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Chapter 27

Chapter 27. Gravimetric and Combustion Analysis. Representative Gravimetric Analyses. Precipitation. AgNO 3 + NaCl -----> AgCl + NaNO 3. Precipitation. AgNO 3 + NaCl -----> ? AgNO 3 + NaCl -----> AgCl + NaNO 3 total ionic equation

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Chapter 27

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  1. Chapter 27 Gravimetric and Combustion Analysis Dr. S. M. Condren

  2. Representative Gravimetric Analyses Dr. S. M. Condren

  3. Dr. S. M. Condren

  4. Dr. S. M. Condren

  5. Precipitation AgNO3 + NaCl -----> AgCl + NaNO3 Dr. S. M. Condren

  6. Precipitation AgNO3 + NaCl -----> ? AgNO3 + NaCl -----> AgCl + NaNO3 total ionic equation Ag+ + NO3- + Na+ + Cl- -----> AgCl + Na+ + NO3- net ionic equation Ag+ + Cl- -----> AgCl Dr. S. M. Condren

  7. Precipitation. WHY? Solubility Rules 1. The common salts of the alkali metals (Group IA) and the ammonium ion (NH4+) are soluble. 2. Salts of nitrate (NO3-), chlorate (ClO3-), perchlorate (ClO4-), and acetate (CH3COO-) anions are soluble. 3. All chlorides, bromides, and iodides are soluble, except those of Ag+, Pb+2, and Hg2+2 (the form that Hg(I) exists in water). Dr. S. M. Condren

  8. Precipitation. WHY? Solubility Rules 4. Sulfates (SO4-2) are soluble except those of Ba+2, Pb+2, Hg+2, and Hg2+2. Dr. S. M. Condren

  9. Precipitation. WHY? Solubility Rules 5. Most metal hydroxides are insoluble. The exceptions are the hydroxides of the alkali metals and the heavier alkaline earth metals (Ca, Sr, Ba). 6. All carbonates (CO3-2) and phosphates (PO4-3) are insoluble, except those of the Group IA metals and NH4+ ions. Dr. S. M. Condren

  10. Crystal Growth nucleation particle growth supersaturated Dr. S. M. Condren

  11. Crystal Growth relative supersaturation = (Q - S)/S where Q => actual concentration S => equilibrium concentration Dr. S. M. Condren

  12. Crystal Growth techniques to promote crystal growth 1. Raising the temperature to increase S and thereby decrease relative supersaturation. 2. Adding precipitant slowly with vigorous mixing, to avoid a local, highly supersaturated condition where the stream of precipitant first enters the analyte. 3. Keeping the volume of solution large so that the concentration of analyte and precipitant are low. Dr. S. M. Condren

  13. Dr. S. M. Condren

  14. Precipitation in the Presence of Electrolyte coagulate => “to change or be changed from a liquid into a thickened mass” Dr. S. M. Condren

  15. Precipitation in the Presence of Electrolyte Dr. S. M. Condren

  16. Precipitation in the Presence of Electrolyte adsorption => attached to surface absorption => penetration beyond surface Dr. S. M. Condren

  17. Digestion process of keeping mixture warm while the size of the crystals increase Dr. S. M. Condren

  18. Purity absorbed impurities inclusions - impurity ions which randomly occupy sites in the crystal lattice occulusions - pockets of impurity trapped inside growing crystal coprecipitaion Dr. S. M. Condren

  19. Purity gathering agent - precipitating agent used to collect trace component (process - gathering) masking agent post-precipitation peptization Dr. S. M. Condren

  20. Product Composition • product must be of known composition • hygroscopic substance • ignition • thermogravimetric Dr. S. M. Condren

  21. Thermogravimetric curve for calcium salicylate Dr. S. M. Condren

  22. Example: What weight of Fe2O3 can be obtained from 1.63 g of Fe3O4? 2 Fe3O4 + [O] -----> 3 Fe2O3 (1.63 g Fe2O3) #g Fe2O3 = -------------------- Dr. S. M. Condren

  23. Example: What weight of Fe2O3 can be obtained from 1.63 g of Fe3O4? 2 Fe3O4 + [O] -----> 3 Fe2O3 (1.63 g Fe2O3)(1 mole Fe3O4) #g Fe2O3 = -------------------------------------- (231.54 g Fe3O4) Dr. S. M. Condren

