*m*_{s} = 8(*A*_{C}) + 8(*A*_{H})
= 8(12.01 g/mol) + 8(1.008 g/mol) = 104.14 g/mol
Now, using Equation 14.7, it is possible to calculate the repeat unit weight of the unknown repeat unit type, *m*_{x}. Thus
Finally, it is necessary to calculate the repeat unit molecular weights for each of the possible other repeat unit types. These are calculated below:
*m*_{ethylene} = 2(*A*_{C}) + 4(*A*_{H}) = 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol
*m*_{propylene} = 3(*A*_{C}) + 6(*A*_{H}) = 3(12.01 g/mol) + 6(1.008 g/mol) = 42.08 g/mol
*m*_{TFE} = 2(*A*_{C}) + 4(*A*_{F}) = 2(12.01 g/mol) + 4(19.00 g/mol) = 100.02 g/mol
*m*_{VC} = 2(*A*_{C}) + 3(*A*_{H}) + (*A*_{Cl}) = 2(12.01 g/mol) + 3(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol
Therefore, propylene is the other repeat unit type since its *m* value is almost the same as the calculated *m*_{x}.
14.19 *(a) Determine the ratio of butadiene to styrene repeat units in a copolymer having a number-average molecular weight of 350,000 g/mol and degree of polymerization of 4425.*
* (b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?*
__Solution__
(a) This portion of the problem asks us to determine the ratio of butadiene to styrene repeat units in a copolymer having a weight-average molecular weight of 350,000 g/mol and a degree of polymerization of 4425. It first becomes necessary to calculate the average repeat unit molecular weight of the copolymer, , using Equation 14.6 as
If we designate *f*_{b} as the chain fraction of butadiene repeat units, since the copolymer consists of only two repeat unit types, the chain fraction of styrene repeat units *f*_{s} is just 1 – *f*_{b}. Now, Equation 14.7 for this copolymer may be written in the form
in which *m*_{b} and *m*_{s} are the repeat unit molecular weights for butadiene and styrene, respectively. These values are calculated as follows:
*m*_{b} = 4(*A*_{C}) + 6(*A*_{H}) = 4(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol
*m*_{s} = 8(*A*_{C}) + 8(*A*_{H}) = 8(12.01 g/mol) + 8(1.008 g/mol) = 104.14 g/mol
Solving for *f*_{b} in the above expression yields
Furthermore, *f*_{s} = 1 – *f*_{b} = 1 – 0.50 = 0.50; or the ratio is just
(b) Of the possible copolymers, the only one for which there is a restriction on the ratio of repeat unit 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 not only alternating, but also random, graft, and block.
14.20 *Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylene may have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both repeat unit types.*
__Solution__
For a copolymer consisting of 60 wt% ethylene and 40 wt% propylene, we are asked to determine the fraction of both repeat unit types.
In 100 g of this material, there are 60 g of ethylene and 40 g of propylene. The ethylene (C_{2}H_{4}) molecular weight is
*m*(ethylene) = 2(*A*_{C}) + 4(*A*_{H})
= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol
The propylene (C_{3}H_{6}) molecular weight is
*m*(propylene) = 3(*A*_{C}) + 6(*A*_{H})
= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol
Therefore, in 100 g of this material, there are
and
Thus, the fraction of the ethylene repeat unit, *f*(ethylene), is just
Likewise,
14.21 *A random poly(isobutylene-isoprene) copolymer has a number-average molecular weight of 200,000 g/mol and a degree of polymerization of 3000. Compute the fraction of isobutylene and isoprene repeat units in this copolymer.*
__Solution__
For a random poly(isobutylene-isoprene) copolymer in which = 200,000^{ }g/mol and *DP* = 3000, we are asked to compute the fractions of isobutylene and isoprene repeat units.
From Table 14.5, the isobutylene repeat unit has four carbon and eight hydrogen atoms. Thus,
*m*_{ib} = (4)(12.01 g/mol) + (8)(1.008 g/mol) = 56.10 g/mol
Also, from Table 14.5, the isoprene repeat unit has five carbon and eight hydrogen atoms, and
*m*_{ip} = (5)(12.01 g/mol) + (8)(1.008 g/mol) = 68.11 g/mol
From Equation 14.7
Now, let *x* = *f*_{ib}, such that
since* f*_{ib} + *f*_{ip} = 1. Also, from Equation 14.6
Or
Solving for *x* leads to *x* = *f*_{ib} = *f*(isobutylene) = 0.12. Also,
*f*(isoprene) = 1 – *x* = 1 – 0.12 = 0.88
**Share with your friends:** |