GATE Polymer Chemistry — Molar Mass Averages, PDI and Kinetics

By Aniket Bhardwaj · 14 September 2026 · GATE Chemistry

Polymer chemistry is one of the friendliest parts of the GATE Chemistry syllabus for a student who prepares it properly, because most of the questions are calculable. You are usually asked for a molar mass average, a polydispersity index, a degree of polymerisation, or the order of a polymerisation reaction with respect to initiator. These are short numericals with fixed methods. This guide gives every formula you need, three fully worked examples with the arithmetic shown, and the specific errors that lose marks.

Before anything else: always check the current official GATE notification and syllabus document for what is examinable in your year. Do not rely on any website — including this one — for the paper pattern, the number of questions or the marking scheme.

Why a polymer needs more than one molar mass

A sample of a small molecule such as benzene has exactly one molar mass. A polymer sample does not. It is a mixture of chains of different lengths, so we quote averages. The two you must know are the number average and the weight average.

M_n = Σ N_iM_i / Σ N_i   (number-average molar mass)
M_w = Σ N_iM_i² / Σ N_iM_i   (weight-average molar mass)
PDI = M_w / M_n   (polydispersity index, also written Đ)

What the symbols mean:

N_i — the number of chains (or moles of chains) that have molar mass M_i.
M_i — the molar mass of that group of chains, in g/mol.
M_n counts every chain equally, so it is what colligative methods (osmometry, freezing-point depression, end-group analysis) measure.
M_w weights each chain by its own mass, so heavy chains dominate. Light scattering measures M_w.
PDI is always ≥ 1. It equals exactly 1 only for a perfectly monodisperse sample, where every chain has the same length.

Worked example 1 — M_n, M_w and PDI

A polymer sample contains 2 mol of chains of molar mass 10,000 g/mol and 3 mol of chains of molar mass 20,000 g/mol. Find M_n, M_w and the PDI.

Step 1 — M_n.
Σ N_iM_i = (2 × 10,000) + (3 × 20,000) = 20,000 + 60,000 = 80,000
Σ N_i = 2 + 3 = 5
M_n = 80,000 ÷ 5 = 16,000 g/mol

Step 2 — M_w.
Σ N_iM_i² = (2 × 10,000²) + (3 × 20,000²) = (2 × 1.0 × 10⁸) + (3 × 4.0 × 10⁸) = 2.0 × 10⁸ + 12.0 × 10⁸ = 1.4 × 10⁹
M_w = 1.4 × 10⁹ ÷ 80,000 = 1.4 × 10⁹ ÷ 8.0 × 10⁴ = 17,500 g/mol

Step 3 — PDI.
PDI = 17,500 ÷ 16,000 = 1.09 (to 2 decimal places)

Sanity check: M_w came out larger than M_n, and the PDI is greater than 1. If your answer breaks either of these, you have made an arithmetic error — the inequality M_w ≥ M_n is mathematically guaranteed.

Degree of polymerisation

The degree of polymerisation is simply how many monomer units the average chain contains.

DP_n = M_n / M₀

Here M₀ is the molar mass of the repeating unit. Note that the repeating unit is not the monomer if a small molecule is lost during the reaction — in a condensation polymerisation, water or HCl leaves, so the repeat unit is lighter than the sum of the monomers.

Worked example 2 — degree of polymerisation of polystyrene

A polystyrene sample has M_n = 1.04 × 10⁵ g/mol. Find DP_n.

Step 1 — molar mass of the repeat unit. The styrene repeat unit is C₈H₈ (addition polymerisation loses nothing).
C: 8 × 12.011 = 96.088
H: 8 × 1.008 = 8.064
M₀ = 96.088 + 8.064 = 104.152 g/mol

Step 2 — divide.
DP_n = 104,000 ÷ 104.152 = 998.5 ≈ 999 units per chain

So the average chain carries about a thousand styrene units. Notice the first step is a plain molar-mass calculation — that is why polymer numericals reward students who are fast and accurate with formula masses.

Step-growth polymerisation and the Carothers equation

For a linear step-growth (condensation) polymerisation with exact stoichiometry, the average chain length depends only on how far the reaction has gone:

DP_n = 1 / (1 − p) where p = fractional extent of reaction of the functional groups

Under the most probable (Flory) distribution, the same reaction also gives PDI = 1 + p, so a step-growth polymer taken to high conversion tends towards a PDI of about 2. This single result explains why condensation polymers are broadly distributed while a well-controlled living polymerisation can approach PDI ≈ 1.

Worked example 3 — Carothers equation for nylon-6,6

A nylon-6,6 polymerisation reaches p = 0.98. Find DP_n and M_n.

