the Mole !

1. It’s time to learn about . . . the Mole !

2. There she blows!!!

3. Stoichiometry: Molar SolutionsAt the conclusion of our time together, you should be able to: Define molarity Determine the molarity of a given solution Make a solution with a given molarity

4. Some Definitions A solution is a homogeneous mixture of 2 or more substances in a single phase. The larger constituent is usually regarded as the SOLVENT and the others as SOLUTES.

5. K+(aq) + MnO4-(aq) IONIC COMPOUNDSCompounds in Aqueous Solution Many reactions involve ionic compounds, in water these are — aqueous solutions. KMnO4 in water

6. What Do You Get From a Pampered Cow? Spoiled Milk. 

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

8. PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl2•6H2O Step 2: Calculate Molarity [NiCl2•6 H2O ] = 0.0841 M

9. moles solute ( M ) Molarity = liters of solution Concentration of Solute Therefore: x both sides by Liters = MV = moles = grams/molar mass

10. x 1 mol NiCl2 237.71 g NiCl2 SAME PROBLEM: Using moles = MV. 5.00 g NiCl2 = M x 0.250 L = 0.0841 M NiCl2*6H2O

11. x 1 mol H2C2O4 90.04 g H2C2O4 USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? moles = M•V X g H2C2O4 = 0.0500 x 0.250 L = 1.13 g H2C2O4

12. x 1 mol NaNO3 85.00 g NaNO3 Page 25:2 moles = M•V X g NaNO3 = 0.50 M x 1.00 L = 42.5 g NaNO3

13. Over-Worked Mouse

14. x 40.00 g NaOH 1 mol NaOH Learning Check the Old Way?!! How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? moles = MV mol = 3.0 M x 0.400 L mol = 1.2 48 g NaOH 1.2 mol NaOH

15. x 1 mol NaOH 40.00 g NaOH The Better Way moles = M•V X g NaOH = 3.0 M x 0.400 L = 48 g NaOH

16. Hopefully, you’re not this lost!!!

17. Preparing Solutions • Determine the mass of solute. • Place in the appropriate volumetric flask. • Add deionized water and swirl until solute is dissolved. • Add water to the mark on the neck of the flask. • Stopper and mix thoroughly.

18. Interesting Vanity Plate…

19. Stoichiometry: Molar SolutionsLet’s see if you can: Define molarity Determine the molarity of a given solution Make a solution with a given molarity

20. moles solute ( M ) Molarity = liters of solution Concentration of Solute Therefore: MV = moles = grams/molar mass

21. Support Bacteria!! • They're the only culture some people have.

22. x 1 mol AgNO3 169.88 g AgNO3 PROBLEM: Mr. T accidentally dropped 15.14 g of silver (I) nitrate into 100.0 mL of deionized water. Rather than throw the solution away, help him by determining its molar concentration so that he can label it and still use it.. 15.14 g AgNO3 = M x 0.100 L = 0.8912 M AgNO3

23. x 1 mol Pb(NO3)2 331.22 g Pb(NO3)2 PROBLEM: Mr. T needs to make 0.500 L of a 0.100 M solution of lead (II) nitrate for a future lab. Please help him by calculating the amount of solute needed and by outlining in detail the steps he would need to take to make this molar solution. X g Pb(NO3)2 = 0.100 M x 0.500 L = 16.6 g Pb(NO3)2

24. Preparing Solutions • Measure 16.6 g Pb(NO3)2. • Place in a 0.500 L volumetric flask. • Add deionized water and swirl until solute is dissolved. • Add water to the mark on the neck of the flask. • Stopper and mix thoroughly.

25. Most Caring Person!! An elderly gentleman had recently lost his wife.Upon seeing the man cry, a little 4 year old neighbor boy went into the old gentleman's yard, climbed onto his lap, and just sat there. When his mother asked him what he had said to the neighbor, the little boy just said, "Nothing, I just helped him cry."

