Presentation on theme: "Unit 12 Chemistry Langley"— Presentation transcript:
1Unit 12 Chemistry Langley STOICHIOMETRY BASICSUnit 12ChemistryLangley**Corresponds to Chapter 12 in the Prentice Hall Chemistry Textbook
2INTRODUCTION Stoichiometry Dervived from two Greek words meaning measured and element4 typesMole-MoleMass-MoleMass-MassVolume-VolumeMathematically basedBalanced equations very important; first place to start
3Interpreting Chemical Reaction From a balanced chemical equation, you can describe a chemical reaction in terms of quantities of products and reactants. These quantities include the number of molecules, atoms, moles, mass, and volume of the compounds used as well as each element used to make the respective compounds
4Interpreting a Chemical Reaction N2 + 3H2 2NH3 (Figure 12.3, pg. 357)Number of atoms: 2 atoms of nitrogen, 6 atoms of hydrogen, 8 atoms of ammoniaNumber of molecules: 1 molecule of nitrogen, 3 molecules, 2 molecules of ammonia (1:3:2 ratio)Number of relative moles: 1 mole of nitrogen, 3 relative moles of hydrogen, 2 relative moles of ammoniaMolecules and moles are not the same thingMass: g of nitrogen, 6.0 g of hydrogen, 34 g of ammoniumVolume: 22.4L of nitrogen, 67.2L of hydrogen, 44.8L of ammoniaBecause 1 mol of any gas at STP occupies a volume of 22.4L
5Molar Mass of Compounds The molar mass (MM) of a compound is determined the same way, except now you 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 = g/mol Ca + (2 X 35.45) g/mol Cl g/mol CaCl220Ca 17 Cl35.45
6Molar Mass Calculation Calculate the Molar Mass of calcium phosphateFormula = (NH4)3(PO4)Masses elements:Molar Mass =
7Atoms 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)
8Everything must go through Moles!!! Calculationsmolar mass Avogadro’s number Grams Moles particlesEverything must go through Moles!!!
9Chocolate 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?
10Recipes and EquationsJust 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
11Chemistry 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?
12Starting Point-Balanced Equation Write 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?
13Mole-Mole Calculations These 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 1 mol Cl22 mol Na= 2.5 moles Cl2
14Mole-Mole Calculations How many moles of ammonia are produced when 0.60 moles of nitrogen reacts with hydrogen?N2(g) + 3H2 (g) 2NH3(g)2.0 mol NH30.6 mol N2= 1.2 mol NH31.0 mol N2
15Mass-Mole Calculations Sometimes you are going to start with mass and will have to 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
16Mass-Mole Calculations Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide
17Mole-Mass Calculations 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 gramsExample: 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 1 mol Cl g Cl22 mol Na 1 mol Cl2= 177g Cl2
18Mole-Mass Calculations Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.
19Mass-Mass Calculations 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
20Mass-Mass Calculations Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.N2 + 3 H2 2 NH32.00g N mol N mol NH g NH328.02g N2 1 mol N mol NH3= 2.4 g NH3
21Mass-Mass Calculations How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?
22Volume-Volume Calculations 1 mole of any substance = 22.4LNitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of STP.
23Volume-Volume Calculations 2NO + O2(g) 2NO2(g)Know: volume of oxygen = 34L O22 mol NO2/1 mol O21 mol O2 = 22.4 L O2 (at STP)1 mole NO2 = 22.4 L NO (at STP)34 L O21 mol O22 mol NO222.4 L NO2= 64 L O222.4 L O21 mol O21 mol NO2
24Limiting Reactants 1 cup butter 1/2 cup white sugar 1 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?
25Limiting ReactantMost of the time 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 This is called the limiting reactant.
26Limiting ReactantTo find the correct answer, we have to try all of the reactants. We have to 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!
27Limiting Reactant: Limiting Reactant 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
28Limiting ReactantWe 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
29Limiting Reactant15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.
30Finding the Amount of Excess By calculating the amount of the excess 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 previous problem?
31Finding Excess15.0 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!
32Limiting Reactant: Recap 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.You can recognize a limiting reactant problem because there is MORE THAN ONE 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!
33Empirical vs. Molecular Formulas Empirical Formula of a compound shows the smalles whole number ration of the atoms in the compoundA compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the emprical formula of the compound?25.9 g N * (1 mol/14.0g) = 1.85 mol N74.1 g O * (1 mol O/16.0g) = 4.63 mol OEmpirical Formula = N2O5
34Empirical vs. Molecular Formulas Molecular Formula tells you the actual number of each kind of atom present in a molecule of the compoundCalculate the molecular formula of a compound whose molar mass is 60.0 g and empirical formula is CH4N.Find empirical formula mass: =30gMolar mass/efm = 60/30 = 2Molecular Formula = C2H8N2
35Percent, Theoretical, and Actual Yield Actual Yield – the amount produced in a reaction (amount of products)Theoretical Yield – the amount that should have been produced in a reactionMaximum amount of product that could be formed from given amount of reactantsPercent Yield – (actual/theoretical) X 100%Often less than the theoretical