Polymer Molecular Weight - Lecture Notes PDF

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This document provides a lecture outline and notes on polymer molecular weight, covering topics like molecular weight determination, distribution, and different types of average molecular weights, with examples and various methods of calculation for both simple compounds and synthetic polymers.

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Subject Code:CH301 Course Title: Polymer Materials Unit-3, Lecture-16,17,18 Polymer molecular weight, and its distribution, method of determination of different molecular weights, ✓ Textbook of Polymer Science/ F.W. Billmeyer/ John Wiley ✓ Polymer Science/ V.R. Gowarikar/ New Age Int...

Subject Code:CH301 Course Title: Polymer Materials Unit-3, Lecture-16,17,18 Polymer molecular weight, and its distribution, method of determination of different molecular weights, ✓ Textbook of Polymer Science/ F.W. Billmeyer/ John Wiley ✓ Polymer Science/ V.R. Gowarikar/ New Age International ✓ Introduction to polymers/ Young and Powell Molecular weight of Polymers Simple compounds : Fixed molecular weight Ex. Acetone: 58 All the molecules present have identical structure Unit of molecular weight: g/mol or Dalton Not dependent how it is made Synthetic polymer : the molecule form have different chain lengths Molecular weight of Polymers The growth and termination of polymer chains are subjected to variation in the production of mixture of chemically identical molecules of different sizes Ex Ethylene : CH =CH M. W= 28 2 2 Molecular weight of polymer samples: viewed statically and Polymerization R-(CH -CH ) -R MW=14000 2 2 500 expressed as some average of R-(CH2-CH2)550-R MW=15400 the molecular weight contribution by individual R-(CH2-CH2)600-R MW=16800 molecules Molecular weight of Polymers Averages Subjects Credits Marks Maths 5 70 English 5 70 Hindi 5 70 Art 2 70 Music 2 70 Physical Education 1 70 Straight Average of Marks= (70+70+70+70+70+70)/6= 70 Weight average of Marks = (5x70 +5 x70+5x70+2x70+2x70+1x70)/20 =70 Molecular weight of Polymers Weight Averages Subjects Credits Student A Student B Marks Marks Maths 5 90 70 English 5 90 70 Hindi 5 90 70 Art 2 70 90 Music 2 70 90 Physical Education 1 70 90 Straight Average for Student A &B = 80 Weight average of Marks for student A = (5x90 +5 x90+5x90+2x70+2x70+1x70)/20 =85 Weight average of Marks for student B= (5x70 +5 x70+5x70+2x90+2x90+1x90)/20 =74 Molecular weight of Polymers Types of average molecular weight in polymers Number Weight Viscosity Z Average Average Average Average Molecular weight of Polymers Vegetable entity Number of Units in each entity, n Weight of each Unit, M (g) Total weight of each entity W= nM Onions 2 10 20 Brinjals 4 20 80 Cabbages 6 100 600 Cauliflowers 3 250 750 15 - 780 Total no. of vegetables contained in the basket= 15 Assumption: weight of a vegetable is same: No. of onions present in the basket = 2 All onion has same weight No. fraction of Onions= 2/15 Number fraction for other vegetables, 4/15, 6/15, 3/15 Contribution made by 2 Onions towards average weight of vegetable in the basket : Number fraction of onions x weight of each onion = (2/15) x 10 = 1.33 g Similarly, Contribution by the other vegetables , 5.33 g, 40 g, 50 g Therefore, Contribution made by each vegetable, number average weight of the total vegetables= 