A polymer sample is not like sugar where every molecule is identical. When you polymerize, chains grow to different lengths — some short, some huge. So there is no single "molecular weight." Instead we report averages . The question is: how do you average a crowd of unequal molecules? The answer depends on whether you count by number of molecules or by mass of molecules — and that single choice gives us two different averages.
There are two honest ways to take an average of these masses.
Intuition Why this averageImagine lining up every single molecule and asking "what is the average mass per molecule ?" Each molecule gets one vote , regardless of how big it is. A tiny chain counts exactly as much as a giant chain.
Derivation (first principles):
Total mass of sample = sum of (number × mass) of each kind:
Total mass = ∑ i N i M i \text{Total mass} = \sum_i N_i M_i Total mass = ∑ i N i M i
Total number of molecules:
Total number = ∑ i N i \text{Total number} = \sum_i N_i Total number = ∑ i N i
Average mass per molecule = total mass ÷ total number:
This is just the ordinary arithmetic mean weighted by how many molecules there are.
Intuition Why this averageNow weight each molecule by how much mass it contributes , not by how many there are. A big molecule "speaks louder" because it carries more mass. Techniques like light scattering physically respond to mass, so they naturally measure this average.
Derivation (first principles):
The weight fraction of species i i i w i = N i M i ∑ j N j M j w_i = \frac{N_i M_i}{\sum_j N_j M_j} w i = ∑ j N j M j N i M i
Average mass weighted by mass-fraction:
M ˉ w = ∑ i w i M i = ∑ i N i M i ⋅ M i ∑ j N j M j \bar{M}_w = \sum_i w_i M_i = \frac{\sum_i N_i M_i \cdot M_i}{\sum_j N_j M_j} M ˉ w = ∑ i w i M i = ∑ j N j M j ∑ i N i M i ⋅ M i
Intuition Why the extra power ofM M M M ˉ n \bar{M}_n M ˉ n M i M_i M i M ˉ w \bar{M}_w M ˉ w N i M i N_iM_i N i M i then average over M i M_i M i M i 2 M_i^2 M i 2
PDI = M ˉ w M ˉ n \text{PDI} = \frac{\bar{M}_w}{\bar{M}_n} PDI = M ˉ n M ˉ w width / spread of the molecular-weight distribution.
Intuition Why PDI ≥ 1 alwaysBecause M ˉ w \bar{M}_w M ˉ w always ≥ M ˉ n \bar{M}_n M ˉ n only when every molecule has the same mass (monodisperse). The more spread out the lengths, the bigger the gap, the larger PDI.
PDI value
Meaning
= 1 =1 = 1 All chains identical (monodisperse ). Natural polymers (proteins, DNA)
≈ 1.5 \approx 1.5 ≈ 1.5 2 2 2 Many addition (free-radical) polymers, depending on termination mode
→ 2 \to 2 → 2 Step-growth/condensation polymers (Flory most-probable distribution, PDI = 1 + p \text{PDI}=1+p PDI = 1 + p
> 2 >2 > 2 Broad distributions, e.g. some coordination/chain-transfer-dominated processes
Worked example Example 1 — a tiny 3-molecule sampleSample: 2 molecules of mass 10, 3 molecules of mass 20.
Step 1 — M ˉ n \bar{M}_n M ˉ n
M ˉ n = ( 2 ) ( 10 ) + ( 3 ) ( 20 ) 2 + 3 = 20 + 60 5 = 80 5 = 16 \bar{M}_n = \frac{(2)(10)+(3)(20)}{2+3} = \frac{20+60}{5} = \frac{80}{5} = 16 M ˉ n = 2 + 3 ( 2 ) ( 10 ) + ( 3 ) ( 20 ) = 5 20 + 60 = 5 80 = 16 Why this step? Total mass over total count — one vote per molecule.
