ChE 412-512 - Spring 2010
Polymer Materials Engineering
Homework Set #2 Ch 3, 4, & 6 (125 pts)
due Thursday, February 4
1. (10 pts) Draw chemical structures of isotactic and syndiotactic oligomers of styrene with five
repeat units; include hydrogen atoms for the end groups.
Repeat unit is -(CH2CH φ)Isotactic
HHHHHHHH H H
H-C-C-C-C-C-C-C-C-C-C-H
φH φH φH φ HφH
HH H φ H H H φ H H
H-C- C- C- C- C- C- C- C- C- C- H
H φ H H H φ H H H φ
Using φ to represent the phenyl groups (C6H5), it
alternates from above to below the backbone in the
syndiotactic polymer
2. (15 pts) Given the following chain lengths for poly(ethylene terephthalate), determine the number
average molecular weight, weight average molecular weight, and polydispersity index.
The system consists of 10 different polymer chains, with repeat unit lengths of 12, 25, 175, 186,
192, 194, 199, 200, 202 and 212.
The repeat unit for PET is shown in row D, page 23. It has the chemical formula: C10O4H8; Mr is 192;
Ignore the end groups. (it would be -OH and -H for an additional 18 amu)
Mn = ΣniMi / Σni = 192 * (12 + 25 + 175 + 186 + 192 + 194 + 199 + 200 + 202 + 212) / 10
Mn = 30,600
Mw = ΣwiMi/ Σwi = ΣniMi2 / ΣniMi = 36,700
PDI = Mw/Mn = 1.20
3. (20) A. (10 pts) Draw a structure to represent the triblock polymer, poly(ethylene oxide-b-propylene
oxide-b-ethylene oxide), PEOm-PPOn-PEOm. This polymer is used as a surfactant and has the
tradename Pluronic®. For your structure, use Pluronic F127, where m = 99 and n = 65.
Ethylene oxide has the structure:
It polymerizes to form:
(this shows 3 2/3 repeat
units, incorrectly)
Propylene oxide is
formed into a
polymer:
(this shows 4 repeat
units, correctly)
Therefore the triblock copolymer of Pluronic F127 would have the structure:
-[CH2-CH2-O]99 - [CH(CH3)CH2O]65 - [CH2-CH2-O]99
B. (10 pts) If you were to stretch out a Pluronic F127 in planar zigzag form, without violating
required bond angles, so that all of the atoms in the polymer backbone were in the same plane,
estimate the polymer chain length within 20%, including the end groups (-H’s), on the
polymer backbone.
Bond
C-C
C=C
C-H
C-O
C-N
C=O
Bond Length (Å)
1.54
1.34
1.20
1.43
1.47
1.23
Bond Type
C-C-C
Bond Angle (o)
109.5
C-C=C
122
C-O-C
O-C-C
108
110
Visualize a single repeat unit for ethylene oxide, then multiply length by 198, and the same for propylene
oxide, with a length of 65.Add in the two end groups (CH on the left, OH on the right).
Only the polymer backbone matters to determine chain length.
Since the backbone is C-C-O, the bond lengths needed are C-O and C-C.
The bond angles of C-O-C and O-C-C are needed.
The backbones of EO & PO are the same, so we can simply assume we have 263 CCO units connected
here.
Draw these out; if you assume the first C-C bond is the direction of the polymer, then the first 3 bonds
repeat over and over again in the structure, with the polymer extending in a linear direction.(approx).
So the length of two repeat units would be approximately (O-C-C-O-C-C-O):
C-C
C-O
O-C
1.54 + 1.43 sin 20 + 1.43 cos 2= 3.46 Å
Thus, 263 CCO units would be approximately 263 * 3.46 Å = 910 Å long, minus the end groups. The H
groups could add as much as 2.4 Å. So the Pluronic molecule stretched out would be ~ 90-100 nm long.
(900- 1000 Å). However, individual polymers are never fully extended like this! (Full credit if within ~ 20
%)
4. (25) A. (15 pts) Many commercially-used polymers are actually blends of two or more different
molecular weights, which gives a net composition with a bimodal molecular weight distribution.
If an equal weight of polymers A, B and C (below) are blended, what are the new Mn, Mw, and
polydispersity of the blend?