1.33+5.33+40+50= 96.66 g Molecular weight of Polymers Vegetable entity Number of Units in each entity, n Weight of each Unit, M (g) Total weight of each entity W= nM Onions 2 10 20 Brinjals 4 20 80 Cabbages 6 100 600 Cauliflowers 3 250 750 15 - 1450 Average weight is based the assumption that the individual vegetable variety contributes to total weight in the proportion not of its number but its weight Total weight of all vegetables in the basket = 1450 g Weight of onion present in the basket= 20 g Weight fraction of Onion= 20/1450 Similarly, weight fraction of other vegetables 80/1450, 600/1450, 750/1450 Contribution made by onion towards average weight of vegetable in the basket : Weight fraction of onion X average weight of onions= (20/1450) x 10 = 0.14 Similarly, for other vegetables contribution 1.10g, 41.38 g, 129.31g Total contribution made by each vegetable, the weight average of the total vegetables = 0.14+1.1+41.38+129.31= 171.93 Molecular weight of Polymers Number Average Molecular Weight Total number of molecules (n) given by 𝒏 = 𝒏𝟏 + 𝒏𝟐 + 𝒏𝟑 + … … … … … … + 𝒏𝒊 = ෍ 𝒏𝒊 Number of molecules in fraction 1= 𝒏𝟏 𝒏𝟏 𝒏𝟏 Number fraction of fraction 1= = σ 𝒏𝒊 𝒏 𝒏𝟏𝑴𝟏 Molecular weight contribution by fraction 1= 𝒏 Similarly molecular weight contribution by other fractions will be 𝒏𝟐𝑴𝟐 𝒏𝟑𝑴𝟑 𝒏𝟒𝑴𝟒 𝒏𝒊𝑴𝒊 , , , ……………… 𝒏 𝒏 𝒏 𝒏 Number average molecular weight of whole polymer will be 𝒏𝟏𝑴𝟐 𝒏𝟐𝑴𝟐 𝒏𝟑𝑴𝟑 𝒏𝟒𝑴𝟒 𝒏𝒊𝑴𝒊 σ 𝒏𝒊𝑴𝒊 + + + + ………………+ = σ 𝒏𝒊 = 𝑴𝒏 𝒏 𝒏 𝒏 𝒏 𝒏 Weight Average Molecular Weight Total weight of Polymer = W=σ 𝒏𝒊𝑴𝒊 Weight of fraction 1 = W1= 𝒏𝟏𝑴𝟏 𝒏𝟏𝑴𝟏 𝒏𝟏 𝑴𝟏 Weight fraction of fraction 1 = = σ 𝒏𝒊𝑴𝒊 𝑾 𝒏𝟏𝑴𝟏𝑴𝟏 𝒏𝟏 𝑴𝟏𝟐 Molecular weight contribution by fraction 1 = σ 𝒏𝒊 𝑴𝒊 = σ 𝒏𝒊𝑴𝒊 Similarly, the molecular weight contribution by the other fractions will be 𝒏𝟐 𝑴𝟐𝑴𝟐 𝒏𝟑 𝑴𝟑𝑴𝟑 𝒏𝟒 𝑴𝟒𝑴𝟒 𝒏𝒊 𝑴𝒊𝟐 σ 𝒏𝒊𝑴𝒊 , σ , σ ………………………………….. σ 𝒏𝒊𝑴𝒊 𝒏𝒊𝑴𝒊 𝒏𝒊𝑴𝒊 Weight average molecular of the whole polymer will be 𝒏𝟏 𝑴𝟏𝟐 𝒏𝟐 𝑴𝟐𝟐 𝒏𝟑 𝑴𝟑𝟐 𝒏𝟒 𝑴𝟒𝟐 𝒏𝒊 𝑴𝒊𝟐 σ 𝒏𝒊𝑴𝒊𝟐 σ 𝒏𝒊𝑴𝒊 +σ + σ 𝒏𝒊𝑴𝒊 + σ 𝒏𝒊𝑴𝒊 …………………………………..+ σ = σ 𝒏𝒊𝑴𝒊 = 𝑴𝒘 𝒏𝒊𝑴𝒊 𝒏𝒊𝑴𝒊 Z Average Molecular Weight Emphasis on the large molecules even more than Mw Useful for some calculation used in mechanical properties Method uses a centrifuge to separate the polymer Useful for melt flow properties of polymer Viscosity average Molecular weight Molecular weight can also be calculated from the viscosity of a polymer solution The principle is a simple one: Bigger polymers molecules make a solution more viscous than small ones do The molecular weight obtained by measuring the viscosity is a different from either the number average or the weight average molecular weight. But it's closer to the weight average than the number average. Degree of Polymerization The number of repeating units in the chains of which a polymer is made up is called degree of a polymerization (n) Degree of Polymerization (P) = M/m; Where, M= Mass of Polymer; m = mass of monomeric unit Average degree of polymerization σ 𝒏𝒊 𝑫𝒑 𝒊 𝑫𝒑 𝒏 = σ 𝒏𝒊 σ 𝒏𝒊 𝑫𝒑 𝒊𝟐 𝑫𝒑 𝒘 = σ 𝒏𝒊 𝑫𝒑 𝒊 Molecular weight Average 𝑴𝒏 = 𝑫𝒑 𝒏. 𝒎 𝑴𝒘 = 𝑫𝒑 𝒘. 