Step 1 — DP_n.
DP_n = 1 ÷ (1 − 0.98) = 1 ÷ 0.02 = 50

Step 2 — mass of one structural unit. The nylon-6,6 repeat unit is C₁₂H₂₂N₂O₂:
C: 12 × 12.011 = 144.132
H: 22 × 1.008 = 22.176
N: 2 × 14.007 = 28.014
O: 2 × 15.999 = 31.998
M(repeat unit) = 144.132 + 22.176 + 28.014 + 31.998 = 226.320 g/mol

Step 3 — mind the convention. Nylon-6,6 is made from two different monomers, so one repeat unit contains two structural units (one diamine residue and one diacid residue). In the standard Carothers treatment DP_n counts structural units, so the average structural-unit mass is 226.320 ÷ 2 = 113.160 g/mol.
M_n = 50 × 113.160 = 5658 g/mol

Convention warning, because textbooks genuinely differ. Some books define DP_n as the number of repeat units, which for this polymer would give M_n = 50 × 226.320 = 11,316 g/mol — twice the value above. Neither book is wrong; they are counting different things. In an exam, read whether the question says "structural units" or "repeat units", and if it gives you M₀ directly, just use the M₀ it gives you.

Chain-growth (free-radical) kinetics — the result GATE asks for

For a free-radical polymerisation with initiator decomposition, propagation and bimolecular termination, applying the steady-state approximation to the radical concentration gives:

R_p = k_p [M] ( f k_d [I] / k_t )^1/2

R_p — rate of polymerisation (rate of monomer consumption).
[M], [I] — monomer and initiator concentrations.
k_p, k_d, k_t — rate constants for propagation, initiator decomposition and termination.
f — initiator efficiency, the fraction of radicals that actually start a chain rather than recombining in the solvent cage.

The examinable consequence: the rate is first order in monomer but only half order in initiator. Doubling [I] multiplies the rate by √2 ≈ 1.41, not by 2. The kinetic chain length ν = R_p/R_i therefore falls as you add more initiator — more initiator gives you a faster reaction but shorter chains. That trade-off is a favourite one-line question.

Classification you should be able to recite

Basis	Categories	Point to remember
Mechanism	Chain-growth (addition) vs step-growth (condensation)	Chain-growth reaches high molar mass early; step-growth needs p very close to 1
Thermal behaviour	Thermoplastic vs thermosetting	Thermoplastics soften and can be remoulded; thermosets are cross-linked and cannot
Tacticity	Isotactic, syndiotactic, atactic	Regular tacticity allows crystallinity; atactic chains are usually amorphous
Structure	Linear, branched, cross-linked, network	Branching lowers density and crystallinity (LDPE vs HDPE)
Copolymer sequence	Random, alternating, block, graft	Reactivity ratios r₁, r₂ decide which one you get

Two transitions are also standard: T_g, the glass transition temperature, is a property of the amorphous regions and is a second-order-like transition; T_m, the melting temperature, belongs to crystalline regions and is a genuine first-order transition. A fully amorphous polymer has a T_g but no T_m. Ziegler–Natta and metallocene catalysts matter because they control tacticity and therefore crystallinity.

Common mistakes that cost marks

Reporting a PDI below 1. Impossible. If you get one, you have swapped M_w and M_n, or squared the wrong term in M_w.
Using ΣN_i in the denominator of M_w. The denominator of M_w is ΣN_iM_i, not ΣN_i.
Taking the monomer mass as the repeat-unit mass in a condensation polymer. A molecule of water (or HCl) is lost per linkage, so the repeat unit is lighter than the monomers you started with.
Making the rate first order in initiator. It is one-half order under steady-state with bimolecular termination — the square root is the whole point.
Confusing T_g with T_m, or expecting a melting point for an atactic, fully amorphous polymer.
Forgetting that the Carothers equation assumes exact stoichiometry. An excess of one monomer, or a monofunctional impurity, caps the chains and limits DP_n no matter how long you run the reaction.

Where polymer chemistry sits in your GATE preparation

Sub-topic	What is typically asked	Preparation priority
Molar mass averages and PDI	Direct numerical from a table of N_i, M_i	High — fast, certain marks
Degree of polymerisation	M_n ÷ M₀, often combined with a formula-mass step	High
Carothers equation	DP_n from p, or p needed for a target DP_n	High
Free-radical kinetics	Order in [I] and [M]; effect on chain length	Medium — conceptual one-liners
Characterisation methods	Which technique gives M_n vs M_w	Medium
Tacticity, T_g/T_m, copolymers	Match-the-following and reasoning statements	Medium

Treat the numerical rows as guaranteed practice: they take under two minutes each once the method is automatic, and the method never changes. For the exact syllabus wording and the current pattern, always consult the official GATE information brochure for your examination year.

Every polymer numerical starts with a formula mass. Before you can divide M_n by M₀, you have to get M₀ right — C₈H₈ for styrene, C₁₂H₂₂N₂O₂ for the nylon-6,6 repeat unit, C₂H₄ for polyethylene. The free Molar Mass & Composition calculator takes any formula and returns the element-wise breakdown, so you can check that first step in seconds instead of losing a whole question to an addition slip.

Open the Molar Mass Calculator →

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