Step 2 — M ˉ w \bar{M}_w M ˉ w
M ˉ w = ( 2 ) ( 10 ) 2 + ( 3 ) ( 20 ) 2 ( 2 ) ( 10 ) + ( 3 ) ( 20 ) = 200 + 1200 80 = 1400 80 = 17.5 \bar{M}_w = \frac{(2)(10)^2+(3)(20)^2}{(2)(10)+(3)(20)} = \frac{200+1200}{80} = \frac{1400}{80} = 17.5 M ˉ w = ( 2 ) ( 10 ) + ( 3 ) ( 20 ) ( 2 ) ( 10 ) 2 + ( 3 ) ( 20 ) 2 = 80 200 + 1200 = 80 1400 = 17.5 Why this step? Numerator uses M 2 M^2 M 2
Step 3 — PDI:
PDI = 17.5 16 = 1.094 \text{PDI} = \frac{17.5}{16} = 1.094 PDI = 16 17.5 = 1.094 Why this step? Slightly above 1 → mild spread. Notice M ˉ w > M ˉ n \bar{M}_w > \bar{M}_n M ˉ w > M ˉ n
Worked example Example 2 — equal numbers, very different sizes1 molecule of mass 100, 1 molecule of mass 10000.
M ˉ n = 100 + 10000 2 = 5050 \bar{M}_n = \frac{100+10000}{2} = 5050 M ˉ n = 2 100 + 10000 = 5050 M ˉ w = 100 2 + 10000 2 100 + 10000 = 10 4 + 10 8 10100 = 100010000 10100 ≈ 9902 \bar{M}_w = \frac{100^2 + 10000^2}{100+10000} = \frac{10^4 + 10^8}{10100} = \frac{100010000}{10100} \approx 9902 M ˉ w = 100 + 10000 10 0 2 + 1000 0 2 = 10100 1 0 4 + 1 0 8 = 10100 100010000 ≈ 9902 PDI = 9902 5050 ≈ 1.96 \text{PDI} = \frac{9902}{5050} \approx 1.96 PDI = 5050 9902 ≈ 1.96 Why this step? Even with equal numbers , the one giant chain holds almost all the mass , so M ˉ w \bar{M}_w M ˉ w M ˉ n \bar{M}_n M ˉ n
Worked example Example 3 — monodisperse check5 molecules, all mass 50.
M ˉ n = 5 ( 50 ) 5 = 50 , M ˉ w = 5 ( 50 ) 2 5 ( 50 ) = 50 2 50 = 50 \bar{M}_n = \frac{5(50)}{5}=50,\quad \bar{M}_w=\frac{5(50)^2}{5(50)}=\frac{50^2}{50}=50 M ˉ n = 5 5 ( 50 ) = 50 , M ˉ w = 5 ( 50 ) 5 ( 50 ) 2 = 50 5 0 2 = 50 PDI = 1 \text{PDI}=1 PDI = 1 Why this step? All identical → both averages collapse to the same number. Confirms PDI = 1 boundary.
Recall Predict before computingSample: 4 chains of mass 1000, 1 chain of mass 5000.
Forecast: Will M ˉ w \bar{M}_w M ˉ w
Verify:
M ˉ n = 4 ( 1000 ) + 5000 5 = 9000 5 = 1800 \bar{M}_n = \frac{4(1000)+5000}{5} = \frac{9000}{5}=1800 M ˉ n = 5 4 ( 1000 ) + 5000 = 5 9000 = 1800 M ˉ w = 4 ( 1000 ) 2 + ( 5000 ) 2 9000 = 4 × 10 6 + 25 × 10 6 9000 = 29 × 10 6 9000 ≈ 3222 \bar{M}_w = \frac{4(1000)^2 + (5000)^2}{9000} = \frac{4\times10^6 + 25\times10^6}{9000}=\frac{29\times10^6}{9000}\approx 3222 M ˉ w = 9000 4 ( 1000 ) 2 + ( 5000 ) 2 = 9000 4 × 1 0 6 + 25 × 1 0 6 = 9000 29 × 1 0 6 ≈ 3222 = 3222 / 1800 ≈ 1.79 = 3222/1800 \approx 1.79 = 3222/1800 ≈ 1.79 M ˉ w \bar M_w M ˉ w
Common mistake "Just average all theM i M_i M i Why it feels right: Averaging usually = add them up, divide by count. So you might do ( 10 + 20 ) / 2 = 15 (10+20)/2 = 15 ( 10 + 20 ) /2 = 15 The fix: You must weight by how many of each (N i N_i N i types . With multiplicities, M ˉ n = 16 \bar{M}_n = 16 M ˉ n = 16
M ˉ w \bar{M}_w M ˉ w M ˉ n \bar{M}_n M ˉ n Why it feels right: Different formulas → maybe sometimes one is bigger, sometimes the other.
