STOICHIOMETRY is. Math-tastic! Let s make some Cookies! 2/21/2015
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1 Math-tastic! Unit 9: Math of Chemistry Part II - Stoichiometry Lesson # 9.4: The Arithmetic of Equations Mr. Mole 87 STOICHIOMETRY is Greek for measuring elements Pronounced stoy-kee-ahm-uhtree Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. Let s make some Cookies! When baking cookies, a recipe is usually used, telling the exact amount of each ingredient. If you need more, you can double or triple the amount Thus, a recipe is much like a balanced equation. There are 4 ways to interpret a balanced chemical equation: 1
2 #1. In terms of Particles An Element is made of atoms A Molecular compound (made of only nonmetals) is made up of molecules (Don t forget the diatomic elements) Ionic Compounds (made of a metal and nonmetal parts) are made of formula units Example: 2H 2 + O 2 2H 2 O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. Another example: 2Al 2 O 3 Al + 3O 2 2 formula units Al 2 O 3 form 4 atoms Al and 3 molecules O 2 Now read this: 2Na + 2H 2 O 2NaOH + H 2 #2. In terms of Moles The coefficients tell us how many moles of each substance 2Al 2 O 3 Al + 3O 2 2Na + 2H 2 O 2NaOH + H 2 Remember: A balanced equation creates a Molar Ratio 2
3 #3. In terms of Mass The Law of Conservation of Mass applies We can check mass by using moles. 2H 2 + O 2 2H 2 O 2 moles H g H 2 1 mole H 2 = 4.04 g H g O 2 1 mole O 2 = g O 1 mole O g H 2 + O 2 + reactants In terms of Mass (for products) 2H 2 + O 2 2H 2 O 2 moles H 2 O g H 2 O 1 mole H 2 O = g H 2 O g H 2 + O 2 = g H 2 O grams reactant = grams product The mass of the reactants must equal the mass of the products. #4. In terms of Volume (@ STP) At STP, 1 mol of any gas = 22.4 L 2H 2 + O 2 2H 2 O (2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 ) (2 x 22.4 L H 2 O) 67.2 Liters of reactant 44.8 Liters of product! NOTE: mass and atoms are ALWAYS conserved - however, molecules, formula units, moles, and volumes will not necessarily be conserved! 3
4 Practice: Show that the following equation follows the Law of Conservation of Mass (show the atoms balance, and the mass on both sides is equal) 2Al 2 O 3 Al + 3O 2 Mole-Mole Ratio A Conversion Factor that shows the mole-tomole ratio between two of the substances in a balanced equation Derived from the coefficients of any two substances in the equation 97 Writing Mole Ratios 4 Fe + 3 O 2 2 Fe 2 O 3 Fe and O 2 4 mol Fe and 3 mol O 2 3 mol O 2 4 mol Fe Fe and Fe 2 O 3 4 mol Fe and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 4 mol Fe O 2 and Fe 2 O 3 3 mol O 2 and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O
5 Learning Check #1 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mol ratio between H 2 and N 2 is 1) 3 mol N 2 2) 1 mol N 2 3) 1 mol N 2 1 mol H 2 3 mol H 2 2 mol H 2 B. A mol ratio between NH 3 and H 2 is 1) 1 mol H 2 2) 3 mol H 2 3) 3 mol H 2 2 mol NH 3 2 mol NH 3 2 mol NH 3 99 Chemical Calculations Mole to Mole Conversions 2Al 2 O 3 Al + 3O 2 each time we use 2 moles of Al 2 O 3 we will also make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 3 mole O 2 2 moles Al 2 O 3 These are the two possible conversion factors to use in the solution of the problem. Mole to Mole conversions How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2Al 2 O 3 Al + 3O 2 3 mol O 3.34 mol x 2 = 5.01 mol O 2 Al 2 O 2 mol Al 2 O 3 3 Conversion factor from balanced equation If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! 5
6 Chemical Calculations 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react? 6.0 mol O 2 x mol Fe 2 O 3 = 4.0 mol Fe 2 O 3 mol O Learning Check 2 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 mol of O 2? 1) 3.00 mol Fe 2) 9.00 mol Fe 3) 16.0 mol Fe 103 Solution #2 4 Fe + 3 O 2 2 Fe 2 O mol O 2 x 4 mol Fe = 16.0 mol Fe 3 mol O
7 Practice: 2C 2 H O 2 4CO H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.6 mol) How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (8.95 mol) If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (4.94 mol) Steps to Calculate Stoichiometric Problems 1. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit. Example Calculation The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? Write the equation H 2 (g) + O 2 (g) H 2 O (g) Balance the equation 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 107 7
8 Organize data mol bridge 2 H 2 (g) + O 2 (g) 2 H 2 O (g)? g 13.1 g Plan g of H 2 O mol H 2 O mol O 2 g of O 2 Setup 13.1 g H 2 O x 1 mol H 2 O x 1 mol O 2 x 32.0 g O g H 2 O 2 mol H 2 O 1 mol O 2 = 11.6 g O Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 2 2Al 2 O g Al 1 mol Al 2 mol Al 2 O g Al 2 O g Al 4 mol Al 1 mol Al 2 O 3 =? g Al 2 O 3 (6.50 x 1 x 2 x ) (26.98 x 4 x 1) = 12.3 g Al 2 O 3 are formed Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4 Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu 8
9 Points to Remember 1. Read an equation in moles 2. Convert given amount to moles 3. Use mole factor to give desired moles 4. Convert moles to grams grams (given moles (given) grams (desired) moles (desired) 111 Learning Check # 3 How many O 2 molecules will react with 505 grams of Na to form Na 2 O? 4 Na + O 2 2 Na 2 O Complete the set up: 505 g Na x 1 mol Na x x 23.0 g Na 112 Solution # 3 4 Na + O 2 2 Na 2 O 505 g Na x 1 mol Na x 1 mol O 2 x 6.02 x g Na 4 mol Na 1 mol O 2 = 3.30 x molelcules 113 9
10 Learning Check # 4 Acetylene gas C 2 H 2 burns in the oxyactylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g of CO 2? 2 C 2 H O 2 4 CO H 2 O 75.0 g CO 2 x x x 114 Solution # 4 2 C 2 H O 2 4 CO + 2 H 2 O 75.0 g CO 2 x 1 mol CO 2 x 2 mol C 2 H 2 x 26.0 g C 2 H g CO 2 4 mol CO 2 1 mol C 2 H 2 = 22.2 g C 2 H Volume-Volume Calculations: How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2? CH 4 + 2O 2 CO 2 + 2H 2 O 17.5 L O 2 1 mol O L O 2 1 mol CH 4 2 mol O L CH 4 1 mol CH 4 = 8.75 L CH 4 Notice anything relating these two steps? 10
11 Avogadro told us: Equal volumes of gas, at the same temperature and pressure contain the same number of particles. Moles are numbers of particles You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same. 1 mole = 22.4 STP Shortcut for Volume-Volume? How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2? CH 4 + 2O 2 CO 2 + 2H 2 O 17.5 L O 2 1 L CH 4 2 L O 2 = 8.75 L CH 4 Note: This only works for Volume- Volume problems STP. Given (A) grams (A) Pathways for Problems Using Equations Needed (B) grams (B) Molar mass Molar mass moles (A) Coefficients moles (B) particles (A) Avogadro s Number particles (B) Avogadro s Number
12 How do you get good at this? 12
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