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Stoichiometry

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**Molar Mass of Compounds**

Molar Mass (MM) = add 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 g/mol CaCl2 20 Ca 17 Cl 35.45

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**Atoms or Molecules Moles Volume (liters) Mass (grams)**

Multiply by 6.02 X 1023 Divide by 6.02 X 1023 Moles Multiply by atomic/molar mass from periodic table Divide 22.4 Multiply by STP Volume (liters) Divide by atomic/molar mass from periodic table Mass (grams)

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**Other Stoichiometric Calculations**

Problem-Solving Approach With your knowledge of conversion factors and this problem-solving approach, you can solve a variety of stoichiometric problems. Identifying What conversion factor is used to convert moles to representative particles?

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**Practice Calculate the Molar Mass of calcium phosphate Ca3(PO4)2**

Formula = Masses elements: Molar Mass = Ca3(PO4)2

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**Everything must go through Moles!!!**

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

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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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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! Example: 2 Na + Cl2 2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

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Mole -Mole These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical EX: 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 1 mol Cl2 2 mol Na = 2.5 moles Cl2

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Volume-Volume How many liters of CH4 at STP are required to completely react with 17.5 L of O2? CH4 + 2O2 ® CO2 + 2H2O 1 L CH4 17.5 L O2 = 8.75 L CH4 2 L O2

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**Mole-Mass Conversions**

Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams EX: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2 2 NaCl 5.00 moles Na mol Cl g Cl2 2 mol Na 1 mol Cl2 = 177g Cl2

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Mass-Mole We can start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water 2 C2H6 + 7 O2 4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 18.0 g H2O 6 mol H20 = mol C2H6

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**Mass-Mass Conversions**

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, mole ratio, and back to grams of compound we are interested in

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Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + 3 H2 2 NH3 the bridge 2.00g N mol N mol NH g NH3 28.02g N2 1 mol N mol NH3 = 2.4 g NH3

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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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Limiting Reactant Sometimes in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

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Limiting Reactant To find the correct answer, try all of the reactants. Calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

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**Limiting Reactant: Example**

10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3 Start with Al: Now Cl2: 10.0 g Al 1 mol Al mol AlCl g AlCl3 27.0 g Al mol Al mol AlCl3 = 49.4g AlCl3 35.0g Cl mol Cl mol AlCl g AlCl3 71.0 g Cl mol Cl mol AlCl3 = 43.9g AlCl3

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**Limiting reaction Example**

We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete

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**Limiting Reactant Practice**

15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

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**Finding the Amount of Excess**

By calculating the amount of the excess reactant needed to completely react with the limiting reactant, subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem?

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**Finding Excess Practice**

15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I mol I mol K g K 254 g I mol I mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

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**Limiting Reactant: Recap**

You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. Convert ALL of the reactants to the SAME product (pick any product you choose.) The lowest answer is the correct answer. The reactant that gave you the lowest answer is the LIMITING REACTANT. The other reactant(s) are in EXCESS. To find the amount of excess, subtract the amount used from the given amount. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

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

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**Mass-Mass Problem: 4Al + 3O2 2Al2O3 12.3 g Al2O3 are formed**

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al O2 2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 g Al2O3 = ? g Al2O3 26.98 g Al 4 mol Al 1 mol Al2O3 12.3 g Al2O3 are formed (6.50 x 1 x 2 x ) ÷ (26.98 x 4 x 1) =

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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 Answer = 17.2 g Cu

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**Volume-Volume Calculations:**

How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ? CH4 + 2O2 ® CO2 + 2H2O 1 mol CH4 17.5 L O2 = 8.75 L CH4 2 mol O2

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**Limiting Reagent & Percent Yield**

OBJECTIVES: Identify the limiting reagent in a reaction. Calculate theoretical yield, percent yield, and the amount of excess reagent that remains unreacted given appropriate information

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

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**Limiting Reagents - Combustion**

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

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

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**Cu is the Limiting Reagent, 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? 2Cu + S ® Cu2S Cu is the Limiting Reagent, since it produced less product. 1 mol Cu 1 mol Cu2S g Cu2S 10.6 g Cu 63.55g Cu 2 mol Cu 1 mol Cu2S = 13.3 g Cu2S 1 mol S 1 mol Cu2S g Cu2S 3.83 g S 32.06g S 1 mol S 1 mol Cu2S = 19.0 g Cu2S

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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 had in Driver’s Education class.

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

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

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