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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
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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?
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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?
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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?
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Get on with it! What does this have to do with CHEMISTRY?
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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 + chips (a synthesis reaction) (177ºC) 1 batch of chocolate chip cookies! coefficient unit substance
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Welcome to STOICHIOMETRY
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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.
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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?
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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
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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
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… 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
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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.
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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
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Mole Ratios 2 H 2 + O 2 2 H 2 O
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Practice For each reaction, write all possible mole ratios.
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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
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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 4 0.5 mol Alx 3 mol CuSO 4 2 mol Al = 0.8 mol CuSO 4
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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.
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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 32.00 g O 2 1 mol O 2 x
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Mass-Mole Practice CuSO 4 AlAl 2 (SO 4 ) 3 Cu323 ++ 1. a.13.5 g Al 1 mol Al 26.98 g Al =x 0.751 mol CuSO 4 b. c. 13.5 g Al 1 mol Al 2 (SO 4 ) 3 2 mol Al = x0.250 mol Al 2 (SO 4 ) 3 13.5 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 26.98 g Al x 1 mol Al 26.98 g Al x
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Mole – Mass Practice Ca + AlCl 3 CaCl 2 + Al3232 0.095 mol AlCl 3 x 3 mol Ca 2 mol AlCl 3 x 40.08 g Ca 1 mol Ca =5.7 g Ca
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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 = 32.00 g O 2 1 mol O 2 xx 18.02 g H 2 O 1 mol H 2 O 54.1 g H 2 O
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Mass-Mass Practice CuSO 4 AlAl 2 (SO 4 ) 3 Cu323 ++ 1. a.8.5 g Al 1 mol Al 26.98 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 26.98 g Al x 1 mol CuSO 4 342.14 g Al 2 (SO 4 ) 3 1 mol Al 2 (SO 4 ) 3 x x 159.61 g CuSO 4
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Mass-Mass Practice c.8.5 g Al 3 mol Cu 2 mol Al = x 30. g Cu 1 mol Al 26.98 g Al xx 63.55 g Cu 1 mol Cu
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