𝒎 Molecular weight of Distribution Polymer Molecular Weight Distribution Polydispersity Index Gives the idea of the lowest and the highest molecular weight polymer chain a well as the distribution pattern of the intermediate molecular weight 𝑴𝒘 Polydispersity Index (PDI) = 𝑴𝒏 Polydispersity arises owing to variation in the degree of polymerization attained by different molecules during the polymerization process Polymer Molecular Weight Distribution PDI values for synthetic polymers obtained by different polymerization techniques Polymerization type Polydispersity index (a) Free radical chain Polymerization (i) Solution, suspension and emulsion with precise 1.5-2 temperature control (ii) Bulk polymerization with moderate temperature control 2-5 (iii) Bulk Polymerization without temperature control (auto 8-10 acceleration) (b) Polycondensation, polyaddition and ring opening 2-3 polymerization (c) Coordination polymerization >10 (d)Polymerization system leading to chain branching >20 Importance of Different Molecular Weights Significant : Processing of polymer : Properties of a polymer end-product The tensile strength is a function of the weight-average molecular weight. The tensile strength is most influenced by the large molecules in the material. Importance of different Molecular Weights Manufacture of plastic product having plastic flexing hinges, where the flex life of the plastic is important. The flex life is directly related to the Z-average molecular weight. The extremely large molecules are apparently most important in determining that property. Importance of different Molecular Weights The brittleness of a polymer directly related to the levels of both the low and the high molecular weight species in a sample of polymer. find a relationship between product quality and the number-average molecular weight Importance of different Molecular Weights Extrudability or the "moldability" of a polymer The ease with which a molten polymer can be made to flow through an orifice from an extruder into a mold and completely fill the mold The viscosity-average molecular weight is important. The viscosity-average can be directly related to the processability Chain Entanglement in Polymers Chain Entanglement in Polymers Threshold molecular weight The attractive force between the chains Identify the chemical structure: chain length and attractive force Increase molecular weight Increase polymer properties Threshold molecular weight Threshold molecular weight : structure of polymer Polyethylene (non polar): low attractive force, high threshold Polyamide (H Bond): high attractive force, low threshold Threshold molecular weight Range between optimum properties and ease of processing Degree of Polymerization: 200 to 2000 MW: 20,000 to 2,00,000 Measurement of molecular weight of polymers End group analysis Osmometry Light Scattering Viscometry Ultra centrifugation 𝑴𝒏 𝑴𝒏 𝑴𝒘 𝑴𝒗 𝑴𝒛 Direct Method Direct Method Direct Method Indirect method Direct Method Size exclusion chromatography or Gel permeation chromatography 𝑴𝒏 𝑴𝒘 𝑴𝒗 𝑴𝒛 Molecular weight distribution curve Indirect method Polymer Molecular weight by end group analysis method End group analysis method mainly applied to condensation polymers Used reliably only for samples consisting of linear molecules with determinable end groups for those obtained by known polymerization mechanism without side reactions. limitation : sensitivity decrease with increasing molecular weight of the polymer, which restricts its use to molecular weight up to 20,000 da. Polymer Molecular weight by end group analysis method The number average molecular weight of certain liner polymers can be found by estimating the number of end groups by chemical analysis (a) Measurement of amino end group in polyamides Nylon (1 gm) Dissolve in 25 ml methanol/phenol mixture Refluxing for 30 min Titrate against 0.02 N HCl using thymol blue as an indicator (color change from blue to yellow) 𝑨−𝑩 𝑿 𝑵𝒐𝒓𝒎𝒂𝒍𝒊𝒕𝒚 𝒐𝒇 𝑯𝑪𝒍 𝑿 𝒆𝒒.𝒘𝒕 𝒐𝒇 𝑯𝑪𝒍 Amount of reagent used in titration = 𝟏𝟎𝟎𝟎 Where: A is the volume in ml of 0.02N HCl for the Sample B is the volume in ml of 0.02 N HCL for the Blank Polymer Molecular weight by end group analysis method (b) Measurement of Carboxyl end groups in PET PET Chips (1 gm) 50 ml of 2:3 Phenol: Chloroform mixture Refluxed for 2 h to dissolve the polymer Titrate against 0.1 N KOH solution in benzoyl alcohol using 0.5 ml tertrabromophenol as an indicator (color change yellow to blue violet) 𝑨−𝑩 𝑿 𝑵𝒐𝒓𝒎𝒂𝒍𝒊𝒕𝒚 𝒐𝒇 𝑲𝑶𝑯 𝑿 𝒆𝒒.𝒘𝒕 𝒐𝒇 𝑲𝑶𝑯 Amount of reagent used in titration = 𝟏𝟎𝟎𝟎 Where: A is the volume in ml of 0.1N KOH for the Sample B is the volume in ml of 0.1 N KOH for the Blank Polymer Molecular weight by end group analysis method The number average molecular weight can be calculated from the end group analysis using 𝒇𝒘𝒆 𝑴𝒏 = 𝒂 Where , f= functionality or reactive groups per molecule in the sample w= weight of the polymer a= amount of reagent used in the titration e = equivalent weight of the reagent Numerical 2.7 gm of a polyester that is known to be linear and to contain acid groups at both ends requires titration with 15 ml 0.1 N alcoholic KOH to reach a phenolphthalein end point. Calculate the 𝑴𝒏 Solution 𝑨−𝑩 𝑿 𝑵𝒐𝒓𝒎𝒂𝒍𝒊𝒕𝒚 𝒐𝒇 𝑲𝑶𝑯 𝑿 𝒆𝒒.𝒘𝒕 𝒐𝒇 𝑲𝑶𝑯 Amount of reagent used in titration = 𝟏𝟎𝟎𝟎 𝟏𝟓 𝒙 𝟎.𝟏 𝒙 𝟓𝟔 Amount of reagent used a= = 0.084 𝟏𝟎𝟎𝟎 𝒇𝒘𝒆 𝑴𝒏 = 𝒂 𝟐 𝒙 𝟐. 𝟕𝒙 𝟓𝟔 𝑴𝒏 = = 𝟑𝟔𝟎𝟎 𝒈𝒎/𝒎𝒐𝒍 𝟎. 𝟎𝟖𝟒 Viscometry Viscosity average molecular weight determine by this method Polymer solution shows very high viscosity which varies with concentration and molecular weight Intrinsic viscosity (Staudinger index or limiting viscosity number )[]: It is strongly dependent on the molecular dimensions of the solute particles. Since molecular dimension depends on molecular weight, suitable calibration curves have been developed which led to well known relation called the Mark Houwink Sakurado equation Where, “k” and “a” are constant  = 𝒌𝑴 a Reported in literature based on the calibrations using molecular weight determined by other methods Viscometry Measurement of Intrinsic viscosity Liquid is flowing through a capillary tube Time required for the liquid of Volume V to pass through the capillary of radius r and length l is related to its absolute viscosity  by Poiseuille equation 𝝅𝑷 𝒓𝟒 𝒕 = P: pressure head under which liquid flow takes place 𝟖 𝑽𝒍 𝟖𝑽𝒍𝜼 t= Equation correlates the flow time with 𝝅𝑷 𝒓𝟒 the absolute viscosity of the liquid Ostwald Viscometer If  = absolute viscosity of a solution t = time of flow of solution through capillary o = absolute viscosity of the pure solvent t0= time of flow of solution through capillary Viscometry Time of flow of solution 𝟖𝑽𝒍𝜼 A t= 𝟒 𝝅𝑷 𝒓 Time of flow of solvent 𝟖 𝑽 𝒍𝜼 𝒐 B to= 𝟒 𝝅𝑷 𝒓 Dividing equation, A by B 𝒕  = 𝒕𝒐 𝒐 𝒕  𝒔𝒑 Relative Viscosity (r): = Reduced viscosity = 𝒕𝒐 𝒐 𝒄 𝒕−𝒕𝒐 −𝒐 𝒓 Specific Viscosity sp : = = r- 1 Inherent viscosity= ln 𝒕𝒐 𝒐 𝒄 𝒔𝒑 𝒕−𝒕𝒐 Intrinsic viscosity []= 𝑪𝟎 = 𝒄 𝟎 𝒄 𝒕𝒐 Viscometry Viscosity data for polymethyl acrylate sample in benzene solution at 30°C is given. Flow time for solvent to= 216 second. Given K = 8.630 x 10-5 dl/gm, a = 0.725. Calculate viscosity average molecular weight. 𝒔𝒑 Concertation (c) gm/dl Flow time t , second (t- to)/to = 𝒔𝒑 𝒄 0.2716 459.8 1.128703704 4.155757377 0.1940 378.2 0.750925926 3.870752195 0.1509 337.9 0.564351852 3.739906242 0.1235 321.8 0.489814815 3.966111861 0.1045 296.4 0.372222222 3.561935141 Viscometry 𝒔𝒑 𝑪𝟎 = []= 3.1976 4.2 𝒄 4.1 y = 3.5199x + 3.1976  = 𝒌𝑴 a 𝒔𝒑 4 R² = 0.9987 𝒄 3.9 ഥ 0.725 3.8 3.1976= 8.630 x 10-5 𝑴 3.7 ഥ 0.725= 3.7033 x 104 3.6 𝑴 3.5 0 0.1 0.2 0.3 ഥ = (37033)1/0.725 𝑴 C ഥ = (37033)1.38 𝑴 ഥ = 2,016,752.83 𝑴 Viscometry Viscosity Molecular weight constants for Polymer/solvent system at 30C Polymer Solvent K x 103 (ml/g) a Cellulose triacetate Chloroform 4.5 0.90 Polyacrylic acid 1, 4 dioxane 76 0.50 Polyacrylonitrile Dimethyl formamide 20.9 0.75 Polyethyl acrylate Acetone 20 0.66 Polyethylene Phenol/tetrachloroeth 22.9 0.73 terephthalate ane (3/5 vol ratio) Polymethylmethacrylate Acetone 7.70 0.70 Polyvinyl alcohol Water 45.3 0.64 Gel Permeation Chromatography Gel Permeation Chromatography Separates polymers based on the size Gel Permeation Chromatography Beads: copolymers of styrene and divinyl benzene Gel Permeation Chromatography Gel Permeation Chromatography Calibration curve with standard Molecular weight Polystyrene standards, solvent THF Gel Permeation Chromatography Gel Permeation Chromatography Gel Permeation Chromatography https://www.youtube.com/watch?v=9trws35gLNM https://www.youtube.com/watch?v=L-xIHihOGKs Numerical Q. 1Size exclusion chromatography data of a new polymer shows the following molecular weight distribution Number of molecules (ni) Mass of each molecules (Mi) 5 10000 3 30000 2 60000 Calculate number average, weight average and z average molecular weight for this sample σ 𝒏𝒊𝑴𝒊 σ 𝒏𝒊𝑴𝒊𝟐 σ 𝒏𝒊𝑴𝒊𝟑 Solution = 𝑴𝒏 = 𝑴𝒘 = 𝑴𝒛 σ 𝒏𝒊 σ 𝒏𝒊𝑴𝒊 σ 𝒏𝒊 𝑴𝒊𝟐 ni Mi niMi niMi* Mi niMi *Mi *Mi 5 10000 50000 500000000 5E+12 3 30000 90000 2700000000 8.1E+13 Mn 26000 2 60000 120000 7200000000 4.32E+14 Mw 40000 10 100000 260000 10400000000 5.18E+14 Mz 49807.69 Numerical Q.2 A polydisperse sample of Polystyrene is prepared by mixing three monodisperse samples in the following proportion Weight of sample (wi=niMi) (gm) Molecular weight (Mi) (gm/mol) 1 10,000 2 50,000 2 100,000 Using this information calculate the number average, weight average molecular weight and PDI of Mixture. σ 𝒘𝒊 σ 𝒏𝒊𝑴𝒊 σ 𝒘𝒊𝑴𝒊 σ 𝒏𝒊𝑴𝒊𝟐 Solution = = 𝑴𝒏 = = 𝑴𝒘 𝒘𝒊 σ 𝒏𝒊 σ 𝒘𝒊 σ 𝒏𝒊𝑴𝒊 σ( ) 𝑴𝒊 wi= niMi Mi wi/Mi wiMi 1 10000 0.0001 10000 2 50000 0.00004 100000 Mn 31250 2 100000 0.00002 200000 Mw 62000 5 160000 0.00016 310000 PDI 1.984 Numerical Q.3 A sample of polystyrene is composed of a series of fraction of different sized molecules Fraction Weight Fraction (Wi or hi) Molecualr weight (Mi) Da A 0.10 12000 B 0.19 21000 C 0.24 35000 D 0.18 49000 E 0.11 73000 F 0.08 102000 G 0.06 122,000 H 0.04 146,000 Calculate number average and weight average molecular weight and Polydispersity index of this polymer σ 𝒘𝒊 σ 𝒏𝒊𝑴𝒊 σ 𝒘𝒊𝑴𝒊 σ 𝒏𝒊𝑴𝒊𝟐 Solution = = 𝑴𝒏 = = 𝑴𝒘 𝒘𝒊 σ 𝒏𝒊 σ 𝒘𝒊 σ 𝒏𝒊𝑴𝒊 σ( ) 𝑴𝒊 wi= niMi Mi wi/Mi wi Mi 0.1 12000 8.33333E-06 1200 0.19 21000 9.04762E-06 3990 0.24 35000 6.85714E-06 8400 0.18 49000 3.67347E-06 8820 0.11 73000 1.50685E-06 8030 0.08 102000 7.84314E-07 8160 0.06 122,000 4.91803E-07 7320 Mn 32290.87 0.04 146,000 2.73973E-07 5840 Mw 51760 1 560000 3.09685E-05 51760 PDI 1.60293

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