Chapter 3 Stoichiometry - Chemistry

Chapter 3!Stoichiometry:Calculations with ChemicalFormulas and EquationsStoichiometry

Anatomy of a Chemical EquationCH4 (g) 2O2 (g)CO2 (g) 2 H2O (g)Stoichiometry

Anatomy of a Chemical EquationCH4 (g) 2 O2 (g)Reactants appear on theleft side of the equation.CO2 (g) 2 H2O (g)Stoichiometry

Anatomy of a Chemical EquationCH4 (g) 2 O2 (g)CO2 (g) 2 H2O (g)Products appear on theright side of the equation.Stoichiometry

Anatomy of a Chemical EquationCH4 (g) 2 O2 (g)CO2 (g) 2 H2O (g)The states of the reactants and productsare written in parentheses to the right ofeach compound.Stoichiometry

Anatomy of a Chemical EquationCH4 (g) 2 O2 (g)Coefficients are inserted tobalance the equation.CO2 (g) 2 H2O (g)Stoichiometry

Subscripts and Coefficients GiveDifferent Information Subscripts tell the number of atoms ofeach element in a moleculeStoichiometry

Subscripts and Coefficients GiveDifferent Information Subscripts tell the number of atoms ofeach element in a molecule Coefficients tell the number ofmolecules (compounds).Stoichiometry

ReactionTypes!Stoichiometry

Combination Reactions Two or moresubstancesreact to formone product Examples:N2 (g) 3 H2 (g) 2 NH3 (g)C3H6 (g) Br2 (l) C3H6Br2 (l)2 Mg (s) O2 (g) 2 MgO (s)Stoichiometry

2 Mg (s) O2 (g) 2 MgO (s)Stoichiometry

Decomposition Reactions One substance breaks down into two or moresubstances Examples:CaCO3 (s) CaO (s) CO2 (g)2 KClO3 (s) 2 KCl (s) O2 (g)2 NaN3 (s) 2 Na (s) 3 N2 (g)Stoichiometry

Combustion Reactions! Rapid reactions thathave oxygen as areactant sometimesproduce a flame Most often involvehydrocarbons reactingwith oxygen in the air toproduce CO2 and H2O. Examples:CH4 (g) 2 O2 (g) CO2 (g) 2 H2O (g)C3H8 (g) 5 O2 (g) 3 CO2 (g) 4 H2O (g)2H2 O2 ------- 2H2OStoichiometry

FormulaWeights!Stoichiometry

The amu unit Defined (since 1961) as: 1/12 mass of the 12C isotope. 12C 12 amuStoichiometry

Formula Weight (FW)! Sum of the atomic weights for the atomsin a chemical formula So, the formula weight of calciumchloride, CaCl2, would beCa: 1(40.1 amu) Cl: 2(35.5 amu)111.1 amu These are generally reported for ioniccompoundsStoichiometry

Molecular Weight (MW) Sum of the atomic weights of the atomsin a molecule For the molecule ethane, C2H6, themolecular weight would beC: 2(12.0 amu) H: 6(1.0 amu)30.0 amuStoichiometry

Percent CompositionOne can find the percentage of the massof a compound that comes from each ofthe elements in the compound by usingthis equation:% element (number of atoms)(atomic weight)(FW of the compound)x 100Stoichiometry

Percent CompositionSo the percentage of carbon andhydrogen in ethane (C2H6, molecularmass 30.0) is:%C %H (2)(12.0 amu)(30.0 amu)(6)(1.01 amu)(30.0 amu) 24.0 amu30.0 amu6.06 amu30.0 amux 100 80.0%x 100 20.0%Stoichiometry

MolesStoichiometry

Atomic mass unit and the mole amu definition: 12C 12 amu. The atomic mass unit is defined thisway. 1 amu 1.6605 x 10-24 g How many 12C atoms weigh 12 g? 6.02x1023 12C weigh 12 g. Avogadroʼs number The moleStoichiometry

