Chapter 12 Stoichiometry - PC\ MAC

Chapter 12“Stoichiometry”Mr. Mole

Let’s make some Cookies! Whenbaking cookies, a recipeis usually used, telling the exactamount of each ingredient. If you need more, you candouble or triple the amount Thus,a recipe is much like abalanced equation.

Stoichiometry is Greek for “measuring elements”Pronounced “stoy kee ahm uh tree” Definedas: calculations of thequantities in chemical reactions,based on a balanced equation.are 4 ways to interpret abalanced chemical equation There

#1. In terms of Particles An Element is made of atoms A Molecular compound (madeof only nonmetals) is made upof molecules (Don’t forget the diatomic elements) Ionic Compounds (made of ametal and nonmetal parts) aremade of formula units

Example: 2H2 O2 2H2O Twomolecules of hydrogen and onemolecule of oxygen form two molecules ofwater. Another example: 2Al2O3 Al 3O22 formula units Al2O3 form 4 atoms Aland 3 molecules O2Now read this: 2Na 2H2O 2NaOH H2

#2. In terms of Moles Thecoefficients tell us howmany moles of each substance2Al2O3 Al 3O22Na 2H2O 2NaOH H2 Remember:A balancedequation is a Molar Ratio

#3. In terms of Mass TheLaw of Conservation of Mass applies We can check mass by using moles.2H2 O2 2H2O2 moles H21 mole O22.02 g H21 mole H232.00 g O21 mole O2 4.04 g H2 32.00 g O236.04 gg HH22 O236.04reactants

In terms of Mass (for products)2H2 O2 2H2O2 moles H2O18.02 g H2O 36.04 g H2O1 mole H2O36.04 g H2 O2 36.04 g H2O36.04 grams reactant 36.04 grams productThe mass of the reactants mustequal the mass of the products.

#4. In terms of Volume AtSTP, 1 mol of any gas 22.4 L2H2 O2 2H2O(2 x 22.4 L H2) (1 x 22.4 L O2) (2 x 22.4 L H2O)67.2 Liters of reactant 44.8 Liters of product!NOTE: mass and atoms areALWAYS conserved - however,molecules, formula units, moles, andvolumes will not necessarily beconserved!

Practice: Showthat the followingequation follows the Law ofConservation of Mass (showthe atoms balance, and themass on both sides is equal)2Al2O3 Al 3O2

Section 12.2Chemical Calculations OBJECTIVES: Construct “mole ratios” frombalanced chemicalequations, and apply theseratios in mole-molestoichiometric calculations.

Chemical Calculations OBJECTIVES: Calculate stoichiometricquantities from balancedchemical equations usingunits of moles, mass,representative particles, andvolumes of gases at STP.

Mole to Mole conversions2Al2O3 Al 3O2 each time we use 2 moles of Al2O3we will also make 3 moles of O22 moles Al2O33 mole O2or3 mole O22 moles Al2O3These are the two possible conversionfactors to use in the solution of the problem.

Mole to Mole conversions Howmany moles of O2 areproduced when 3.34 moles ofAl2O3 decompose?2Al2O3 Al 3O23.34 mol Al2O33 mol O22 mol Al2O3 5.01 mol O2Conversion factor from balanced equationIf you know the amount of ANY chemical in the reaction,you can find the amount of ALL the other chemicals!

Practice:2C2H2 5 O2 4CO2 2 H2O If 3.84 moles of C2H2 are burned, howmany moles of O2 are needed?(9.6 mol) How many moles of C2H2 are needed toproduce 8.95 mole of H2O? (8.95 mol) If 2.47 moles of C2H2 are burned, howmany moles of CO2 are formed? (4.94 mol)

How do you get good at this?

Steps to CalculateStoichiometric Problems1. Correctly balance the equation.2. Convert the given amount intomoles.3. Set up mole ratios.4. Use mole ratios to calculate molesof desired chemical.5. Convert moles back into final unit.

