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

Stoichiometry

Stoichiometry : Mole Ratios to Stoichiometry : Mole Ratios to Determining Grams of Product At the conclusion of our time together, you should be able to: Review the conversion of particles or grams to moles Determine mole ratios from a balanced chemical equation Determine the amount of product produced when given the amount of reactants

Review the Molar Mass of Compounds The molar mass (MM) of a compound is determined by adding up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl  110.98 g/mol CaCl2 20 Ca  40.08 17 Cl 35.45

Practice Calculate the Molar Mass of calcium phosphate Formula = Masses elements: Molar Mass = 310.18 g Ca3(PO4)2

Calculations molar mass Avogadro’s number Grams Moles Particles Everything must go through Moles!!!

Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023 Multiply by 6.02 X 1023 Multiply by atomic/molar mass from periodic table Moles Divide by atomic/molar mass from periodic table Mass (grams)

“Mole Relationships” Lab – Page 9 Lead (II) nitrate and sodium iodide  ???? Which test tube will produce the max amount? Test #1 Test #2 Test #3 Test #4 Test #5 1.5:4.5 2:4 3:3 4:2 4.5:1.5

Chocolate Chip Cookies!! 1 cup butter 2 eggs 1/2 cup white sugar 1 teaspoon salt 1 cup packed brown sugar 1 teaspoon vanilla extract 2 1/2 cups all-purpose flour 1 teaspoon baking soda 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How much brown sugar would I need if I had 1½ cups white sugar?

Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction/equations Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Are obtained by comparing the coefficients in a balanced chemical equation.

Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! _SiO2 + _HF  _SiF4 + _H2O Mole Ratios SiO2 to SiF4 SiF4 to H2O H2O to HF SiO2 to HF

25 moles of SiO2 converts to: Practice – Page 11 25 moles of SiO2 converts to: Moles of SiF4? Grams of SiO2? Molecules of SiF4?

5.0 moles of Ag converts to: Practice – Page 11 _Ag + _Cl2 +  _AgCl 5.0 moles of Ag converts to: Moles of Cl2? Grams of Cl2?

Mass-Mass Conversions – Page 13 Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, to mole ratios, and back to grams of the compound we are interested in.

Mass-Mass Conversion – Page 13 Calculate how many grams of iron are required to make 86.7g of FeS? __Fe + __S8  __FeS

Practice Mass-Mole-Mass – Page 13 #2 If 17.6g of carbon disulfide react with an excess of chlorine, how many grams of disulfur dichloride will result?

Stoichiometry : Mole Ratios to Stoichiometry : Mole Ratios to Determining Grams of Product Let’s see if you can: Review the conversion of particles or grams to moles Determine mole ratios from a balanced chemical equation Determine the amount of product produced when given the amount of reactants

Mass-Mole-Molecules: Determine the number of molecules in 73 g of water # H2O molecules = 73 g H2O x 1 mol H2O 18.02 g H2O x 6.02x1023 molecules 1 mol H2O = 2.4 x 1024 molecules H2O

Try this one: Calculate the mass in grams of iodine required to react completely with 0.50 moles of aluminum. Al + I2  AlI3 2 Al + 3 I2  2 AlI3 x 3 mol I2 2 mol Al 0.50 mol Al x 253.80 g I2 1 mol I2 = 190 g I2

Try this one: Calculate the mass in grams of iodine required to react completely with 0.50 g of aluminum. Al + I2  AlI3 2 Al + 3 I2  2 AlI3 x 1 mol Al 26.98 g Al 0.50 g Al x 3 mol I2 2 mol Al x 253.80 g I2 1 mol I2 = 7.1 g I2