2 Molar Mass of Compounds Molar Mass (MM) = add up all the atomic masses for the molecule (or compound)Ex. Molar mass of CaCl2Avg. Atomic mass of Calcium = 40.08gAvg. Atomic mass of Chlorine = 35.45gMolar Mass of calcium chloride =40.08 g/mol Ca + (2 X 35.45) g/mol Cl g/mol CaCl220Ca 17 Cl 35.45
3 Atoms or Molecules Moles Volume (liters) Mass (grams) Multiply by 6.02 X 1023Divide by 6.02 X 1023MolesMultiply by atomic/molar mass from periodic tableDivide 22.4Multiply by STPVolume (liters)Divide by atomic/molar mass from periodic tableMass (grams)
4 Other Stoichiometric Calculations Problem-Solving ApproachWith 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?
5 Practice Calculate the Molar Mass of calcium phosphate Ca3(PO4)2 Formula =Masses elements:Molar Mass =Ca3(PO4)2
6 Everything must go through Moles!!! Calculationsmolar mass Avogadro’s number Grams Moles particlesEverything must go through Moles!!!
7 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?
8 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
9 Chemistry RecipesLooking 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 NaClThis reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chlorideWhat if we wanted 4 moles of NaCl? 10 moles? 50 moles?
10 Mole -MoleThese mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemicalEX: 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 1 mol Cl22 mol Na= 2.5 moles Cl2
11 Volume-VolumeHow many liters of CH4 at STP are required to completely react with 17.5 L of O2?CH4 + 2O2 ® CO2 + 2H2O1 L CH417.5 L O2= 8.75 L CH42 L O2
12 Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of molesWe still go through moles and use the mole ratio, but now we also use molar mass to get to gramsEX: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride?2 Na + Cl2 2 NaCl5.00 moles Na mol Cl g Cl22 mol Na 1 mol Cl2= 177g Cl2
13 Mass-MoleWe can start with mass and convert to moles of product or another reactantWe use molar mass and the mole ratio to get to moles of the compound of interestCalculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water2 C2H6 + 7 O2 4 CO2 + 6 H2010.0 g H2O 1 mol H2O 2 mol C2H618.0 g H2O 6 mol H20= mol C2H6
14 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) orhow 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
15 Mass-Mass ConversionEx. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.N2 + 3 H2 2 NH3the bridge2.00g N mol N mol NH g NH328.02g N2 1 mol N mol NH3= 2.4 g NH3
16 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?
17 Limiting ReactantSometimes 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.
18 Limiting ReactantTo 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!
19 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 AlCl3Start with Al:Now Cl2:10.0 g Al 1 mol Al mol AlCl g AlCl327.0 g Al mol Al mol AlCl3= 49.4g AlCl335.0g Cl mol Cl mol AlCl g AlCl371.0 g Cl mol Cl mol AlCl3= 43.9g AlCl3
20 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
21 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.
22 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?
23 Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KIWe 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 K254 g I mol I mol K= 4.62 g K USED!15.0 g K – 4.62 g K = g K EXCESSGiven amount of excess reactantAmount of excess reactant actually usedNote that we started with the limiting reactant! Once you determine the LR, you should only start with it!
24 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!
26 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 2Al2O36.50 g Al1 mol Al2 mol Al2O3g Al2O3=? g Al2O326.98 g Al4 mol Al1 mol Al2O312.3 g Al2O3are formed(6.50 x 1 x 2 x ) ÷ (26.98 x 4 x 1) =
27 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 + 3CuAnswer = 17.2 g Cu
28 Volume-Volume Calculations: How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?CH4 + 2O2 ® CO2 + 2H2O1 mol CH417.5 L O2= 8.75 L CH42 mol O2
29 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
30 “Limiting” ReagentIf 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
32 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 chemicalDo two stoichiometry problems, one for each reagent you are given.
33 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 ® Cu2SCu is the Limiting Reagent, since it produced less product.1 mol Cu1 mol Cu2Sg Cu2S10.6 g Cu63.55g Cu2 mol Cu1 mol Cu2S= 13.3 g Cu2S1 mol S1 mol Cu2Sg Cu2S3.83 g S32.06g S1 mol S1 mol Cu2S= 19.0 g Cu2S
34 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
35 The Concept of:A little different type of yield than you had in Driver’s Education class.
36 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
37 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 %
38 Percent yield tells us how “efficient” a reaction is. Details on YieldPercent 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