  24. Example: What weight of Fe2O3 can be obtained from 1.63 g of Fe3O4? 2 Fe3O4 + [O] -----> 3 Fe2O3 (1.63)(1 mole Fe3O4)(3 mole Fe2O3) #g Fe2O3 = ------------------------------------------ (231.54)(2 mole Fe3O4) Dr. S. M. Condren

  25. Example: What weight of Fe2O3 can be obtained from 1.63 g of Fe3O4? 2 Fe3O4 + [O] -----> 3 Fe2O3 (1.63)(1)(3moleFe2O3)(159.69gFe2O3) #g Fe2O3 = ------------------------------------------------ (231.54)(2) (1 mol Fe2O3) Dr. S. M. Condren

  26. Example: What weight of Fe2O3 can be obtained from 1.63 g of Fe3O4? 2 Fe3O4 + [O] -----> 3 Fe2O3 (1.63)(1)(3)(159.69gFe2O3) #g Fe2O3 = ----------------------------------- = 1.69 g (231.54)(2)(1) Dr. S. M. Condren

  27. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 Dr. S. M. Condren

  28. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 if use all of the 2.85 g Pb(NO3)2 Dr. S. M. Condren

  29. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 if use all of the 2.85 g Pb(NO3)2 (2.85 g Pb(NO3)2 #g PbI2 = ------------------------ Dr. S. M. Condren

  30. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 if use all of the 2.85 g Pb(NO3)2 (2.85 g Pb(NO3)2)(1 mol Pb(NO3)2 #g PbI2 = -------------------------------------------- (331 g Pb(NO3)2) Dr. S. M. Condren

  31. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 if use all of the 2.85 g Pb(NO3)2 (2.85 g Pb(NO3)2)(1 mol Pb(NO3)2 #g PbI2 = -------------------------------------------- (331 g Pb(NO3)2) Dr. S. M. Condren

  32. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 if use all of the 2.85 g Pb(NO3)2 (2.85)(1 mol Pb(NO3)2)(1mol PbI2) #g PbI2 = -------------------------------------------- (331) (1 mol Pb(NO3)2) Dr. S. M. Condren

  33. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 if use all of the 2.85 g Pb(NO3)2 (2.85)(1)(1 mol PbI2)(461 g PbI2) #g PbI2 = -------------------------------------------- (331) (1) (1 mol PbI2) = 3.97 g PbI2 if use all of the 225 mL of 0.0550 M KI (225 mL KI)(1 L KI)(0.0550 mol KI)(1 mol PbI2)(461 g PbI2) #g PbI2 = -------------------------------------------------------- (1000 mL KI)(1 L KI)(2 mol KI) (1 mol PbI2) = 2.85 g PbI2 Since 2.85 g is the lesser amount of PbI2, the KI must be the limiiting reactant, as it limits the amount of PbI2 that is produced in the reaction. Dr. S. M. Condren

  34. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 if use all of the 2.85 g Pb(NO3)2 (2.85)(1)(1 mol PbI2)(461 g PbI2) #g PbI2 = -------------------------------------------- (331) (1) (1 mol PbI2) = 3.97 g PbI2 if use all of the 225 mL of 0.0550 M KI (225 mL KI)(1 L KI)(0.0550 mol KI)(1 mol PbI2)(461 g PbI2) #g PbI2 = -------------------------------------------------------- (1000 mL KI)(1 L KI)(2 mol KI) (1 mol PbI2) = 2.85 g PbI2 Since 2.85 g is the lesser amount of PbI2, the KI must be the limiiting reactant, as it limits the amount of PbI2 that is produced in the reaction. Dr. S. M. Condren

  35. EXAMPLE: What mass of PbI2 will precipitate if 2.85 g Pb(NO3)2 is added to 225 mL of 0.0550 M KI(aq)? Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3 if use all of the 2.85 g Pb(NO3)2 (2.85)(1)(1)(461 g PbI2) #g PbI2 = ------------------------------- = 3.97 g (331)(1)(1) if use all of the 225 mL of 0.0550 M KI (225 mL KI)(1 L KI)(0.0550 mol KI)(1 mol PbI2)(461 g PbI2) #g PbI2 = -------------------------------------------------------- (1000 mL KI)(1 L KI)(2 mol KI) (1 mol PbI2) = 2.85 g PbI2 Since 2.85 g is the lesser amount of PbI2, the KI must be the limiiting reactant, as it limits the amount of PbI2 that is produced in the reaction. Dr. S. M. Condren

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