2. Copyright Sautter 2003

3. STOICHIOMETRY “Measuring elements” Determining the Results of A Chemical Reaction

4. PREDICTING HOW MUCH OF A SUBSTANCE CAN BE MADE BY A CHEMICAL REACTION BEFORE IT IS CARRIED OUT!! STEP I ALWAYS WRITE THE EQUATION USING CHEMICAL FORMULAE AND BALANCE IT. (REMEMBER TO BE SURE YOU USE THE CORRECT SUBSCRIPTS FOR THE FORMULAE AND CHANGE ONLY THE COEFFICIENTS WHEN BALANCING THE EQUATION) EXAMPLE: CALCIUM CARBONATE (LIMESTONE) WHEN HEATED GIVES CALCIUM OXIDE (LIME) AND CARBON DIOXIDE CALCIUM = Ca (+2) CARBONATE = CO 3 (-2) CALCIUM CARBONATE = CaCO 3 OXIDE = O (-2) CALCIUM OXIDE = CaO CARBON DIOXIDE = CO 2 CaCO 3(S)  CaO (s) + CO 2(g) THERE IS ONE Ca ON EACH SIDE OF THE EQUATION, ONE C ON EACH SIDE AND THREE O ON EACH SIDE. THE EQUATION IS BALANCED!

5. #1. In terms of Particles u An Element is made of atoms u A Molecular compound (made of only nonmetals) is made up of molecules (Don’t forget the diatomic elements) u Ionic Compounds (made of a metal and nonmetal parts) are made of formula units

6. 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 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 Now read this: 2Na + 2H 2 O  2NaOH + H 2

7. #2. In terms of Moles u The coefficients tell us how many moles of each substance 2Al 2 O 3  Al + 3O 2 2Na + 2H 2 O  2NaOH + H 2 u Remember: A balanced equation is a Molar Ratio

8. #3. In terms of Mass u The Law of Conservation of Mass applies u We can check mass by using moles. 2H 2 + O 2   2H 2 O 2 moles H 2 2.02 g H 2 1 mole H 2 = 4.04 g H 2 1 mole O 2 32.00 g O 2 1 mole O 2 = 32.00 g O 2 36.04 g H 2 + O 2 + reactants

9. In terms of Mass (for products) 2H 2 + O 2   2H 2 O 2 moles H 2 O 18.02 g H 2 O 1 mole H 2 O = 36.04 g H 2 O 36.04 g H 2 + O 2 = 36.04 g H 2 O The mass of the reactants must equal the mass of the products. 36.04 grams reactant = 36.04 grams product

10. #4. In terms of Volume u 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) NOTE: mass and atoms are ALWAYS conserved - however, molecules, formula units, moles, and volumes will not necessarily be conserved! 67.2 Liters of reactant ≠ 44.8 Liters of product!

11. Practice: u 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

12. Chemical Calculations u OBJECTIVES: Construct “mole ratios” from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations.

13. 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 2 moles Al 2 O 3 3 mole O 2 These are the two possible conversion factors to use in the solution of the problem.

14. Mole to Mole conversions u 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 3.34 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 = 5.01 mol O 2 If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! Conversion factor from balanced equation

15. Practice: 2C 2 H 2 + 5 O 2  4CO 2 + 2 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)

16. How do you get good at this?

17. 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.

18. 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 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 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = 12.3 g Al 2 O 3 are formed

19. Another example: u 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

20. Volume-Volume Calculations: u 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 22.4 L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4 Notice anything relating these two steps?

21. Avogadro told us: u Equal volumes of gas, at the same temperature and pressure contain the same number of particles. u Moles are numbers of particles u 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 L @ STP

22. Shortcut for Volume-Volume? u 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 O2O2 2 L O 2 1 L CH 4 = 8.75 L CH 4 Note: This only works for Volume-Volume problems.

