1. 1 A BALANCED EQUATION THE HEART OF STOICHIOMETRY

2. 2 Proportional Relationships  Stoichiometry  mass relationships between substances in a chemical reaction  based on the mole ratio  Mole Ratio  indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

3. 3 Proportional Relationships  I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 3 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs 3 doz. 2 eggs = 7.5 dozen cookies Ratio of eggs to cookies

4. 4 HOW TO SOLVE PROBLEMS  There are four basic types of problems: problems:  Mole to mole  Mole to gram  Gram to mole  Gram to gram

5. 5 MOLE TO MOLE PROBLEMS   X Moles A is the amount given in the problem   Moles A to Moles B is the mole ratio from your equation.   The coefficients are the mole ratio! X mol Amol B mol A = mol B

6. 6 2H2 + O2  2H2O  If you have 12 moles of oxygen, how many moles of water will you make? 2 mol H 2 O 1 mol O 2 = 24 mol H 2 O (from problem) 12 mol O 2 (from equation)

7. 7 You try one Write and balance: sodium + fluorine  sodium fluoride sodium + fluorine  sodium fluoride If you have 5.50 moles of sodium, how many moles of fluorine do you need? If you have 5.50 moles of sodium, how many moles of fluorine do you need?

8. 8 2Na + F 2  2NaF  If you have 5.50 moles of sodium, how many moles of fluorine do you need? 5.50 mol Na1 mol F 2 2 mol Na = 2.75 mol F 2 (from problem) (from equation)

9. 9 MOLE TO GRAM PROBLEMS  You have to find the molar mass of compounds to solve these problems. X mol Amol B mol A (from problem)(from equation) grams B 1 mol B (from periodic table) = grams B

10. 10 2Na + F 2  2NaF  If you have 18 moles of fluorine, how many grams of NaF will you make?  First, find the g/mol of your compound  Then, set up your problem.  You must use labels in the problem!!  Check your sig figs!

11. 11 Find the mass of NaF, then set up the problem 23g Na + 19g F = 42g/mol NaF 18 mol F 2 (from problem) (from equation)(from periodic table) 2 mol NaF 1 mol F 2 42 g NaF 1 mol NaF = 1500 g NaF

12. 12 2Na + F 2  2NaF  If you have 25.0 moles of Na, how many grams of fluorine will you need? 25.0 mol Na 1 mol F 2 38g F 2 25.0 mol Na 1 mol F 2 38g F 2 2 mol Na 1 mol F 2 2 mol Na 1 mol F 2 (from problem) (from equation) (from periodic table) = 475 g F 2

13. 13 GRAM TO MOLE PROBLEMS  You have to find molar mass of compounds to solve these problems, too. gram A 1 mol A mol B gram A mol A gram A mol A (from problem) (from periodic table) (from equation) (from problem) (from periodic table) (from equation) = mol B

14. 14 4K + O 2  2K 2 O  If you have 245g K, how many moles of K 2 O will you produce? 245g K 1 mol K 39 g K 2 mol K 2 O 4 mol K = 3.14 mol K 2 O

15. 15 GRAM TO GRAM PROBLEMS  These are the longest problems of all.  You have to find the molar mass of several compounds to solve the problems.  You must have labels in every part of your equation!

16. 16 GRAM TO GRAM PROBLEMS  Now is not the time to take shortcuts! I will mark them wrong!  Do not combine any equalities or skip any equalities.  My way is the only way!

17. 17 GRAM TO GRAM PROBLEMS X grams A 1 mol A grams A (from problem)(from P.T.) mol B mol A (from equation) grams B 1 mol B (from P.T.) = g B

18. 18 2NaOH + H 2 SO 4  Na 2 SO 4 + 2H 2 O  If you have 95 grams of sodium hydroxide, how many grams of sulfuric acid will you need?  First, find g/mol of both sodium hydroxide and sulfuric acid. Na = 23 1 O = 16 1 H = 1 40g/mol 2H = 2 1 S = 32 4 O = 64 98g/mol

19. 19 The set-up 95 g NaOH1 mol NaOH 40 g NaOH 1 mol H 2 SO 4 2 mol NaOH 98 g H 2 SO 4 1 mol H 2 SO 4 = 116 g H 2 SO 4

20. 20 Use the same information and equation to solve the following: How many grams of each product will be produced?  First, find the molar mass of each product. Na = 2(23) = 46 S = 1(32) = 32 S = 1(32) = 32 O = 4(16) = 64 O = 4(16) = 64 142 g/mol 142 g/mol H = 2 (1) = 2 O = 1(16) = 16 O = 1(16) = 16 18g/mol 18g/mol

21. 21 2NaOH + H 2 SO 4  Na 2 SO 4 + 2H 2 O 95g NaOH 1 mol NaOH 1 mol Na 2 SO 4 142 g Na 2 SO 4 = g Na 2 SO 4 40g NaOH 2 mol NaOH 1 mol Na 2 SO 4 40g NaOH 2 mol NaOH 1 mol Na 2 SO 4 95g NaOH 1 mol NaOH 2 mol H 2 O 18 g H 2 O = g H 2 O 40g NaOH 2 mol NaOH 1 mol H 2 O 40g NaOH 2 mol NaOH 1 mol H 2 O

22. 22 Solution  168.63 g Na 2 SO 4  42.75 g H 2 O