 / 0.07 / 0.11
Solution
(a) From the tabulated data, we are asked to compute , 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
______
(b) From the tabulated data, we are asked to compute , the weight-average molecular weight.
Molecular wt.
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
______
(c) Now we are asked to compute the degree of polymerization, which is possible using Equation 14.6. For polypropylene, the repeat unit molecular weight is just
m = 3(AC) + 6(AH)
= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol
And
14.6 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 14.3 is this polymer? Why?
Range g/mol / xi / wi
15,000–30,000 / 0.04 / 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 , 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
______
(b) From the tabulated data, we are asked to compute , 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
______
(c) We are now asked if the degree of polymerization is 710, which of the polymers in Table 14.3 is this material? It is necessary to compute m in Equation 14.6 as
The repeat unit molecular weights of the polymers listed in Table 14.3 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.
14.7 Is it possible to have a poly(methyl methacrylate) homopolymer with the following molecular weight data and a of polymerization of 527? Why or why not?
Range (g/mol) / wi / xi
8,000–20,000 / 0.02 / 0.05
20,000–32,000 / 0.08 / 0.15
32,000–44,000 / 0.17 / 0.21
44,000–56,000 / 0.29 / 0.28
56,000–68,000 / 0.23 / 0.18
68,000–80,000 / 0.16 / 0.10
80,000–92,000 / 0.05 / 0.03
Solution
This problem asks if it is possible to have a poly(methyl methacrylate) homopolymer with the given molecular weight data and a degree of polymerization of 527. The appropriate data are given below along with a computation of the number-average molecular weight.
Molecular wt.
Range Mean Mi xi xiMi
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 7400
80,000-92,000 86,000 0.03 2580
______
For PMMA, from Table 14.3, each repeat 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
Now, we will compute the degree of polymerization using Equation 14.6 as
Thus, such a homopolymer is not possible since the calculated degree of polymerization is 477 (and not 527).
14.8 High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen.
(a) Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 5% of all the original hydrogen atoms.
(b) In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)?
Solution
(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 determined using a modified form of Equation 4.3 as
(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
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)
Furthermore, the average end-to-end distance for a series of polymer molecules r in Figure 14.6 is equal to
(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
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 q = 109° (Section 14.4); therefore, from Equation 14.11
It is now possible to calculate the average chain end-to-end distance, r, using Equation 14.12 as
14.10 Using the definitions for total chain molecule length, L (Equation 14.11) and average chain end-to-end distance r (Equation 14.12), for a linear polyethylene determine:
(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 14.11 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 q = 109°. Thus
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 using Equation 14.6, 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
(b) Next, we are to determine the number-average molecular weight for r = 20 nm. Solving for N from Equation 14.12 leads to
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 14.6
Molecular Configurations
14.11 Sketch portions of a linear polystyrene molecule that are (a) syndiotactic, (b) atactic, and (c) isotactic. Use two-dimensional schematics per footnote 8 of this chapter.
Solution
We are asked to sketch portions of a linear polystyrene molecule for different configurations (using two-dimensional schematic sketches).
(a) Syndiotactic polystyrene
(b) Atactic polystyrene
(c) Isotactic polystyrene
14.12 Sketch cis and trans structures for (a) butadiene, and (b) chloroprene. Use two-dimensional schematics per footnote 11 of this chapter.
Solution
This problem asks for us to sketch cis and trans structures for butadiene and chloroprene.
(a) The structure for cis polybutadiene (Table 14.5) is
The structure of trans butadiene is
(b) The structure of cis chloroprene (Table 14.5) is
The structure of trans chloroprene is
Thermoplastic and Thermosetting Polymers
14.13 Make comparisons of thermoplastic and thermosetting polymers (a) on the basis of mechanical characteristics upon heating, and (b) according to possible molecular structures.
Solution
(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.
14.14 (a) Is it possible to grind up and reuse phenol-formaldehyde? Why or why not?
(b) Is it possible to grind up and reuse polypropylene? Why or why not?
Solution
(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
14.15 Sketch the repeat structure for each of the following alternating copolymers: (a) poly(butadiene-chloroprene), (b) poly(styrene-methyl methacrylate), and (c) poly(acrylonitrile-vinyl chloride).
Solution
This problem asks for sketches of the repeat unit structures for several alternating copolymers.
(a) For poly(butadiene-chloroprene)
(b) For poly(styrene-methyl methacrylate)
(c) For poly(acrylonitrile-vinyl chloride)
14.16 The number-average molecular weight of a poly(styrene-butadiene) alternating copolymer is 1,350,000 g/mol; determine the average number of styrene and butadiene repeat units per molecule.
Solution
Since it is an alternating copolymer, the number of both types of repeat units will be the same. Therefore, consider them as a single repeat unit, and determine the number-average degree of polymerization. For the styrene repeat unit, there are eight carbon atoms and eight hydrogen atoms, while the butadiene repeat consists of four carbon atoms and six hydrogen atoms. Therefore, the styrene-butadiene combined repeat unit weight is just
m = 12(AC) + 14(AH)
= (12)(12.01 g/mol) + (14)(1.008 g/mol) = 158.23 g/mol
From Equation 14.6, the degree of polymerization is just
Thus, there is an average of 8530 of both repeat unit types per molecule.
14.17 Calculate the number-average molecular weight of a random nitrile rubber [poly(acrylonitrile-butadiene) copolymer] in which the fraction of butadiene repeat units is 0.30; assume that this concentration corresponds to a degree of polymerization of 2000.