Polymer A:
Polymer B:
Polymer C:
Mn
200,000
350,000
150,000
Mw
600,000
400,000
150,000
PDI
3.00
1.14
1.00
See Chapter 6, example 2. Assume there is 1 gram of each polymer.
Mn = ΣniMi / Σni = Σwi / Σwi/Mni = 3 g / (1/200,000 + 1/350,000 + 1/150,000) = 207,000
Mw = ΣwiMwi/ Σwi = {1 g (600,000) + 1 g (400,000) + 1 g (150,000)} / 3g = 383,000
PDI = 1.85
B. (10 pts) Draw (sketch) a graph to represent the results of running this blend through a gel
permeation chromatograph. The y-axis is an indicator of intensity (refractive index or
absorbance) and is proportional to the number of polymers that are leaving the GPC column.
Label the x- and y- axes and label the parts of the chromatograph to show the different polymers
in the blend (A, B & C). Consider that the A, B & C peaks may overlap.
The y-axis should be intensity, and the x-axis should be either time or elution volume. Remember that the
large molecular weight material comes out first.
the peak for polymer A would be very broad since it has a high PDI. It will be centered around the low
Mn of 200,000, but will extend throughout the peaks for B & C polymers, too.
The peak for polymer B will be much more narrow. It will come out after A has started coming out, but
before A peaks.
Polymer C will come out of the chromatographic column in a single bolus, as a large peak at 150,000.
(or the longest time. However, peak A will continue even longer than polymer C.
Intensity
The area under each peak should be the same, so the peak for polymer C should be extremely high
compared to the rest of the chromatogram. Although the figure shows 3 different peaks, the GPC has no
way of knowing the difference between A, B & C, so the actual graph would have only a single line to
trace out the peaks.
C
B
A
Time
5. (15 pts) An osmometry experiment was done to estimate the molecular weight of a sample of
polystyrene. The concentration of polymer in toluene was varied and the height difference
between the solution and the solvent was measured. The temperature was 27 oC, where the
specific gravity of toluene is 0.8667.
Based on the following data, determine the molecular weight of PS. Is this Mw or Mn?
Concentration, CA
g/mL
0.0320
0.0660
0.100
0.140
0.190
Osmotic Pressure, πA,
cm of solvent
0.700
1.82
3.10
5.44
9.30
Osmometry determines Mn
lim (c→0) π/c = RT/Mn
Osmotic
Pressure/Concentration, cm
solv/(g/mL)
So, need to plot π /c vs. c and find the intercept.
Watch units!!
Using given units,
intercept = 1.579 cm
solvent/(g/mL)
60
y = 169.17x + 15.787
R2 = 0.9884
50
40
30
20
10
0
0
0.05
0.1
0.15
Concentration of PS g/mL
0.2
Convert cm solvent to
energy
15.79 cm toluene/(g/mL)
* {0.8667 g/mL} * 9.8
m/s2 * 100 cm/m= 13410
cm2/s2
Mn = RT/intercept
R = 8.314 J/K-mol
T = 300 K (absolute temp!)
Mn
= (8.314 J/K-mol) (300 K) / {(13410 cm2/s2) *}
= 18.60 J/mol s2/cm2 * {10000 cm2/m2} {note: 1J = 1 Nm = 1 kg-m/s2-m}
Mn = 1.86 * 106 g/gmol
6. (10 pts) End Group Analysis
6.50 moles of ethanediamine (H2N-CH2CH2-NH2) were reacted with 6.50 moles of adipic
acid (HOOC-CH2-CH2-CH2-CH2-COOH). After the reaction, a titration was performed using
NaOH (which reacts with COOH) and an indicating dye to determine the number of acid groups
remaining free after the polymerization. If 73.0 mL of 3.0 M NaOH were required to neutralize
the polymer, what is the average molecular weight (Mn) of the polymer?
1 mole of acid reacts with one mole of NaOH, thus the amount of NaOH added is equal to the number of
moles of COOH end groups.
0.073 L * 3.0 mol/L = 0.219 moles. (thus, there are 0.219 moles of polymer chains)
Thus the remainder of the adipic acid reacted.
There are still 6.5 moles of adipic acid, but they are now part of polymer chains of various lengths.
6.5 moles of monomer are distributed among 0.219 moles of polymer, thus
6.5/0.219 =29.68 monomer repeat units per polymer, on average.
An alternate way to arrive at the same answer is to use the conversion equation:
Xn = 1/(1-p)
Here p = # of COOH reacted / initial # of COOH groups.