23. IN ORDER TO PREDICT THE RESULTS OF A CHEMICAL REACTION WE MUST BE GIVEN THE QUANTITIES OF MATERIALS THAT ARE TO BE USED IN THE REACTION SUPPOSE WE ARE GIVEN 200 GRAMS OF CALCIUM CARBONATE AND ASKED TO DECIDE HOW MUCH CALCIUM OXIDE AND CARBON DIOXIDE CAN BE MADE?? WE ALREADY HAVE THE BALANCED EQUATION BUT WHAT DO WE DO NEXT?? WELL, SINCE BALANCED EQUATIONS SHOW THE NUMBER OF ATOMS AND MOLECULES INVOLVED, WE MUST WORK WITH NUMBERS OF ATOMS AND MOLECULES. REMEMBER WE COUNT ATOMS AND MOLECULES WITH MOLES. (6.02 X 10 23 = 1 MOLE)

24. **REMEMBER ** TO CONVERT GRAMS (MASS) TO MOLES WE DIVIDE THE WEIGHT OF ONE MOLE FROM THE PERIODIC TABLE INTO THE GIVEN NUMBER OF GRAMS FOR EXAMPLE: SINCE CaCO 3 CONTAINS 1Ca FROM THE PERIODIC TABLE WE USE 40grams x 1 1 C FROM THE PERIODIC TABLE WE USE 12 grams x1 3 O FROM THE PERIODIC TABLE WE USE 16 grams x 3 (1x 40) + (1x 12) + (3 x 16) = 100 THE MASS OF ONE MOLE OF CALCIUM CARBONATE IS 100 grams IN OUR PROBLEM WE ARE USING 200 grams of CaCO 3 DIVIDING 200grams by 100grams per mole of CaCO 3 WE FIND THAT WE HAVE EXACTLY 2 MOLES OF THE CALCIUM CARBONATE REACTANT.

25. STEP II IN SOLVING STIOCHIOMETRY PROBLEMS THEN IS TO CONVERT THE GIVEN NUMBER OF GRAMS TO MOLES..

26. CaCO 3(S)  CaO (s) + CO 2(g) THE EQUATION SAYS THAT ONE CALCIUM CARBONATE MAKES ONE CALCIUM OXIDE. HOW MANY CALCIUM OXIDES WOULD TEN CALCIUM CARBONATES MAKES? HOW ABOUT TEN? WHAT ABOUT ONE HUNDRED CALCIUM CARBONATES? WOULDN’T THEY MAKE ONE HUNDRED CALCIUM OXIDES? OF COURSE !! THEN WHAT ABOUT A MOLE OF CALCIUM CARBONATE? WOULDN’T THEY BE EXPECTED TO MAKE A MOLE OF CALCIUM OXIDE? AND OF COURSE THEN TWO MOLES WOULD MAKE TWO MOLES!

27. STEP III IS PROBABLY THE MOST DIFFICULT ONE! IT REQUIRES US TO PUT TOGETHER THE BALANCED EQUATION FROM STEP I AND THE MOLES THAT WE CALCULATED IN STEP II !! ** REMEMBER ** HERE’S OUR BALANCED EQUATION CaCO 3(S)  CaO (s) + CO 2(g) AND HERE’S THE MOLES WE CALCULATED DIVIDING 200grams by 100grams per mole of CaCO 3 WE FIND THAT WE HAVE EXACTLY 2 MOLES OF THE CALCIUM CARBONATE REACTANT

28. CaCO 3(S)  CaO (s) + CO 2(g) WHAT ABOUT THE CARBON DIOXIDE? THE BALANCED EQUATION ALSO SAYS THAT ONE CALCIUM CARBONATE MAKES ONE CARBON DIOXIDE TOO. SO USING THE SAME LOGIC THAT WE APPLIED TO THE CALCIUM OXIDE, IT IS OBVIOUS THAT TWO MOLES OF CALCIUM CARBONATE WILL PRODUCE EXACTLY TWO MOLES OF CARBON DIOXIDE ALSO. STEP III – USING THE BALANCED EQUATION RATIOS FOUND IN STEP I AND THE MOLES DETERMINED IN STEP II, FIND THE MOLES OF EACH PRODUCT MATERIAL FORMED.

