Unformatted text preview:

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 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 polymethyl methacrylate, from Table 14.3, 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 14.3, 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 polyethylene terephthalate, from Table 14.3, 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 14.9 (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 justmC = 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 CCl = 177.25 g600.5 g + 95.76 g + 177.25 g x 100 = 20.3 wt% (b) Chlorinated polyethylene differs from polyvinyl 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. 14.11 This problem first of all asks for us to calculate, using Equation (14.11), 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 (14.4a). For polyethylene, from Table 14.3, 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 nn = M nm = 300,000 g/mol28.05 g/mol = 10,695 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 14.4); therefore, from Equation (14.11)L = Nd sin θ2      = (21, 390)(0.154 nm) sin 109°2            = 2682 nm 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) 21,390 = 22.5 nm 14.23 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 g28.05 g / mol= 2.14 mol of ethylene and 40 g42.08 g / mol= 0.95 mol of propylene Thus, the fraction of the ethylene mer, f(ethylene), is justf(ethylene) =2.14 mol2.14 mol+ 0.95 mol= 0.69 Likewise, f(propylene) =0.95 mol2.14 mol+ 0.95 mol=


View Full Document

UTK MSE 201 - Chapter 14 Notes

Download Chapter 14 Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Chapter 14 Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Chapter 14 Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?