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10/3/09 Stoichiometry Stoichiometry Chemistry Preparatory Topics Lecture 5 – Dr. Gondran Composi?on (Chemical Formula) Stoichiometry: The propor?ons of elements in stoichiometric compounds. In H2O, hydrogen and oxygen is 2:1. Reac?on (Chemical Equa?on) Stoichiometry: The stoichiometric propor?ons by which species combine or are produced in chemical equa?ons. 2 H2 + O2 2 H20 Mass‐to‐Mass Stoichiometry: The mass stoichiometric propor?ons in which species combine or are produced. (use formula masses) 4 grams of H2 and 32 grams of O2 make 36 grams of H2O Formula Stoichiometry Calcula1ng Molecular Mass Formula or composi?on stoichiometry gives the number of atoms of each element that are in a given compound or molecule Molar mass of any compound or molecule is the sum of the masses of the cons?tuent atoms Example: Water, H2O has 2 atoms of H and one atom of O Example: Glucose, C6H12O6 has 6 atoms of C, 12 atoms of H, and 6 atoms of O Stoichiometry Composi?on (Chemical Formula) Stoichiometry: The propor?ons of elements in stoichiometric compounds. In H2O, hydrogen and oxygen is 2:1. Reac?on (Chemical Equa?on) Stoichiometry: The stoichiometric propor?ons by which species combine or are produced in chemical equa?ons. 2 H2 + O2 2 H20 Mass‐to‐Mass Stoichiometry: The mass stoichiometric propor?ons in which species combine or are produced. (use formula masses) 4 grams of H2 and 32 grams of O2 make 36 grams of H2O Example: Water, H2O has 2 atoms of H and one atom of O 2 ( 1 g/mole) + 16 g/ mole = 18 g/mole Example: Glucose, C6H12O6 has 6 atoms of C, 12 atoms of H, and 6 atoms of O 6 (12 g/mole) + 12 (1 g/mole) + 6 (16 g/mole) = (72 + 12 + 96) g/mole = 180. g/mole Chemical Reac1ons Chemical reac?on occur when chemical bonds are broken and new ones are formed to change the molecules present. In the process reactants are transformed into products. The number and type of atoms does not change Example: 2 H2 + O2 2 H2O reactants product(s) 1 10/3/09 Types of Reac1ons Combina?on (or synthesis) A + X → AX Decomposi?on AX → A + X Single replacement A + BX → AX + B Double replacement AX + BY → AY + BX and Combus?on CxHy + y/4 O2 → x CO2 + y/2 H2O Balancing Equa1ons Do not change the iden??es (formulas) of any of the reactants or products. Determine what coefficients are necessary to ensure that the same number of each type of atom appears on both reactant and product sides. 1) 2) 3) Balancing Equa1ons: Example 1) 2) 3) Make a first guess star?ng with 1 for the most complicated molecule(s). Balance the number of atoms for each element in the star?ng formula (compound), first those that only occur in one other compound then those that occur several ?me. Save atoms that occur in a homogeneous molecule for last. Mul?ply through to reduce to eliminate frac?ons with the lowest whole number coefficients possible Make a first guess star?ng with 1 for the most complicated molecule(s). Balance the number of atoms for each element in the star?ng formula (compound), first those that only occur in one other compound then those that occur several ?me. Save atoms that occur in a homogeneous molecule for last. Mul?ply through to eliminate frac?ons obtaining the lowest whole number coefficients possible Balancing Equa1ons: Example 1) 2) 3) Make a first guess star?ng with 1 for the most complicated molecule(s). Balance the number of atoms for each element in the star?ng formula (compound), first those that only occur in one other compound then those that occur several ?me. Save atoms that occur in a homogeneous molecule for last. Mul?ply through to reduce to eliminate frac?ons with the lowest whole number coefficients possible __ (NH4) 2Cr2O7(S) __ N2(g) + __ H2O(g) + __ Cr2O3(s) __CaSiO3(aq) +__HF(aq) __SiF4(g) +__CaF2(s) + __H2O(l) 1 (NH4) 2Cr2O7(S) 1 N2(g) + 4 H2O(g) + 1 Cr2O3(s) 1 CaSiO3(aq) + 6 HF(aq) 1 SiF4(g) + 1 CaF2(s) + 3 H2O(l) Balancing Equa1ons: Example Moles to Moles 1) 2) 3) Make