Question 16.5: Number and Weight Average Molecular Weights We have a polyet......
Number and Weight Average Molecular Weights
We have a polyethylene sample containing 4000 chains with molecular weights between 0 and 5000 g/mol, 8000 chains with molecular weights between 5000 and 10,000 g/mol, 7000 chains with molecular weights between 10,000 and 15,000 g/mol, and 2000 chains with molecular weights between 15,000 and 20,000 g/mol. Determine both the number and weight average molecular weights.
First we need to determine the number fraction x_{i} and weight fraction f_{i} for each of the four ranges. For x_{i}, we simply divide the number in each range by 21,000, which is the total number of chains. To find f_{i}, we first multiply the number of chains by the mean molecular weight of the chains in each range, giving the “weight” of each group, then find f_{i} by dividing by the total weight of 1.925 × 10^{8} g/mol. We can then use Equations 16-2 and 16-3 to find the molecular weights.
\overline{M_{n}} = ∑x_{i}M_{i} = 9160 \ g/mol
\overline{M_{w}} =∑ f_{i}M_{i} = 11,331 \ g/mol
The weight average molecular weight is larger than the number average molecular weight.
| Number of Chains | Mean M per Chain (g/mol) | x_{i} | x_{i}M_{i} (g/mol) | Weight (g/mol) | f_{i} | f_{i}M_{i} (g/mol) |
| 4000 | 2500 | 0.191 | 477.5 | 1.0 × 10^7 | 0.0519 | 129.75 |
| 8000 | 7500 | 0.381 | 2857.5 | 6.0 × 10^7 | 0.3118 | 2338.50 |
| 7000 | 12,500 | 0.333 | 4162.5 | 8.75 × 10^7 | 0.4545 | 5681.25 |
| 2000 | 17,500 | 0.095 | 1662.5 | 3.5 × 10^7 | 0.1818 | 3181.50 |
| ∑ = 21,000 | ∑ = 1 | ∑ = 9160 | ∑ = 1.925 × 10^8 | ∑ = 1 | ∑ = 11.331 |