Stoichiometry. Stoichiometry. Calcula1ng Molecular Mass. Formula Stoichiometry. Chemical Reac1ons. Stoichiometry 10/3/09

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1 Chemistry Preparatory Topics Lecture 5 Dr. Gondran Composi?on (Chemical Formula) Stoichiometry: The propor?ons of elements in stoichiometric compounds. In H 2 O, 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 H 2 + O 2 2 H 2 0 Mass to Mass Stoichiometry: The mass stoichiometric propor?ons in which species combine or are produced. (use formula masses) 4 grams of H 2 and 32 grams of O 2 make 36 grams of H 2 O Formula Formula or composi?on stoichiometry gives the number of atoms of each element that are in a given compound or molecule Example: Water, H 2 O has 2 atoms of H and one atom of O Example: Glucose, C 6 H 12 O 6 has 6 atoms of C, 12 atoms of H, and 6 atoms of O Calcula1ng Molecular Mass Molar mass of any compound or molecule is the sum of the masses of the cons?tuent atoms Example: Water, H 2 O has 2 atoms of H and one atom of O 2 ( 1 g/mole) + 16 g/ mole = 18 g/mole Example: Glucose, C 6 H 12 O 6 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) = ( ) g/mole = 180. g/mole Composi?on (Chemical Formula) Stoichiometry: The propor?ons of elements in stoichiometric compounds. In H 2 O, 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 H 2 + O 2 2 H 2 0 Mass to Mass Stoichiometry: The mass stoichiometric propor?ons in which species combine or are produced. (use formula masses) 4 grams of H 2 and 32 grams of O 2 make 36 grams of H 2 O 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 H 2 + O 2 2 H 2 O reactants product(s) 1

2 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 C x H y + y/4 O 2 x CO 2 + y/2 H 2 O 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. (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. 3) Mul?ply through to eliminate frac?ons obtaining the lowest whole number coefficients possible Balancing Equa1ons: Example (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. 3) Mul?ply through to reduce to eliminate frac?ons with the lowest whole number coefficients possible Balancing Equa1ons: Example (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. 3) Mul?ply through to reduce to eliminate frac?ons with the lowest whole number coefficients possible (NH 4 ) 2 Cr 2 O 7(S) N 2(g) + H 2 O (g) + Cr 2 O 3(s) CaSiO 3(aq) + HF (aq) SiF 4(g) + CaF 2(s) + H 2 O (l) 1 (NH 4 ) 2 Cr 2 O 7(S) 1 N 2(g) + 4 H 2 O (g) + 1 Cr 2 O 3(s) 1 CaSiO 3(aq) + 6 HF (aq) 1 SiF 4(g) + 1 CaF 2(s) + 3 H 2 O (l) Balancing Equa1ons: Example (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. 3) Mul?ply through to reduce to eliminate frac?ons with the lowest whole number coefficients possible NaHCO 3(s) Na 2 CO 3(s) + H 2 O (g) + CO 2 (g) 1 NaHCO 3(s) 1/2 Na 2 CO 3(s) + 1/2 H 2 O (g) + 1/2 CO 2 (g) 2 NaHCO 3(s) 1 Na 2 CO 3(s) + 1 H 2 O (g) + 1 CO 2 (g) Moles to Moles Each balanced chemical equa?on contains the informa?on about the ra?os of all of the components CaSiO 3(aq) + 6 HF (aq) SiF 4(g) + CaF 2(s) + 3 H 2 O (l) 1 mole of CaSiO 3 1 mole of CaSiO 3 6 moles of HF 1 mole of SiF 4 = 1 1 mole of CaSiO 3 1 mole of CaSiO 3 1 mole of CaF 2 3 mole of H 2 O 2

3 Moles to Moles: Example SiC is made from sand, SiO 2 and elemental C with CO as a by product. How many moles of CO are produced when 1/3 mole of C is used? SiO 2 + C SiC + CO 1 SiO 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 is a Mole? The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C; Avogadro s number. One mole represents 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 What Does a Mole Look Like? Molar mass of any compound or molecule is the sum of the masses of the cons?tuent atoms Example: Water, H 2 O, molar mass = 18 g/mole density 1 g/ ml one mole is 18 ml ~ 1.2 tbsp Example: Glucose, C 6 H 12 O 6, m. m.= 180 g/mole density 1.5 g/ ml one mole is 120 ml Composi?on (Chemical Formula) Stoichiometry: The propor?ons of elements in stoichiometric compounds. In H 2 O, 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 H 2 + O 2 2 H 2 0 Mass to Mass Stoichiometry: The mass stoichiometric propor?ons in which species combine or are produced. (use formula masses) 4 grams of H 2 and 32 grams of O 2 make 36 grams of H 2 O Mass to Mass 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 1) Amount of reactant Moles of reactant 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 3) Gas Laws 3

