Polymer Properties
Exercise 1
Structure
a)
b)
c)
d)
CH2=CH–CH3
CH2=C(CH3)2
CH3–CH=CH–C2H5
CH3–CH=CH–CH=CH2
O
e)
f)
CH3
CH CH C2H5
Draw the different stereoregular
polymer structures that can be
obtained.
1a)
n CH2
CH2
CH CH3
CH
CH3
n
Stereoregularity of PP:
isotaktinen PP
n
syndiotaktinen PP
n
1b)
CH3
CH2
C(CH3)2
CH2
C
CH3
n
No stereoregularity differences in polymer.
1c)
CH3
CH3
CH CH C2H5
CH CH
C2H5
n
C2H5 C2H5 C2H5
C2H5 C2H5 C2H5
C2H5 C2H5 C2H5
C2H5 C2H5 C2H5
C2H5 C2H5 C2H5
n
n
n
n
C2H5 C2H5 C2H5
n
n
1d)
3,4-polymerization
1,2-polymerization
Stereoregular polymers in the case of 3,4-polymerization form in the
same way as in c). In case of the 1,2-polymerization can produce as
in a) both isotactic or syndiotactic polymers as well as cis-trans
isomers:
CH3
H
H
H
C C
C C
CH
H
CH
CH3
CH2
CH2
n
n
cis
trans
1e)
O
CH3
CH CH C2H5
CH3
CH3
CH CH O
O CH CH
C2H5
C2H5
n
n
Stereoregular polymers form in the same way as in c).
1f)
CH3
n
CH3
CH3
CH3
(CH2)4
C
n
(CH2)4 CH C
CH
n
n
n
CH3
(CH2)4
C
cis
C
CH3
H
n
(CH2)4
H
C
C
H
trans
n
(CH2)4
C
C
H
CH3
(CH2)4
cis
n
C
C
CH3
trans
n
Ring opening polymerization can occur or the ring may remain intact. In the case
the ring does not open only few of the possible structures have been shown.
Avarage molecular weights
Equations for number average and weight average molecular weights as well as polydispersity index are defined as follows:
Mn
nM


n
i
w


n
i
i
i
Mw 
 wi M i
w
i
where
i
Mi
ni
wi

2
n
M
i i
n M
i
Mw
PD 
Mn
i
molecular weight of molecules i
number of molecules with molecular weight i
mass of the molecules with molecular weight i
2
A sample of polystyrene is
composed of a series of
fractions of different sized
molecules.
a) Calculate the number
average and weight
average molecular weights
of this sample as well as
the PDI.
b) How would adding styrene
oligomer change the
average molecular
weights? Added amount is
5wt.% of polymer mass
and M=1000g/mol.
Table 1. PS fractions.
Fraction
weight fraction
Molecular weight
[g/mol]
A
0.130
11000
B
0.300
14000
C
0.400
17000
D
0.170
21000
2a)
• Determine the number of moles in each fraction. Assume that
the sample is 10 g in the beginning. Number of moles of the
fraction is ni = wi / Mi.
fractio
wi
Mi
ni
n
[g]
[g/mol]
[mmol]
A
1.30
11000
0.118
B
3.00
14000
0.214
C
4.00
17000
0.235
D
1.70
21000
0.0810
wi = 10.0 g
ni = 0.648 mmol
2a)
• Number average molecular weight:
Mn
n M w



n n
i
i
i
i
i
10.0 g
g

 15400 mol
0.648mmol
• Weight average molecular weight:
Mw
wM


w
i
i
i
g
(1.3 11000  3.0 14000  4.0 17000  1.7  21000) g  mol
g

 16000 mol
10.0 g
• Polydispersity index:
g
M w 16000 mol
PD 

 1.04
g
M n 15400 mol
2b)
• When 5.0 wt-% of styrene oligomer (fraction E) is added, the
total mass and number of moles increase as follow:
fraction
wi
Mi
ni
[g]
[g/mol]
[mmol]
A
1.30
11000
0.118
B
3.00
14000
0.214
C
4.00
17000
0.235
D
1.70
21000
0.0810
E
0.50
1000
0.50
wi = 10.5 g
ni = 1.10 mmol
2b)
• number average molecular weight
Mn
nM


