Chocolate Chip Cookies 2.25 cups flour 8 Tbsp butter 0.5 cups shortening 0.75 cups sugar 0.75 cups brown sugar 1 tsp salt 1 tsp baking soda 1 tsp vanilla 0.5 cups Egg Beaters 1 bag chocolate chips

Chocolate Chip Cookies 2.25 cups flour 8 Tbsp butter 0.5 cups shortening 0.75 cups sugar 0.75 cups brown sugar 1 tsp salt 1 tsp baking soda 1 tsp vanilla 0.5 cups Egg Beaters 1 bag chocolate chips How much? What units? Of what?

Chocolate Chip Cookies 2.25 flour 8 butter 0.5 shortening 0.75 sugar 0.75 brown sugar 1 salt 1 baking soda 1 vanilla 0.5 Egg Beaters 1 chocolate chips How much? Of what?

Chocolate Chip Cookies 2.25 cups flour 8 Tbsp butter 0.5 cups shortening 0.75 cups sugar 0.75 cups brown sugar 1 tsp salt 1 tsp baking soda 1 tsp vanilla 0.5 cups Egg Beaters 1 bag chocolate chips How much? What units? Of what?

Get on with it! What does this have to do with CHEMISTRY?

2.25 cups flour + 8 Tbsp butter cups shortening cups sugar cups brown sugar + 1 tsp salt + 1 tsp baking soda + 1 tsp vanilla cups Egg Beaters + chips (a synthesis reaction) (177ºC) 1 batch of chocolate chip cookies! coefficient unit substance

Welcome to STOICHIOMETRY

What is stoichiometry? Composition stoichiometry deals with the mass relationships of elements in compounds. Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction. All reaction stoichiometry calculations start with a balanced chemical equation.

Let’s Revisit the Cookies… 2.25 cups flour 8 Tbsp butter 0.5 cups shortening 0.75 cups sugar 0.75 cups brown sugar 1 tsp salt 1 tsp baking soda 1 tsp vanilla 0.5 cups Egg Beaters For 1 batch: The Egg Beaters I have are close to expiring! I’d like to use the rest of them in this recipe. I have 1.5 cups of Egg Beaters. How many batches of cookies can I make with that many Egg Beaters?

Let’s Revisit the Cookies… 2.25 cups flour 8 Tbsp butter 0.5 cups shortening 0.75 cups sugar 0.75 cups brown sugar 1 tsp salt 1 tsp baking soda 1 tsp vanilla 0.5 cups Egg Beaters For 1 batch: I have 1.5 cups of Egg Beaters. How many batches of cookies can I make with that many Egg Beaters? 1.5 cups E.B. x 1 batch cookies 0.5 cups E.B. = 3.0 batches of cookies

Let’s Revisit the Cookies… 2.25 cups flour 8 Tbsp butter 0.5 cups shortening 0.75 cups sugar 0.75 cups brown sugar 1 tsp salt 1 tsp baking soda 1 tsp vanilla 0.5 cups Egg Beaters For 1 batch: I have 1.5 cups of Egg Beaters. How much butter do I need to deplete (use up) the Egg Beaters? 1.5 cups E.B. x 8 Tbsp butter 0.5 cups E.B. = 24 Tablespoons of butter

… Back to Chemistry There are three types of stoichiometry problems: – Mole-Mole problems (1 conversion) – Mass-Mole problems (2 conversions) – Mass-Mass problems (3 conversions) given required

Stoichiometry Problems Stoichiometric problems are solved by using ratios from balanced chemical equations to convert the given quantity. A mole ratio is a conversion factor that relates the amount in moles of any two substances involved in a chemical reaction. This information is obtained from the balanced chemical equation.

Mole Ratios Example: The relationships between product and reactants or reactants can be expressed in the following mole ratios: 2 H 2 + O 2 2 H 2 O

Mole Ratios 2 H 2 + O 2 2 H 2 O

Practice For each reaction, write all possible mole ratios.

Mole-Mole Problems Example: 2 H 2 + O 2 2 H 2 O How many moles of water can be formed from 0.5 mol H 2 ? 0.5 mol H 2 x 2 mol H 2 2 mol H 2 O = 0.5 mol H 2 O

Mole-Mole Practice 3CuSO 4 +2AlAl 2 (SO 4 ) 3 +3Cu 1. Convert 0.5 mol CuSO 4 to mol Cu 0.5 mol CuSO 4 x3 mol Cu 3 mol CuSO 4 = 0.5 mol Cu 2. Convert 0.5 mol Al to mol CuSO mol Alx 3 mol CuSO 4 2 mol Al = 0.8 mol CuSO 4

Mass – Mole Problems Step 1: Write a BALANCED EQUATION. Step 2: Calculate the molar mass of your given substance and convert from mass to moles. Step 3: Determine the mole ratio from the coefficients in the balanced equation. Step 4: Set up the conversion and solve.

Mass-Mole Problems Example: 2 H 2 + O 2 2 H 2 O How many moles of water can be formed from 48.0 g O 2 ? 48.0 g O 2 x 1 mol O 2 2 mol H 2 O = 3.00 mol H 2 O g O 2 1 mol O 2 x

Mass-Mole Practice CuSO 4 AlAl 2 (SO 4 ) 3 Cu a.13.5 g Al 1 mol Al g Al =x mol CuSO 4 b. c g Al 1 mol Al 2 (SO 4 ) 3 2 mol Al = x0.250 mol Al 2 (SO 4 ) g Al 3 mol Cu 2 mol Al = x0.751 mol Cu Mole ratio x 3 mol CuSO 4 2 mol Al 1 mol Al g Al x 1 mol Al g Al x

Mole – Mass Practice Ca + AlCl 3 CaCl 2 + Al mol AlCl 3 x 3 mol Ca 2 mol AlCl 3 x g Ca 1 mol Ca =5.7 g Ca

Mass-Mass Problems Example: 2 H 2 + O 2 2 H 2 O How many grams of water can be formed from 48.0 g O 2 ? 48.0 g O 2 x 1 mol O 2 2 mol H 2 O = g O 2 1 mol O 2 xx g H 2 O 1 mol H 2 O 54.1 g H 2 O

Mass-Mass Practice CuSO 4 AlAl 2 (SO 4 ) 3 Cu a.8.5 g Al 1 mol Al g Al = x 75 g CuSO 4 b.8.5 g Al 1 mol Al 2 (SO 4 ) 3 2 mol Al = x 54 g Al 2 (SO 4 ) 3 Mole ratio x 3 mol CuSO 4 2 mol Al 1 mol Al g Al x 1 mol CuSO g Al 2 (SO 4 ) 3 1 mol Al 2 (SO 4 ) 3 x x g CuSO 4

Mass-Mass Practice c.8.5 g Al 3 mol Cu 2 mol Al = x 30. g Cu 1 mol Al g Al xx g Cu 1 mol Cu