MOLAR MASS AND DEGREE OF POLYMERIZATION

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MOLAR MASS AND DEGREE OF POLYMERIZATION

• Many properties of polymer show a strong
  dependence upon the size of the polymer chains.
• So it is essential to characterize their
  dimensions.
• This is normally done by measuring the molar mass
  (M) of a polymer.
• Molar mass defined as the mass of 1 mole of the
  polymer it is quoted in gmol-1 or kg mol-1.
• For network polymer we use the molar mass of
  chains existing between junction points. The
  molar mass of the network is infinite ?.
• Usually the molar mass of a homopolymer is
  related to the degree of polymerization ( X)
  which is the number of repeat units in the
  polymer chain by the simple relation.
• M XMo 1.1a
• where
• Mo ? molar mass of the repeat unit
• example (C2H4)1200 X 1200
• For copolymer the sum of the products ?Mo for
  each type of R.U. is required to define the
  molar mass.
• M ?Xi Moi 1.1b

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Molar Mass Distribution

• A polymer sample is a mixture of molecules with
  different degree of polymerization.
• In general a polymer contain a range of molar
  masses.
• Fig. 1.1 shows a typical molar mass distribution
  curve
• A sample of polymer has molecules of different
  molar mass
• M1, M2, M3, M4 ........ Mi
• N1, N2, N3, N4 ....... Ni

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Molar Mass Averages

• It is some times convenient to use a molar mass
  average in place of a complete distribution.
• The number - average molar mass Mn
• Definition ? the sum of the products of the molar
  mass of each fraction multiplied by its mole
  fraction.
• Mn ?XiMi 1.2
• where
• Xi is the mole fraction of the molecules of
  molar mass Mi

Ni Xi _______ N number of
moles ?Ni
?NiMi ?Wi
W ?Mn ___________ ________ _____
?Ni
?Ni ?Ni 1.3
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Sometimes weight fraction are used in place of
mole fraction. The weight fraction Wi is
defined as the mass of molecules of molar mass
Mi divided by the total mass of all the
molecules present.  
NiMi Wi __________ 1.4
?NiMi
Wi ? Ni ? ______
________ 1.5 Mi
?NiMi
Combining equations 1.3 1.5 gives Mn in terms
of weight fraction.   1 Wi
1 Mn _____________
or ? _______ _______
1.6 ?Wi/Mi Mi Mn
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The weight -average molar mass Mw   Definition ?
the sum of the products of the molar mass of each
fraction multiplied by its weight fraction.
?WiMi Mw _________ 1.7
?W i
Combine equation 1.4 and 1.7  ?NiMi2
?WiMi ?WiMi Mw
____________ __________
___________ ?NiMi ?W i W
1.8
We use Mw rather than Mn for polymeric materials
where some of the physical properties the
contribution of the high Mi species dominates the
behavior.   Because of the squared Mi term in the
numerator, Mw is sensitive to the presence of
high Mi species.
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Example1.1

• Consider a blend of mass 2g formed from 1g of
  each of the two paraffin's one C95H192 and the
  other C105H212, What are Mn and Mw of the blend ?
• Solution
• The two molar masses are
• M95 9512192 1332 g/mol
• M105 10512212 1472 g/mol
• It then follows that in the 2g specimen the
  number of moles present is
• n95 1 / 1332 7.5110-4 mol
• n105 1 / 1472 6.7910-4 mol
• The average molar masses are then, From eqn. 1.3
• Mn 11/ (7.516.79) 10-4 1399 g/ mol
• Mw 11332 11472 / 2 1402 g/ mol
• In this case the two averages are almost the
  same.

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Example1.2

• Consider a blend of mass 2g formed from 1g of
  each of the two paraffin's on C10H22 and the
  other C190H382, What are Mn and Mw of the blend ?
• Solution
• The two molar masses are
• M10 101222 142 g/mol
• M190 19012382 2662 g/mol
• It then follows that in the 2g specimen the
  number of moles present is
• n10 1 / 142 70.4210-4 mol
• n190 1 / 2662 3.7610-4 mol
• The average molar masses are then, From eqn. 1.3
• Mn 11/ (70.423.76) 10-4 270 g/ mol
• Mw 1142 12662 / 2 1402 g/ mol
• In this case the two averages are not the same.

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Example1.3

• Consider a blend of mass 2g formed from 1g of
  each of the two paraffin's on C10H22 and the
  other C1000H2002, What are Mn and Mw of the blend
  ?
• Solution
• The two molar masses are
• M10 101222 142 g/mol
• M1000 1000122002 14002 g/mol
• It then follows that in the 2g specimen the
  number of moles present is
• n10 1 / 142 70.4210-4 mol
• n1000 1 / 14002 .7110-4 mol
• The average molar masses are then, From eqn. 1.3
• Mn 11/ (70.42.71) 10-4 281 g/ mol
• Mw 1142 114002 / 2 7072 g/ mol
• In this case the two averages are not the same.

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Comments on Example 1.1, 1.2 and 1.3

• The result of the example illustrate the
  following points. 
• Mn is sensitive to the mixture of molecules of
  low molecular mass.
• Mn for 10/190 and 10/1000 are much lower than the
  95/105 blend
• Mw is sensitive to the mixture of molecules of
  high molecular mass.
• Mw for 10/1000 blend greatly exceeds that for the
  other two (where they are equal 1402)
• Mw always higher than Mn
• The ratio Mw/Mn is a measure of the range of
  molecular sizes in the specimen it is 1.00, 5.19
  and 25.17 it is normally in the ranges of 2 - 100
  some polymer has very small or very high value
  of polydispersity index.
• This ratio is known as polydispersity or
  heterogeneity index.
• Monodisperse polymer would have Mw/Mn 1.00

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Z-average Molar Mass (Mz)

• quoted for higher molar mass averages. 

?NiMi3 ?WiMi2 Mz
_____________ ____________
1.9 ?NiMi2 ?WiMi
The number average and weight average degree of
polymerization For a homopolymer are given by
Xn Mn/Mo 1-10 and
Xw Mw/Mo 1-11
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Homework

• Solve the last three example for the Z-average
  molar mass ?
• Comment on the results ?

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Thank You

• See You Next Lecture