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Stoichiometry Molar Mass of Compounds Molar Mass (MM) = add up all the atomic masses for the molecule (or compound) Molar Mass (MM) = add up all the.

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Presentation on theme: "Stoichiometry Molar Mass of Compounds Molar Mass (MM) = add up all the atomic masses for the molecule (or compound) Molar Mass (MM) = add up all the."— Presentation transcript:

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2 Stoichiometry

3 Molar Mass of Compounds Molar Mass (MM) = add up all the atomic masses for the molecule (or compound) Molar Mass (MM) = add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl 2 Ex. Molar mass of CaCl 2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = Molar Mass of calcium chloride = g/mol Ca + (2 X 35.45) g/mol Cl g/mol Ca + (2 X 35.45) g/mol Cl g/mol CaCl 2 20 Ca 40.08a 17 Cl Cl

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

5 Other Stoichiometric Calculations Problem-Solving Approach

6 Practice Calculate the Molar Mass of calcium phosphate Calculate the Molar Mass of calcium phosphate Formula = Formula = Masses elements: Masses elements: Molar Mass = Molar Mass = Ca 3 (PO 4 ) 2

7 molar mass Avogadros number Grams Moles particles molar mass Avogadros number Grams Moles particles Everything must go through Moles!!! Calculations

8 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?

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

10 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) 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! Be sure you have a balanced reaction before you start! Example: Example: 2 Na + Cl 2 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 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? What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

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

12 Volume-Volume How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2O 2 CO 2 + 2H 2 O CH 4 + 2O 2 CO 2 + 2H 2 O 17.5 L O2O2 2 L O 2 1 L CH 4 = 8.75 L CH 4

13 Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles 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 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? EX: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2 2 NaCl 5.00 moles Na 1 mol Cl g Cl 2 2 mol Na 1 mol Cl 2 = 177g Cl 2

14 Mass-Mole We can start with mass and convert to moles of product or another reactant 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 We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water 2 C 2 H O 2 4 CO H C 2 H O 2 4 CO H g H 2 O 1 mol H 2 O 2 mol C 2 H g H 2 O 6 mol H 2 0 = mol C 2 H 6

15 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 ) or 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 ( how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from, mole ratio, and of compound we are interested in Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

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

17 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?

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

19 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. 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 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! 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 cant compare to see which is greater and which is lower unless the product is the same! Be sure to pick a product! You cant compare to see which is greater and which is lower unless the product is the same!

20 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? 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 Cl 2 2 AlCl 3 Start with Al: Start with Al: Now Cl 2 : Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl g AlCl g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl g Cl 2 1 mol Cl 2 2 mol AlCl g AlCl g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

21 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. 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. Once the 35.0g of chlorine is used up, the reaction comes to a complete. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete.

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

23 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. 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? Can we find the amount of excess potassium in the previous problem?

24 Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2 2 KI 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2 2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 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!

25 Limiting Reactant: Recap 1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2. Convert ALL of the reactants to the SAME product (pick any product you choose.) 3. The lowest answer is the correct answer. 4. The reactant that gave you the lowest answer is the LIMITING REACTANT. 5. The other reactant(s) are in EXCESS. 6. To find the amount of excess, subtract the amount used from the given amount. 7. If you have to find more than one product, be sure to start with the limiting reactant. You dont have to determine which is the LR over and over again!

26 How do you get good at this?

27 Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 2 2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O g Al 2 O 3 (6.50 x 1 x 2 x ) ÷ (26.98 x 4 x 1) = 12.3 g Al 2 O 3 are formed

28 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? If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4 Fe 2 (SO 4 ) 3 + 3Cu 2Fe + 3CuSO 4 Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu

29 Volume-Volume Calculations: How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2O 2 CO 2 + 2H 2 O CH 4 + 2O 2 CO 2 + 2H 2 O 17.5 L O 2 2 mol O2O2 1 CH 4 = 8.75 L CH 4

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

31 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? 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 limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make The limiting reagent determines how much product you can make

32 Limiting Reagents - Combustion

33 How do you find out which is limited? The chemical that makes the least amount of product is the limiting reagent. 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 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. Do two stoichiometry problems, one for each reagent you are given.

34 If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed? If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed? 2Cu + S Cu 2 S 2Cu + S Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S g Cu 2 S = 19.0 g Cu 2 S Cu is the Limiting Reagent, since it produced less product.

35 Another example: If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? 2Al + 3CuSO 4 3Cu + Al 2 (SO 4 ) 3 the CuSO 4 is limited, so Cu = 20.6 g the CuSO 4 is limited, so Cu = 20.6 g How much excess reagent will remain? How much excess reagent will remain? Excess = 4.47 grams

36 The Concept of: A little different type of yield than you had in Drivers Education class.

37 What is Yield? Yield is the amount of product made in a chemical reaction. Yield is the amount of product made in a chemical reaction. There are three types: 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

38 Example: 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4 Al 2 (SO 4 ) 3 + 3Cu 2Al + 3 CuSO 4 Al 2 (SO 4 ) 3 + 3Cu What is the actual yield? What is the actual yield? What is the theoretical yield? What is the theoretical yield? What is the percent yield? What is the percent yield? = 6.78 g Cu = 13.8 g Cu = 49.1 %

39 Details on Yield Percent yield tells us how efficient a reaction is. Percent yield tells us how efficient a reaction is. Percent yield can not be bigger than 100 %. Percent yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! 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 Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring

40 How do you get good at this?


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