2Molar Mass of Compounds Molar mass (MM) of a compound - determined by adding up the atomic masses of each elementEx. Molar mass of CaCl2MM of Calcium = g/molMM of Chlorine = 35.45g g/mol x 2 = 70.9 g/molMolar Mass of CaCl2= g/mol Ca g/mol Cl = g/mol CaCl220Ca 17 Cl 35.45
3Flowchart 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)
4Practice Calculate the Molar Mass of calcium phosphide Formula = Masses elements:Ca = g/mol * 3P = g/mol * 2Molar Mass = g/molCa3P2
5Calculationsmolar mass Avogadro’s number Grams Moles particles 22.4 L Volume Everything must go through Moles!!!
6Chocolate Chip Cookies!! 1 cup butter 1 cup packed brown sugar 2 eggs 2 1/2 cups all-purpose flour 2 cups semisweet chocolate chips Makes 3 dozenHow many eggs are needed to make 3 dozen cookies?How many eggs to make 9 dozen cookies?How much brown sugar would I need if I had 4 eggs?262
7Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes called equationsInstead of using cups and teaspoons, we use molesLastly, instead of eggs, butter, sugar, etc. we use compounds
8Chemistry RecipesBalanced reactions tell us how much reactant will react to get a productlike the cookie recipeBe sure you have a balanced equation before you start!Ex: 2 Na + Cl2 2 NaClReaction tells us 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?
9PracticeWrite 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 H2, 1 O24 H2, 2 O26 H2, 6 H2O25 O2, 50 H2O
10Stoichiometry is… Greek for “measuring elements” Pronounced “stoy kee ah muh 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
11#1. In terms of Particles An Element is made of atoms A Molecular compound (made of only nonmetals) is made up of moleculesIonic Compounds (made of a metal and nonmetal parts) are made of formula units
12Example: 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
13#2. In terms of MolesCoefficients tell us how many moles of each substance2Al2O3 ® 4Al + 3O22 mol Al2O3, 4 mol Al, 3 mol O22Na + 2H2O ® 2NaOH + H2Remember: A balanced equation is a Molar Ratio
14#3. In terms of Mass Be + 2F BeF2 The Law of Conservation of Mass appliesWe can check mass by using moles.Be + 2F BeF21 mole Be9.01 g Be=9.01 g Be1 mole Be+19.00 g F2 mole F38.00 g F=1 mole F47.01 g Be + 2F36.04 g H2 + O2reactants
15In terms of Mass (for products) Be + 2F BeF21 moles BeF247.01 g BeF2=47.01 g BeF21 mole BeF247.01 g Be + 2F=47.01 g BeF2reactant = productMass of reactants and products. must be equal
1667.2 Liters of reactant ≠ 44.8 Liters of product! #4. In terms of VolumeAt STP, 1 mol of any gas = 22.4 L2H O ® H2O(2 x 22.4 L H2) + (1 x 22.4 L O2) ® (2 x 22.4 L H2O)**Mass and atoms are ALWAYS conservedhowever, molecules, formula units, moles, and volumes will not necessarily be conserved!67.2 Liters of reactant ≠ 44.8 Liters of product!
17Practice: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 + 3O2Atoms: 4 Al and 6 O = Al and 6 OMass: Reactant: g/mol = Products: 108 g/mol g/mol
18Practice Atoms: 2 N and 6 H = 2 N and 6 H Moles: 1 N2 and 3 H2 ≠ 2 NH3 1). Balance the equation and interpret it in terms of atoms, moles and mass. Show that the law of conservation is observed.N H2--> NH3Atoms: 2 N and 6 H = N and 6 HMoles: 1 N2 and 3 H ≠ NH3Mass: Reactants: g/mol g/mol = Products: g/mol
20Mole to Mole conversions 2Al2O3 ® 4Al + 3O2each time we use 2 moles of Al2O3 we will also make 3 moles of O22 moles Al2O33 mole O2or3 mole O22 moles Al2O3This is how we can convert from mols to mols.This is why we need balanced equations.
21Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?2Al2O3 ® 4Al + 3O23.34 mol Al2O33 mol O2=5.01 mol O22 mol Al2O3Conversion factor from balanced equationIf you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals!
22Practice: 2C2H2 + 5O2 ® 4CO2 + 2H2O 8.95 mol C2H2 9.6 mol O2 If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?3.84 moles C2H25 moles O29.6 mol O22 moles C2H2How many moles of C2H2 are needed to produce 8.95 mole of H2O?2 moles C2H28.95 moles H2O8.95 mol C2H22 moles H2O
23Steps to Calculate Stoichiometric Problems 1. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit.
25Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of molesWe still use the mole ratio, but now we also use molar mass to get to gramsExample: How many grams of chlorine are required to react with 5.00 moles of sodium to produce sodium chloride?2 Na + Cl2 2 NaCl5.00 moles Na 1 mol Cl g Cl22 mol Na 1 mol Cl2= 177g Cl2
26PracticeCalculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.Al + 3I AlI3g I0.50 mol Al3 mol I= g I1 mol I1 mol Al
27Mass-Mole We can also start with mass and convert to moles We 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
28PracticeCalculate how many moles of oxygen are required to make 10.0 g of aluminum oxide2Al + 3O Al2O310.0 g Al2O31 mol Al2O33 mol O2g Al2O31 mol Al2O30.29 g O2
29Mass-Mass Problem: 4Al + 3O2 2Al2O3 12.3 g Al2O3 are formed 6.50 grams of aluminum reacts with 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) =
30Another 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 + 3Cu10.1 g Fe1 mol Fe3 mol Cu63.55 g Cu55.85 g Fe2 mol Fe1 mol CuAnswer = 17.2 g Cu
31Volume-Volume Calculations: How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?CH4 + 2O2 ® CO2 + 2H2O22.4 L O21 mol O21 mol CH41 mol CH422.4 L CH41 mol O222.4 L CH417.5 L O222.4 L O22 mol O21 mol CH4= 8.75 L CH4
33Section 12.3 Limiting Reagent & Percent Yield OBJECTIVES:Identify the limiting reagent in a reaction.
34“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
35How do you find out which is limited? The chemical that makes the least amount of product is the “limiting reagent”.Limiting reagent problems will give you 2 amounts of chemicalsYou must do two stoichiometry problems; one for each reagent given.If 2 products are given, pick one and use it for both calculations
36Cu is the Limiting Reagent, since it produced less product. 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 ® 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
37Finding the Amount of Excess By calculating the amount of the reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.Can we find the amount of excess potassium in the next problem?
38Finding 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.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!
39Another 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)3CuSO4 limiting reactant10.3 g Al1 mol Al3 mol Cu63.55 g Cu= g Cu26.98 g Al2 mol Al1 mol Cu51.7 g CuSO41 mol CuSO43 mol Cu63.55 g Cug CuSO43 mol CuSO41 mol Cu= g Cu
40How much excess reactant do we have from the last problem? CuSO4 limiting reactantExcess Al = g – 5.83 g =51.7 g CuSO41 mol CuSO426.98 g Al2 mol Alg CuSO43 mol CuSO41 mol Al= 5.83 g Al actually used4.47 g Al excess
41Limiting 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 limiting reactantThe 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!
42The Concept of:A little different type of yield than you had in Driver’s Education class.
43What is Yield?Yield is the amount of product made in a chemical reaction.There are three types:1. Theoretical yield- what the balanced equation tells should be made2. Actual yield- what you actually get in the lab when the chemicals are mixed3. Percent yield = Actual Theoreticalx 100%
44Example: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 Cu3.92 g Al1 mol Al3 mol Cu63.55 mol Cu= 13.8 g Cu26.98 g Al2 mol Al1 mol Cu6.78 g CuX 100= 49.1 %13.8 g Cu
45Details 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