a first guess star?ng with 1 for the most complicated molecule(s). Balance the number of atoms for each element in the star?ng formula (compound), first those that only occur in one other compound then those that occur several ?me. Save atoms that occur in a homogeneous molecule for last. Mul?ply through to reduce to eliminate frac?ons with the lowest whole number coefficients possible __ NaHCO3(s) __ Na2CO3(s) + __ H2O(g) + __ CO2 (g) 1 NaHCO3(s) 1/2 Na2CO3(s) + 1/2 H2O(g) + 1/2 CO2 (g) Each balanced chemical equa?on contains the informa?on about the ra?os of all of the components CaSiO3(aq) + 6 HF(aq) SiF4(g) + CaF2(s) + 3 H2O(l) 1 mole of CaSiO3 1 mole of CaSiO3 6 moles of HF 1 mole of SiF4 2 NaHCO3(s) 1 Na2CO3(s) + 1 H2O(g) + 1 CO2 (g) = 1 1 mole of CaSiO3 1 mole of CaSiO3 1 mole of CaF2 3 mole of H2O 2 10/3/09 Moles to Moles: Example SiC is made from sand, SiO2 and elemental C with CO as a by‐product. How many moles of CO are produced when 1/3 mole of C is used? __ SiO2 + __ C __ SiC + __ CO 1 SiO2 + 3 C 1 SiC + 2 CO 1/3 mole of C x 2 moles of CO = 2/9 moles of CO 3 moles of C What Does a Mole Look Like? What is a Mole? The number equal to the number of carbon atoms in exactly 12 grams of pure 12C; Avogadro’s number. One mole represents 6.022 × 1023 units. 1) Molar mass and the mass of the samples 2) Molarity and volume of a solu?on 3) Gas laws knowing temperature, pressure and volume Stoichiometry Molar mass of any compound or molecule is the sum of the masses of the cons?tuent atoms Composi?on (Chemical Formula) Stoichiometry: The propor?ons of elements in stoichiometric compounds. In H2O, hydrogen and oxygen is 2:1. Example: Water, H2O, molar mass = 18 g/mole density 1 g/ ml one mole is 18 ml ~ 1.2 tbsp Reac?on (Chemical Equa?on) Stoichiometry: The stoichiometric propor?ons by which species combine or are produced in chemical equa?ons. 2 H2 + O2 2 H20 Example: Glucose, C6H12O6, m. m.= 180 g/mole density 1.5 g/ ml one mole is 120 ml Mass‐to‐Mass Stoichiometry: The mass stoichiometric propor?ons in which species combine or are produced. (use formula masses) 4 grams of H2 and 32 grams of O2 make 36 grams of H2O Mass‐to‐Mass Stoichiometry A balanced chemical equa?on is a quan?ta?ve descrip?on of the reac?on and it tells you how many moles of each reactant combine to form how many moles of each product. We can not directly measure moles We need to measure masses, volumes, concentra?ons Then convert to moles so we can compare the molar propor?ons in the balances chemical equa?on Finally we need to convert back to masses, volumes, concentra?ons … something we will be able to measure Mass‐to‐Mass Stoichiometry 1) Amount of reactant Moles of reactant 1) Molar mass 2) Solu?on molarity and volume 3) Gas laws 2) Moles of reactant Moles of product Stoichiometric coefficients in the balanced chemical equa?on 3) Moles of product Amount of product 1) Molar mass 2) Solu?on molarity and volume 3) Gas Laws 3 10/3/09 Calcula1ng Moles with Molarity How many moles of HCl are in 0.25 liters of a 0.05 M solu?on? moles = volume x concentra?on moles = 0.25 l x 0.05 moles/ l = 0.0125 moles If the volume remained the same when excess solid NaOH was added and what would the concentra?on of NaCl solu?on? molarity (concenta?on) = moles / volume molarity = 0.0125 moles / 0.25 l = 0.05 M Mass‐to‐Mass Stoichiometry 1) Amount of reactant Moles of reactant 1) Molar mass 2) Solu?on molarity and volume 3) Gas laws 2) Moles of reactant Moles of product Stoichiometric coefficients in the balanced chemical equa?on 3) Moles of product Amount of product 1) Molar mass 2) Solu?on molarity and volume 3) Gas Laws Limi1ng Reactants Using the stoichiometric rela?onship given in the balanced chemical equa?on or mass‐to‐mass stoichiometry, it is possible to determine which (if any) reactant is present in excess (more is present than will react with all of the other component) and which reactant will be consumed completely and thus limit how much stuff can react, the “limi?ng reactant”. Calcula1ng Moles from Molar Mass How many moles are in 28 gm of C3H6O? 3 carbon atoms 6 hydrogen atoms and 1 oxygen atom molar mass = 3 x 12 + 6 x 1.0 + 1 x 16 = 58 gm/mole number of moles = 28 gm x 1 mole = 0.48 moles 58 gm How many grams dose 1.45 moles of CO2 weigh? 