1. 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) 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

2. 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, M Mi
N1, N2, N3, N Ni
6000
3000
5000
4000
2000
1000
No. of molecules of mass Mi
Mi gmol-1 (105)
Mn = 100,000
Mw = 200,000
Mz = 300,000
Figure 1.1

3. 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
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

4. Combining equations 1.3 & 1.5 gives Mn in terms of weight fraction.
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 = __________
NiMi
Wi  Ni
 ______ = ________
Mi NiMi
Combining equations 1.3 & 1.5 gives Mn in terms of weight fraction. Wi
Mn = _____________ or  _______ = _______
Wi/Mi Mi Mn

5. 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 = _________
W i
Combine equation and 1.7 
NiMi WiMi WiMi
Mw = ____________ = __________ = ___________
NiMi W i W
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.

6. 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 = 95* = 1332 g/mol
M105 = 105* = 1472 g/mol
It then follows that in the 2g specimen the number of moles present is
n95 = 1 / 1332 = 7.51*10-4 mol
n105 = 1 / 1472 = 6.79*10-4 mol
The average molar masses are then, From eqn. 1.3
Mn = 1+1/ ( ) *10-4 = 1399 g/ mol
Mw = 1* *1472 / 2 = 1402 g/ mol
In this case the two averages are almost the same.

7. 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 = 10*12+22 = 142 g/mol
M190 = 190* = 2662 g/mol
It then follows that in the 2g specimen the number of moles present is
n10 = 1 / 142 = 70.42*10-4 mol
n190 = 1 / 2662 = 3.76*10-4 mol
The average molar masses are then, From eqn. 1.3
Mn = 1+1/ ( ) *10-4 = 270 g/ mol
Mw = 1* *2662 / 2 = 1402 g/ mol
In this case the two averages are not the same.

8. 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 = 10*12+22 = 142 g/mol
M1000 = 1000* = g/mol
It then follows that in the 2g specimen the number of moles present is
n10 = 1 / 142 = 70.42*10-4 mol
n1000 = 1 / = .71*10-4 mol
The average molar masses are then, From eqn. 1.3
Mn = 1+1/ ( ) *10-4 = 281 g/ mol
Mw = 1* *14002 / 2 = 7072 g/ mol
In this case the two averages are not the same.

9. 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 it is normally in the ranges of 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

10. Z-average Molar Mass (Mz)
quoted for higher molar mass averages. 
NiMi WiMi2
Mz = _____________ = ____________
NiMi WiMi
The number average and weight average degree of polymerization
For a homopolymer are given by
Xn = Mn/Mo
and Xw = Mw/Mo

11. Homework Solve the last three example for the Z-average molar mass ?
Comment on the results ?

12. Thank You
See You Next Lecture