2.  Objective: Understand molecular formulas and balancing equations.  Before: Introduction to molecular formulas  During: Discuss molecular formulas and balancing equations  After: Review molecular formulas and balancing equations 1

3. The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

4. The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 2. Divide the molecular mass by the mass given by the emipirical formula.

5. The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

6. Chemical change involves a reorganization of the atoms in one or more substances. C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O reactantsproducts 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water When the equation is balanced it has quantitative significance:

7. Balancing Equations In order to do stoichiometric problems a balance chemical equation is needed

8.  Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4  Reactants: Ca 2+ – 3, PO 4 3- - 2, H + – 2, SO 4 2+ - 1  Products: Ca 2+ - 1, SO 4 2- - 1, H + - 3, PO 4 3- - 1 7

9. ____C 3 H 8 (g) + _____ O 2 (g) ----> _____CO 2 (g) + _____ H 2 O(g) ____B 4 H 10 (g) + _____ O 2 (g) ----> ___ B 2 O 3 (g) + _____ H 2 O(g)

10. 1. Balance the equation. 2. Convert masses to moles. 3. Determine which reactant is limiting. 4. Use moles of limiting reactant and mole ratios to find moles of desired product. 5. Convert from moles to grams.

11. 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. Al+O2O2 Al 2 O 3 b. What are the reactants? a. Every reaction needs a yield sign! c. What are the products? d. What are the balanced coefficients? 43 2

12. 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =12.3 g Al 2 O 3

13.  Objective: Understand limiting reactant(s) in a chemical reaction.  Before: Introduction to limiting reactants  During: Discuss limiting reactants and chemical equations  After: Review limiting reactants and chemical equations 12

14. The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.

15. LIMITING REACTANT The reactant that gives the least number of product moles “limits” the reaction. To understand this concept, let’s suppose you were an elf working for Santa Claus and your job was to make candy canes. You take one red stick and one white stick then twist them around to make one candy cane. The ratio of red to white is 1:1. Time is ticking and you find that you have 24 red stick and 17 white sticks left. What is the maximum number of candy canes you can make? *MOLES* The answer is 17 candy canes! The white sticks “limit” the amount of product you could make. In chemistry we do not use sticks but *MOLES* to determine which starting material will limit the maximum amount of product that can be produced in a chemical reaction.

16. LIMITING REACTANT a) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? 4 Na (s) + O 2(g)  2 Na 2 O (s) Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 6.74 g of Na 2 O 5.00g Na ( 1 mole Na ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 6.74 g of Na 2 O 23 g Na 4 mole Na 1 mol Na 2 O 19.38 g of Na 2 O 5.00g O 2 ( 1 mole O 2 ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 19.38 g of Na 2 O 32 g O 2 1 mole O 2 1 mol Na 2 O Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide. Wrong answer

17. LIMITING REACTANT b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4 Na (s) + O 2(g)  2 Na 2 O (s) There are two methods used to answer this question. The Law of Conservation of mass and Stoichiometry. The amount of O 2 used to make 6.74 g of Na 2 O is calculated by: 1.74 g of O 2 was used 5.00g Na ( 1 mole Na ) ( 1 mole O 2 )( 32 g O 2 ) = 1.74 g of O 2 was used 23 g Na 4 mole Na 1 mol O 2 Or use the Law of Conservation of mass: Mass of product (6.74 g) – mass of limiting reactant (5.00 g) = mass of other reactant, in this case oxygen (1.74 g). The amount of oxygen (O 2 ) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

18. PRACTICE PROBLEM #19 1. How much AgCl product will be produced if 100.00 g of BaCl 2 reacted with excess AgNO 3 ? 2. How many moles of carbon dioxide could be produced from 220.0 g of C 2 H 2 and 545.0 g of O 2 ? 3. How many grams of CO 2 can be produced by the reaction of 35.5 grams of C 2 H 2 and 45.9 grams of O 2 ? 4. In the reaction between CH 4 and O 2, if 25.0 g of CO 2 are produced, what is the minimum amount of each reactant needed? 5. Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag. When 10.0 g of copper was reacted with 60.0 g of silver nitrate solution, 30.0 g of silver was obtained. What is the percent yield of silver obtained? 137.57 g 50.5 g 13.63 mol 9.09 g of CH 4 & 36.4 g of O 2 88.3%