Atomic mass unit and the mole amu definition: 12C 12 amu.1 amu 1.6605 x 10-24 gHow many 12C atoms weigh 12 g?6.02x1023 12C weigh 12 g.Avogadroʼs numberThe mole #atoms (1 atom/12 amu)(1 amu/1.66x10-24 g)(12g) 6.02x1023 12C weigh 12 gStoichiometry

Therefore:Any 6.02 x 1023 1 mole of 12C has amass of 12 gStoichiometry

The mole The mole is just a number of things1 dozen 12 things1 pair 2 things1 mole 6.022141x1023 thingsStoichiometry

Molar MassThe trick: By definition, this is the mass of 1 mol ofa substance (i.e., g/mol)– The molar mass of an element is the massnumber for the element that we find on theperiodic table– The formula weight (in amuʼs) will be thesame number as the molar mass (in g/mol)Stoichiometry

Using MolesMoles provide a bridge from the molecular scale to thereal-world scaleThe number of moles correspond to the number ofmolecules. 1 mole of any substance has the samenumber of molecules.Stoichiometry

Mole Relationships One mole of atoms, ions, or molecules containsAvogadroʼs number of those particles One mole of molecules or formula units containsAvogadroʼs number times the number of atoms orions of each element in the compoundStoichiometry

FindingEmpiricalFormulasStoichiometry

Combustion Analysisgives % compositionCnHnOn O2nCO2 1/2nH2O Compounds containing C, H and O are routinelyanalyzed through combustion in a chamber like this– %C is determined from the mass of CO2 produced– %H is determined from the mass of H2O produced– %O is determined by difference after the C and H haveStoichiometrybeen determined

Calculating Empirical FormulasOne can calculate the empirical formula fromthe percent compositionStoichiometry

Calculating Empirical FormulasThe compound para-aminobenzoic acid (you may haveseen it listed as PABA on your bottle of sunscreen) iscomposed of carbon (61.31%), hydrogen (5.14%),nitrogen (10.21%), and oxygen (23.33%). Find theempirical formula of PABA.Stoichiometry

Calculating Empirical FormulasAssuming 100.00 g of para-aminobenzoic acid,C:H:N:O:1 mol12.01 g1 mol5.14 g x1.01 g1 mol10.21 g x14.01 g1 mol23.33 g x16.00 g61.31 g x 5.105 mol C 5.09 mol H 0.7288 mol N 1.456 mol OStoichiometry

Calculating Empirical FormulasCalculate the mole ratio by dividing by the smallest numberof moles:C:5.105 mol0.7288 mol 7.005 7H:5.09 mol0.7288 mol 6.984 7N:0.7288 mol0.7288 mol 1.000O:1.458 mol0.7288 mol 2.001 2Stoichiometry

Calculating Empirical FormulasThese are the subscripts for the empirical formula:C7H7NO2OH2NO-Stoichiometry

Elemental AnalysesCompoundscontaining otherelements areanalyzed usingmethods analogousto those used for C,H and OStoichiometry

Stoichiometric CalculationsThe coefficients in the balanced equation givethe ratio of moles of reactants and productsStoichiometry

Stoichiometric CalculationsFrom the mass ofSubstance A you canuse the ratio of thecoefficients of A andB to calculate themass of Substance Bformed (if itʼs aproduct) or used (ifitʼs a reactant)Stoichiometry

Stoichiometric CalculationsExample: 10 grams of glucose (C6H12O6) react in acombustion reaction. How many grams of each product areproduced?C6H12O6(s) 6 O2(g) 6 CO2(g) 6 H2O(l)10.g? ?Starting with 10. g of C6H12O6 we calculate the moles of C6H12O6 use the coefficients to find the moles of H2O & CO2and then turn the moles to gramsStoichiometry

Stoichiometric calculationsC6H12O6 10.gMW: 180g/mol6O2 6CO2? 44 g/mol6H2O?18g/mol#mol: 10.g(1mol/180g)0.055 055mol)18g/mol5.9 gStoichiometry

LimitingReactantsStoichiometry

How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredientsOnce you run out of sugar, you will stop making cookiesStoichiometry

