1
KE-100.3410 Polymer properties
Calculations 1.
Structure and average molecular weights
Exercise 1.1
Draw the different stereoregular polymer structures that can be obtained from the following
monomers
a)
b)
c)
CH2=CH–CH3
CH2=C(CH3)2
CH3–CH=CH–C2H5
d)
CH3–CH=CH–CH=CH2
O
CH3 CH CH C2H5
e)
f)
CH3
Solution 1.1
a)
n CH2
CH2
CH CH3
CH
CH3
n
Stereoregular PPs:
isotaktinen PP
n
syndiotaktinen PP
b)
CH3
CH2
C(CH3)2
CH2
C
CH3
n
No stereoregularity differences in polymer.
n
2
c)
CH3
CH3
CH CH C2H5
CH CH
C2H5
n
Stereoregular polymers:
C2H5 C2H5 C2H5
C2H5 C2H5 C2H5
C2H5 C2H5 C2H5
n
n
n
C2H5 C2H5 C2H5
C2H5 C2H5 C2H5
n
C2H5 C2H5 C2H5
n
n
d)
3,4-polymerization
1,2-polymerization
Stereoregular polymers in the case of 3,4-polymerization form in the same way as in c). In the case
of 1,2 and 1,4-polymerization, they can produce structures as in a) both isotactic or syndiotactic
polymers as well as cis-trans isomers in sidegroups (1,2-polymerization) or in backchain (1,4polymerization):
H
H
H
C C
CH
CH3
C C
CH
CH3
H
CH2
CH2
cis
n
trans
n
3
e)
O
CH CH C2H5
CH3
CH3
CH3
CH CH O
O CH CH
C2H5
C2H5
n
n
Stereoregular polymers form in the same way as in c).
f) 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.
CH3
n
CH3
CH3
CH3
(CH2)4
n
(CH2)4 CH C
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
4
AVERAGE MOLECULAR WEIGHTS
Equations for number average and weight average molecular weights as well as polydispersity
index are defined as follows:
Mn 
n M
n
i
i
i
where

w
n
Mw 
i
i
Mi
ni
wi
w M
w
i
i
i

n M
n M
i
2
i
i
i
PD 
Mw
Mn
molecular weight of molecules i
number of molecules with molecular weight i
mass of the molecules with molecular weight i
Solution 1.2
a)
Determine the number of moles in each fraction. Assume that the sample is 10g in the beginning.
Number of moles of the fraction is ni = wi / Mi.
fraction
wi
[g]
Mi
[g/mol]
ni
[mmol]
A
B
C
D
1.30
3.00
4.00
1.70
11000
14000
17000
21000
0.118
0.214
0.235
0.0810
wi = 10.0 g
ni = 0.648 mmol
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 
w M
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
PD 
g
M w 16000 mol

 1.04
g
M n 15400 mol
5
b)
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
B
C
D
E
1.30
3.00
4.00
1.70
0.50
11000
14000
17000
21000
1000
wi = 10.5 g
0.118
0.214
0.235
0.0810
0.50
ni = 1.10 mmol
thus number average molecular weight
Mn 
n M
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
10.5g
Polydispersity index
PD 
g
M w 15300 mol

 1, 67
g
Mn
9150 mol
Viscosity average molecular weight
relative viscosity
r 
 t

0 t 0
Relative viscosity increment (or specific viscosity) is the ratio of difference in viscosities (or efflux
times) to solvent viscosity (or solvent efflux time)
sp 
  0 t  t 0

0
t0
Reduced viscosity (or viscosity number)
red 
sp
c
6
Inherent viscosity
inh 
ln r
c
Mark-Houwink equation
   k  M va
Intrinsic viscosity [] can be defined:
 sp 

