In one line
Make a table of Ni, Mi, NiMi and NiMi2, then use Mn = ΣNiMi/ΣNi and Mw = ΣNiMi2/ΣNiMi.
Explained simply
The method is mechanical, like filling a bill. Each row is one group of molecules. Multiply, add the columns, divide. The only real trap is reading whether the question gives numbers of molecules or masses of molecules.
How it works
- Read the data: if "number of molecules", "mole %" or "% of molecules" is given, these are Ni. If "mass" or "weight %" is given, these are wi; convert using Ni = wi/Mi if needed.
- Build the table: Ni, Mi, NiMi, NiMi2.
- Add each column.
- Compute Mn, Mw and, if asked, PDI = Mw/Mn.
- Check: Mw ≥ Mn.
Key formula
\( \bar{M}_n = \dfrac{\sum N_i M_i}{\sum N_i} \qquad \bar{M}_w = \dfrac{\sum N_i M_i^2}{\sum N_i M_i} \)
If mass data wi are given: \( \bar{M}_n = \dfrac{\sum w_i}{\sum (w_i/M_i)} \) and \( \bar{M}_w = \dfrac{\sum w_i M_i}{\sum w_i} \).
Quick example
A polymer sample contains 30% molecules of molecular weight 20,000, 40% of 30,000 and 30% of 60,000. Find Mn, Mw and PDI.
| Ni | Mi | NiMi | NiMi2 |
|---|---|---|---|
| 30 | 20,000 | 6.0 × 105 | 1.20 × 1010 |
| 40 | 30,000 | 1.2 × 106 | 3.60 × 1010 |
| 30 | 60,000 | 1.8 × 106 | 1.08 × 1011 |
| Σ = 100 | 3.6 × 106 | 1.56 × 1011 |
- Mn = 3.6 × 106/100 = 36,000 g/mol
- Mw = 1.56 × 1011/3.6 × 106 = 43,333 g/mol
- PDI = 43,333/36,000 = 1.20
Common mistakes
- Treating weight-percent data as number-percent data (gives the wrong averages).
- Arithmetic slips with powers of ten; write each column in scientific notation.
- Stopping at Mn when the question also asks for Mw or PDI.
How MU asks it
"A polymer sample contains... Calculate the number average and weight average molecular weights" (3–5 marks). For full marks write both formulas first, present a clear table with column totals, give answers with units (g/mol), and comment that Mw > Mn shows a polydisperse sample.