1. Unit 4 Solutions and Stoichiometry

2. Outline of Topics Solutions Solutions Molarity Molarity Dilution Dilution Introduction to Chemical Reactions Introduction to Chemical Reactions Stoichiometry Stoichiometry Competing Reactions Competing Reactions Limiting Reactants Limiting Reactants

3. Solutions Common homogeneous mixture Common homogeneous mixture Components Components Solvent Solvent Most commonly a liquid Most commonly a liquid Solute Solute May be solid, liquid or gas May be solid, liquid or gas Seawater Seawater Water is the solvent Water is the solvent Solutes may be one of a variety of salts Solutes may be one of a variety of salts

4. Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

5. Polar versus NonPolar Polar substances: Such as Ionic Compounds and a few Covalent Compounds have a positive side and a negative side. Polar substances: Such as Ionic Compounds and a few Covalent Compounds have a positive side and a negative side. Nonpolar substances: Most Covalent Compounds do not have a positive side or a negative side. Nonpolar substances: Most Covalent Compounds do not have a positive side or a negative side.

6. Like Dissolves Like Polar substances are dissolved by Polar Substances Polar substances are dissolved by Polar Substances Nonpolar substances are dissolved by Nonpolar Substances. Nonpolar substances are dissolved by Nonpolar Substances. Polar does not dissolve Nonpolar. Polar does not dissolve Nonpolar. Nonpolar does not dissolve Nonpolar. Nonpolar does not dissolve Nonpolar.

7. Nonpolar will not dissolve in Polar Iodine is NonPolar, water is Polar, so it won’t dissolve Iodine is NonPolar, water is Polar, so it won’t dissolve Iodine will dissolve in Carbon Tetra Chloride which is NonPolar. Iodine will dissolve in Carbon Tetra Chloride which is NonPolar.

8. Solute Concentrations - Molarity Definition of molarity Definition of molarity Molarity = moles of solute/liters of solution Molarity = moles of solute/liters of solution Symbol is M Symbol is M Square brackets are used to indicate concentration in M Square brackets are used to indicate concentration in M [Na + ] = 1.0 M [Na + ] = 1.0 M Consider a solution prepared from 1.20 mol of substance A, diluted to a total volume of 2.50 L Consider a solution prepared from 1.20 mol of substance A, diluted to a total volume of 2.50 L Concentration is 1.20 mol/2.50 L or 0.480 M Concentration is 1.20 mol/2.50 L or 0.480 M

9. Concentration of Solute The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution

10. PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate Molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = 0.0841 M

11. Step 1: Change mL to L. 250 mL * 1L/1000mL = 0.250 L Step 2: Calculate. Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles Step 3: Convert moles to grams. (0.0125 mol)(90.00 g/mol) = 1.13 g USING MOLARITY moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution?

12. Colligative Properties On adding a solute to a solvent, the properties of the solvent are modified. Vapor pressure decreases Vapor pressure decreases Melting point decreases Melting point decreases Boiling point increases Boiling point increases Osmosis is possible (osmotic pressure) Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

13. Additivity Masses are additive; volumes are not Masses are additive; volumes are not The total mass of a solution is the sum of the mass of the solute and the solvent The total mass of a solution is the sum of the mass of the solute and the solvent The total volume of a solution is not the sum of the volumes of the solute and solvent The total volume of a solution is not the sum of the volumes of the solute and solvent

14. Molarity as a Conversion Factor The molarity can be used to calculate The molarity can be used to calculate The number of moles of solute in a given volume of solution The number of moles of solute in a given volume of solution The volume of solution containing a given number of moles of solute The volume of solution containing a given number of moles of solute

15. Example 3.2

16. Example 3.2, (Cont’d)

18. Dissolving Ionic Solids When an ionic solid is dissolved in a solvent, the ions separate from each other When an ionic solid is dissolved in a solvent, the ions separate from each other MgCl 2 (s) → Mg 2+ (aq) + 2 Cl - (aq) MgCl 2 (s) → Mg 2+ (aq) + 2 Cl - (aq) The concentrations of ions are related to each other by the formula of the compound: The concentrations of ions are related to each other by the formula of the compound: Molarity MgCl 2 of = molarity of Mg 2+ Molarity MgCl 2 of = molarity of Mg 2+ Molarity of Cl - = 2 X molarity of MgCl 2 Molarity of Cl - = 2 X molarity of MgCl 2 Total number of moles of ions per mole of MgCl 2 is 3 Total number of moles of ions per mole of MgCl 2 is 3

19. Example 3.3

20. Example 3.3, (Cont’d)

22. Dilution The process of reducing the Molarity of a solution. The process of reducing the Molarity of a solution. Stock Solutions: Solutions at a high molarity. Stock Solutions: Solutions at a high molarity. Equation: M 1 V 1 = M 2 V 2 Equation: M 1 V 1 = M 2 V 2 Solvent is added to the existing solution until the desired Molarity is achieved. Solvent is added to the existing solution until the desired Molarity is achieved.