 c 
   limc0 
ln r 

 c 
   limc0 
By plotting viscosities as a function of concentration the intrinsic viscosity can be estimated by
extrapolation of polymer solution to zero concentration.
the intrinsic viscosity is expressed by the Huggins equation (Fried 3.119):
red  [ ]  k H [ ]2 c
where kH is dimensionless parameter whose value depends on temperature as well as the specific
polymer/solvent combination.
Exercise 1.3
The 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 M v . 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
[mg/ml]
efflux time
[t/s]
0
5.0
10.0
15.0
20.0
25.0
110.0
123.5
138.0
153.6
170.2
187.9
7
Solution 1.3
Calculate the required viscosity parameters:
c
efflux time
r
[mg/ml]
[t/s]
= t/t0
0
5.0
10.0
15.0
20.0
25.0
110.0
123.5
138.0
153.6
170.2
187.9
1.123
1.255
1.396
1.547
1.708
sp
inh
red
= (t-t0)/t0
=ln(r)/c
= sp/c
0.123
0.255
0.396
0.547
0.708
0.0232
0.0227
0.0222
0.0218
0.0214
0.0246
0.0255
0.0264
0.0274
0.0283
Draw inh and red as function of concentration.
red = 0.0002x + 0.0237
R2 = 0.9997
0,03
0,025
0,02
inh = -0.00009x + 0.0236
R2 = 0.9966
0,015
0,01
0
10
20
30
concentration [mg/ml]
  is obtained from the plot from the y-axis intercept:
 sp
 c
   limc0 

  0.0237 ml/mg =23.7 ml/g

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:
8
   k  M va  M va 
 
k
ml

1
23.65

a
   
g
 Mv  
 
 0.00848 ml
 k 

g

1
 0.748


 40205



Viscosity average molecular weight M v  40000
g
.
mol
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  the accuracy of the calculation is not particularly
good.
Exercise 1.4
Low-Angle Laser Light-Scattering = LALLS (2o-10o) can be used to determine the molecular
weight of polymer particles even from very dilute solutions using the Debye equation:
Kc
1

 2 A2 c
R( ) M w
where R() is Rayleigh ratio, Mw weight average molecular weight, c particle concentration (g/dm3).
A2 is second virial coefficient and K is a function of the refractive index.
2 2 no2  dn 
K
 
N A 4  dc 
2
where no is the refractive index of the pure solvent. NA = 6.0231023 mol-1 Avogadro´s number. 
wavelength. dn/dc specific refractive increment of the dilute polymer solution.
For cellulose acetate, the Rayleigh ratio R() in dioxane with LALLS measurement with different
concentrations is:
Error! Not a valid link.
Refractive index for dioxane is no = 1.4199. Cellulose acetate solution has dn/dc = 6.29710-2 cm3/g
and the wavelength is  = 6328 Å. Calculate the weight average molecular weight and the second
virial coefficient (A2).
Solution 1.4
9
Both weight average molecular weight Mw and second virial coefficient A2 can be determined from
graph when Kc/R() is plotted as a function of concentration
Kc
1

 2 A2 c
R( ) M w
1/Mw is the y-axis intercept and A2 is half of the linear coefficient.
2
3

8 m 
2


1,
4199

6,
297

10
2


2
g 
2 2 no2  dn 

12 mol  m
K


1,
633

10
 
N A 4  dc  6, 023 1023 mol 1  (6328 1010 m)4
g2
2
c (g / m3)
5,034E+02
1,007E+03
1,510E+03
2,014E+03
2,517E+03
R() (m-1)
2,390E-04
4,400E-04
6,060E-04
7,900E-04
9,020E-04
2
Kc / R()
3,440E-06
3,737E-06
4,071E-06
4,163E-06
4,558E-06
5,E-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
mol
g
= 3,20 10-6
 M w  313000
Mw
g
mol
And second virial coefficient:
2 A2  5, 29 1010
3
mol  m3
10 mol  m

A

2,
64

10
2
g2
g2
2000
2500
10
Exercise 1.5*
Polymers A and B are monodisperse polystyrenes. Molecular weight of Polymer A is three times
the molecular weight of polymer B. Polymer C is polydisperse PS with Mw = 2.0105 g/mol. A
mixture containing 25 g of polymer A, 50 g of polymer B and 25 g of polymer C was measured
with light scattering, and molecular weight obtained was 112500 g/mol. With osmotic pressure, the
molecular weight was determined to be 60000 g/mol. Estimate the number average molecular
weight Mn of the polymer C.
Solution 1.5*
Polymers A and B are monodisperse and thus the number average and weight average
molecular weights are equal Mn=Mw. Additionally it is known that M B 
MA
. Light scattering
3
measurement reveals the weight average molecular weight which Mw = 112500g/mol. Osmotic
pressure is used to measure number average molecular weight and thus for the mixture Mn =
60000g/mol.
Molecular weight for polymer A can be calculated since the weight average molecular weight of the
mixture is known:
Mw 
wM
w
i
i

