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“Stoichiometry” Mr. Mole

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First… A bit of review

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**The Arithmetic of Equations**

OBJECTIVES: Interpret balanced chemical equations in terms of: a) moles b) representative particles c) mass

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**Remember Avocado, I mean, AVOGADRO?**

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**Remember Avocado, I mean, AVOGADRO?**

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**Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023**

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

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**Remember calculating Molar Mass?**

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Molar Mass Molar mass is determined by adding the atomic masses for the atoms, molecules or compounds you are working with Ex. Molar mass of CaCl2 Avg. Atomic mass of Calcium = 40g Avg. Atomic mass of Chlorine = 35.5g Molar Mass of calcium chloride = 40 g/mol Ca + (2 X 35.5) g/mol Cl 111 g/mol CaCl2 20 Ca 40 17 Cl 35.5

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**Practice 310.3 g/mol Calculate the Molar Mass of calcium phosphate**

Formula = Masses elements: Ca: 3 Ca’s X 40 = P: 2 P’s X 31 = O: 8 O’s X 16 = Molar Mass = Ca3(PO4)2 120 g 62 g 128 g 120g Ca + 62g P + 128g O = 310.3 g/mol

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**Let’s make some Cookies!**

When baking cookies, a recipe is usually used, telling the exact amount of each ingredient. If you need more, you can double or triple the amount Thus, a recipe is much like a balanced equation.

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**Chocolate Chip Cookies!!**

1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 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 many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

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**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

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Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H2 + O2 2 H2O How many moles of reactants are needed? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? 2 mol H2 1 mol O2 4 mol H2 2 mol O2 6 mol H2, 6 mol H2O 25 mol O2, 50 mol H2O

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**Stoichiometry is… Greek for “measuring elements”**

Pronounced “stoy kee ahm uh tree” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There are 4 ways to interpret a balanced chemical equation

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**#1. In terms of Particles An Element is made of atoms**

A Molecular compound (made of only nonmetals) is made up of molecules (Don’t forget the diatomic elements) Ionic Compounds (made of a metal and nonmetal parts) are made of formula units

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Example: 2H2 + O2 → 2H2O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. Another example: 2Al2O3 ® 4Al + 3O2 2 formula units Al2O3 form 4 atoms Al and 3 molecules O2 Now read this: 2Na + 2H2O ® 2NaOH + H2

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**Remember: A balanced equation is a Molar Ratio**

#2. In terms of Moles The coefficients tell us how many moles of each substance 2Al2O3 ® 4Al + 3O2 2Na + 2H2O ® 2NaOH + H2 Remember: A balanced equation is a Molar Ratio

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Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl2 2 NaCl 5 moles Na ? mol Cl2 X 2 mol Na 1 mol Cl2 = 2.5 moles Cl2

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**#3. In terms of Mass 2H2 + O2 ® 2H2O**

The Law of Conservation of Mass applies We can check mass by using moles. 2H2 + O2 ® 2H2O 2 g H2 2 moles H2 = 4 g H2 1 mole H2 + 32 g O2 1 mole O2 = 32 g O2 1 mole O2 36 g H2 + O2 36.04 g H2 + O2 reactants

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**In terms of Mass (for products)**

2H2 + O2 ® 2H2O 18.02 g H2O = 36.04 g H2O 2 moles H2O 1 mole H2O 36.04 g H2 + O2 = 36.04 g H2O 36.04 grams reactant = grams product The mass of the reactants must equal the mass of the products.

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Practice: Show that the following equation follows the Law of Conservation of Mass (show the atoms balance, and the mass on both sides is equal) 2Al2O3 ® 4Al + 3O2

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**Chemical Calculations**

OBJECTIVES: Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, and representative particles

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**Chemical Calculations**

Basic equation for all of these: Given amount Unknown amount X Recipe amount Recipe amount Cross multiply and you got this!

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**Mole to Mole conversions**

How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2Al2O3 ® 4Al + 3O2 3.34 mol Al2O ? O2 X 2 mol Al2O mol O2 = 5.01 mol O2 If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals!

