6 Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023 Multiply by 6.02 X 1023MolesMultiply by atomic/molar mass from periodic tableDivide by atomic/molar mass from periodic tableMass (grams)
8 Molar MassMolar mass is determined by adding the atomic masses for the atoms, molecules or compounds you are working withEx. Molar mass of CaCl2Avg. Atomic mass of Calcium = 40gAvg. Atomic mass of Chlorine = 35.5gMolar Mass of calcium chloride = 40 g/mol Ca + (2 X 35.5) g/mol Cl 111 g/mol CaCl220Ca 4017 Cl 35.5
9 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)2120 g62 g128 g120g Ca + 62g P + 128g O =310.3 g/mol
10 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 amountThus, a recipe is much like a balanced equation.
11 Chocolate Chip Cookies!! 1 cup butter1/2 cup white sugar1 cup packed brown sugar1 teaspoon vanilla extract2 eggs2 1/2 cups all-purpose flour1 teaspoon baking soda1 teaspoon salt2 cups semisweet chocolate chipsMakes 3 dozenHow 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?
12 Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as wellInstead of calling them recipes, we call them reaction equationsFurthermore, instead of using cups and teaspoons, we use molesLastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients
13 PracticeWrite the balanced reaction for hydrogen gas reacting with oxygen gas.2 H2 + O2 2 H2OHow 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 O24 mol H2 2 mol O26 mol H2, 6 mol H2O25 mol O2, 50 mol H2O
15 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
16 #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
17 Example: 2H2 + O2 → 2H2OTwo molecules of hydrogen and one molecule of oxygen form two molecules of water.Another example: 2Al2O3 ® 4Al + 3O22formula unitsAl2O3form4atomsAland3moleculesO2Now read this: 2Na + 2H2O ® 2NaOH + H2
18 Remember: A balanced equation is a Molar Ratio #2. In terms of MolesThe coefficients tell us how many moles of each substance2Al2O3 ® 4Al + 3O22Na + 2H2O ® 2NaOH + H2Remember: A balanced equation is a Molar Ratio
19 Mole RatiosThese mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemicalExample: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)?2 Na + Cl2 2 NaCl5 moles Na ? mol Cl2X2 mol Na 1 mol Cl2= 2.5 moles Cl2
20 #3. In terms of Mass 2H2 + O2 ® 2H2O The Law of Conservation of Mass appliesWe can check mass by using moles.2H2 + O2 ® 2H2O2 g H22 moles H2=4 g H21 mole H2+32 g O21 mole O2=32 g O21 mole O236 g H2 + O236.04 g H2 + O2reactants
21 In terms of Mass (for products) 2H2 + O2 ® 2H2O18.02 g H2O=36.04 g H2O2 moles H2O1 mole H2O36.04 g H2 + O2=36.04 g H2O36.04 grams reactant = grams productThe mass of the reactants must equal the mass of the products.
22 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
23 Chemical Calculations OBJECTIVES:Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, and representative particles
24 Chemical Calculations Basic equation for all of these:Given amount Unknown amountXRecipe amount Recipe amountCross multiply and you got this!
25 Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?2Al2O3 ® 4Al + 3O23.34 mol Al2O ? O2X2 mol Al2O mol O2= 5.01 mol O2If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals!
27 Practice:2C2H2 + 5 O2 ® 4CO2 + 2 H2OIf 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?
28 Practice:2C2H2 + 5 O2 ® 4CO2 + 2 H2OIf 3.84 moles of C2H2 are burned, how many moles of O2 are needed?
29 Practice:2C2H2 + 5 O2 ® 4CO2 + 2 H2OIf 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?
30 Practice:2C2H2 + 5 O2 ® 4CO2 + 2 H2OIf 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)
32 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
33 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
34 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?
35 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 + 3CuAnswer = 17.2 g Cu
36 Section 12.3 Limiting Reactant & Percent Yield OBJECTIVES:Identify the limiting reactant in a reaction.
37 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.
38 “Limiting” ReactantIf 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
39 Limiting Reactant: Cookies 1 cup butter1/2 cup white sugar1 cup packed brown sugar1 teaspoon vanilla extract2 eggs2 1/2 cups all-purpose flour1 teaspoon baking soda1 teaspoon salt2 cups semisweet chocolate chipsMakes 3 dozenIf 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?
40 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 chemicalDo two stoichiometry problems, one for each reactant you are given.
41 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
42 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
43 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
44 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…
45 Another example:20.6g of Cu was produced in the following reaction. How much Al was required?2Al + 3CuSO4 → 3Cu + Al2(SO4)3
46 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?
47 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)3the CuSO4 is limited, so Cu = 20.6 gHow much excess reagent will remain?Excess = 4.47 grams
48 The Concept of:A little different type of yield than you learned in Driver’s Ed…
49 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 mixed2. Theoretical yield- what the balanced equation tells should be made3. Percent yield = Actual Theoreticalx 100
50 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 + 3CuWhat is the actual yield?What is the theoretical yield?What is the percent yield?= 6.78 g Cu= 13.8 g Cu= 49.1 %
51 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
52 Since you now know all you need to do, GO DO!!!
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