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 Mole-Based 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 100-g 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 Non-stoichiometric 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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