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226
Applying this formula (or the graph) for the unknown sample with [
η
] = 46 ml/g
yields M = 38.831 g/mol. Which type of molecular weight average is obtained?
a)
If the sample turns out to be monodisperse (by supplementary
measurements, e.g. chromatography): No problem. M is M!
b)
The sample is polydisperse:
In this case the value 38.831 g/mol represents the viscosity average, defined
as:
M M c
c
v
i
a
i
i
a
= ∑
∑
⎛
⎝
⎜
⎞
⎠
⎟
1
Note that M
v
= M
w
for the special case a = 1.
Case 2: Polydisperse standards
Again, consider the following example (another polymer):
Sample
M
w
M
n
[
η
]
Std 1
22 000 11 000
31
Std 2
54 000 27 000
52
Std 3
120 000 60 000
82
Unknown
?
?
62
y = 0.1004x
0.5734
y = 0.1493x
0.5734
10
100
1 000
10 000 100 000 1 000 000
M (g/mol)
[
η
] (ml/g)
Note that M
w
/M
n
= 2.0 for the standards, suggesting (but not strictly proven!)
that the standards are linear polymers obtained by random depolymerisation.
In this case we obtain two different MHS-relations, one for M
w
and one for M
n
: