Gravimetric analysis, P K MANI

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Gravimetric analysis

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  • Absolutely brilliant presentation. As a bit of a novice teacher I am excited to have found such a fantastic reference on gravimetric analysis. The worked examples are very helpful and I will be returning to these slides when I teach analytical chemistry in the near future. Thank you Professor Mani ! Reginald Crosswell.
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Gravimetric analysis, P K MANI

  1. 1. Gravimetric Analysis and Precipitation Equilibria Analytical Chemistry, ACSS-501 Class – I to VI Dr. Pabitra Kumar Mani Door meten tot weten “Through measurement to knowledge.” Heike Kamerlingh Onnes, NL “gentleman of absolute zero”
  2. 2. A History of Gravimetric Analysis Jons Jacob Berzelius (1779 - 1848), considered the leading chemist of his time, developed much of the apparatus and many of the techniques of 19th century analytical chemistry. Examples include the use of ashless filter paper in gravimetry, the use of hydrofluoric acid to decompose silicates, and the use of the metric system in weight determinations. He performed thousands of analyses of pure compounds to determine the atomic weights of most of the elements known then. Berzelius also developed our present system of symbols for elements and compounds.
  3. 3. A History of Gravimetric Analysis Theodore W. Richards (1868 - 1928) …and his graduate students at Harvard developed or refined many of the techniques of gravimetric analysis of silver and chlorine. These techniques were used to determine the atomic weights of 25 of the elements by preparing pure samples of the chlorides of the elements, decomposing known weights of the compounds, and determining the chloride content by gravimetric methods. From this work Richards became the first American to receive the Nobel Prize in Chemistry in 1914.
  4. 4. Accuracy vs. Precision • Accuracy: getting the correct answer • Precision: being able to get the same answer more than once • We can have precision without accuracy and accuracy without precision • Both are important for valuable results Chemistry 215 Copyright D Sharma 4
  5. 5. Chemistry 215 Copyright D Sharma 5 http://www.chem.ualberta.ca/courses/Chem211/Harrison/
  6. 6. Analytes: The constituents of interest in a sample. Matrix: All other constituents in a sample except for the analytes.
  7. 7. How to Perform a Successful Gravimetric Analysis • What steps are needed? 1. 2. 3. 4. 5. 6. 7. 8. 9. Sampled dried, triplicate portions weighed Preparation of the solution Precipitation Digestion Filtration Washing Drying or igniting Weighing Calculation
  8. 8. Gravimetric Analysis • Gravimetric Analysis – one of the most accurate and precise methods of macro-quantitative analysis. • Analyte selectively converted to an insoluble form. • Measurement of mass of material • Correlate with chemical composition • Why? • Simple • Often required for high precision Accuracy: A measure of the agreement between an experimental result and its expected value. Precision: An indication of the reproducibility of a measurement or result.
  9. 9. Common Desiccants Mechanism of Action Hydration ANHYDRONE® (Magnesium Perchlorate anhydrous), CaCl2, MgO, MgSO4, K2CO3, KOH, Drierite, Na2SO4 (anhydrous), H2SO4, ZnCl2 Absorption and/ or Adsorption BaO, CaSO4, Molecular Sieve, H3PO4, NaOH Pellets Chemisorption CaO, P2O5
  10. 10. • Silica gel goes from blue to • Anhydrous sodium sulfate pink as it absorbs moisture gets clumpy as it absorbs Can be regenerated in oven water More Information about desiccants including common interferents and regeneration temperature can be found at: http://www.jtbaker.com/techlib/documents/3045.html
  11. 11. Principles of gravimetric Analysis (i)Evolution or Volatalization method (ii) Precipitation method (iii) Miscellaneous method : Magnetic separation, Differential separation, Electro gravimetry
  12. 12. Volatilization Gravimetry (i)Direct Method (ii)Indirect method Direct Method The desired constituent is directly weighed eg. estimation of water in minerals i) Talc H Mg (SiO ) Δ water (absorbed in a suitable absorbant, 2 3 3 4 Anhydrous Mg (ClO4)2, P2O5 ii) Carbonates, bicarbonates Δ or H+ CO2 ignited iii) (Borates, Boric Acid ) + H2SO4 + CH3OH (BOCH3)3 (weighed) Methyl borate Ca(BO3)2 Ca-Borate (ester) absorbed in
  13. 13. Indirect Method A) N in Nitrate, KNO3 2KNO3 + SiO2 heat K2SiO3 + N2O5 (more volatile) (residue in the crucible) W1-W2 = wt of N2O5 ----- lost heat B) SiO2(Fe, Cu, Pb) + HF Pt crucible SiF4 + residue (impurities) W1-W2 = wt of pure silica
  14. 14. For carbonate: carbon dioxide Solid adsorbent: Ascarite II - NaOH on a nonfibrous silicate
  15. 15. Precipitation Technique: To get a substance, sparingly soluble or very little soluble by double decomposition reaction, in solution state AB + CD AD + BC (soln) Reagent Sparingly soln soluble Due care should taken for selection of a suitable reagent for the desired constituent to be analysed. The reagent must (i)Produce a precipitate of low solubility product (ii)Be selective or better specific for the desired constituent. Other constituents should not produce a similar low solubility product with the reagent
  16. 16. Desired ion Reagent Precipitate form Weighable form Cl- Ag+ as AgNO3 soln AgCl AgCl Ag+ Cl- as KCL soln AgCl AgCl SO4-2 Ba+2 as BaCl2 or BaNO3 BaSO4 BaSO4 Ba+2 SO4-2 BaSO4 BaSO4 Mg+2, Zn+2, Mn+2 HPO4-2 as (NH4)2HPO4 soln. M(NH4)PO4.6H2O M2P2O7 pyrophosphate HPO4-2 , PO4-3 pH 5-6.5, MgCl2 soln NH4Cl soln Mg(NH4)PO4.6H2O Mg2P2O7 pyrophosphate K+ HClO4 KClO4 KClO4 Na+ [(C6H5)4 B]- K [(C6H5)4 B] K [(C6H5)4 B] Na3 [ Co(NO2)6 ] K2Na [ Co(NO2)6 ] K2Na [ Co(NO2)6 ]
  17. 17. Desired ion Reagent Precipitate form Ni+2 Dimethyl glyoxime (DMG) Ni(DMG)2 Co+2 α-nitroso β-napthol CoR3 N O Co3O4 N OH Al+3 pH 4.5-6.0 Mg+2 pH 7.0 8 hydroxy quinoline at Weighable form O + Co O AlR3
  18. 18. Desired ion Reagent Be+2, Cu+2, Fe+3 N-nitroso phenyl hydroxyl amine (Cupferron) Precipitate form Weighable form BeR2 CuR2 Weighable CuO FeR3 N N Ca+2 O Fe2O3 3+ Fe O CaC2O4.H2O CaC2O4.H2O Low heat <400°C >600°C C2O42- CaO Cu+2 CNS- or KCNS CuCNS Directly weighable F- PbCl2 satd. Soln. PbFCl Directly weighable
  19. 19. Organic precipitating agents are chelating agents. Organic precipitating agents are chelating agents. They form insoluble metal chelates. . They form insoluble metal chelates ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)
  20. 20. Effect of particle size Solubility of a precipitate increases with decreasing particle size. For having diameter in the range less than 1-2 µ, the solubility RT ln S r 2σ = M S dr Sr =solubility of particles having radius r S= solubilty of macroparticles, solubility M= molecular weight σ = surface tension, d= density of the particle For different substances, however, depending upon the nature of bond and molecular structure, the ratio Sr/S becomes diffferent For BaSO4 ppts, S0.02/S = 931 (at 25°C) Ag2CrO4, S0.02/S = 4 (at 25°C) The particle size during pptn is influenced by the relative super saturation of the medium. The initial velocity of formation of particles, Velocity of formation, V α (Q-S)/S where, Q=total concentration of the substance that is to precipitate S= equilibrium solubility, (Q-S) denote the supersaturation at the moment pptn commences.
  21. 21. This condition applies when the pptn. begins, by changing the value of relative supersaturation, the particle size can be chnage. With very low value of it, initially fewer no. of nuclei (particles) are formed, which later grow in size slowly to produce larger particles of not only low solubility but also of perfect crystalline set and easy filterability. The rate of growth of initially precipitated particles is given by Noyes-Nernst Equation V= [ (D/l) x A] (QS) D= diffusion coefficent l= diffusion path, A= surface area at a given temp. and stirring rate (D/l) x A= const, K V= K(Q-S) Particle size increases with decreasing super saturation. This is due to two effects in combine . (i) The Aggregation velocity (ii) The Orientation velocity The final form of the precipitate particle depends upon the competition of these two effects. On digestion and aging particularly at higher temp.(hot soln.) and dil. soln (i.e. under condition of increasing solubility) , the crystals of precipitates become stable and of perfect shape and size
  22. 22. Sometimes very finely dispersed particles of colloidal nature can be readily coagulated to larger size by adding very small amount of an organic substance behaving as a lyophillic colloid i.e., a dilute soln of gelatin (1%) or gum arabic sol. The process is called sensitization. Thus by adding a few ml of gelatin soln. quantitative pptn of colloidal silica is obtained through digestion and evaporation of colloidal silicic acid.
  23. 23. Important Factors for Gravimetric Analysis • Nucleation – Individual ions/atoms/molecules coalesce to form “nuclei” • Particle Growth – Condensation of ions/atoms/molecules with existing “nuclei” forming larger particles which settle out • Colloidal Suspension – Colloidal particles remain suspended due to adsorbed ions giving a net + or - charge Coagulation, agglomeration Suspended colloidal particles coalesce to form larger filterable particles (inert electrolyte allows closer approach) Peptization Re-dissolution of coagulated colloids by washing and removing inert electrolyte
  24. 24. Schematic model of the solid–solution interface at a particle of AgCl in a solution containing excess AgNO3.
  25. 25. Mechanism of Precipitation Digestion • Heating the precipitate within the mother liquor (or solution from which it precipitated) for a certain period of time to encourage densification of nuclei. – During digestion, small particles dissolve and larger ones grow (Ostwald ripening). This process helps produce larger crystals that are more easily filtered from solution ∆T 35
  26. 26. Conditions for Analytical Precipitation • Precipitation from hot solution – The molar solubility (S) of precipitates increases with an increase in temperature – An increase in S decreases the supersaturation and increases the size of the particle. • Precipitation from dilute solution – This keeps the molar concentration of the mixed reagents low. Slow addition of precipitating reagent and thorough stirring keeps Q low. (Uniform stirring prevents high local concentrations of the precipitating agent.) 36
  27. 27. Conditions for Analytical Precipitation • Precipitation at a pH near the acidic end of the pH range in which the precipitate is quantitative. – Many precipitates are more soluble at the lower (more acidic) pH values and so the rate of precipitation is slower. • Digestion of the precipitate. – The digestion period can lead to improvements in the organization of atoms within the crystalline nuclei, such as expulsion of foreign atoms (or other impurities). 37
  28. 28. During digestion at elevated temperature: During digestion at elevated temperature: Small particles tend to dissolve and reprecipitate on larger ones. Small particles tend to dissolve and reprecipitate on larger ones. Individual particles agglomerate. Individual particles agglomerate. Adsorbed impurities tend to go into solution. Adsorbed impurities tend to go into solution. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 10.1. Ostwald ripening.
  29. 29. Cl- -adsorbs on the particles when in excess (primary layer). Cl adsorbs on the particles when in excess (primary layer). A counter layer of cations forms. A counter layer of cations forms. The neutral double layer causes the colloidal particles to The neutral double layer causes the colloidal particles to coagulate. coagulate. Washing with water will dilute the counter layer and the primary layer charge Washing with water will dilute the counter layer and the primary layer charge causes the particles to revert to the colloidal state (peptization). causes the particles to revert to the colloidal state (peptization). So we wash with an electrolyte So we wash with an electrolyte that can be volatilized on heating (HNO3). that can be volatilized on heating (HNO ). 3 ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Representation of silver chloride colloidal particle and adsorptive layers when Cl is in excess.
  30. 30. Class-III
  31. 31. Effect of particle size Solubility of a precipitate increases with decreasing particle size. For having diameter in the range less than 1-2 µ, the solubility RT ln S r 2σ = M S dr Sr =solubility of particles having radius r S= solubilty of macroparticles, solubility M= molecular weight σ = surface tension, d= density of the particle For different substances, however, depending upon the nature of bond and molecular structure, the ratio Sr/S becomes diffferent For BaSO4 ppts, S0.02/S = 931 (at 25°C) Ag2CrO4, S0.02/S = 4 (at 25°C) The particle size during pptn is influenced by the relative super saturation of the medium. The initial velocity of formation of particles, Velocity of formation, V α (Q-S)/S where, Q=total concentration of the substance that is to precipitate S= equilibrium solubility, (Q-S) denote the supersaturation at the moment pptn commences.
  32. 32. This condition applies when the pptn. begins, by changing the value of relative supersaturation, the particle size can be chnage. With very low value of it, initially fewer no. of nuclei (particles) are formed, which later grow in size slowly to produce larger particles of not only low solubility but also of perfect crystalline set and easy filterability. The rate of growth of initially precipitated particles is given by Noyes-Nernst Equation V= [ (D/l) x A] (Q-S) D= diffusion coefficent l= diffusion path, A= surface area at a given temp. and stirring rate (D/l) x A= const, K V= K(Q-S) Particle size increases with decreasing super saturation. This is due to two effects in combine . (i) The Aggregation velocity (ii) The Orientation velocity The final form of the precipitate particle depends upon the competition of these two effects. On digestion and aging particularly at higher temp.(hot soln.) and dil. soln (i.e. under condition of increasing solubility) , the crystals of precipitates become stable and of perfect shape and size
  33. 33. Sometimes very finely dispersed particles of colloidal nature can be readily coagulated to larger size by adding very small amount of an organic substance behaving as a lyophillic colloid i.e., a dilute soln of gelatin (1%) or gum arabic sol. The process is called sensitization. Thus by adding a few ml of gelatin soln. quantitative pptn of colloidal silica is obtained through digestion and evaporation of colloidal silicic acid.
  34. 34. Solubility product and Factors affecting solubility of a ppt. BmAn (solid) ⇋ BmAn (soln.) ⇋ m Bx+ + n AySparingly soluble almost complete dissociation since the resulting electrolyte substance [ B ] = ms moles/L [ A ] = ns moles/L soln. is very dilute Solubility = s moles/L (having ionic lattice) [B ] [ A ] Ke = x+ m y− n [ Bm An ] [ ] [A ] S=B x+ m y− n x+ y− But, [BmAn] is constant because the soln is in contact with large excess of solid BmAn Ke X [BmAn] = S (Solubility product ) = (ms)m (ns)n
  35. 35. Precipitation Equilibria:The Solubility Product • Solubility of Slightly Soluble Salts: • AgCl(s)(AgCl)(aq) Ag+ + Cl• Solubility Product KSP = ion product • KSP = [Ag+][Cl-] • Ag2CrO4(s)  2 Ag+ + CrO42• KSP = [Ag+]2[CrO42-]
  36. 36. The molar solubility depends on the stoichiometry of the salt. The molar solubility depends on the stoichiometry of the salt. A 1:1 salt is less soluble than a nonsymmetric salt with the same K sp.. A 1:1 salt is less soluble than a nonsymmetric salt with the same K sp ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)
  37. 37. Let’s consider silver chloride AgCl (s)  Ag+ + Cls s s • What is the equilibrium constant? = s2 Ksp = (Ag+) (Cl-) • Now what about the value of the Ksp? Ksp = 1.6 x 10-10 • Can we calculate 1.3 xsolubility, s? s = (Ksp)1/2 = the 10-5 M PGCC CHM 102 Sinex
  38. 38. What is the pH of a bottle of Milk of Magnesia, a common antacid? Milk of Magnesia is Mg(OH)2 with Ksp = 8.9 x 10-12 Mg(OH)2 (s)  Mg++ + 2OHs s 2s Ksp = (Mg++)(OH-)2 = s(2s)2 = 4s3 s = (Ksp/4)1/3 = 1.3 x 10-4 M (OH-) = 2s = 2.6 10-4 M PGCC CHM 102 Sinex pOH = 3.58 pH = 10.42
  39. 39. Problem: A saturated solution of silver chromate was to found contain 0.022 g/L of Ag2CrO4. Find Ksp Eq. Expression: Ag2CrO4 (s) ⇔ 2Ag+ + CrO42– Ksp = [Ag+]2[CrO42–] So we must find the concentrations of each ion and then solve for Ksp. 52
  40. 40. Problem: A saturated solution of silver chromate was to contain 0.022 g/L of Ag2CrO4. Find Ksp Eq. Expression: Ag2CrO4 (s) ⇔ 2Ag+ + CrO42– Ksp = [Ag+]2[CrO42–] Ag+: 0.022 g Ag2CrO4 L 2 Ag + mol 332g 1 Ag 2CrO 4 2 1 CrO 40.022g Ag2CrO4 mol CrO4 : L 332 g 1 AgCrO 4 –2 [ K sp = 1.33 x 10 mol Ag + = 1.33 x 10–4 L = 6.63 x 10–5 2 mol CrO 4L ] [6.63 x 10 ] = 1.16 x 10 -4 2 -5 -12 53
  41. 41. Problems working from Ksp values. Given: Ksp for MgF2 is 6.4 x 10–9 @ 25 oC Find: solubility in mol/L and in g/L Ksp = [Mg2+][F–]2 MgF2(s) ⇔ Mg2+ + 2F– I. C. E. N/A N/A N/A 0 +x +x 0 +2x +2x Ksp= [x][2x]2 = 4x3 6.4 x 10–9 = 4x3 [ ] 6.4 x 10 -9 x=3 = 1.2 x 10-3 = Mg 2+ = [ MgF2 ] 4 now for g/L: 1.2 x 10-3 mol MgF2 L 62.3 g mol = 7.3 x 10–2 g MgF2 L 54
  42. 42. The common ion effect: A salt is generally less soluble in a solution containing an ion which is the same as one of the constituent ions of that salt. This is known as the COMMON ION EFFECT. We can see that this must necessarily occur if we apply Le Chatelier's Principle to an equilibrium such as Adding an excess Cl- (or Na+) to a saturated solution of NaCl imposes a stress on the equilibrium, which will adjust in order to oppose the stress. A shift to the left will use up Na+ or Cl- to form solid NaCl. Le Chatelier's Principle "If a system in equilibrium is subjected to a stress the equilibrium will shift in the direction which tends to relieve that stress."
  43. 43. The degree of ionization of an electrolyte is suppressed by the addition of a strong electrolyte containing common ion. This effect is known common ion effect. The phenomenon of lowering the degree of ionization of a weak electrolyte by adding a solution of a strong electrolyte having a common ion is called common ion effect. APPLICATION OF COMMON ION EFFECT Knowledge of common ion effect is very useful in analytical chemistry. It is frequently applied in qualitative analysis. An electrolyte is precipitated only when the concentration of its ions exceeds the solubility product (KSP).. PRECIPITATION OF THE CATIONS OF GROUP II Sulphides of basic radicals of groups II are precipitated by passing H2S gas through the acidified solution by HCl. Ionization of H2S: H2S ⇋ 2H+ + S-2 Here HCl provides common ion H+ which shifts the above equilibrium to the left as given by Le-Chatelier's principle. HCl ⇋ H+ + ClAddition of HCl suppresses the ionization of H2S and lowers the concentration of S-2 ions, just enough to exceeds the KSP of II group sulphides. In this way only cations of gr. II are precipitated as CuS, PbS, CdS etc. but precipitation of the sulphides of gr. IV is prevented because they have high KSP values as compared to the sulphides of gr. II.
  44. 44. PRECIPITATION OF THE CATIONS OF GROUP IV (Co,Ni, Zn) Cations of groups IV are precipitated as sulphides by passing H2S gas through the solution in the presence of NH4OH. Ionization of NH4OH: NH4OH ⇋ NH4+ + OH- In this analysis NH4OH provides OH- ions which combines with H+ ions of H2S to form H2O. H2S ⇋ 2H+ + S-2 : H+ + OH- ⇋ H2 O Removal of H+ ions from product side shifts the equilibrium to right and the concentration of S-2 increases which is enough to exceed the KSP of the sulphides of group IV. In this way CoS, NiS or ZnS can easily be precipitated. PRECIPITATION OF THE CATIONS OF GROUP III (Al,Fe,Cr) Cations of groups III are precipitated as hydroxides by passing NH4OH in the presence of NH4Cl. Here, NH4Cl provides common ion NH4+ which suppresses the ionization of NH4OH. NH4OH ⇋ NH4+ + OH- : NH4Cl ⇋ NH4+ + Cl- Common ion NH4+ shifts the equilibrium to left side and the concentration of OH- ions decreases. Under these circumstances, the KIP of the hydroxides of Al, Fe and Cr is only exceeded and they are precipitated as Al (OH)3, Fe (OH)3 and Cr (OH)3 but the
  45. 45. COMMON ION EFFECT Consider a simple sparingly soluble precipitate like (BA) e.g. AgCl (1:1 electrolyte) B+ or A- is a common ion with ppt BA. In a soln. Containing ion as B+ as its soluble salt in C moles/L BA ⇋ B+ + As = solubility S = [B+] [A- ] = s x s = s2 In terms of common ion B+ , the solubilty of BA changes to and becomes s′ ∴ [B+] = s′ + C [A- ] = s′ ∴ S = (s′ + C )(s′) = s′ 2 + s′C ∴ s′ < s Hence, solubilty of a ppt or a sparingly soluble electrolyte decreases due to common ion effect. This is why during pptn from a soln the reagent is added in a little excess of the theoretical value
  46. 46. The common ion effect “Le Chatelier” Why is AgCl less soluble in sea water than in fresh water? AgCl(s) ⇔ Ag+ + Cl– Seawater contains NaCl 60
  47. 47. Problem: The solubility of AgCl in pure water is 1.3 x 10–5 M. What is its solubility in seawater where the [Cl–] = 0.55 M? (Ksp of AgCl = 1.8 x 10–10) I. C. E. AgCl(s) ⇔ Ag+ + Cl– N/A 0 0.55 N/A +x +x N/A +x 0.55 + x Ksp= [x][0.55 + x] Ksp= [Ag+][Cl–] try dropping this x Ksp = 0.55x 1.8 x 10–10 = 0.55x x = 3.3 x 10–10 = [Ag+]=[AgCl] “AgCl is much less soluble in seawater” 61
  48. 48. more Common ion effect: a. What is the solubility of CaF2 in 0.010 M Ca(NO3)2? Ksp(CaF2) = 3.9 x 10–11 CaF2(s) ⇔ Ca2+ + 2F– [Ca2+] [F–] I. 0.010 0 C. +x +2x 2x E. 0.010 + x Ksp= [0.010 + x][2x]2 Ksp=[Ca2+][F-]2 ≅ [0.010][2x]2 = 0.010(4x2) 3.9 x 10–11 = 0.010(4x2) x = 3.1 x 10–5 M Ca2+ from CaF2 so = M of CaF2 Now YOU determine the solubility of CaF2 in 0.010 M NaF. 62
  49. 49. Answer: 3.9 x 10–7 M Ca2+ CaF2(s) ⇔ Ca2+ + 2F– 0 0.010 +x 2x x 0.010 + 2x Ksp = [x][0.010 + 2x]2 ≅ x(0.010)2 3.9 x 10-11 =x(0.010)2 x = 3.9 x 10-7 What does x tell us 63
  50. 50. Effect of Temperature on solubility Raising the temp of the soln. usually increases the solubility of a ppt. or of the sparingly soluble substance. Thus a ppt is usually more soluble in hot soln than in the cold. But difft. ppts show different trend in respect of increase in solubility with temp. ppt 10ºC 50ºC 100ºC AgCl 0.89 5.23 21.1 BaSO4 2.2 3.36 3.9 Presence of a little excess of the precipitant (peptising reagent), however, diminishes the solubility of the ppt to such an extent that the temp. effect on solubility often becomes negligible
  51. 51. Factors affecting solubility product Analytical Chemistry, ACSS-501 Class –IV Dr. Pabitra Kumar Mani Door meten tot weten “Through measurement to knowledge.” Heike Kamerlingh Onnes , NL “gentleman of absolute zero”
  52. 52. Effect of solvent: As a rule solubility of inorganic precipitates in water is decreased by addition of organic solvents like methanol, ethanol, acetones etc. CaSO4 ---- a so much insoluble in 50% ethanol that quantitative separation of Ca or ppt is possible in their way. comps like LiCl----- is soluble in alcohols whereas NaCl or KCl is insoluble in alcohol. LiCl---- Li low ionic size and covalent character Thus, by using Amyl alcohol(C5-alcohol), Li can be separated from Na+ or K+ in the form of their chlorides. In the organic solvents LiCl is completely soluble. Solubility (at 25ºC , g/100g) Water KClO4 NaClO4 Methanol Ethanol N-Butanol Ethyl acetate 2.02 0.105 0.012 0.0045 0.0015 Soluble in all the solvents Separation of KClO4 from NaClO4 is possible by using ethyl acetate or n-butanol
  53. 53. Effect of Electrolytes: Presence of other ions added in the form of electrolytes or already present in the soln. as foreign ions increase the ionic strength of the medium. Ionic strength, μ = ½ ∑ Ci Zi2 Ci = concn of ion, Zi= charge or valence of ion As ionic strength increases activity coeffcicent of an ion including that of the ion derived from ppt itself decreases μ = -A logfi A = constant , fi = activity coefficient ai = ci.fi a= activity of any ion, ci = concn of ion, From thermodynamics it is found that the solubility pdt as defined in terms of molar concn is not a trur constant. Instead the activity product is true constant S = aB+ . aA a - = (cB+ x cA-) (fB+ x fA-) = Sc.(f±)2 Thus, in-order to keep the activity pdt const at a particular temp. as the activity coeffcient value decreases, the molar concn must increase i.e., the substance(ppt) becomes more soluble. Hence, solubility of a precipitae increases by adding large amount of electrolytes in the solution. Sa = activity pdt., Sc = solubilty pdt in terms of molar concn , f± = mean activity coeffcient
  54. 54. Influence of complex ion formation Many ppts show a tendency to form a complex ion with an excess of their own ion resulting in the increase in solublty of the precipitate Ag+ KCN AgCN KCN [Ag(CN)2]Complex ion formation AgCl HCl H[AgCl2] S AgCN = 4 x 10-12 , when CN- is added in equivalent amount to Ag+ in soln [CN- ] = √4 x 10-12 = 2 x10-6 g-ion/ L Complex formation rreaction. AgCN + CN- ⇋ [Ag(CN)2]- K⇋ [AgCN] + [CN-] [Ag(CN) ]2 long as. AgCN As ∴ [CN-] [Ag(CN) ]2 = K= dissociation const /instability constant remains in solid phase, [AgCN]= const K′ ...........10-10 or, [CN-] = 10-10 [Ag(CN)2]- eq.....1 Now, suppose excess of [CN-] is added corresponding to 0.01 molar soln (10-2) [CN-] + [Ag(CN)2]- = 10-2 eq........2 free complexed ion
  55. 55. [Ag(CN)2]- {10-10 +1 } = 10-2 or, [Ag(CN)2]- ≈ 10-2 Hence, practically all the [CN-] added is transformed into the complex ion and it can be said that KCN dissolves an equiv amount of AgCN . Complex formation as above is quite common during pptn analysis particularly when the resulting complex ion has a very high stability. Hg+2 + I- → HgI2 HgI2 + 2I- → [HgI4]-2 ( insoluble) (highly stable) As a rule, solubility for many precipitates at first decreases due to common ion effect and a min. solubilty may be found at a certain excess of the common ion and the solubility then increases with further addition of the reagent providing the common ion but forming the complex ion at same time The complex ions may also be formed with constituents other than the common – ion present in the soln. Solubilty of AgOH or Cu(OH) 2 in ammonia solution. AgOH + 2NH3 ⇋ [Ag(NH3)2]+ + OHCu(OH)2 + 4NH3 ⇋ [Cu(NH3)4]+2 + 2OH-
  56. 56. Influence of [H+] or pH If the ppt. is salt of a weak acid, strong acids will have a solvent effect on it CaF2 + 2HCl → Ca+2 + 2Cl- + 2HF Salt strong acid weak acid Thus all carbonates dissolved in mineral acid. Phosphates, Aresnates, Sulphides and Cyanide except AgCN which is in effect the salt of relatively strong acid H[Ag(CN)2] Ag+ Ag[Ag(CN)2] ≡ 2AgCN HNO3 Insoluble Argento cyanic acid (strong acid) The solvent effect of acid is of importance even incase of slightly soluble sulfate. Among the hydroxides and hydrous oxides precipitates pH ha s agreat role on their formation. Fractional precipitation of hydrous oxide can be carried out by controlling the pH of the medium. pH Metal ions Pptd as Hydrous oxides 3 4 5 6 7 8 9 11 Fe+3 Th+4 Al+3 Be+2 Fe+2 Co+2 Ag+ Mg+2 U+4 Cu+2 Ni+2 Mn+2 Zn+2 Cr+3 U+6 Cd+2 Hg+2 Zr+4 Sn+2 H+R- + M+n → MRn + H+ pH has effect in all cases of pptn pptant ppt where reagent is an weak acid
  57. 57. I- I-
  58. 58. Co-precipitation and Post precipitation Co-precipitation means incorporation of foreign ions into a ppts having a common ion or an ion of identical size or charge with any ion of the ppt. Co-pptn may take place due to mainly 3 reasons (i)Due to adsorption at the surface of the particles exposed to the soln. having ionic lattice. Colloidal particles like Silver halide of hydrated oxides are highly contaminated in this way. (ii)Due to adsorption at the interior or occlusion of foreign ions during growth of primary particles. In this case ions may or may not fit into the lattice and even solvent molecules are often held in the ppt in this way. (iii)Due to formation of chemical comp. with the unsatisfied valency of any ions of the lattice. e.g. Ba Cl- When SO4-2 is pptd with BaCl2 soln Ba + SO4 SO4 Ba+ Cl- Ba + Ba+ K+ When Ba+2 is pptd by K2SO4 soln SO4 Ba+ K+ lattice
  59. 59. The degree of contamination of a ppt. by Co-pptn depends on specific adsorption, specific surface, decreasing solubility of the contaminant, increasing concn of the contaminant, higher valence, and smaller size of the foreign ion, lower temp. and greater no. of particles/ unit mass of the ppts
  60. 60. Co-pptn can be minimised by the following techniques i) By pptn from hot soln so that initially a few particles are formed and relative supersaturation is increased ii) For ppts having appreciable solubility, however the soln must be cooled and the relative super saturation is increased by the control of pH or any other device iii) By pptn from dilute soln to decrease the concn. of foreign ions iv) By adding reagent as a solution , slowly with stirring in order to reduce the local concn. (Q) v) By prior removal of the foreign ion as ppt by rn with a difft. Reagent or by conversion of the foreign ion to a difft. valency state or sometimes to a complex spp. vi) E.g. Fe+2 during BaSO4 pptn N N [Fe(O-pn)3] Fe(OH)3. xH2O Phenanthroline i) By allowing the ppt. to satnd (digestion and aging) for perfect growth of crystals ii) By thorough and proper washing of the ppt. iii) By double pptn. +2
  61. 61. Ist ppt. Fe+3, Ca+2 Fe(OH)3.xH2O (Ca+2) HCl + Ca+2 in soln Included as co-ppt. Fe+3, Ca+2 (less) OHFe(OH)3.xH2O + Soln (Ca+2) 2nd ppt.
  62. 62. Three examples of impurities that may form during precipitation. The cubic frame represents the precipitate and the blue marks are impurities: (a) inclusions, (b) occlusions, and (c) surface adsorbates. Inclusions are randomly distributed throughout the precipitate. Occlusions are localized within the interior of the precipitate and surface adsorbates are localized on the precipitate’s exterior. For ease of viewing, in (c) adsorption is shown on only one surface.
  63. 63. Post precipitation Post –pptn involves incorporation of impurities when the supernatant liquid is super saturated w.r.t some phase which crystalline slowly. Eg. thus in case of metal sulfide pptn from 0.1N acid medium (HCl), ZnS is not normally precipitated A soln saturated with H2S, however, remains supersaturated with ZnS, which slowly separates as a ppt which is accelerted in presence of the fine particles of HgS or CuS. This is known as post precipitation of ZnS  Similar situation arises when Mg remains in a solution containing large excess of oxalate after Ca-Oxalate pptn. Mg2C2O4 precipitate slowly from the supernatant liquid by post-pptn
  64. 64. Post precipitation differs from Co-pptn in the following respects Contamination increases with aging or digestion (in case of post pptn), but the reverse is true in case of Co-pptn Contamination increases with faster agitation or stirring of the supernatant liquid in case of post pptn., but the reverse is true for co-pptn. Degree of contamination or its magnitude is usually much greater due to post pptn in comparison to Co-pptn. To avoid post –pptn, a shorter period of digestion particularly under warm condition is to be carried out and the ppt. should be filtered as quickly as possible so that the ppt does not remain in contact with a mother–liquor for a long time
  65. 65. Washing of the precipitate The precipitate after filtration must be washed thoroughly with a proper liquid.  The washed liquid should dissolve out the foreign substances only and it should have no peptising or disperse action on the ppt.  It should not form any volatile or insoluble pdt. with the ppt and should be itself easily removed during drying.  It should contain no substance which interferes for the determination of any constituents in the filtrate.  The washed liquid should not interact chemically with the ppt during drying or ignition.
  66. 66. Nature of the wash liquid To prevent peptisation the wash liquid must contain an electrolyte. Usually dilute ammonia, dil. Acids and some salts particularly ammonium salts are used as electrolytes. The wash liquid often contains a common ion with the ppt to decrease the solubility during washing. A few drops of reagent solution is usually added to washed liquid for this purpose During AgCl pptn wahing with dil HNO3 For Hydrous oxide pptn…. dil NH3 containing small amount of NH4NO3 is used For washing of PbSO4 ppt…… a saturated soln of PbSO4 is used as first wash liquid, the ppt was then washed with 40% alcohol The washing must be carried out by using a very small amount of wash liquid at a time and the number of washing must be as large as possible.
  67. 67. The common ion effect is used to decrease the solubility. The common ion effect is used to decrease the solubility. Sulfate concentration is the amount in equilibrium and is equal to the Sulfate concentration is the amount in equilibrium and is equal to the BaSO4 solubility. BaSO solubility. 4 In absence of excess barium ion, solubility is 10-5 M. In absence of excess barium ion, solubility is 10-5 M. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 10.3. Predicted effect of excess barium ion on solubility of BaSO 4.
  68. 68. Ksp = Ksp0/fAg+fSO42Ksp = Ksp0/fAg+fSO42Solubility increases with increasing ionic strength as activity coefficients decrease. Solubility increases with increasing ionic strength as activity coefficients decrease. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Predicted effect of increased ionic strength on solubility of BaSO4. Solubility at zero ionic strength is 1.0 x 10-5 M.
  69. 69. Gravimetric Calculation Analytical Chemistry, ACSS-501 Class –V Dr. Pabitra Kumar Mani Door meten tot weten “Through measurement to knowledge.” Heike Kamerlingh Onnes “gentleman of absolute zero”
  70. 70. Find moles of precipitate. Find moles of sought substance(Analyte) Find weight of sought substance (Analyte). mol ratio=mol analyte/ mol ppt Calculate the percentage of the sample that is the sought substance
  71. 71. Gravimetric Calculations   The point here is to find the weight of analyte from the weight of precipitate. We can use the concepts discussed previously in stoichiometric  calculations but let us learn something else. Assume Cl 2 is to be precipitated as  AgCl, then we can write a stoichiometric factor reading as follows: one mole of Cl 2  gives 2 moles of AgCl. This is in fact what is called the gravimetric factor (GF)  where we can substitute for the number of moles by grams to get:   GF for Cl2 = 1 mol Cl2/2 mol AgCl = FW Cl2/2 FW AgCl = x g analyte/y g precipitate Weight of substance sought = weight of precipitate x GF One can also consider the problem by looking at the number of mmoles of analyte in  terms of the mmoles of the precipitate where for the precipitation of Cl 2 as AgCl, we  can write (analyte) Cl2 = 2 AgCl (ppt) mmol Cl2 = 1/2 mmol AgCl (mg Cl2/FW Cl2) = 1/2 (mg AgCl/FW AgCl)