= (13 - 0.219)moles / 13 moles = 0.983
Thus Xn = 59.4 (total monomer residues, or 29.7 AAs and 29.7 BB’s, or better said 29.7 AABB’s, as
calculated above)
To find molecular weight (Mn), the molecular weight of the repeat unit must be found:
H-(HNCH2CH2-NH-OC-CH2-CH2-CH2-CH2-CO-)OH
The repeat unit is between the parenthesis: C8N2O2H14
Mr = 170
Thus Mn = 29.68 * 170 = 5,045 g/gmol
7. (20 points) Five students have taken viscometry data for a monodisperse sample of atactic polystyrene
at five different concentrations in toluene. Unfortunately, the techniques were not all identical, so they
reported different versions of viscosity. The density of toluene at 30 oC is 0.8665 g/mL; 0.8656 g/mL at
40 oC. The viscosity of toluene is 0.5260 cP at 30 oC; 0.4710 cP at 40 oC.
Given the data below, resolve the experiments so that you can determine the viscosity-average
molecular weight of the polymer.
Expt 1: Toluene Concentration = 0.1000 g/dL; inherent viscosity = 0.053672 dL/g, at 30 oC.
Expt 2: C = 0.0500 g/dL; used an Ostwald viscometer at 30 oC to determine a flow time of 4
minutes, 37.01 seconds; in the viscometer calibration equation η/ρ = at + b/t,
a = 0.001340; and b = 65.760 (where time is in seconds, density in g/mL and viscosity in cP)
Expt 3: C = 0.1100 g/dL; solution viscosity = 0.47360 cP at 40 oC.
Expt 4: C = 0.0240 g/dL; used a cone-and-plate viscometer at 30 oC which returned a viscosity of
0.526620 cP
Expt 5: C = 8.0 g/dL; determined specific viscosity was 10.896 at 30 oC
Mark-Houwink constants for atactic polystyrene at 30 oC are: K = 0.2 * 10-3 mL/g ; a = 0.72.
(Note that the experiments were done by students and some data may need to be thrown out! If you throw
out any data, give an explanation why that data may not be valid.)
First, take a look at the experiments. Two of them are not valid for estimating molecular weight
by viscometry: Experiment #2 was done at a different temperature (and it’s got a fairly high
concentration), and Expt 5 is NOT a dilute solution!
Therefore, only 3 experiments were done that are valid for finding viscosity-average MW.
Mv = ([η]/K)(1/a)
Find [η] by plotting η red vs. c.
Only data from experiments 1,3 and 4 are valid.
cP
cP
Experiment g/dL
visc
visc solv relat
specif
reduced
#
n
ns
nsp
nred
conc
dL/g
nr
inherent
4
0.024 0.526620
0.526 1.001179 0.001179 0.049113
3
0.05 0.527339
0.526 1.002546 0.002546 0.050927
1
0.1
a
b
0.00134
t
65.76
277
0.526 1.005382 0.005382 0.053816 0.053672
A Plot of η red vs. c yields an intercept of 0.047728 dL/g (or 4.7728 mL/g)
{watch units on K vs. [η]}
Mv = 1,203,000 g/gmol
8. (10 pts) Sketch (qualitatively) a SEC chromatogram of weight vs. elution volume for a blend
containing a, b and c, below.
(a) 10 g of monodisperse polystyrene with Mn = 50,000
(b) 5 g of polydisperse polystyrene (PI = 1.2) with Mw = 100,000
(c) 25 g of polydisperse polystyrene (PI = 4) with Mw = 500,000
Show the relative peak positions and peak heights, being careful to consider potential overlapping.
This graph should be quantity (or chain weight) vs. elution volume (or time)
This solution should give you an idea of what to expect before going into the lab to run an experiment, but
the best you can do with the information given is come up with a qualitative curve.
The polymers will be detected in order of decreasing molecular weight (c,b, then a)
The first peak should be a broad, short peak representing sample (c), centered at 500,000 MW; the area
under this peak should be 5 times as large as sample (b).
The second peak (b) should begin before the first one returns to zero, and should be narrower, with a
similar height to the first peak.
The third peak (a) should be nearly a Dirac delta function, isolated from the other 2 peaks, or perhaps
slightly overlapped with peak b. It should be the highest peak of the three.