MA
 wC  M w,C
g
3
 112500
wA  wB  wC
mol
wA  M A  wB 
i
M w  ( wA  wB  wC )  wC  M w,C
w
wA  B
3
g
g
112500
 (25 g  50 g  25 g )  25 g  2 105
mol
mol  150000 g

50 g
mol
25 g 
3
 MA 
And then molecular weight for polymer B can be calculated:
M
MB  A 
3
g
mol  50000 g
3
mol
150000
Number average molecular weight for polymer C can be obtained:
Mn 
n M
n
i
i
i

w
n
i
i

wA  wB  wC
g
 60000
w
wA wB
mol

 C
M A M B M n ,C
11
 M n ,C 

wC
wA  wB  wC wA wB


Mn
MA MB
25 g
25 g
25 g  50 g  25 g
50 g


g
g
g
60000
150000
50000
mol
mol
mol
 50000
g
mol
Membrane osmometry
The osmotic pressure,  of a polymer solution can be obtained from the chemical potential, 1, or
equivalently from the activity, a1, of the solvent through the basic relationship (Fried 3.96):
1  RT ln a1  V1
where V1 is the molar volume of the solvent. Substitution of the Flory-Huggins expression for
solvent activity into the above equation and rearrangement gives (Fried 3.97):

RT
ln(1  2 )  2  1222 
V1 
where 2 is lattice volume fraction of second component (polymer) and 12 is interaction parameter.
The Van’t Hoff equation for the osmotic pressure of an ideal, dilute solution when 12=0.5:
 RT

c
M
In more general form for number average molecular weight (Fried 3.101):
 1

  RTc 
 A2 c  A3c 2  ... 
 Mn

where A2 and A3 virial coefficients. In dilute solutions (c<1g/dL) can the second and higher
concentrations be ignored. Then when plotting /(RTc) as a function of concentration, a linear
1
correlation is obtained where the slope is A2 and the cross point of y-axis M n .
The osmotic pressure is calculated from the difference in between the height of solvent in the
solvent capillary and the height of solution in the opposite capillary at equilibrium as (Fried 3.104):
   gh
Exercise 1. 6*
The following measurements have been obtained for a polymer solution at 25oC:
c (g/dL) h (cm of solvent)
0,32
0,70
0,66
1,82
1,00
3,10
1,40
5,44
1,90
9,30
12
Density of the solvent is 0.85 g/cm3.
a) Plot /(RTc) as a function of concentration c.
b) Determine the average molecular weight for the polymer and the second virial coefficient.
Solution 1.6*
a)
Osmotic pressure (Fried 3.104):
   gh
when the density of the solvent is.85 g/cm3 ja constant g=9.80665 m/s2. R = 8.3145 J/(K mol) ja
temperature T = 298.15K.
c (g/m3)
3200
6600
10000
14000
19000
h (m)

0,007 58,3495675
0,0182 151,7088755
0,031 258,4052275
0,0544 453,459496
0,093 775,2156825
RTc) (mol/g)
7,35558E-06
9,27248E-06
1,04239E-05
1,30659E-05
1,64588E-05
 /(RTc) vs. c
 /(RTc) (mol/g)
2,0E-05
1,5E-05
y = 5,69E-10x + 5,31E-06
R2 = 9,88E-01
1,0E-05
5,0E-06
0,0E+00
5,0E+03
1,0E+04
1,5E+04
2,0E+04
3
c (g/m )
b)
Slope gives us A2 =5.710-10 (molm3)/g2 and the y-axis intercept 1
Mn 
1
g
 188000
.
6
5,3110
mol
Mn