Moles, Formulas, Reactions & Stoichiometry Lecture 3 Topics Brown, chapter 3 1. Moles 3.4 Molecular mass (aka molecular weight) 3.3 2. Molar converstions Percent composition 3.3 Empirical formulas 3.5 3. Stoichiometry: balancing chemical equations 3.1 4. Patterns of Chemical Reactivity 3.2 Combination Decomposition Combustion Exchange 5. Stoichiometry & Conversions 3.6 Limiting reactants 3.7 Theoretical & percent yield

Stoichiometric conversions Balanced equations can predict yields. One reactant often limits the… …theoretical yield of product. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

Stoichiometry is not about mass! How do we ‘interpret’ the meaning of chemical equations? 2Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq) There are several correct interpretations: 2 atoms Na + 2 molecules water become 1 molecule H2 gas + 2 molecules NaOH 2 dozen atoms + 2 dozen molecules become 1 dozen molecules + 2 dozen molecules 2 cases atoms + 2 cases molecules become 1 case molecules + 2 case molecules 2 MOLES atoms + 2 MOLES molecules become 1 MOLE molec. + 2 MOLES molec. And an INCORRECT interpretation: Stoichiometry counts moles & molecules. Not mass! 2 g Na + 2 g water become 1 g H2 gas + 2 g NaOH How do you convert to grams? 2 moles Na 23.22 g = 46 g 1 mole 1 mole H2 2.02 g = 2.0 g 1 mole 46 g + 36 g --> 2 g + 80 g 2 moles H2O 18.01 g = 36 g 1 mole 2 moles NaOH 40.00 g = 80.0 g 1 mole p.98-99

Stoichiometry “mall map” atoms X molecules A subsc Av# For the reaction: A  B Where A contains element X, and B contains element Y. mass A moles A MW Stoic mass B moles B MW Av# atoms Y subsc molecules B p.149-50

Stoichiometric conversions Combust C2H6 to produce water. How much ethane do you need to burn to produce 2.34 moles of water? Steps: A balanced chemical equation Go to moles. Use coefficients to change one reactant or product to another. 2C2H6 + 7O2  4CO2 + 6H2O 2.34 moles 2.34 mol H2O 2 mol C2H6 = 0.78 mol C2H6 6 mol H2O How many grams of CO2 are produced by burning 155.0 g of ethane? Same balanced chemical equation! (see above) 155.0 g C2H6 1 mol C2H6 4 mol CO2 44.0 g CO2 = 455.0 g CO2 30 g 2 mol C2H6 1 mol CO2 p.100-2

Limiting reactants: used up or left over? You want to make grilled cheese sandwiches and find 10 slices of bread & 4 slices of cheese. How many sandwiches can you make? Each sandwich requires 2 slices of bread & 1 slice of Swiss: 2:1 ratio. (B2C) So for 10 slices of bread you’d need 10/2 = 5 slices of Swiss… …but there are only 4 slices. So cheese is limiting (sandwiches produced) & bread is in excess. Since the product still includes O2, O2 was in excess & H2 was limiting O2 + 2H2  2H2O p.102-105

Limiting reactants strategy Keys to solving limiting reactant problems: 1. Start with a balanced chemical equation. 2. Determine which reactant is limiting (in terms of moles). 3. Base moles of product yield on moles of limiting reactant. 4. Run a race in which the two reactants compete to make product. The reactant that makes less product is limiting. How much H2 gas can be made from 30.0 g of Na and 35.0 g of H2O? 2Na + 2H2O --> H2 + 2NaOH 30.o g 35.0 g 30.0 g 1 mol Na 1 mol H2 2.021 g H2 = 1.31 g H2 22.99 g 2 mol Na 1 mol  limiting Na H2O H2 Moles 1.30 1.94 0 Change -1.30 -1.30 +.65 Left 0 0.64 0.65 Grams 0 11.5 1.31 30.5 g 1 mol H2O 1 mol H2 2.02 g H2 = 1.71 g H2 18.02 g 2 mol Na 1 mol p.102-105

Examples: limiting reactants How much water can be made from 150 g of H2 and 1500 g of O2? 2H2 + O2 --> 2H2O 150 g 1500 g 150 g 1 mol H2 2 mol H2O 18.01 g H2O = 1.3 x 103 g H2O made 2.02 g 2 mol H2 1 mol So H2 is limiting & dictates yield 1500 g 1 mol O2 2 mol H2O 18.02 g H2O = 1.7 x 103 g H2O made 32.00 g 1 mol O2 1 mol How much excess reactant is leftover? 1.3 x 103 g H2O 1 mole H2O 1 mole O2 32.00 g = 1200 g O2 used 18.02 g 2 mole H2O 1 mole Leftover O2 = 1500 g given – 1200 g used = 300 g leftover p.102-105

Percent yield Percent yield compares actual yield to theoretical yield. It’s a measure of how good the chemist’s yield was. We’ve just learned to calculate theoretical yield in limiting reactions. % yield = (actual/theoretical)(100) How much Fe can be made from 1.000 kg of rust and 445.2 g CO? Rust is Fe2O3, MW 159.7 g/mole Fe2O3 + 3CO --> 2Fe + 3CO2 103 g 445.2 g ? 103 g Fe2O3 1 mol Fe2O3 3 mol CO 28.00 g = 526.0 g CO required so CO limiting 159.70 g 1 mol Fe2O3 1 mol CO 445.2 g CO 1 mol CO 2 mol Fe 55.85 g = 592.0 g Fe produced 28.00 g 3 mol CO 1 mol Fe If the actual yield was 500.00 g of Fe, what’s the percent yield? 500.00 g (100) = 84.45% 592.0 g p.105-106

Putting it all together! How many grams of ethane does it take to make 1.00 kg of water if the process is 85% efficient? (C2H6; MW = 33.11 g/mol) You need to make 1.00 kg of water, but the rxn is only 85% efficient, so: 1.oo kg X kg --> 1.18 kg must be made = 1.18x103 g H2O 85% 100% 2C2H6 + 7O2 --> 4CO2 + 6H2O ? g 1180 g 1.18E3 g H2O 1 mol H2O 2 mol C2H6 33.11 g = 723 g of ethane (C2H6) 18.01 g 6 mol H2O 1 mol C2H6