Question

Molecular weight data for some polymer are tabulated. Determine:
a) The number-average molecular weight
b) The weight-average molecular weight
c) What is the number-average degree of polymerization for this polymer?
d) What is the weight-average degree of polymerization for this polymer?

Side note: Please consider the monomer C3H6

Molecular weight data for some polymer are tabulated. Determine: The number-average molecular weight The weight-average molecular weight What is the number-average degree of polymerization for this polymer? What is the weight-average degree of polymerization for this polymer?

Answer 1

Given:

              

Molecular Weight Range (g/mol)	xi	wi
8,000 – 16,000	0.05	0.02
16,000 – 24,000	0.16	0.10
24,000 – 32,000	0.24	0.20
32,000 – 40,000	0.28	0.30
40,000 – 48,000	0.20	0.27
48,000 – 56,000	0.07	0.11

Solution:

           (a) From the tabulated data, let us find out Mn, the number-average molecular weight.

Molecular Weight Range (g/mol)	Mean Mi	xi	xiMi
8,000 – 16,000	12,000	0.05	600
16,000 – 24,000	20,000	0.16	3,200
24,000 – 32,000	28,000	0.24	6,720
32,000 – 40,000	36,000	0.28	10,080
40,000 – 48,000	44,000	0.20	8,800
48,000 – 56,000	52,000	0.07	3,640

      Mn = $\sum$xiMi= 33,040 g/mol

             (b) From the tabulated data, let us find out Mw, the weight-average molecular weight.

Molecular Weight Range (g/mol)	Mean Mi	wi	wiMi
8,000 – 16,000	12,000	0.02	240
16,000 – 24,000	20,000	0.10	2,000
24,000 – 32,000	28,000	0.20	5,600
32,000 – 40,000	36,000	0.30	10,800
40,000 – 48,000	44,000	0.27	11,880
48,000 – 56,000	52,000	0.11	5,720

      Mw = $\sum$xiMi= 36,240 g/mol

   We can define the average degree of polymerization (# of monomer units per polymer molecule) according to the number-average (n_n) or weight-average (n_n):

                             n_n = Mn / m    or

                             n_w = Mn / m

where m is the molecular weight of the monomer unit (in this case, C₃H₆).

(c) For number - average degree of polymerization:

                 Molecular weight of the monomer (m) = 3*12.011 + 6*1.00794

                                                                                 = 42.08 g/mol

                 Now, n_n = Mn / m   

                 For which Mn = 33,040 g/mol

                 Now, calculating n_n, we get,

                              

                           n_n = 33,040 / 42.08 = 785

                           

(d) For weight - average degree of polymerization:

                 Molecular weight of the monomer (m) = 3*12.011 + 6*1.00794

                                                                                 = 42.08 g/mol

                 Now, n_w = Mn / m   

                 For which Mn = 36,240 g/mol

                 Now, calculating n_w, we get,

                              

                           n_w = 36,240 / 42.08 = 861