  72. 72. Q1.  Calculate the grams analyte to mg precipitate for the following:            P (at wt =30.97) in Ag3PO4 (FW = 711.22), Bi2S3 (FW 514.15) in BaSO4  (FW = 233.40)   Solution: 1   P = Ag3PO4 mmol p = mmol Ag3PO4 Mg P/30.97 = mg Ag3PO4/711.22 Mg P/mg Ag3PO4 = 30.97/711.22 = 0.04354 (GF)   Bi2S3 = 3 BaSO4 mmol Bi2S3 = 1/3 mmol BaSO4 mg Bi2S3/FW Bi2S3 = 1/3 mg BaSO4/FW BaSO4 mg Bi2S3/514.15 = 1/3 mg BaSO4/233.40 mg Bi2S3/ BaSO4 = 1/3 (514.15/ 233.40) =  0.73429 (GF)
  73. 73. (NH4)2PO4.12 MoO3  (FW  =  1876.5).  Find  percentage  P  (at  wt  =  30.97)  and  percentage P2O5 (FW = 141.95) in the sample.   Solution: 2.  First we set the mol relationship between analyte and precipitate  P = (NH4)2PO4.12 MoO3  mmol P = mmol (NH4)2PO4.12 MoO3   mg P/at wt P = mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3 mg P = at wt P x (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3)   mg P = 30.97 (1.1682x103 / 1876.5) = 19.280 mg % P = (19.280/271.1) x 100 = 7.111%     The same procedure is applied for finding the percentage of P 2O5   P2O5 = 2 (NH4)2PO4.12 MoO3 mmol P2O5 = 1/2 (NH4)2PO4.12 MoO3 mg P2O5/FW P2O5 = 1/2 (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3) mg P2O5 = 1/2 x FW P2O5 x (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3) mg P2O5 = 1/2 x 141.95 (1.1682x103/1876.5) = 44.185 mg
  74. 74. Q.3. Manganese  in  a  1.52 g sample was  precipitated  as  Mn3O4  (FW  =  228.8)  weighing 0.126 g. Find percentage Mn2O3 (FW = 157.9) and Mn (at wt = 54.94) in  the sample.   Solution. 3 (analyte)  3 Mn2O3 = 2 Mn3O4 (ppt) mmol Mn2O3 = 3/2 mmol Mn3O4 mg Mn2O3 / FW Mn2O3 = 3/2 (mg Mn3O4/FW Mn3O4) mg Mn2O3 = 3/2 FW Mn2O3 (mg Mn3O4/FW Mn3O4) mg Mn2O3 = 3/2 x 157.9 ( 126/228.8) = 130 mg % Mn2O3 = (130/1520) x 100 = 8.58%   The same idea is applied for the determination of Mn in the sample 3 Mn = Mn3O4 mmol Mn = 3 mmol Mn3O4 mg Mn / at wt Mn = 3 (mg Mn3O4/FW Mn3O4) mg Mn = 3 at wt Mn (mg Mn3O4/FW Mn3O4) mg Mn2O3 = 3 x 54.94 ( 126/228.8) = 90.8 mg   % Mn = (90.8/1520) x 100 = 5.97%
  75. 75.   Q.4. A  mixture  containing  only  FeCl3  (FW  =  162.2)  and  AlCl3  (FW  =  133.34)  weighs 5.95 g. The chlorides are converted to hydroxides and ignited to Fe2O3 (FW  = 159.7) and Al2O3 (FW = 101.96). The oxide mixture weighs 2.26 g. Calculate the  percentage Fe (at wt = 55.85) and Al (at wt = 26.98) in the sample.   Solution 4. Fe = FeCl3 1 mol Fe = 1 mol FeCl3 g Fe/at wt Fe = g FeCl3/ FW FeCl3 Rearrangement gives g FeCl3 = g Fe (FW FeCl3/at wt Fe) In the same manner g AlCl3 = g Al ( FW AlCl3/at wt Al) g FeCl3 + g AlCl3 = 5.95 g Fe (FW FeCl3/at wt Fe) + g Al ( FW AlCl3/at wt Al) = 5.95                    assume         g Fe = x, g Al = y then: x (FW FeCl3/at wt Fe) + y ( FW AlCl3/at wt Al) = 5.95                   x (162.2/55.85) + y (133.34/26.98) = 5.95 2.90 x + 4.94 y = 5.95   ................ (1)
  76. 76. Soln 4:  The same treatment with the oxides gives 2 Fe =Fe2O3 mol Fe = 2 mol Fe2O3 g Fe/at wt Fe = 2 (g Fe2O3/FW Fe2O3) g Fe2O3 = 1/2 g Fe (FW Fe2O3/at wt Fe) In the same manner g Al2O3 = 1/2 g Al (FW Al2O3/at wt Al)               g Fe2O3 + g Al2O3 = 2.26 1/2g Fe(FW Fe2O3/at wt Fe) + 1/2 g Al (FW Al2O3/at wt Al) = 2.26 1/2 x (159.7/55.85) + 1/2 y (101.96/26.98) = 2.26 1.43 x + 1.89 y = 2.26 ...............2) from (1) and (2) we get  x = 1.07 y = 0.58 % Fe = (1.07/5.95) x 100 = 18.0% % Al = (0.58/5.95) x 100 = 9.8%  
  77. 77. Q :  A 0.4960 g sample of a CaCO3 is dissolved in an acidic solution.  The calcium is precipitated as CaC2O4.H2O and the dry precipitate is  found to weigh 0.6186 g. What is the percentage of CaO in the sample? Unknown: % CaO in the 0.4960 g sample of CaCO3 Knowns: wt of CaCO3 sample = 0.4960 g, wt of CaC2O4.H2O = 0.6186 g Concepts: gravimetric factor, moles, mole ratio, FW wt of CaO = 0.2374 g
  78. 78. x= 1.07g, y=0.58 g %Fe = (1.07/5.95) x100     =18.0% %Al = (0.58/5.95) x100  =9.8.0%
  79. 79. Solubility in Presence of Diverse Ions Find  the  solubility  of  AgCl (ksp = 1.0 x 10-10) in 0.1 M NaNO3.  The  activity  coefficients for silver and chloride are 0.75 and 0.76, respectively.     Solution AgCl(s) = Ag+ + ClWe  can  no  longer  use  the  thermodynamic  equilibrium  constant  (i.e  in  absence  of  diverse  ions ) and we have to consider the concentration equilibrium constant or use activities instead  of concentration : Ksp = aAg+ aClKsp = [Ag+] fAg+ [Cl-] fCl1.0x10-10 = s x 0.75 x s x 0.76 s = 1.3x10-5 M We  have  calculated  the  solubility  of  AgCl in pure water to  be  1.0x10-5 M,  if  we  compare the this value to that obtained in presence of diverse ions we see % increase in solubility = {(1.3x10-5 – 1.0x10-5)/1.0x10-5}x 100 = 30% Therefore, once again we have an evidence for an increase in dissociation or a shift of equilibrium to right in presence of diverse ions .
  80. 80. Q. A certain barium halide exists as the hydrated salt BaX2.2H2O, where X is the halogen. The barium content of the salt can be determined by gravimetric methods. A sample of the halide (0.2650 g) was dissolved in water (200 cm3) and excess sulfuric acid added. The mixture was then heated and held at boiling for 45 minutes. The precipitate (barium sulfate) was filtered off, washed and dried. Mass of precipitate obtained = 0.2533 g. Determine the identity of X.
