Weight-Average Molar Mass

Introduction

Weight-average molar mass gives greater influence to heavy polymer chains because they contribute more of the sample's total mass. It is different from number average even though both describe the same molecules. The difference is especially visible when a small number of very long chains is present.

Core explanation

For classes containing N i chains of mass M i, define the mass fraction w i = N iM i/ΣN iM i. Then M w = Σw iM i = ΣN iM i²/ΣN iM i. The squared M i in the numerator explains why a high-mass tail can raise M w strongly. Consider three chains of 10 kg mol⁻¹ and one of 30 kg mol⁻¹. The total count-weighted mass is 60, while ΣN iM i² = 3×100 + 1×900 = 1200 in consistent squared units; M w = 1200/60 = 20 kg mol⁻¹. The same sample had Mₙ = 15 kg mol⁻¹. M w is never below Mₙ for a nonnegative distribution of polymer masses; equality occurs only when all chains have the same mass. Static light scattering often yields a weight-average molar mass because scattering intensity gives stronger weight to larger molecules, although data reduction must account for concentration and interactions. Melt viscosity and entanglement can also be sensitive to longer chains, but no single average completely predicts material properties. Do not confuse a weight fraction with a number fraction. To compute M w from a table, use either the squared-mass formula with counts or the simpler Σw iM i if true mass fractions are supplied.

Step-by-step reasoning

Determine whether the data give chain counts N i or mass fractions w i. From counts, calculate ΣN iM i² and ΣN iM i, then divide. From mass fractions, calculate Σw iM i. Check that M w lies between the smallest and largest M i and is at least Mₙ.

Visual explanation

Draw three short chains and one long chain. Make each chain's bar width proportional to its mass. On a mass-weighted plot, the long chain occupies half of the material in the example despite being only a quarter of the molecules.

Real-world analogy

At a vote weighted by the amount invested, one large shareholder carries more influence than one small shareholder. M w similarly weights each chain according to how much sample mass it represents.

Real-world example

Static light scattering is commonly used to determine M w in a suitable polymer solution. A few large chains can contribute substantial scattering, so sample preparation must avoid dust and aggregates that imitate a high-mass tail.

Why?

M w averages chain mass by sample mass share, not by molecule count. Heavy chains contribute more material and therefore receive larger weight. Algebraically, this places M i² in the count-based numerator.

Common misconception

M w is not simply the arithmetic mean of the listed class masses unless their mass fractions happen to be equal. Likewise, it is not the mass of the single largest polymer molecule.

Worked example

Question: A sample contains two chains of 20 kg mol⁻¹ and two of 40 kg mol⁻¹. Find M w and compare it with Mₙ. Reasoning: Numerator is 2×20² + 2×40² = 800 + 3200 = 4000; denominator is 2×20 + 2×40 = 120. The number average is 120/4 = 30 kg mol⁻¹. Answer: M w = 4000/120 ≈ 33.3 kg mol⁻¹, above Mₙ = 30 kg mol⁻¹ because the heavier chains carry two thirds of the sample mass.

Quick check

1. Write M w in terms of N i and M i. Answer: M w = ΣN iM i²/ΣN iM i.

Exam focus

Show intermediate sums and keep powers of mass straight. Compare M w with Mₙ as an error check: if your result is smaller, revisit the arithmetic or the weighting.

Advanced insight

The inequality M w ≥ Mₙ follows from the nonnegative variance of molar mass under the number distribution. Equality only for a single mass is a mathematical expression of a perfectly uniform chain population.

Summary

Weight-average molar mass is ΣN iM i²/ΣN iM i or Σw iM i. It emphasises long chains more than Mₙ, so a high-mass tail can shift it markedly. M w cannot fall below Mₙ for the same sample and equals it only for a monodisperse distribution.

Practice questions

1. Calculate M w for equal numbers of 10 and 30 kg mol⁻¹ chains. Answer: (10² + 30²)/(10 + 30) = 1000/40 = 25 kg mol⁻¹.

2. Why does a long chain affect M w strongly? Answer: Its mass appears in the sample mass fraction and squared in the count-based numerator.

3. Can M w be smaller than Mₙ for a valid mass distribution? Answer: No. The mass weighting gives M w ≥ Mₙ.

4. Which common method is associated with M w? Answer: Static light scattering under appropriate solution and analysis conditions.