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**How do you get good at this?**

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Practice: 2C2H2 + 5 O2 ® 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? How many moles of C2H2 are needed to produce 8.95 mole of H2O? If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

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Practice: 2C2H2 + 5 O2 ® 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?

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Practice: 2C2H2 + 5 O2 ® 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? How many moles of C2H2 are needed to produce 8.95 mole of H2O? If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

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Practice: 2C2H2 + 5 O2 ® 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? (9.6 mol) How many moles of C2H2 are needed to produce 8.95 mole of H2O? (8.95 mol) If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? (4.94 mol)

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**How do you get good at this?**

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**Steps to Calculate Stoichiometric Problems**

Correctly balance the equation. Place the given amount over the recipe amount. Place the unknown amount over the recipe amount. Cross multiply to solve

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Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu

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Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form?

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**Another example: Answer = 17.2 g Cu 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu**

If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu Answer = 17.2 g Cu

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**Section 12.3 Limiting Reactant & Percent Yield**

OBJECTIVES: Identify the limiting reactant in a reaction.

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**Section 12.3 Limiting Reactant & Percent Yield**

OBJECTIVES: Calculate theoretical yield, percent yield, and the amount of excess reactant that remains unreacted given appropriate information.

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“Limiting” Reactant If you are given one dozen loaves of bread and three pieces of salami, how many salami sandwiches can you make? The limiting reactant is the reactant you run out of first. The excess reactant is the one you have left over. The limiting reactant determines how much product you can make

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**Limiting Reactant: Cookies**

1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

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**How do you find out which is limited?**

The chemical that makes the least amount of product is the “limiting reactant”. You can recognize limiting reactant problems because they will give you 2 amounts of chemical Do two stoichiometry problems, one for each reactant you are given.

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**Cu is the Limiting Reactant, since it produced less product.**

If 10.6 g of copper reacts with g sulfur, how many grams of the product (copper (I) sulfide) will be formed? Cu is the Limiting Reactant, since it produced less product. = 13.3 g Cu2S = 13.3 g Cu2S = 19.0 g Cu2S

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**Another example: Al + CuSO4 → Cu + Al2(SO4)3**

If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper (grams) will be produced? Al + CuSO4 → Cu + Al2(SO4)3

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**Another example: 2Al + 3CuSO4 → 3Cu + Al2(SO4)3**

If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper (grams) will be produced? 2Al + 3CuSO4 → 3Cu + Al2(SO4)3

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Another example: In the preceding reaction, how much of the excess reactant will remain? Remember: the CuSO4 is limited, so the Cu produced was 20.6 g. Flip the equation around to see how much Al was used to make this much Cu. In other words…

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Another example: 20.6g of Cu was produced in the following reaction. How much Al was required? 2Al + 3CuSO4 → 3Cu + Al2(SO4)3

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Another example: Next, subtract the needed amount from the given amount from the first problem. If 10.3 g of aluminum are reacted with 51.7g of CuSO4 how much copper (grams) will be produced?

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**How much excess reagent will remain?**

Another example: If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper (grams) will be produced? 2Al + 3CuSO4 → 3Cu + Al2(SO4)3 the CuSO4 is limited, so Cu = 20.6 g How much excess reagent will remain? Excess = 4.47 grams

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The Concept of: A little different type of yield than you learned in Driver’s Ed…

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What is Yield? Yield is the amount of product made in a chemical reaction. There are three types: 1. Actual yield- what you actually get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield = Actual Theoretical x 100

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Example: 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield? = 6.78 g Cu = 13.8 g Cu = 49.1 %

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**Details on Yield Percent yield tells us how “efficient” a reaction is.**

Percent yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring

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**Since you now know all you need to do,**

GO DO!!!

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Chapter 12 Stoichiometry Mr. Mole. Molar Mass of Compounds Molar mass (MM) of a compound - determined by up the atomic masses of – Ex. Molar mass of CaCl.

Chapter 12 Stoichiometry Mr. Mole. Molar Mass of Compounds Molar mass (MM) of a compound - determined by up the atomic masses of – Ex. Molar mass of CaCl.

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