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

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 6000 3000 5000 4000 2000 1000 No. of molecules of mass Mi 1 2 3 4 5 6 7 Mi gmol-1 (105) Mn = 100,000 Mw = 200,000 Mz = 300,000 Figure 1.1

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

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 = __________ 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

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.

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*12+192 = 1332 g/mol M105 = 105*12+212 = 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/ (7.51+6.79) *10-4 = 1399 g/ mol Mw = 1*1332 + 1*1472 / 2 = 1402 g/ mol In this case the two averages are almost the same.

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*12+382 = 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/ (70.42+3.76) *10-4 = 270 g/ mol Mw = 1*142 + 1*2662 / 2 = 1402 g/ mol In this case the two averages are not the same.

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*12+2002 = 14002 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 / 14002 = .71*10-4 mol The average molar masses are then, From eqn. 1.3 Mn = 1+1/ (70.42+.71) *10-4 = 281 g/ mol Mw = 1*142 + 1*14002 / 2 = 7072 g/ mol In this case the two averages are not the same.

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

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

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

Thank You See You Next Lecture