Chapter 1 Assignment Solution Keys (10 pt each, total 50 pt)
2.
Average number of backbone carbon between ethyl side chains
= (total number of carbons on the backbone)/ (total number of ethyl group)
= (44.9% x 4 + 47.2% x 4 + 7.9% x 2) / (7.9% x 1)
= 48.6
(Each 1,4 addition contributes 4 carbons and each 1,2 addition contributes 2 carbons to the backbone
carbons; each 1, 2 addition contributes 1 ethyl group)
5.
(a)
If the star polymer is formed with a full conversion, there will be no B end group left, as shown in
eq.1. However, a full conversion is never possible. Each unreacted B will start a linear polymer and
linear polymer bearing B end group will present to some degree (eq. 2).
R is the multifunctional unit with multiple A’ functional groups from which the branches grow.
b is the number of branches that grow out of the multifunctional unit.
y is the average number of monomer unit subtended by an A’ group of multi-functional unit.
(b)
“Q = the number of equivalent of multifunctional group that react per mole of monomer
L = the number of equivalent of unreacted end groups (B) per mole of monomer”
Assign B = the total number of AB monomer.
the total number of functional group (A) of the star center = QB
the total number of branches that subtended from a multi-functional unit = QB
the total number of unreacted end group (B) = the total number of unreacted monomer = the total
number of linear polymer that formed = LB
Because of equal reactivity of all functional groups, the average number of monomer unit subtended by
an A’ group of multi-functional unit will equal the number attached to the B terminal unit of the linear
polymer, therefore
<y> = (total number of reacted monomers) / (total number of polymer chains)
= (total number of monomer-total number of unreacted monomer) / (total number of linear chain +
total number of branches that formed from the star center)
= (B-LB) / (LB+QB)
= (1-L) / (L+Q)
1
(c)
The number average molecular weight of star polymer (Mn)
= the molecular weight of the multi-functional unit + (the number average molecular weight of the
branches) x (the number of branch)
= the molecular weight of the multi-functional unit + (the number average degree of polymerization) x
(repeating unit molecular weight) x (the number of branch)
= Mmultifunctional unit + <y>M0b
= b(Mmultifunctional unit/b + <y>M0)
= b[Mb + (1-L)/(Q+L)M0]
M0 is a constant and Mb is related to the molecular weight of the multi-functional unit (Mmultifunctional
unit) by a factor of 1/b.
(d) Using parameters for the following multi-functional units and monomer and evaluate the Mn.
________________________________________________________________________
b
4
8
Q
0.2169
0.134
L
0.0018
0.00093
M(multifunctional
unit)
458.46
770.77
Mo
113.16
113.16
Mn = M(multifunctional unit) +
(1-L)/(Q+L)M0b
2524.4
7473.8
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6.
Mn = ∑Mini / ∑ni = ∑Mi(gi/Mi) / ∑(gi/Mi) = ∑gi / ∑(gi/Mi)
Mw = ∑Miwi = ∑(Migi)/∑gi
Mz = ∑Mi3ni / ∑Mi2ni = ∑Mi3(gi/Mi) / ∑Mi2(gi/Mi) = ∑(Mi2gi) / ∑(Migi)
fraction
mass gi
(g)
1
2
3
4
5
6
7
8
9
1.15
0.73
0.415
0.35
0.51
0.34
1.78
0.1
0.94
Mix10-4
1.25
2.05
2.4
3.2
3.9
4.5
6.35
4.1
9.4
gi/Mi
0.92
0.36
0.17
0.11
0.13
0.08
0.28
0.02
0.10
Migi
1.44
1.50
1.00
1.12
1.99
1.53
11.30
0.41
8.84
2
Mi gi
1.80
3.07
2.39
3.58
7.76
6.89
71.77
1.68
83.06
Mn
Mw
Mz
2.91
4.61
6.25
2
12.
15.
Mn1 = ∑gi1 / ∑(gi1/Mi1) = 70k, Mw1 = ∑(Mi1gi1)/∑gi1 =100k, ∑gi = 10g
Mn2 = ∑gi1 / ∑(gi1/Mi1) = 20k, Mw2 = ∑(Mi2gi2)/∑gi2 = 60k, ∑gi = 20g
Mn = ∑gi/∑gi/Mi) = (∑gi1+∑gi2)/[∑(gi1/Mi1)+∑(gi2/Mi2)] = (∑gi1 +∑gi2)/(∑gi1/Mn1+∑gi2 /Mn2)
= (10 + 20)/(10/70 + 20/20) = 26.25 k
Mw1 = ∑(Migi)/∑gi = [∑(Mi1gi1)+∑(Mi1gi1)]/(∑gi1+∑gi2) = (Mw1∑gi1+Mw2∑gi2)/(∑gi1 + ∑gi2)
= (100x10 + 60x20)/(10 + 20) = 73.33 k
3