Mass-Mass Problem:6.50 grams of aluminum reacts with an excess ofoxygen. How many grams of aluminum oxide areformed?4Al 3O2 2Al2O36.50 g Al1 mol Al2 mol Al2O3 101.96 g Al2O326.98 g Al4 mol Al1 mol Al2O3(6.50 x 1 x 2 x 101.96) (26.98 x 4 x 1) ? g Al2O312.3 g Al2O3are formed

Another example: If10.1 g of Fe are added to asolution of Copper (II) Sulfate, howmany grams of solid copper wouldform?2Fe 3CuSO4 Fe2(SO4)3 3CuAnswer 17.2 g Cu

Volume-Volume Calculations: Howmany liters of CH4 at STP arerequired to completely react with 17.5 Lof O2 ?CH4 2O2 CO2 2H2O1 mol O2 1 mol CH4 22.4 L CH417.5 L O222.4 L O2 2 mol O2 1 mol CH4 8.75 L CH4Notice anything relating these two steps?

Avogadro told us: Equalvolumes of gas, at the sametemperature and pressure containthe same number of particles. Moles are numbers of particles You can treat reactions as if theyhappen liters at a time, as long asyou keep the temperature andpressure the same. 1 mole 22.4 L @ STP

Shortcut for Volume-Volume? Howmany liters of CH4 at STP arerequired to completely react with 17.5 Lof O2?CH4 2O2 CO2 2H2O17.5 L O21 L CH42 L O2 8.75 L CH4Note: This only works forVolume-Volume problems.

Section 12.3Limiting Reagent & Percent Yield OBJECTIVES: Identify the limiting reagent ina reaction.

Section 12.3Limiting Reagent & Percent Yield OBJECTIVES: Calculate theoretical yield,percent yield, and the amountof excess reagent that remainsunreacted given appropriateinformation.

“Limiting” Reagent Ifyou are given one dozen loaves ofbread, a gallon of mustard, and threepieces of salami, how many salamisandwiches can you make? The limiting reagent is the reactant you runout of first. The excess reagent is the one you haveleft over. The limiting reagent determines how muchproduct you can make

Limiting Reagents - Combustion

How do you find out which is limited? Thechemical that makes the leastamount of product is the “limitingreagent”. You can recognize limiting reagentproblems because they will giveyou 2 amounts of chemical Do two stoichiometry problems,one for each reagent you are given.

If10.6 g of copper reacts withCugramsis the of the3.83 g sulfur, how manyLimitingproduct (copper (I) sulfide)will be formed?Reagent,2Cu S Cu2Ssince it1 mol Cu2S 159.16 g Cu2S1molCuproduced less10.6 g CuCu63.55g Cu 2 mol1 mol Cu2Sproduct. 13.3 g Cu2S1molS3.83 g S32.06g S1 mol Cu2S 159.16 g Cu2S1 mol S1 mol Cu2S 19.0 g Cu2S

Another example: If 10.3 g of aluminum arereacted with 51.7 g of CuSO4how much copper (grams) willbe produced?2Al 3CuSO4 3Cu Al2(SO4)3the CuSO4 is limited, so Cu 20.6 g Howmuch excess reagent willremain? Excess 4.47 grams

The Concept of:A little different type of yield thanyou had in Driver’s Educationclass.

What is Yield? Yieldis the amount of product made ina chemical reaction. There are three types:1. Actual yield- what you actually get inthe lab when the chemicals are mixed2. Theoretical yield- what the balancedequation tells should be made3. Percent yield Actualx 100Theoretical

Example: 6.78g of copper is produced when3.92 g of Al are reacted with excesscopper (II) sulfate.2Al 3 CuSO4 Al2(SO4)3 3Cu Whatis the actual yield? Whatis the theoretical yield? 13.8 g Cu Whatis the percent yield? 6.78 g Cu 49.1 %

Details on Yield Percentyield tells us how “efficient” areaction is.yield can not be biggerthan 100 %. Theoretical yield will always be largerthan actual yield! Percent Why? Due to impure reactants; competingside reactions; loss of product in filtering ortransferring between containers; measuring

Chapter 12 "Stoichiometry" . 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. Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of