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 C2HOCl C14H9Cl5 + H2O
chlorobenzene chloral DDT

3. Example: DDT is an insecticide prepared by the following reaction:
2 C6H5Cl C2HOCl C14H9Cl5 + H2O
chlorobenzene chloral DDT
If g of chlorobenzene reacts with g of chloral, calculate (a) the theoretical yield of DDT in grams, and (b) the percentage yield if g of DDT is actually isolated.

4. Example: DDT is an insecticide prepared by the following reaction:
2 C6H5Cl C2HOCl C14H9Cl5 + H2O
chlorobenzene chloral DDT
If g of chlorobenzene reacts with g of chloral, calculate (a) the theoretical yield of DDT in grams, and (b) the percentage yield if g of DDT is actually isolated.
molar mass of chlorobenzene = g/mol
molar mass of chloral = g/mol

5. moles of chlorobenzene =
= mol chlorobenzene

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

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

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

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

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

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

13. = %

14. Concentration of Solutions

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

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

17. Example: A 2. 60 molar sodium chloride solution, expressed as 2
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: l of solution = 2.60 mol of NaCl.

18. Problem example: A 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 = g/mol

20. Moles of C12H22O11 =
= mol sucrose

21. Moles of C12H22O11 =
= mol sucrose
molarity of C12H22O11 =
= M

22. Problem example: How many grams of NaCl are present in 50. 0 ml of a 2
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 g/mol

23. Problem example: How many grams of NaCl are present in 50. 0 ml of a 2
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 g/mol
First find the number of moles of NaCl present in 50.0 ml of solution.

24. Problem example: How many grams of NaCl are present in 50. 0 ml of a 2
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 g/mol
First find the number of moles of NaCl present in 50.0 ml of solution.
moles of NaCl =
= moles

25. Problem example: How many grams of NaCl are present in 50. 0 ml of a 2
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 g/mol
First find the number of moles of NaCl present in 50.0 ml of solution.
moles of NaCl =
= moles
grams of NaCl =
= 7.19 g NaCl

26. molar mass of Na2SO4 = 142 g/mol
Problem example: How would you prepare 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

27. molar mass of Na2SO4 = 142 g/mol moles of Na2SO4 = = 0.0250 mol Na2SO4
Problem example: How would you prepare 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 =
= mol Na2SO4

28. molar mass of Na2SO4 = 142 g/mol moles of Na2SO4 = = 0.0250 mol Na2SO4
Problem example: How would you prepare 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 =
= mol Na2SO4
grams of Na2SO4 =
= g Na2SO4

29. molar mass of Na2SO4 = 142 g/mol moles of Na2SO4 = = 0.0250 mol Na2SO4
Problem example: How would you prepare 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 =
= mol Na2SO4
grams of Na2SO4 =
= g Na2SO4
Therefore, dissolve 3.55 g of Na2SO4 in sufficient water to make ml of solution.

30. Titration: Acids and Bases

31. 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.

32. 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.

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

34. 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.

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:

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)

37. 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.

38. 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.

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

40. 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 colorless in acidic solution and pink in basic solution.

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

42. Problem example: ml of M NaOH solution is needed to completely neutralize 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)

43. Step 2
moles of NaOH =
= mol NaOH

44. Step 2
moles of NaOH =
= mol NaOH
Step 3
moles of HCl =
= mol HCl

45. Step 4
The molarity of the HCl solution = =
= M

46. Problem example: In a titration experiment, 17. 8 ml of 0
Problem example: In a titration experiment, 17.8 ml of M H2SO4 solution is required to completely neutralize 20.0 ml of a KOH solution. Calculate the concentration of the KOH solution.

47. Problem example: In a titration experiment, 17. 8 ml of 0
Problem example: In a titration experiment, 17.8 ml of 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)

48. Step 2
moles of H2SO4 =
= mol H2SO4

49. Step 2
moles of H2SO4 =
= mol H2SO4
Step 3
moles of KOH =
= mol KOH

50. The molarity of the KOH solution =

51. If you are clear on the above steps, then you may shorten the calculation to:
molarity =
______________________________________
= M

52. Percentage Concentration

53. Percentage Concentration
weight/weight percent (better would be mass/mass percent): The number of grams of solute per 100 g of solution.

54. Percentage Concentration
weight/weight percent (better would be mass/mass percent): The number of grams of solute per 100 g of solution. Example: 0.25 % w/w NaOH means that there is a ratio of 0.25 g of NaOH in 100 g of NaOH solution.

55. volume/volume percent : When both the solute and the solvent are liquids, it is sometimes convenient to describe concentrations as percent by volume.

56. volume/volume percent : When both the solute and the solvent are liquids, it is sometimes convenient to describe concentrations as percent by volume. This is the number of volumes of solute in 100 volumes of solution.

57. volume/volume percent : When both the solute and the solvent are liquids, it is sometimes convenient to describe concentrations as percent by volume. This is the number of volumes of solute in 100 volumes of solution. Volume can be any convenient volume unit – provided the same volume is used for both the solute and the solution.

58. volume/volume percent : When both the solute and the solvent are liquids, it is sometimes convenient to describe concentrations as percent by volume. This is the number of volumes of solute in 100 volumes of solution. Volume can be any convenient volume unit – provided the same volume is used for both the solute and the solution. Example: 10% v/v ethanol has 10 volumes of ethanol in 100 volumes of ethanol/water solution.

59. Making dilute solutions from concentrated solutions.

60. Making dilute solutions from concentrated solutions.
Assume the concentration units are molarity. Use the abbreviations dil for dilute and conc for concentrated.