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THE HEART OF STOICHIOMETRY

THE HEART OF STOICHIOMETRY. A BALANCED EQUATION. 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. Ratio of eggs to cookies.

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THE HEART OF STOICHIOMETRY

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  1. THE HEART OF STOICHIOMETRY A BALANCED EQUATION

  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 + O2 2 MgO

  3. Ratio of eggs to cookies Proportional Relationships 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 • I have 5 eggs. How many cookies can I make? 5 eggs 3 doz. 2 eggs = 7.5 dozen cookies

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

  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 A mol B = mol B mol A

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

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

  8. 2Na + F2 2NaF • If you have 5.50 moles of sodium, how many moles of fluorine do you need? 5.50 mol Na 1 mol F2 = 2.75 mol F2 2 mol Na (from problem) (from equation)

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

  10. 2Na + F2 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. Find the mass of NaF, then set up the problem 23g Na + 19g F = 42g/mol NaF 18 mol F2 2 mol NaF 1 mol F2 42 g NaF 1 mol NaF = 1500 g NaF (from problem) (from equation) (from periodic table)

  12. 2Na + F2 2NaF • If you have 25.0 moles of Na, how many grams of fluorine will you need? 25.0 mol Na 1 mol F2 38g F2 2 mol Na 1 mol F2 (from problem) (from equation) (from periodic table) = 475 g F2

  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 (from problem) (from periodic table) (from equation) = mol B

  14. 4K + O2 2K2O • If you have 245g K, how many moles of K2O will you produce? 1 mol K 39 g K 245g K 2 mol K2O 4 mol K = 3.14 mol K2O

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

  18. 2NaOH + H2SO4 Na2SO4 + 2H2O • 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. The set-up 95 g NaOH 1 mol NaOH 40 g NaOH 1 mol H2SO4 2 mol NaOH 98 g H2SO4 1 mol H2SO4 = 116 g H2SO4

  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 O = 4(16) = 64 142 g/mol H = 2 (1) = 2 O = 1(16) = 16 18g/mol

  21. 2NaOH + H2SO4 Na2SO4 + 2H2O 95g NaOH 1 mol NaOH 1 mol Na2SO4 142 g Na2SO4 = g Na2SO4 40g NaOH 2 mol NaOH 1 mol Na2SO4 95g NaOH 1 mol NaOH 2 mol H2O 18 g H2O = g H2O 40g NaOH 2 mol NaOH 1 mol H2O

  22. Solution • 168.63 g Na2SO4 • 42.75 g H2O

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