23. Example Problem A 250 mL of 2.0 M HCl must be prepared from a 16.0 M Stock Solution. How much of the stock solution must be used, and how much water must be added? A 250 mL of 2.0 M HCl must be prepared from a 16.0 M Stock Solution. How much of the stock solution must be used, and how much water must be added? Determine knowns and unknowns Determine knowns and unknowns M 1 = 16 M M 1 = 16 M V 1 = X V 1 = X M 2 = 2.0 M M 2 = 2.0 M V 2 = 250 mL = 0.250 L V 2 = 250 mL = 0.250 L

24. Example problem continued Determine Formula Determine Formula M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 V 2 M 1 V 1 = V 2 M 1 V 1 = V 2 M 2 M 2 Plug in Numbers Plug in Numbers 2.0 M x 0.250 L = V 2 2.0 M x 0.250 L = V 2 16.0 M 16.0 M Solve Solve V 2 = 0.0312 L is the beginning volume of 16.0 M HCl. V 2 = 0.0312 L is the beginning volume of 16.0 M HCl. You must add (0.250L – 0.0312L) = 0.229 L of water to the original stock solution. You must add (0.250L – 0.0312L) = 0.229 L of water to the original stock solution.

25. Mass Relations in Reactions Chemical equations represent chemical reactions Chemical equations represent chemical reactions Reactants appear on the left Reactants appear on the left Products appear on the right Products appear on the right Equation must be balanced Equation must be balanced Number of atoms of each element on the left … Number of atoms of each element on the left … …equals the number of atoms of each element on the right …equals the number of atoms of each element on the right

26. How are Equations Written? We must know the reactants and the products for a reaction for which an equation is to be written We must know the reactants and the products for a reaction for which an equation is to be written It is often necessary to do an experiment and an analysis to determine the products of a reaction It is often necessary to do an experiment and an analysis to determine the products of a reaction Determining the products is often time consuming and difficult Determining the products is often time consuming and difficult

27. Writing Chemical Equations 1. Write a skeleton equation for the reaction. 2. Indicate the physical state of each reactant and product. 3. Balance the equation Only the coefficients can be changed; subscripts are fixed by chemical nature of the reactants and products Only the coefficients can be changed; subscripts are fixed by chemical nature of the reactants and products It is best to balance atoms that appear only once on each side of the equation first It is best to balance atoms that appear only once on each side of the equation first

28. Example 3.8

29. Balancing Chemical Reactions Matter cannot be created or destroyed in a chemical reaction. Matter cannot be created or destroyed in a chemical reaction. The total number of atoms of each element in a chemical reactions must be the same in the reactants (start) and the products (end) The total number of atoms of each element in a chemical reactions must be the same in the reactants (start) and the products (end) Coefficients are used to balance an equation. Coefficients are used to balance an equation.

30. Steps in Balancing 1. Write the skeleton equation. 2. Determine the number of atoms on each side of the equation. 3. Add coefficients to each reactant and product until the number of atoms on each side is equal. 4. Hint: As long as a polyatomic ion remains unchanged, you can treat it as a single unit.

31. Example Copper metal is added to Aqueous Silver Nitrate making Aqueous Copper (II) Nitrate and Silver metal. Copper metal is added to Aqueous Silver Nitrate making Aqueous Copper (II) Nitrate and Silver metal. 1. Write the skeletal equation: Cu (s) + AgNO 3 (aq)  Cu(NO 3 ) 2 (aq) + Ag (s) 2. Determine the number of each species on the reactant and product side. 1Cu1 1Ag1 1 NO 3 - 2

32. Example Continued 3. Working one species at a time, add coefficents. Cu (s) + 2 AgNO 3 (aq)  Cu(NO 3 ) 2 (aq) + Ag (s) 1Cu1 2Ag1 2 NO 3 - 2

