Stoichiometry and the Mole Chapter 9 Stoichiometry and the Mole

9.2 The Mole The mole Relates number of atoms/molecules to grams Conversion unit that is necessary due to small size of individual atoms and molecules Equal to a set number of particles

9.2 The Mole (Continued) Consider the chemical equation : H2 + I2  2HI This equation expresses amounts in molecules and atoms: In terms of moles:

9.2 The Mole (Continued) Molar mass Allows conversion between mass and moles Simply replace atomic mass amu with grams to find the mass of one mole.

9.2 The Mole (Continued) Suppose you obtain 393.934 g of gold. (a) What is the molar mass of gold? (b) How many moles of gold do you have? (c) How many atoms of gold do you have?

Molar mass of molecules 9.2 The Mole (Continued) Molar mass of molecules Number of atoms = number of moles Consider CO2…

Conversions with molar mass 9.2 The Mole (Continued) Conversions with molar mass 1 mole of water has a mass of 18.015 g/mol. Moles= How many moles of water are in 250.0 g of H2O? How many grams of water are in 13.88 moles of water?

9.2 The Mole (Continued) Stoichiometry Conversions

Part 2 9.3 Reaction Stoichiometry According to the equation below… H2 + I2  2HI “One mole of H2 reacts with one mole of I2 to form 2 moles of HI.” What do we say if there are 2 moles of each reactant?

9.3: Reaction Stoichiometry Theoretical yield Maximum amount of product that can be produced from a certain reactant amount

9.3 Reaction Stoichiometry Actual yield Amount actually produced in a chemical reaction Often is less than theoretical yield due to side reactions going on simultaneously Is directly measured rather than derived (the mass of the ppt from lab is the actual yield!)

9.3 Reaction Stoichiometry Percent yield Percentage of theoretical yield that was measured What would be the percent yield if 225.10 g of HI was actually produced?

9.3 Reaction Stoichiometry Practice Calculating Actual Yield from a Reaction- STEP BY STEP! 4 Al + 3 O2  2 Al2O3 Step 1: Determine the molar ratio “__moles of Al react with__ moles of oxygen to produce ___moles of aluminum oxide”

9.3 Reaction Stoichiometry Calculating Yield from a reaction: 4 Al + 3 O2  2 Al2O3 Step 2: Convert moles to masses Al: 4 moles x ______ g/mol = O2 : 3 mol x ______ g/mol= Aluminum Oxide 3mol x _______g/mol=

9.3 Reaction Stoichiometry Calculating Yield from a reaction: 4 Al + 3 O2  2 Al2O3 Therefore… 4 mol of Al require 3mol O2 to completely react ANOTHER INTERPRETATION… 4 mol of Al react with 3mol O2 to produce 2mol Al2O3

9.3 Reaction Stoichiometry (Continued) Calculating Yield from a reaction: 4 Al + 3 O2  2 Al2O3 Step 3: Calculate % Yield The theoretical yield = 203.86g Al2O3 After the reaction is complete, the actual yield =200.3g Calculate the % yield

9.4 Dealing with a Limiting Reactant Reactant that is in short supply “Limits” the amount of product that can be made Will be completely used up in a reaction The reactants that is consumed (runs out) first Excess reactant Present in excess in a reaction Will be left over after the reaction stops Is done sometimes to make sure a reaction goes to completion (to make sure there’s enough)

9.4 Dealing with a Limiting Reactant Finding the limiting reactant Write out the balanced equation Calculate the number of moles of each reactant Divide moles by the coefficient in the balanced equation. The smallest mole-to-coefficient ratio is the limiting reactant! SIMPLE!!

9.4 Dealing with a Limiting Reactant 6 moles of propane are available to react with 29 moles of oxygen…What is the limiting reactant? Balanced Equation: C3H8 + O2  CO2 + H2O

9.4 Limiting Reactant Look Familiar? 4 Al + 3 O2  2 Al2O3 Find the limiting reactant if 13.3mol of Al react with 11.8mol O2

9.4 Limiting Reactant Look Familiar? 4 Al + 3 O2  2 Al2O3 Find the limiting reactant if 13.3mol of Al react with 11.8mol O2 Al: O2:

9.4 Limiting Reactant What happens with yield when there’s a limiting reactant? Yield is calculated based on a revised molar ratio… 4 Al + 3 O2  2 Al2O3 Molar ratio = 4: 3: 2 IF 13.3mol of Al reacts (which makes it limiting) (see previous page)… Theoretical to actual molar ratio of Al is proportional to theoretical to actual molar ratio of Al2O3…

9.4 Limiting Reactant What happens with yield when there’s a limiting reactant? Yield is calculated based on a revised molar ratio… 4 Al + 3 O2  2 Al2O3 Molar ratio = 4: 3: 2 IF 13.3mol of Al reacts, 6.65 moles of Al2O3 are produced

9.4 Limiting Reactant What happens with yield when there’s a limiting reactant? Yield is calculated based on a revised molar ratio… 4 Al + 3 O2  2 Al2O3 The actual yield must be measured in the lab! If the actual yield is only 612g of Al2O3 what is the % yield of Al2O3?