26. x 1 mol Ca(OH)2 74.10 g Ca(OH)2 PROBLEM Page 26-4: Grams to make 100. mL of 0.100 M calcium hydroxide. X g Ca(OH)2 = 0.100 M x 0.100 L = 0.741 g Ca(OH)2

27. 4.00 mol HNO3 PROBLEM Page 26-5: 4.00 moles of nitric acid in 1.50 L of solution = what M. = X M x 1.50 L = 2.67 M HNO3

28. Entrance Quiz #1 • 1. Calculate the molarity if 0.75 mol of NaCl in placed in 300.0 mL of water. • 2. Calculate the number of moles and grams of HCl if there is 12.2 mL of 2.45 M HCl solution.

29. 0.75 mol NaCl X mol HCl x 36.46 g HCl 1 mol HCl = X M x 0.3000 L • 1. Calculate the molarity if 0.75 mol of NaCl in placed in 300.0 mL of water. • 2. Calculate the number of moles and grams of HCl if there is 12.2 mL of 2.45 M HCl solution. = 2.5 M NaCl = 2.45 M x 0.0122 L = 0.0299 mol HCl 0.0299 mol HCl = 1.09 g HCl

30. Entrance Quiz #2 • 1. Calculate the molarity if 0.50 mol of KBr in placed in 750 mL of water. • 2. Calculate the number of moles and molarity of HCl if there is 12.2 mL with 2.45 g of HCl in solution.

31. 0.50 mol KBr 0.0672 mol HCl x 1 mol HCl 36.46 g HCl = X M x 0.75 L • 1. Calculate the molarity if 0.50 mol of KBr in placed in 750 mL of water. • 2. Calculate the number of moles and molarity of HCl if there is 12.2 mL with 2.45 g of HCl. = 0.67 M KBr 2.45 g HCl = 0.0672 mol HCl = X M x 0.0122 L = 5.51 M HCl

32. Stoichiometry: Molar SolutionsAt the conclusion of our time together, you should be able to: Define molarity Determine the molarity of a given solution Make a solution with a given molarity Dilute a given solution to a new molarity

33. Kid’s Letters to God:

34. moles solute ( M ) Molarity = liters of solution For Dilution: the Amount (moles) of Solute #1 = #2 M1V1 = moles and M2V2 = moles Therefore if moles of solute are constant: M1V1 = M2V2 And Using the Formula:

35. The Scientific Methodbegins with Questions about the World Around You.Ever Wonder Why?... there are flotation devices under plane seats instead of parachutes?

36. PROBLEM Page 30-1: Using M1V1 = M2V2 0.150 M NaOH x 0.125 L = M x 0.150 L = 0.125 M NaOH

37. PROBLEM Page 30-5: Using M1V1 = M2V2 2.40 M KCl x 0.500 L = 1.00 M KCl x X L = 1.20 L Therefore: 0.700 L needs to be added

38. How We Doing?? Need Help????

39. Stoichiometry: Molar SolutionsLet’s see if you can: Define molarity Determine the molarity of a given solution Make a solution with a given molarity Dilute a given solution to a new molarity

40. I knew I shouldn’t have done that!!

41. How much water do I need to add to 250 mL of 3.0 M HCl to dilute it to 1.0 M HCl? 3.0 M HCl x 0.250 L = 1.0 M HCl x L = 0.75 L Total, therefore 0.50 L

42. Name ________Class Period _____ Clicker Number Name ________Class Period _____ Clicker Number "Making Molar Solutions A1" (10 points) Make 50.00 mL of a 0.100M BaCl22H2O solution. • __________ grams mass of solute needed • __________ Instructor initials (one point)

43. What are the products of the reaction? Balance the equation. Write the balanced molecular, complete ionic and net ionic equations below. Place the precipitate on the 3rd line of the first row of lines. (5 points) ________ + ________  ________ + ________ ___ ____ + ___ ____  ______ + ___ ____ •   ____ + ___  _____ • What is this the best molar ratio based on the stoichiometry? Should you have put all 50.00 mL of each reactant together to form the most product? Circle yes or no. (2 points) ____________ : ____________ Yes No

44. Preparing Solutions • Determine the mass of solute. • Place in the appropriate volumetric flask. • Add deionized water and swirl until solute is dissolved. • Add water to the mark on the neck of the flask. • Stopper and mix thoroughly.