Example 1
It has been determined that polyvinyl acetate (Table 1) has the molecular size distribution shown in Figure. What are the "weightaverage" molecular weight and the average degree of polymerization?
Table 1. Molecular weight distribution
Range
of 
(MW)_{i}, 
W_{i}, 
(W_{i})(MW)_{i}, 
5,000
 10,000 
7,500 
0.12 
900 
10,000
 15,000 
12,500 
0.18 
2,250 
15,000
 20,000 
17,500 
0.26 
4,550 
20,000
 25,000 
22,500 
0.21 
4,725 
25,000
 30,000 
27,500 
0.14 
3,850 
30,000
 35,000 
32,500 
0.09 
2,925 


Sum
= 1.00 
Sum
= 19,200 
Figure. Polymer size distribution
Solution:
On the basis of 100 g polyvinyl acetate,
M_{w} = {Sum [(W_{i})(MW)_{i}]}/{Sum W_{i}} = 19,200 g/(ave mol wt)
Mer weight of vinyl acetate:
C_{4}H_{6}O_{2}
= 48 + 6 + 32 = 86 g/mer wt
DP = 19200/86 = 224
Units: (g/ave mol wt mers )/(g/mer wt ave mol wt ) *
Example 2
Determine the "numberaverage" molecular weight for the polymer of Example 1.
Solution:
On the basis of 100 g polyvinyl acetate.
Table 2
Range
of molecular weights,
gm/mol wt 
Molecules 
(X_{i})
(MW)_{i} 
5,000
 10,000 
(12)(AN)*/7,500 
(12)
(AN) 
10,000
 15,000 
(18)(AN)/12,500

(18)
(AN) 
15,000
 20,000 
(26)(AN)/17,500 
(26)
(AN) 
20,000
 25,000 
(21)(AN)/22,500 
(2
1) (AN) 
25,000
 30,000' 
(14)(AN)/27,500 
(14)(AN) 
30,000
 35,000 
(9)(AN)/32,500 
(9)
(AN) 

(0.00624)
(AN) 
(100)
(AN) 

100(AN)/
(0.00624)(AN)= 16,010 gm/mole 

As seen from the above example, the numberaverage molecular weight is 16,010, while the weightaverage molecular, determined before, was 19,200. Whenever there is a distribution in sizes, the "numberaverage" molecular weight will be less than the "weightaverage" molecular weight, because of the large numbers of small molecules in the smaller weight fractions. If all the molecules were the same size, the two averages would be identical.