n
i
i
i
10.5 g
g

 9150 mol
1.10mmol
• weight average molecular weight
Mw
w M

w
i
i
i
g
(1.3 11000  3.0 14000  4.0 17000  1.7  21000  0.5 1000) g  mol
g

 15300 mol
10.5 g
• Polydispersity index
g
M w 15300 mol
PD 

 1, 67
g
Mn
9150 mol
3 Viscosity
• relative viscosity:
 t
r  
0 t 0
  0 t  t 0



sp
• Relative viscosity increment (or specific viscosity):
0
t0
sp
• Reduced viscosity (or viscosity number): red 
c
ln r
• Inherent viscosity: inh 
c
• Mark-Houwink equation:    k  M va
• Intrinsic viscosity [] can be defined:
 sp 
   limc0  
 c 
 ln r 

 c 
   limc0 
3)
• Viscosity of atactic polystyrene was measured in dilute
solutions and the results are presented in table 2.
• Determine the viscosity average molecular weight for the
sample . Mark-Houwink constants are k = 0.00848 ml/g and a
= 0.748.
Table 2. Efflux times for polystyrene samples. Solvent toluene. T =25°C.
Polystyrene concentration
efflux time
[mg/ml]
[t/s]
0
110.0
5.0
123.5
10.0
138.0
15.0
153.6
20.0
170.2
25.0
187.9
3)
• Calculate the required viscosity parameters:
c
efflux time
r
sp
inh
[mg/ml]
[t/s]
= t/t0
= (t-t0)/t0 =ln(r)/c = sp/c
0
110.0
5.0
123.5
1.123
0.123
0.0232
0.0246
10.0
138.0
1.255
0.255
0.0227
0.0255
15.0
153.6
1.396
0.396
0.0222
0.0264
20.0
170.2
1.547
0.547
0.0218
0.0274
25.0
187.9
1.708
0.708
0.0214
0.0283
• Draw inh and red as function of concentration.
red
3)
•
[] is obtained from the plot from the
crossing of y-axis:
 sp 
   limc0    0.0237 ml/mg =23.7 ml/g
 c 
ln r
 c
   limc0 
•
•

 = 0.0236 ml/mg = 23.6 ml/g

and the average from these is [] =
23.65 ml/g.
Viscosity average molecular weight
from Mark-Houwink equation:
   k  M va  M va 
Note! Due to empirical coefficients k ja a.
the equation gives the molecular weight
without unit. In literature k = 0.007…0.01
and a = 0.69…0.78  accuracy of the
calculation is not particularly good.
 
k
ml

1
23.65
    a 
g
 Mv  
 
 0.00848 ml
 k 

g







1
0.748
 40205
M v  40000
g
mol
4) Light scattering
• Both weight average molecular weight Mw and second virial
coefficient A2 can be determined from graph when Kc/R(q) is
plotted as a function of concentration:
Kc
1

 2 A2 c
R( ) M w
• 1/Mw is the cross point on y-axis and A2 is half of the linear
coefficient.
2
2
2 
8 m 
2


1,
4199

6,
297

10
2


2
g
2 2 no2  dn 


12 mol  m
K
 1, 633 10
 
4 
23
1
10
4
N A  dc  6, 023 10 mol  (6328 10 m)
g2
3
5,E-06
4)
3
c (g / m )
5,034E+02
1,007E+03
1,510E+03
2,014E+03
2,517E+03
-1
R( ) (m )
2,390E-04
4,400E-04
6,060E-04
7,900E-04
9,020E-04
Kc / R( )
3,440E-06
3,737E-06
4,071E-06
4,163E-06
4,558E-06
Kc/R mol/g
4,E-06
3,E-06
y = 5,29E-10x + 3,20E-06
2
R = 9,77E-01
2,E-06
1,E-06
0,E+00
0
500
1000
1500
c / g/m3
• From the plot:
1
g
-6 mol
= 3,20 10
 M w  313000
Mw
g
mol
• And second virial coefficient:
2 A2  5, 29 1010
3
mol  m3
mol

m
10

A

2,64

10
2
2
g
g2
2000
2500