1 carbon atom and 2 oxygen atoms molar mass = 1 x 12 + 2 x 16 = 44 gm/mole grams of CO2 = 1.45 moles x 44 gm /mole = 63.8 grams Mass‐to‐Mass: Example How much CO2 is released when you burn off 19 grams of sugar? C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Reactant ‐‐ grams to moles ‐‐ moles to moles ‐‐ moles to grams 19 gm sugar x 1 mole sugar x 6 moles CO2 x 44. gm CO2 = 180 gm sugar 1 mole sugar 1 mole CO2 28 gm CO2 Limi1ng Reactant: Baking Analogy We are trying to make as many chocolate chip cookies as possible using a specific recipe and the ingredients I have on hand. The recipe calls for: recipe 1c buoer I have 2c buoer 1c sugar 3c sugar 2 eggs + other stuff 48 cookies 3 eggs If I divided the amounts I have by the amounts needed per batch I have enough buoer sugar eggs to make a double triple 1.5 batch If I try to make a triple batch it won’t come out right! The most I can make using the correct propor?ons is a 1.5 batch. Use 1.5 c buoer 1.5 c sugar 3 eggs to make 72 cookies Leaving 0.5 c buoer 1.5 c sugar 0 eggs 4 10/3/09 Limi1ng Reactant: by Moles 1 CaSiO3(aq)+ 6 HF(aq)1 SiF4(g)+ 1 CaF2(s)+ 3 H2O(l) How much CaF2 is formed if you start with 2 moles of CaSiO3 and 3 moles of HF? 2 moles CaSiO3 x 1 mole CaF2 = 2 moles of CaF2 1 mole CaSiO3 3 moles HF x 1 mole CaF2 = ½ mole of CaF2 6 mole HF Limi1ng Reactant Limi1ng Reactant: by Mass Mass‐to‐Mass Stoichiometry 1) Amount of reactant Moles of reactant 1) Molar mass 2) Solu?on molarity and volume 3) Gas laws 2) Moles of reactant Moles of product Stoichiometric coefficients in the balanced chemical equa?on 3) Moles of product Amount of product 1) Molar mass 2) Solu?on molarity and volume 3) Gas Laws Theore1cal, Actual and Percent Yield How many grams of CO2 are produced when 45 gm of C2H6 are burned in 600 gms of O2? 2 C2H6 + 7 O2 4 CO2 + 6 H20 The amount of the limi?ng reactant will determine the maximum theore?cal yield for the reac?on 45 gm C2H6 x 1 mole C2H6 x 4 moles CO2 = 3 moles CO2 30 gm C2H6 2 moles C2H6 600 gm O2 x 1 mole O2 x 4 moles CO2 = 11 moles CO2 32 gm O2 7 moles O2 The amount of CO2 is limited by the amount of C2H6 grams of CO2 = 3 moles CO2 x 44 gm/mole = 132 gm percent yield = 100% x (actual yield / theore?cal yield) Percent Yield Problem Summary 51 gm SiC are produced from 100 gm of sand, SiO2 and excess elemental C. What is the percent yield for the process? SiO2 + 3 C SiC+ 2 CO How much SiC should be produced from 100 g of sand? 100 gm SiO2 x 1 mole SiO2 x 1 mole SiC x 40 g SiC = 67 gm 60 gms 1 mole SiO2 1 mole SiC Percent yield = actual yield x 100% = 51 gm = 76 % theore?cal yield 67 gm In prac?ce, the actual yield will be some what lower. Example: If we expected to produce 17.64 g of ZnCl and we only produced or only collected 16.05 g of ZnCl The percent yield = 100% x (16.05 g / 17.64 g) = 91.17% Defined stoichiometry Composi?on stoichiometry ‐ Molar mass Reac?on stoichiometry ‐ Balancing equa?ons Mass‐to‐mass stoichiometry Moles from mass or molarity Limi?ng reactants Theore?cal, actual and percent yields 5 10/3/09 One More … ? The Haber process for making ammonia from nitrogen in air is given by the equa?on: N2 + 3 H2 ‐> 2 NH3 What the mass of hydrogen must be supplied to make 500. kg of ammonia in a system that has an 88.8% yield? Solu1on N2 + 3 H2 ‐> 2 NH3 What the mass of H2 must be supplied to make 500. kg of NH3 in a system that has an 88.8% yield? Percent yield = 100% x actual yield/theore?cal yield Theore?cal yield = 100% x actual yield/percent yield = 100% x 500. kg of NH3 / 88.8% = 563 kg of NH3 563,000 gm NH3 x 1 mole NH3 x 3 moles H2 x 2.0 gm H2 17 gm NH3 2 moles NH3 mole H2 = 99,000 gm H2 = 99 kg H2 6