The fix: Mathematically M ˉ w ≥ M ˉ n \bar{M}_w \ge \bar{M}_n M ˉ w ≥ M ˉ n always (it follows from Cauchy–Schwarz / variance ≥ 0). PDI < 1 < 1 < 1 impossible — if you get it, you made an arithmetic error.
Common mistake "Condensation polymers have low PDI (~1.5)."Why it feels right: People half-remember "1.5" as a typical small number.
The fix: Ideal step-growth follows the Flory most-probable distribution with PDI = 1 + p \text{PDI}=1+p PDI = 1 + p p → 1 p\to1 p → 1 → \to → 2 . It is the addition (radical) polymers that often sit around 1.5–2.
Common mistake "PDI = 1 means a small molecule."Why it feels right: "1" sounds like a low/light value.
The fix: PDI is a ratio of averages , dimensionless. PDI = 1 means uniform length , nothing about the size itself. A monodisperse protein of mass 60000 still has PDI = 1.
Common mistake Forgetting to squareM M M M ˉ w \bar M_w M ˉ w Why it feels right: Symmetry tempts you to reuse the M ˉ n \bar M_n M ˉ n The fix: M ˉ w \bar M_w M ˉ w ∑ N i M i 2 \sum N_iM_i^2 ∑ N i M i 2 M i M_i M i
In words, what does each average "vote" by?
Why is M ˉ w ≥ M ˉ n \bar{M}_w \ge \bar{M}_n M ˉ w ≥ M ˉ n
What experimental method gives M ˉ w \bar{M}_w M ˉ w M ˉ n \bar{M}_n M ˉ n
What PDI tells you about a sample of identical chains?
What PDI does an ideal condensation polymer approach at high conversion, and via what formula?
Recall Feynman: explain to a 12-year-oldImagine a school bus full of kids of very different weights. If you ask "what's the average kid's weight?" you add all weights and divide by number of kids — every kid counts once. That's number-average .
But now imagine you're carrying the kids and you ask "the average kid I'm carrying weight from " — the heavy kids feel way more, so they count more. That's weight-average , and it's always a bigger number whenever the kids aren't all the same weight. The ratio of these two tells you how mixed up the weights are: if all kids weigh the same, the ratio is exactly 1.
Mnemonic Remember the formulas"n counts heads, w counts kilos."
n n n ∑ N \sum N ∑ N w w w extra power of M M M on top: ∑ N M 2 \sum N M^2 ∑ N M 2 W eight-average is always the W inner (larger). PDI = w/n ≥ 1. For condensation: "1 + p 1+p 1 + p
Addition vs Condensation Polymerization — condensation follows Flory: PDI = 1 + p → 2 =1+p\to2 = 1 + p → 2 Gel Permeation Chromatography — separates by size, can extract the full distribution & both averages.Light Scattering and Osmometry — light scattering → M ˉ w \bar M_w M ˉ w M ˉ n \bar M_n M ˉ n Colligative Properties — depend on number of particles, hence yield M ˉ n \bar M_n M ˉ n Degree of Polymerization — X ˉ n = M ˉ n / M 0 \bar X_n = \bar M_n / M_0 X ˉ n = M ˉ n / M 0 Variance and Standard Deviation — M ˉ w / M ˉ n − 1 \bar M_w/\bar M_n - 1 M ˉ w / M ˉ n − 1
Number-average molecular weight formula M ˉ n = ∑ N i M i ∑ N i \bar{M}_n = \dfrac{\sum N_i M_i}{\sum N_i} M ˉ n = ∑ N i ∑ N i M i — total mass / total number of molecules
