Molecular weight and molecular weight distribution In a
Molecular weight and molecular weight distribution. In a polymer sample the chains hardly ever have the same lengt They have an average length. Exceptions are for example proteins.
Number average molecular weight. Number of chains with length i Mole fraction ni = Total number of chains = Σn = 1 i N ΣN ·M Mn = =Σ( ) ·M = Σn ·M ΣN ΣN i i i i Ni ΣN i
Weight average molecular weight. Weight of chains with length i Weight fraction wi = = Total weight of all chains ni·Mi Σn ·M i Σw = 1 i Σn ·M Mw =Σw ·M = Σn ·M i Compare: 2 i i i ΣN ·M Mn = Σn ·M = ΣN i i i = i ni·Mi Mn
An example: City Population Amsterdam 1, 000 Groningen 200, 000 Schiermonnikoog 1, 000 Average population: 1/3 x 1, 000 1/3 x 200, 000 1/3 x 1, 000 Mn = Σni·Mi = 400, 333 We can say that the average person lives in a city of about 400, 000.
An example: Weighted average: This is an average that would account for the fact that a large city like Amsterdam holds a larger percentage of the total population of the three cities than Schiermonnikoog. Population City Amsterdam 1, 000 Groningen 200, 000 Schiermonnikoog 1, 000 Mw =Σw ·M i 1, 000 x 1, 000 1, 201, 000 = 1, 000 x 0. 8326 = 832, 600 200, 000 x 200, 000 1, 201, 000 = 200, 000 x 0. 1665 = 1, 000 x 1, 000 1, 201, 000 = 1, 000 x 0. 0008 = 33, 300 0. 8 + 865, 900. 8 We can say that the average person lives in a city of about 866, 000. i
Calculating Mn & Mw. M 1 = 100 M 2 = 10 n 1 = ½ n 2 = ½ Mn = Σn ·M = ½ · 100 + ½ · 10 i i = 50 + 5 = 55 w 1 = n 1·M 1 = ½ · 100 / 55 = 10/11 Mn w 2 = n 2·M 2 = ½ · 10 / 55 = 1/11 Mn Mw =Σw ·M = 10/11 · 100 + 1/11 · 10 i i = 90. 9 + 0. 9 = 91. 8
Calculating Mn & Mw. M 1 = 100 M 2 = 10 n 1 = 1/11 n 2 = 10/11 Mn = Σn ·M = 1/11 · 100 + 10/11 · 10 = 18. 2 w 1 = i i n 1·M 1 Mn w 2 = n 2·M 2 Mn = 1/11 · 100 / 18. 2 = ½ = 10/11 · 10 / 18. 2 = ½ Mw =Σw ·M = ½ · 100 + ½ · 10 = 50 + 5 = 55 i i
ΣN ·M Mn =Σn ·M = ΣN i i M 1 = 100 M 2 = 10 i n 1 = 1/11 i Mn = Σni·Mi = 1/11 · 100 + 10/11 · 10 = 18. 2 i wi = ni·Mi Σn ·M i = i ni·Mi Mn = Σw /M i Σn ·M Mw =Σw ·M = Σn ·M i w 1 = wi/Mi ni i w 2 = = wi·Mi 2 i i z 1 = i Σw ·M i = i ni·Mi Mw Mn n 2·M 2 Mn = 1/11 · 100 / 18. 2 = ½ = 10/11 · 10 / 18. 2 = ½ = Σz /M i w 1·M 1 Mw = ½ · 100 / 55 = 10/11 w 2·M 2 = ½ · 10 / 55 = 1/11 Mw Mz = Σzi·Mi = 10/11 · 100 + 1/11 · 10 = 91. 8 zi/Mi wi n 1·M 1 Mw = Σwi·Mi = ½ · 100 + ½ · 10 = 50 + 5 = 55 z 2 = zi n 2 = 10/11 i Mw Mz = Σz ·M i i Σn ·M = Σn ·M Mn 3 i i 2 Nr of Chains Mz Higher Av. Mw Mw
Consequences of the Schultz Flory Distribution: Dispersity (D): Mw / Mw = 2 Mw = weight average = wi. Mi Mn = number average = ni. Mi , wi = 1 , ni = 1
Schulz-Flory distribution: • Chain growth and chain transfer independent of chain length
Weight fractions Probability for chain transfer: 0. 3 0. 1 0. 001
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