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Theoretical, actual, and percentage yields

Theoretical, actual, and percentage yields. Theoretical yield : The amount of product obtained when all the limiting reagent has reacted. Actual yield : Amount of product actually obtained by experiment from a chemical reaction.

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Theoretical, actual, and percentage yields

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  1. Theoretical, actual, and percentage yields Theoretical yield: The amount of product obtained when all the limiting reagent has reacted. Actual yield: Amount of product actually obtained by experiment from a chemical reaction. Percentage yield: The ratio of the actual yield to the theoretical yield multiplied by 100.

  2. Example: DDT is an insecticide prepared by the following reaction: 2 C6H5Cl + C2HOCl3 C14H9Cl5+ H2O chlorobenzene chloral DDT

  3. Example: DDT is an insecticide prepared by the following reaction: 2 C6H5Cl + C2HOCl3 C14H9Cl5+ H2O chlorobenzene chloral DDT If 450.74 g of chlorobenzene reacts with 899.68 g of chloral, calculate (a) the theoretical yield of DDT in grams, and (b) the percentage yield if 654.13 g of DDT is actually isolated.

  4. Example: DDT is an insecticide prepared by the following reaction: 2 C6H5Cl + C2HOCl3 C14H9Cl5+ H2O chlorobenzene chloral DDT If 450.74 g of chlorobenzene reacts with 899.68 g of chloral, calculate (a) the theoretical yield of DDT in grams, and (b) the percentage yield if 654.13 g of DDT is actually isolated. molar mass of chlorobenzene = 112.56 g/mol molar mass of chloral = 147.39 g/mol

  5. moles of chlorobenzene = = 4.0044 mol chlorobenzene

  6. moles of chlorobenzene = = 4.0044 mol chlorobenzene moles of chloral = = 6.1041 mol chloral.

  7. moles of chlorobenzene = = 4.0044 mol chlorobenzene moles of chloral = = 6.1041 mol chloral. Now determine the limiting reagent.

  8. 4.0044 mol of chlorobenzene would require 2.0022 mol of chloral for reaction (see balanced equation), but we have 6.1041 mol of chloral – which means we have excess chloral. Therefore chlorobenzeneis the limiting reagent.

  9. 4.0044 mol of chlorobenzene would require 2.0022 mol of chloral for reaction (see balanced equation), but we have 6.1041 mol of chloral – which means we have excess chloral. Therefore chlorobenzeneis the limiting reagent. The molar mass of DDT = 354.49 g/mol. mass of DDT =

  10. 4.0044 mol of chlorobenzene would require 2.0022 mol of chloral for reaction (see balanced equation), but we have 6.1041 mol of chloral – which means we have excess chloral. Therefore chlorobenzeneis the limiting reagent. The molar mass of DDT = 354.49 g/mol. mass of DDT = = 709.58 g of DDT

  11. 4.0044 mol of chlorobenzene would require 2.0022 mol of chloral for reaction (see balanced equation), but we have 6.1041 mol of chloral – which means we have excess chloral. Therefore chlorobenzeneis the limiting reagent. The molar mass of DDT = 354.49 g/mol. mass of DDT = = 709.58 g of DDT This is the theoretical yield.

  12. = 92.186 %

  13. Concentration of Solutions

  14. Concentration of Solutions Molar Concentration: This is the most commonly employed concentration used in chemistry.

  15. Concentration of Solutions Molar Concentration: This is the most commonly employed concentration used in chemistry. The unit for molarity, mol/liter, is abbreviated M

  16. Example: A 2.60 molar sodium chloride solution, expressed as 2.60 M NaCl, contains 2.60 moles of NaCl in one liter of the sodium chloride solution. That is: 1 l of solution = 2.60 mol of NaCl.

  17. Problem example: A 6.9846 g sample of sucrose (C12H22O11) is dissolved in enough water to form 67.8 ml of solution. What is the molarity of the solution? molar mass of sucrose = 342.299 g/mol

  18. Moles of C12H22O11 = = 0.020405 mol sucrose

  19. Moles of C12H22O11 = = 0.020405 mol sucrose molarityof C12H22O11 = = 0.302 M

  20. Problem example: How many grams of NaCl are present in 50.0 ml of a 2.45 M NaCl solution? The molar mass of NaCl is 58.443 g/mol

  21. Problem example: How many grams of NaCl are present in 50.0 ml of a 2.45 M NaCl solution? The molar mass of NaCl is 58.443 g/mol First find the number of moles of NaCl present in 50.0 ml of solution.