Weight-average molecular weight formula M ˉ w = ∑ N i M i 2 ∑ N i M i \bar{M}_w = \dfrac{\sum N_i M_i^2}{\sum N_i M_i} M ˉ w = ∑ N i M i ∑ N i M i 2 — mass-fraction weighted average
What does M ˉ n \bar{M}_n M ˉ n Number (one vote per molecule, regardless of size)
What does M ˉ w \bar{M}_w M ˉ w Its mass contribution (big chains count more)
Definition of polydispersity index PDI = M ˉ w / M ˉ n \text{PDI} = \bar{M}_w / \bar{M}_n PDI = M ˉ w / M ˉ n , the spread of the molecular-weight distribution
Why is PDI always ≥ 1? M ˉ w ≥ M ˉ n \bar{M}_w \ge \bar{M}_n M ˉ w ≥ M ˉ n always (follows from variance ≥ 0); equal only if all chains identical
What PDI value means monodisperse? PDI = 1 (all molecules same molar mass, e.g. proteins, DNA)
PDI of an ideal condensation (step-growth) polymer Follows Flory:
PDI = 1 + p \text{PDI}=1+p PDI = 1 + p ; approaches 2 as conversion
p → 1 p\to1 p → 1 Typical PDI for addition (radical) polymers About 1.5–2, depending on termination mode
Which technique gives M ˉ w \bar{M}_w M ˉ w Light scattering (responds to mass)
Which technique gives M ˉ n \bar{M}_n M ˉ n Colligative methods like osmometry (count number of particles)
For a sample of 2 chains mass 10 and 3 chains mass 20, find M ˉ n \bar M_n M ˉ n ( 20 + 60 ) / 5 = 16 (20+60)/5 = 16 ( 20 + 60 ) /5 = 16 Same sample, find M ˉ w \bar M_w M ˉ w ( 200 + 1200 ) / 80 = 17.5 (200+1200)/80 = 17.5 ( 200 + 1200 ) /80 = 17.5 The common error in M ˉ w \bar M_w M ˉ w Forgetting to square
M M M ; it must be
∑ N i M i 2 \sum N_iM_i^2 ∑ N i M i 2 Chains of different lengths
Intuition Hinglish mein samjhoDekho, ek polymer sample mein saare molecules ek jaise lambe nahi hote — kuch chote chains, kuch bahut bade. Isliye ek single "molecular weight" nahi bata sakte; hum average lete hain. Lekin average lene ke do tareeke hain. Agar har molecule ko ek vote do (chahe chota ho ya bada), to milta hai number-average M ˉ n = ∑ N i M i ∑ N i \bar M_n = \frac{\sum N_iM_i}{\sum N_i} M ˉ n = ∑ N i ∑ N i M i
Doosra tareeka: molecule ko uske mass ke hisaab se weight do — bada chain zyada "bolta" hai. Isse milta hai weight-average M ˉ w = ∑ N i M i 2 ∑ N i M i \bar M_w = \frac{\sum N_iM_i^2}{\sum N_iM_i} M ˉ w = ∑ N i M i ∑ N i M i 2 M M M M ˉ w \bar M_w M ˉ w M ˉ n \bar M_n M ˉ n bada ya barabar hota hai.
In dono ka ratio hai PDI = M ˉ w / M ˉ n \bar M_w/\bar M_n M ˉ w / M ˉ n 1 . Important point: ideal condensation (step-growth) polymer Flory distribution follow karta hai, jahan PDI = 1 + p \text{PDI}=1+p PDI = 1 + p p → 1 p\to1 p → 1 2 ki taraf jaata hai — 1.5 nahi! Jabki bahut saare addition (radical) polymers ka PDI typically 1.5–2 range mein hota hai.
Yeh important kyun hai? Kyunki polymer ke properties — strength, melting, viscosity — distribution par depend karte hain. Aur experimentally: light scattering mass ke saath respond karta hai to M ˉ w \bar M_w M ˉ w osmometry (colligative property, particle count) M ˉ n \bar M_n M ˉ n M ˉ w \bar M_w M ˉ w M M M