4 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 = 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 = moles / 0.25 l = 0.05 M Calcula1ng Moles from Molar Mass How many moles are in 28 gm of C 3 H 6 O? 3 carbon atoms 6 hydrogen atoms and 1 oxygen atom molar mass = 3 x x 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 CO 2 weigh? 1 carbon atom and 2 oxygen atoms molar mass = 1 x x 16 = 44 gm/mole grams of CO 2 = 1.45 moles x 44 gm /mole = 63.8 grams Mass to Mass 1) Amount of reactant Moles of reactant 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 3) Gas Laws Mass to Mass: Example How much CO 2 is released when you burn off 19 grams of sugar? C 6 H 12 O O 2 6 CO H 2 O Reactant grams to moles moles to moles moles to grams 19 gm sugar x 1 mole sugar x 6 moles CO 2 x 44. gm CO 2 = 180 gm sugar 1 mole sugar 1 mole CO 2 28 gm CO 2 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. 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 1c sugar 2 eggs + other stuff 48 cookies I have 2c buoer 3c sugar 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

5 Limi1ng Reactant: by Moles 1 CaSiO 3(aq) + 6 HF (aq) 1 SiF 4(g) + 1 CaF 2(s) + 3 H 2 O (l) How much CaF 2 is formed if you start with 2 moles of CaSiO 3 and 3 moles of HF? 2 moles CaSiO 3 x 1 mole CaF 2 = 2 moles of CaF 2 1 mole CaSiO 3 3 moles HF x 1 mole CaF 2 = ½ mole of CaF 2 6 mole HF Limi1ng Reactant Mass to Mass 1) Amount of reactant Moles of reactant 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 3) Gas Laws Limi1ng Reactant: by Mass How many grams of CO 2 are produced when 45 gm of C 2 H 6 are burned in 600 gms of O 2? 2 C 2 H O 2 4 CO H gm C 2 H 6 x 1 mole C 2 H 6 x 4 moles CO 2 = 3 moles CO 2 30 gm C 2 H 6 2 moles C 2 H gm O 2 x 1 mole O 2 x 4 moles CO 2 = 11 moles CO 2 32 gm O 2 7 moles O 2 The amount of CO 2 is limited by the amount of C 2 H 6 grams of CO 2 = 3 moles CO 2 x 44 gm/mole = 132 gm Theore1cal, Actual and Percent Yield The amount of the limi?ng reactant will determine the maximum theore?cal yield for the reac?on In prac?ce, the actual yield will be some what lower. percent yield = 100% x (actual yield / theore?cal yield) Example: If we expected to produce g of ZnCl and we only produced or only collected g of ZnCl The percent yield = 100% x (16.05 g / g) = 91.17% Percent Yield Problem 51 gm SiC are produced from 100 gm of sand, SiO 2 and excess elemental C. What is the percent yield for the process? SiO C SiC+ 2 CO How much SiC should be produced from 100 g of sand? 100 gm SiO 2 x 1 mole SiO 2 x 1 mole SiC x 40 g SiC = 67 gm 60 gms 1 mole SiO 2 1 mole SiC Percent yield = actual yield x 100% = 51 gm = 76 % theore?cal yield 67 gm Summary 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

6 One More? The Haber process for making ammonia from nitrogen in air is given by the equa?on: N H 2 > 2 NH 3 What the mass of hydrogen must be supplied to make 500. kg of ammonia in a system that has an 88.8% yield? Solu1on N H 2 > 2 NH 3 What the mass of H 2 must be supplied to make 500. kg of NH 3 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 NH 3 / 88.8% = 563 kg of NH 3 563,000 gm NH 3 x 1 mole NH 3 x 3 moles H 2 x 2.0 gm H 2 17 gm NH 3 2 moles NH 3 mole H 2 = 99,000 gm H 2 = 99 kg H 2 6