Chem 31 Fall Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Writing and Balancing Chemical Equations


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1 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
2 Example One type of rocket fuel reacts hydrazine and dinitrogen tetroxide and produces nitrogen gas and water 1. hydrazine + dinitrogen tetroxide nitrogen + water 2. N 2 H 4 + N 2 O 4 N 2 + H 2 O Formulas 3. 2N 2 H 4 + N 2 O 4 3N 2 + 4H 2 O Balanced 4. 2N 2 H 4 (l) + N 2 O 4 (l) 3N 2 (g) + 4H 2 O (l) 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 + AgNO 3 AgCl + NaNO 3 Formulas 3. NaCl + AgNO 3 AgCl + NaNO 3 Balanced 4. NaCl (aq) + AgNO 3 (aq) AgCl (s) + NaNO 3 (aq) Na + (aq) +Cl  (aq) + Ag + (aq) + NO 3 (aq) AgCl (s) + Na + (aq) + NO 3 (aq) Ag + (aq) +Cl  (aq) AgCl (s) Net Ionic Equation 4
3 Let s Get Quantitative Recall: 1 mole = 6.02 x 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% 35 Cl % 37 Cl (0.755)(34.97) + (0.245)(36.97) = g/mol This is a weighted average; 1 mol of Cl will have a mass of grams 6
4 General Strategy Be Careful! moles of: atoms? molecules? 7 MoleBased Calculations How many grams of Phosphorous are there in mol P 2 O 5? Strategy: mol P 2 O 5 mol P g P mol P 2 O 5 x 2 mol P x g P = g P 1 mol P 2 O 5 1 mol P Round to: 0.62 g Phosphorous 8
5 How Many Atoms? How many Phosphorous atoms are there in mol P 2 O 5? Strategy: mol P 2 O 5 mol P #P atoms mol P 2 O 5 x 2 mol P x x P atoms = 1 mol P 2 O 5 1 mol P = x P atoms = 1.2 x P atoms 9 Strategy: Empirical Formula from Percent Composition 10
6 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 100g sample: 56.4 g O x 1 mol O = mol O g O 43.6 g P x 1 mol P = mol P g P 11 Empirical Formula  continued This gives: P O Dividing: PO 2.50 P 2 O 5 What about a MOLECULAR formula? need the molar mass (MW) of the compound Example: MW of P 2 O 5 cmpd is 284 g/mol Empirical Formula Mass 2x31 + 5x16 = 142 g MW/Emp Form Mass = 284/142 = 2 So: 2 x P 2 O 5 = P 4 O 10 12
7 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: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) 13 Reactant Quantities How many moles of CH 4 will react with 3.62 mol O 2? convert: mol O 2 mol CH 4 from rxn we know: 2 mol O 2 react with 1 mol CH mol O 2 x 1 mol CH 4 = 1.81 mol CH 4 2 mol O 2 14
8 Product from Reactant Moles How many moles of CO 2 are produced when 1.24 mol O 2 react? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) 1.24 mol O 2 x 1 mol CO 2 = 0.62 mol CO 2 2 mol O 2 15 Product Mass from Reactant Moles How many grams of CO 2 are produced when 1.24 mol O 2 are reacted? mol O 2 mol CO 2 g CO mol O 2 x 1 mol CO 2 x g CO 2 = 2 mol O 2 mol CO 2 = x 10 1 g CO 2 = 2.73x 10 1 g CO 2 16
9 Product Mass from Reactant Mass How many grams of CO 2 are produced when 3.47 grams of O 2 react? g O 2 mol O 2 mol CO 2 g CO g O 2 x 1 mol O 2 x 1 mol CO 2 x g CO 2 = g O 2 2 mol O 2 mol CO 2 = g CO 2 = 2.39 g CO 2 17 The Big Picture Strategy 18
10 What about Nonstoichiometric Amounts? Huh? EXAMPLE: 3Fe + 4H 2 O Fe 3 O 4 + 4H 2 How many moles of H 2 can be prepared from 4.00 mol Fe and 5.00 mol H 2 O? BUT: Fe and H 2 Oreact 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! Define: reagent that is completely consumed before any other In our example: 4.00 mol Fe = 1.33 > 1.25 = 5.00 mol H 2 O 3 mol Fe 4 mol H 2 O So, H 2 Ois our limiting reagent: 5.00 mol H 2 Ox 4 mol H 2 = 5.00 mol H 2 4 mol H 2 O 20
11 Ok, Let s Try Another One How many grams of N 2 F 4 can be prepared from 4.00 g NH 3 and 14.0 g F 2? 2NH 3 + 5F 2 N 2 F 4 + 6HF Which one is limiting reagent? 4.00 g NH 3 x 1 mol NH 3 = mol NH g NH g F 2 x 1 mol F 2 = mol F g F 2 So: mol NH 3 = > = mol F 2 2mol NH 3 5 mol F 2 *F 2 is the Limiting Reagent* 21 On to the answer: Strategy: gf 2 mol F 2 mol N 2 F 4 g N 2 F mol F 2 x 1 mol N 2 F 4 x g N 2 F 4 = 5mol F 2 1 mol N 2 F 4 = g N 2 F 4 = 7.66 g N 2 F 4 22
12 Percent Yield Suppose the previous reaction was performed, but only 4.80 g of N 2 F 4 were produced? Calculate the percent yield of the reaction. %yield = Actual (exptl) Yield x 100 Theoretical Yield = 4.80 g N 2 F 4 x 100 = 62.7% g N 2 F 4 23
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