Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chem 31 Fall 2002 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Writing and Balancing Chemical Equations 1. Write Equation in Words -you cannot write an equation unless you know what happens in the reaction! 2. Substitute Chemical Formulas for Words 3. Balance the Equation -Conservation of Mass 4. Indicate Physical States -solid (s), liquid (l), gas (g), aqueous (aq) 2 1 Example One type of rocket fuel reacts hydrazine and dinitrogen tetroxide and produces nitrogen gas and water 1. hydrazine + dinitrogen tetroxide → nitrogen + water 2. N2H4 + N2O4 → N2 + H2O 3. 2N2H4 + N2O4 → 3N2 + 4H2O 4. 2N2H4 (l) + N2O4 (l) → 3N2 (g) + 4H2O (l) Formulas Balanced Done! 3 Another Example A solution of sodium chloride was added to a solution of silver nitrate, forming a precipitate of silver chloride 1. sodium choride + silver nitrate → silver chloride + sodium nitrate 2. NaCl + AgNO3 → AgCl + NaNO3 Formulas 3. NaCl + AgNO3 → AgCl + NaNO3 Balanced 4. NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq) Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) → AgCl (s) + Na+ (aq) + NO3- (aq) Ag+ (aq) + Cl- (aq) → AgCl (s) Net Ionic Equation 4 2 Let’s Get Quantitative Recall: 1 mole = 6.02 x 1023 particles Atomic Masses (Weights): mass/atom (amu) Molecular/Formula Weights: mass/compound (amu) Molar Masses: mass of a mole of atoms or compound (g) 5 Why Fractional Molar Masses? Need to consider the natural abundances of isotopes Example: Chlorine 75.5% 35Cl + 24.5% 37Cl (0.755)(34.97) + (0.245)(36.97) = 35.45 g/mol -This is a weighted average; 1 mol of Cl will have a mass of 35.45 grams 6 3 General Strategy Be Careful! moles of: atoms? molecules? 7 Mole-Based Calculations How many grams of Phosphorous are there in 0.010 mol P2O5? Strategy: mol P2O5 → mol P → g P 0.010 mol P2O5 x 2 mol P x 30.974 g P = 0.61948 g P 1 mol P2O5 1 mol P Round to: 0.62 g Phosphorous 8 4 How Many Atoms? How many Phosphorous atoms are there in 0.010 mol P2O5? Strategy: mol P2O5 → mol P → #P atoms 0.010 mol P2O5 x 2 mol P x 6.022 x 1023 P atoms 1 mol P2O5 1 mol P = = 1.20440 x 1022 P atoms = 1.2 x 1022 P atoms 9 Strategy: Empirical Formula from Percent Composition 10 5 Empirical Formula from Percent Composition What is the empirical formula for a binary compound which is found to be: 56.4% Oxygen (by mass) 43.6% Phosphorous (by mass)? Strategy: % → grams → mol (% is a relative measure, so DEFINE a sample size (100 g)) In a 100-g sample: 56.4 g O x 1 mol O 43.6 g P 15.999 g O x 1 mol P 30.974 g P = 3.525 mol O = 1.4076 mol P 11 Empirical Formula continued This gives: P1.4076O3.525 Dividing: PO2.50 → P2O5 •What about a MOLECULAR formula? -need the molar mass (MW) of the compound Example: MW of P2O5 cmpd is 284 g/mol Empirical Formula Mass ≈ 2x31 + 5x16 = 142 g MW/Emp Form Mass = 284/142 = 2 So: 2 x P2O5 = P4O10 12 6 Quantifying Reaction Chemistry With a balanced equation, we can relate amounts of reactants and products via molar relationships Let’s look at this combustion reaction: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) 13 Reactant Quantities How many moles of CH4 will react with 3.62 mol O2? convert: mol O2 → mol CH4 from rxn we know: 2 mol O2 react with 1 mol CH4 3.62 mol O2 x 1 mol CH4 = 1.81 mol CH4 2 mol O2 14 7 Product from Reactant Moles How many moles of CO2 are produced when 1.24 mol O2 react? CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ↑ ↑ 1.24 mol O2 x 1 mol CO2 = 0.62 mol CO2 2 mol O2 15 Product Mass from Reactant Moles How many grams of CO2 are produced when 1.24 mol O2 are reacted? mol O2 → mol CO2 → g CO2 1.24 mol O2 x 1 mol CO2 x 44.01 g CO2 = 2 mol O2 mol CO2 = 2.72862 x 101 g CO2 = 2.73x 101 g CO2 16 8 Product Mass from Reactant Mass How many grams of CO2 are produced when 3.47 grams of O2 react? g O2 → mol O2 → mol CO2 → g CO2 3.47 g O2 x 1 mol O2 x 1 mol CO2 x 44.01 g CO2 = 32.00 g O2 2 mol O2 mol CO2 = 2.38616719 g CO2 = 2.39 g CO2 17 The Big Picture Strategy 18 9 What about Non-stoichiometric Amounts? ¾ Huh? ¾ EXAMPLE: 3Fe + 4H2O → Fe3O4 + 4H2 -How many moles of H2 can be prepared from 4.00 mol Fe and 5.00 mol H2O? -BUT: Fe and H2O react in a 3:4 ratio but they are available in a 4:5 ratio We will run out of ONE of the reactants before the other is completely reacted. 19 Limiting Reagent! ¾Define: reagent that is completely consumed before any other In our example: 4.00 mol Fe = 1.33 > 1.25 = 5.00 mol H2O 3 mol Fe 4 mol H2O So, H2O is our limiting reagent: 5.00 mol H2O x 4 mol H2 = 5.00 mol H2 4 mol H2O 20 10 Ok, Let’s Try Another One ¾ How many grams of N2F4 can be prepared from 4.00 g NH3 and 14.0 g F2? 2NH3 + 5F2 → N2F4 + 6HF Which one is limiting reagent? 4.00 g NH3 x 1 mol NH3 = 0.23488 mol NH3 17.03 g NH3 14.0 g F2 x 1 mol F2 = 0.36845 mol F2 37.997 g F2 So: 0.23488 mol NH3 = 0.117 > 0.0737 = 0.36845 mol F2 2 mol NH3 5 mol F2 *F2 is the Limiting Reagent* 21 On to the answer: Strategy: g F2 → mol F2 → mol N2F4 → g N2F4 0.36845 mol F2 x 1 mol N2F4 x 104.01 g N2F4 = 5 mol F2 1 mol N2F4 = 7.6644998 g N2F4 = 7.66 g N2F4 22 11 Percent Yield Suppose the previous reaction was performed, but only 4.80 g of N2F4 were produced? Calculate the percent yield of the reaction. %-yield = Actual (exptl) Yield x 100 Theoretical Yield = 4.80 g N2F4 x 100 = 62.7% 7.6645 g N2F4 23 12