33. Example Continued 4. Cu (s) + 2 AgNO 3 (aq)  Cu(NO 3 ) 2 (aq) + 2 Ag (s) 1Cu1 2Ag2 2 NO 3 - 2

34. Practice Balancing Examples _Mg + _N 2  _Mg 3 N 2 _P + _O 2  _P 4 O 10 _Na + _H 2 O  _H 2 + _NaOH _CH 4 + _O 2  _CO 2 + _H 2 O

35. Writing Chemical Equations 1. Write a skeleton equation for the reaction. 2. Indicate the physical state of each reactant and product. 3. Balance the equation Only the coefficients can be changed; subscripts are fixed by chemical nature of the reactants and products Only the coefficients can be changed; subscripts are fixed by chemical nature of the reactants and products It is best to balance atoms that appear only once on each side of the equation first It is best to balance atoms that appear only once on each side of the equation first

36. Mass Relations from Equations The coefficients of a balanced equation represent the numbers of moles of reactants and products The coefficients of a balanced equation represent the numbers of moles of reactants and products 2 N 2 H 4 (l) + N 2 O 4 (l) → 3 N 2 (g) + 4 H 2 O (l) 2 N 2 H 4 (l) + N 2 O 4 (l) → 3 N 2 (g) + 4 H 2 O (l) 2 mol N 2 H 4 + 1 mol N 2 O 4 → 3 mol N 2 + 4 mol H 2 O 2 mol N 2 H 4 + 1 mol N 2 O 4 → 3 mol N 2 + 4 mol H 2 O

37. Stoichiometry

38. What is Stoichiometry? a branch of chemistry that deals with the quantitative relationships that exist between the reactants and products in chemical reactions. a branch of chemistry that deals with the quantitative relationships that exist between the reactants and products in chemical reactions.chemistry reactantsproductschemistry reactantsproducts Stoichiometry can be used to calculate quantities such as the amount of products that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product). Stoichiometry can be used to calculate quantities such as the amount of products that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product).yield

39. In Other words How much product do I get when I mix certain amounts of Reactants together, How much product do I get when I mix certain amounts of Reactants together, Or, How much do I need to make what I want. Or, How much do I need to make what I want.

40. Steps in Stoichmetry 1. Write the Balanced Chemical Equation. 2. Convert the grams given into moles. 3. Multiply by the mole ratio from the Balanced Chemical Equation. (THIS IS THE ONLY STEP YOU USE THE COEFFICENT IN) 4. Convert the new number of moles to grams.

41. Figure 3.8: Flowchart for Mole- Mass Calculations

42. Railroad Track Stoichiometry

43. The best way to learn is to do it So here we go: You have 5 grams of Lead (II) Nitrate mixed with an excess of Sodium Chloride, how many grams of Lead (II) Chloride will you produce? So here we go: You have 5 grams of Lead (II) Nitrate mixed with an excess of Sodium Chloride, how many grams of Lead (II) Chloride will you produce? Step 1, write the balanced equation: Step 1, write the balanced equation: Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 + 2 NaNO 3 Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 + 2 NaNO 3

44. Step 2: Convert grams to moles DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER. DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER.

45. Step 3: Multiply by mole ratio THIS IS THE ONLY STEP YOU USE THE COEFFIECENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!! THIS IS THE ONLY STEP YOU USE THE COEFFIECENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

46. Step 4: Convert back to grams DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER. DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER.

47. And the Answer is: Solve the problem Solve the problem

48. One More time You have 5 grams of Lead (II) Nitrate mixed with an excess of Sodium Chloride, how many grams of Sodium Nitrate will you produce? You have 5 grams of Lead (II) Nitrate mixed with an excess of Sodium Chloride, how many grams of Sodium Nitrate will you produce? Step 1, write the balanced equation: Step 1, write the balanced equation: Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 + 2 NaNO 3 Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 + 2 NaNO 3

49. Step 2: Convert grams to moles DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER. DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER.

50. Step 3: Multiply by mole ratio THIS IS THE ONLY STEP YOU USE THE COEFFIECENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!! THIS IS THE ONLY STEP YOU USE THE COEFFIECENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

51. Step 4: Multiply by New Molar Mass DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER. DO NOT EVER USE THE COEFFIECENT IN THIS STEP, NEVER, NEVER, NEVER.