  22. Problem example: How many grams of NaCl are present in 50.0 ml of a 2.45 M NaCl solution? The molar mass of NaCl is 58.443 g/mol First find the number of moles of NaCl present in 50.0 ml of solution. moles of NaCl = = 0.123 moles

  23. Problem example: How many grams of NaCl are present in 50.0 ml of a 2.45 M NaCl solution? The molar mass of NaCl is 58.443 g/mol First find the number of moles of NaCl present in 50.0 ml of solution. moles of NaCl = = 0.123 moles grams of NaCl = = 7.19 g NaCl

  24. Problem example: How would you prepare 250.0 ml of 0.100 M Na2SO4 solution? (The question is requesting the mass of Na2SO4 to be used to make the solution). molar mass of Na2SO4 = 142 g/mol

  25. Problem example: How would you prepare 250.0 ml of 0.100 M Na2SO4 solution? (The question is requesting the mass of Na2SO4 to be used to make the solution). molar mass of Na2SO4 = 142 g/mol moles of Na2SO4 = = 0.0250 mol Na2SO4

  26. Problem example: How would you prepare 250.0 ml of 0.100 M Na2SO4 solution? (The question is requesting the mass of Na2SO4 to be used to make the solution). molar mass of Na2SO4 = 142 g/mol moles of Na2SO4 = = 0.0250 mol Na2SO4 grams of Na2SO4 = = 3.55 g Na2SO4

  27. Problem example: How would you prepare 250.0 ml of 0.100 M Na2SO4 solution? (The question is requesting the mass of Na2SO4 to be used to make the solution). molar mass of Na2SO4 = 142 g/mol moles of Na2SO4 = = 0.0250 mol Na2SO4 grams of Na2SO4 = = 3.55 g Na2SO4 Therefore, dissolve 3.55 g of Na2SO4 in sufficient water to make 250.0 ml of solution.

  28. Titration: Acids and Bases

  29. Titration: Acids and Bases Titration: The gradual addition of a solution of accurately known concentration to another solution of unknown concentration until the chemical reaction between the two solutes is complete.

  30. Titration: Acids and Bases Titration: The gradual addition of a solution of accurately known concentration to another solution of unknown concentration until the chemical reaction between the two solutes is complete. Note: Sometimes the solution of unknown concentration is added to the solution of known concentration.

  31. Primary standard: A solution of accurately known concentration or a substance of accurately known purity.

  32. Primary standard: A solution of accurately known concentration or a substance of accurately known purity. Standard solution: A solution of accurately known concentration – usually determined by reaction with a primary standard.

  33. Example: A common example of a titration experiment is the reaction of NaOH of accurately known concentration, with a solution of HCl of unknown concentration:

  34. Example: A common example of a titration experiment is the reaction of NaOH of accurately known concentration, with a solution of HCl of unknown concentration: NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l)

  35. Example: A common example of a titration experiment is the reaction of NaOH of accurately known concentration, with a solution of HCl of unknown concentration: NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l) Equivalence point: Point at which the reaction in a titration is complete.

  36. Example: A common example of a titration experiment is the reaction of NaOH of accurately known concentration, with a solution of HCl of unknown concentration: NaOH(aq) + HCl(aq)NaCl(aq) + H2O(l) Equivalence point: Point at which the reaction in a titration is complete. Endpoint: The point just beyond the equivalence point – the point where the indicator changes color.

  37. Indicator: A substance that is used to indicate the completion of a reaction (usually in a titration) by a sharp change in its color.

  38. Indicator: A substance that is used to indicate the completion of a reaction (usually in a titration) by a sharp change in its color. A common indicator used in many acid-base titrations is phenolphthalein, which is colorlessin acidic solution and pink in basic solution.

  39. Problem example: 37.42 ml of 0.1078 M NaOH solution is needed to completely neutralize 25.00 ml of a HCl solution. What is the concentration of the acid solution?

  40. Problem example: 37.42 ml of 0.1078 M NaOH solution is needed to completely neutralize 25.00 ml of a HCl solution. What is the concentration of the acid solution? Step 1: Write the balanced equation for the reaction: NaOH(aq) + HCl(aq)NaCl(aq)+ H2O(l)

  41. Step 2 moles of NaOH = = 0.004034 mol NaOH

  42. Step 2 moles of NaOH = = 0.004034 mol NaOH Step 3 moles of HCl = = 0.004034 mol HCl

  43. Step 4 The molarity of the HCl solution = = = 0.1614 M

  44. Problem example: In a titration experiment, 17.8 ml of 0.344 M H2SO4 solution is required to completely neutralize 20.0 ml of a KOH solution. Calculate the concentration of the KOH solution.

  45. Problem example: In a titration experiment, 17.8 ml of 0.344 M H2SO4 solution is required to completely neutralize 20.0 ml of a KOH solution. Calculate the concentration of the KOH solution. The balanced equation is (step 1): H2SO4(aq) + 2 KOH(aq)K2SO4(aq) + 2 H2O(l)

  46. Step 2 moles of H2SO4 = = 0.0.00612 mol H2SO4

  47. Step 2 moles of H2SO4 = = 0.0.00612 mol H2SO4 Step 3 moles of KOH = = 0.0122 mol KOH

  48. The molarity of the KOH solution = = 0.610 M

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