How Many Cookies Can I Make? In this example the sugar would be the limiting reactant,because it will limit the amount of cookies you can makeStoichiometry

Limiting Reactants The limiting reactant is the reactant present inthe smallest stoichiometric amount#molesLeft:2H214100 O2 -------- 7522H2O1010Stoichiometry

Limiting ReactantsIn the example below, the O2 would be theexcess reagentStoichiometry

Limiting reagent, example:Soda fizz comes from sodium bicarbonate and citric acid (H3C6H5O7)reacting to make carbon dioxide, sodium citrate (Na3C6H5O7) and water.If 1.0 g of sodium bicarbonate and 1.0g citric acid are reacted, which islimiting? How much carbon dioxide is produced?3NaHCO3(aq) H3C6H5O7(aq) ------ 1.0g1.0g84g/mol192g/mol1.0g(1mol/84g) 1.0(1mol/192g)0.012 mol0.0052 mol3CO2(g) 3H2O(l) Na3C6H5O7(aq)44g/mol(if citrate limiting)0.0052(3) 0.016 0.0052 molSo bicarbonate limiting:0.012 mol0.012(1/3) .0040mol.0052-.0040 .0012mol left0.0012 mol(192 g/mol) 0.023 g left.0.012 moles CO244g/mol(0.012mol) 0.53g CO2Stoichiometry

Theoretical Yield The theoretical yield is the amount ofproduct that can be made– In other words itʼs the amount of productpossible from stoichiometry. The “perfectreaction.” This is different from the actual yield,the amount one actually produces andmeasuresStoichiometry

Percent YieldA comparison of the amount actuallyobtained to the amount it was possibleto makeActual YieldPercent Yield x 100Theoretical YieldStoichiometry

ExampleBenzene (C6H6) reacts with Bromine to producebromobenzene (C6H6Br) and hydrobromic acid. If 30. g ofbenzene reacts with 65 g of bromine and produces 56.7 g ofbromobenzene, what is the percent yield of the reaction?C 6H 6 30.g78g/mol30.g(1mol/78g)0.38 mol(If Br2 limiting)0.41 mol(If C6H6 limiting)0.38 molBr2 ------ C6H5Br HBr65 g56.7 g160.g/mol157g/mol65g(1mol/160g)0.41 mol0.41 mol0.38 mol0.38mol(157g/1mol) 60.gStoichiometry56.7g/60.g(100) 94.5% 95%

Example, one moreReact 1.5 g of NH3 with 2.75 g of O2. How much NOand H2O is produced? What is left?4NH3 5O2-------- 1.5g2.75g17g/mol32g/mol1.5g(1mol/17g) 2.75g(1mol/32g) .088mol.086(If NH3 limiting):.088mol.088(5/4) .11O2 limiting:.086(4/5) .086 mol.069mol.069mol(17g/mol)1.2g2.75g4NO?30.g/mol 6H2O?18g/mol.086 mol(4/5) .086(6/5) .069 mol.10mol.069mol(30.g/mol) .10mol(18g/mol)2.1 g1.8gStoichiometry

Stoichiometry

Gun powder reaction 10KNO3(s) 3S(s) 8C(s) ---- 2K2CO3(s) 3K2SO4(s) 6CO2(g) 5N2(g)Salt peter sulfur charcoalAnd heat.What is interesting about this reaction?What kind of reaction is it?What do you think makes it so powerful?Stoichiometry

Gun powder reactionOxidizingagent Oxidizing Reducingagentagent10KNO3(s) 3S(s) 8C(s) ---- 2K2CO3(s) 3K2SO4(s) 6CO2(g) 5N2(g)Salt peter sulfur charcoalAnd heat.What is interesting about this reaction?Lots of energy, no oxygenWhat kind of reaction is it?Oxidation reductionWhat do you think makes it so powerful and explosive?Makes a lot of gas!!!!Stoichiometry

White phosphorous and Oxygen under waterStoichiometry

The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the . – In other words itʼs the amount of product possible from stoichiometry. The “perfect