  81. 81. Answer: The precipitate is barium sulfate. The first stage is to determine the number of moles of barium sulfate produced, this will, in turn give us the number of moles of barium in the original sample. Relative Molecular Mass(RMM) of barium sulfate = 137.34 (Ba) + 32.06 (S) + (4 x 16.00) (4 x O) = 233.40 Number of moles = mass / RMM = 0.2533 / 233.40 = 1.09 x 10 -3 This is the number of moles of barium present in the precipitate and, therefore, the number of moles of barium in the original sample. Given the formula of the halide, (i.e. it contains one barium per formula unit), this must also be the number of moles of the halide. From this information we can deduce the relative molecular mass of the original halide salt: RMM = mass / number of moles = 0.2650 / 1.09 x 10-3 = 244.18 The relative atomic mass (RAM) of 2 X will be given by the RMM of the whole salt - that of the remaining components; So RAM of 2 X = 244.18 - 173.37 = 70.81 2 X = 70.81, so X = 35.41. The RAM of chlorine is 35.45 which is in good agreement with the result obtained and hence the halide salt is hydrated barium chloride and X = Chlorine
  82. 82. Other Gravimetric Techniques Gravimetric Combustion Analysis Determines the carbon and hydrogen content of organic compounds burned in excess O2. CO2 trap Guard tube prevents H2O and CO2 entering from the air H2O trap 103
  83. 83. Combustion Analysis, or Elemental Analysis C, H, N, S  → CO (g) + H O(g) + N (g) + SO (g) + SO (g)  1050   C 2 2 2 2 3 Determines the C, H, N, and S content in a single operation by using GC and thermal conductivity measurements 104
  84. 84. Measuring Principle The sample under test is weighed in using a tin capsule. The required amount is 2 to  3 mg of organic material and can hardly exceed 10 mg, if inorganic matter with little  carbon content is investigated. After folding the capsule (looking rather like wrapped  tin foil) the sample is placed in the autosampler. The tin capsule enclosing the sample falls into the reactor chamber where  excess oxygen is introduced before. At about 990 °C the material is "mineralized".  Formation of CO is probable at this temperature even under these conditions of excess  O2 . The complete oxidation is reached at a tungsten trioxide (WO3) catalyst  which is passed by the gaseous reaction products. The resulting mixture should thus  consist of CO2, H2O und NOx. But also some excess O2 passes the catalyst. The product gas mixture flows through a silica tube packed with copper granules. In this zone held at about 500 °C remaining oxygen is bound and  nitric/nitrous oxides are reduced. The leaving gas stream includes the analytically  important species CO2, H2O und N2. Eventually included SO2 or hydrohalogenides are absorbed at appropiate traps. High purity helium (Quality 5.0) is used as carrier gas. Finally the gas  mixture is brought to a defined pressure/volume state and is passed to a gas chromatographic system. Separation of the species is done by so called zone chromatography. In this technique a staircase type signal is registrated. Step height
  85. 85. Working Range Mineralization and detection covers every species of the analyte elements  in the sample. Beside purely organic samples various metal organic  compunds and even inorganic samples as carbides and nitrides have  successfully been characterized. The detection limit for carbon and nitrogen at sample amounts of 2 to 3 mg was found to be at about 0,05 w-% (500 ppm) in what case the uncertainty  stays at about 0,02 w-%. Problems and Interferences The weighing of oily or fluid substances is impossible using the thin walled  tin capsules. For this purpose alumina pans with a lid are available. These  pans are tightly closed by cold welding to prevent loss of sample by  spillage and evaporation. As the blank value for nitrogen is dramatically  increased be the enclosed volume of air the determination limit for N in  liquid samples is increased to 0,1 to 0,2 w-%. With highly viscous or even glassy materials elemental analysis is even  impossible with the above method.
  86. 86. It has long been known that phosphorus can interfere in the mineralization  of organic material. In literature the formation of glassy P2O5 xH2O yC has  been described. This effect can be controlled by the addition of vanadium pentoxide (V2O5). Fluorine is mineralized to form HF which reacts at the wall of the silica tubes which form the main part of the reaction zone. The gaseous products, such as SiF4 and relatives, can cause systematic errors which rarely become significant with  respect to the 0,3 w-% tolerance. The mineralization of metal containing samples can also be affected by  interferences. By modification of the method most of these can be compensated  for. Percentage Composition CxHyOz (9.83 mg) + excess O2 x CO2 (23.26 mg) + y/2 H2O (9.52 mg) millimoles of CO2 = 23.26 mg/ 44.01mg/mmol = 0.5285 mmoles of CO2  mmoles of CO2 = mmoles of C in original sample (0.5285 mmoles of C)(12.01mg/mmol C) = 6.35 mg of C in original sample CxHyOz (9.83 mg) + excess O2 x CO2 (23.26 mg) + y/2 H2O (9.52 mg) mmoles of H2O = 9.52 mg/ 18.02 mg/mmol = 0.528 mmoles of H2O  mmoles of H2O = 1/2 mmoles of H in original sample (0.528 mmoles of H)(2)(1.008mg/mmol H) = 1.06 mg of H in original sample
  87. 87. Weight Percentage Composition CxHyOz (9.83 mg) + excess O2 x CO2 (23.26 mg) + y/2 H2O (9.52 mg) %C = 6.35 mg/9.83 mg x 100 = 64.6%  %H = 1.06 mg/9.83 mg x 100 = 10.8%  %O = 100-(64.6 + 10.8) = 24.6% Calculation of Empirical Formula assume for example a 100g sample  64.6% of C: 64.6 g/12.01 g/mol = 5.38 moles of C  10.8% of H: 10.8 g/1.008 g/mol = 10.7 moles of H  24.6% of O: 24.6 g/16.0 g/mol = 1.54 moles of O  Thus: C5.38 H10.7 O1.54  converting to simplest ratio: C5.38/1.54;  H10.7/1.54;  O1.54/1.54  C3.50 H7.00 O1.00 = C7H14O2 What are possible structures?
  88. 88. Other Gravimetric Techniques Thermal Gravimetric Analysis (TGA) • Precisely monitoring weight loss of a sample in a given atmosphere as a function of temperature and/or time • Atmospheres: N2, O2, air, or He • Temperature: ambient to 1000 °C • Records the first derivative of the mass loss • Evaluate the thermal decomposition and stability of materials – Polymers, resins, rubbers, explosives • Information on bulk composition of compounds – Thermal oxidation, heat resistance – Residual water or solvents – Compositional analysis – Ash content in a sample – Quantity of inorganic filler in a polymer 109
  89. 89. Other Gravimetric Techniques TGA cont. • The percent weight loss of a test sample is recorded while the sample is being heated at a uniform rate in an appropriate environment. • The loss in weight over specific temperature ranges provides an indication of the composition of the sample, including volatiles and inert filler, as well as indications of thermal stability. • The gas environment is pre-selected for either a thermal decomposition (inert – He or N2 gas), an oxidative decomposition (air or O2), or a combination therein. Chemistry 215 Copyright D  Sharma 110
  90. 90. Thermogravimetry One method for determining the products of a thermal decomposition is to  monitor the sample’s mass as a function of temperature, a process called  thermogravimetry. 
  91. 91. From 100–250 oC the sample loses 17.61 mg – 15.44 mg, or 2.17 mg,  which is of the sample’s original mass. In terms of CaC2O4•H2O, this corresponds  to a loss of The product’s molar mass and the temperature range for the  decomposition, suggest that this is a loss of H2O(g), leaving a residue of CaC2O4. The loss of 3.38 mg from 350–550 oC is a 19.2% decrease  in the sample’s original mass, or a loss of which is consistent with the loss of CO(g) and a residue of CaCO3. Finally, the loss of 5.30 mg from 600-800 oC is a 30.1% decrease in  the sample’s original mass, or a loss of This loss in molar mass is consistent with the release of CO2(g), leaving a final residue of CaO.
  92. 92. TGA cont. TGA tests may be run in a heating mode at some controlled heating rate, or isothermally. Typical weight loss profiles are analyzed for the amount or percent of weight loss at any given temperature, the amount or percent of noncombusted residue at some final temperature, and the temperatures of various sample degradation processes.
  93. 93. TGA cont. H2O, C7H3O6, C6H4, CO2 are consecutively lost. Chemistry 215 Copyright D  Sharma 114

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