Chap 14 Solns-6E

Chap 14 Solns-6E - CHAPTER 14 POLYMER STRUCTURES PROBLEM...

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CHAPTER 14 POLYMER STRUCTURES PROBLEM SOLUTIONS 14.1 Polymorphism is when two or more crystal structures are possible for a material of given composition. Isomerism is when two or more polymer molecules or mer units have the same composition, but different atomic arrangements. 14.2 The mer structures called for are sketched below. (a) Polyvinyl fluoride (b) Polychlorotrifluoroethylene (c) Polyvinyl alcohol 107
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14.3 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 14.3). If A C and A F represent the atomic weights of carbon and fluorine, respectively, then m = 2(A C ) + 4(A F ) = (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol (b) For polymethyl methacrylate, from Table 14.3, each mer unit has five carbons, eight hydrogens, and two oxygens. Thus, m = 5(A C ) + 8(A H ) + 2(A O ) = (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 14.3, each mer unit has twelve carbons, twenty-two hydrogens, two nitrogens, and two oxygens. Thus, m = 12(A C ) + 22(A H ) + 2(A N ) + 2(A O ) = (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 polyethylene terephthalate, from Table 14.3, each mer unit has ten carbons, eight hydrogens, and four oxygens. Thus, m = 10(A C ) + 8(A H ) + 4(A O ) = (10)(12.01 g/mol) + (8)(1.008 g/mol) + (4)(16.00 g/mol) = 192.16 g/mol 14.4 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 108
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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
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(b) From the tabulated data, we are asked to compute M w , the weight- average molecular weight.
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Chap 14 Solns-6E - CHAPTER 14 POLYMER STRUCTURES PROBLEM...

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