52. And the Answer is:……… Solve the problem Solve the problem

53. Moles instead of grams??? In case of this, you can skip either step 2 or step 4, YEAH!!!!!!!!!!!! In case of this, you can skip either step 2 or step 4, YEAH!!!!!!!!!!!! How many moles of Lead (II) Nitrate are needed to produce 6 grams of Sodium Nitrate? How many moles of Lead (II) Nitrate are needed to produce 6 grams of Sodium Nitrate? Back to Step 1: Back to Step 1: Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 + 2 NaNO 3 Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 + 2 NaNO 3

54. Starting in grams, so We still need Step 2: We still need Step 2:

55. Still have to do Step 3: Moles to Moles: Moles to Moles:

56. Example 3.9

57. Example 3.9, (Cont'd)

60. Limiting Reagent Problems Chemical Reactions will continue until one of the Reagents (Reactants) is used up. Chemical Reactions will continue until one of the Reagents (Reactants) is used up. Once one of the Reagents is used up, the reaction stops! Once one of the Reagents is used up, the reaction stops! It doesn’t matter how much of the other one (Excess) is still left, the reactions STOPS. It doesn’t matter how much of the other one (Excess) is still left, the reactions STOPS.

61. Limiting Reactants require TWO complete Stoichiometry Equations One for each Reactant: One for each Reactant: SAME FOUR STEPS, Two times……….. SAME FOUR STEPS, Two times……….. If 5.00 grams of Lead (II) Nitrate is reacted with 5.00 grams of Sodium Chloride, what is the Limiting Reactant (Reagent), and how much of the solid product is produced? If 5.00 grams of Lead (II) Nitrate is reacted with 5.00 grams of Sodium Chloride, what is the Limiting Reactant (Reagent), and how much of the solid product is produced? Same STEP 1: Same STEP 1: Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 (s) + 2 NaNO 3 Pb(NO 3 ) 2 + 2 NaCl  PbCl 2 (s) + 2 NaNO 3

62. It is a gram to gram problem, so… You still have to do all the steps, but twice. You still have to do all the steps, but twice. Once for each reactant: Once for each reactant:

63. Continuing:

64. And……. 5.00 g Lead (II) Nitrate makes 4.20 g of Lead (II) Chloride. 5.00 g Lead (II) Nitrate makes 4.20 g of Lead (II) Chloride. 5.00 g Sodium Chloride makes 9.81 grams of Lead (II) Chloride. 5.00 g Sodium Chloride makes 9.81 grams of Lead (II) Chloride. So, Lead (II) Nitrate is the Limiting Reagent because it makes the least Product. So, Lead (II) Nitrate is the Limiting Reagent because it makes the least Product. Sodium Chloride is the Excess Reagent, it makes the most Product. Sodium Chloride is the Excess Reagent, it makes the most Product. Only the Least Amount can be made, so only 4.20 g of Lead (II) Chloride is produced. Only the Least Amount can be made, so only 4.20 g of Lead (II) Chloride is produced.

65. YOU CAN ONLY MAKE THE … LEAST AMOUNT OF PRODUCT. LEAST AMOUNT OF PRODUCT. A Car can only go as far as the Gasoline lets it go!!!!!!!!!!! A Car can only go as far as the Gasoline lets it go!!!!!!!!!!!

66. Percent Yield Stoichiometry gives us a perfect answer in a perfect world. Stoichiometry gives us a perfect answer in a perfect world. In the Real World not every atom reacts. In the Real World not every atom reacts. If you run your car out of gas, you can still smell gasoline in the gas tank, some it still left, but the car won’t run. If you run your car out of gas, you can still smell gasoline in the gas tank, some it still left, but the car won’t run.

67. Percent Yield Formula

68. Example In a chemical reaction between Lead (II) Nitrate and Sodium Chloride the Stoichiometry determined that 4.20 grams should be produced. When the Experiment was actually performed, 3.95 grams was produced. What is the Percent Yield? In a chemical reaction between Lead (II) Nitrate and Sodium Chloride the Stoichiometry determined that 4.20 grams should be produced. When the Experiment was actually performed, 3.95 grams was produced. What is the Percent Yield? % yield = 3.95 g actual X 100 = 94.05 % % yield = 3.95 g actual X 100 = 94.05 % 4.20 g predicted

69. Terms Theoretical Yield: The answer to the stoichiometry problem. The amount produced if every atom reacts to form product. Theoretical Yield: The answer to the stoichiometry problem. The amount produced if every atom reacts to form product. Actual Yield (experimental yield): The amount actually produced when a reaction occurs. Actual Yield (experimental yield): The amount actually produced when a reaction occurs.