29. STEP IV – CONVERT THE MOLES FOUND IN STEP III TO GRAMS (MASS) ** REMEMBER** TO CONVERT MOLES TO GRAMS, MULTIPLY THE MASS OF ONE MOLE FROM THE PERIODIC TABLE BY THE NUMBER OF MOLES. EXAMPLE: CaO CONTAINS 1 Ca (1 X 40grams from the Periodic Table) and 1 O (1 x 16 from the Periodic Table) THE MASS OF ONE MOLE OF CaO IS (1 x 40) + (1 x 16) = 56 grams per mole TWO MOLES OF CaO ARE FORMED THEREFORE, 2 MOLES x 56 grams per mole = 112 gram of CaO ARE FORMED IN THE REACTION CO 2 CONTAINS 1 C (1 x 12grams from the Periodic Table) and 2 O (2 x 16 grams from the Periodic Table) THE MASS OF ONE MOLE OF CO 2 IS (1 x 12) + (2 x 16) = 44 grams per moles TWO MOLES OF CO 2 ARE FORMED THEREFORE, 2 MOLES x 44 grams per mole = 88 grams of CO 2 ARE FORMED IN THE REACTION

30. CaCO 3(S)  CaO (s) + CO 2(g) 200 grams 0 grams 0 grams 2 moles 0 moles 0 moles before reaction occurs 0 grams 112 grams 88 grams 0 moles 2moles 2moles after reaction occurs STARTING TOTAL MASS = 200 +0 +0 = 200 GRAMS FINAL TOTAL MASS = 0 + 112 + 88 = 200 GRAMS (CONSERVATION OF MASS) QUESTIONS ?????

31. NOW FOR A HARDER ONE! HYDROGEN GAS REACTS WITH OXYGEN GAS TO FORM WATER. HOW MUCH HYDROGEN AND OXYGEN MUST BE COMBINED TO MAKE 45 GRAMS OF WATER? STEP I – WRITE AND BALANCE THE EQUATION HYDROGEN = H 2 (DIATOMIC ELEMENT) OXYGEN = O 2 (DIATOMIC ELEMENT) WATER = H 2 O 2 H 2(g) + 1 O 2(g)  2 H 2 O (g) THERE ARE 4 ATOMS OF HYDROGEN ON EACH SIDE AND TWO ATOMS OF OXYGEN ON EACH SIDE. THE EQUATION IS BALANCED !!

32. STEP II – CONVERT THE GIVEN GRAMS TO MOLES WATER HAS A MOLAR MASS (MASS OF ONE MOLE) FROM THE PERIODIC TABLE IS 18 GRAMS. { (2 x 1) FOR HYDROGEN AND (1 x 16) FOR OXYGEN } 45 grams of water DIVIDED BY 18 grams per mole = 2.5 moles We want to make 2.5 moles of water in our reaction.

33. STEP III – USE THE BALANCED EQUATION TO FIND THE MOLES OF REACTANT OR PRODUCT REQUIRED Here’s our balanced equation from STEP I 2 H 2(g) + 1 O 2(g)  2 H 2 O (g) TWO MOLES OF WATER ARE MADE FROM TWO MOLES OF HYDROGEN. HOW MANY MOLES OF HYDROGEN WOULD BE NEEDED TO MAKE 2.5 MOLES OF WATER? (A ONE FOR ONE RATIO) HOW ABOUT 2.5 ?? NOW FOR THE OXYGEN ! IT TAKES ONLY 1 MOLE OF OXYGEN TO MAKE 2 MOLES OF WATER, HALF AS MUCH ! (A TWO FOR ONE RATIO) THEREFORE TO MAKE 2.5 MOLES OF WATER NEEDS ONLY ½ OF 2.5 MOLES OF OXYGEN! THEREFORE 1.25 MOLES OF OXYGEN IS REQUIRED.

34. STEP IV – CONVERT MOLES OF UNKNOWN TO GRAMS SINCE WE NOW KNOW THAT 2.5 MOLES OF HYDROGEN IS REQUIRED, WE CAN MULTIPLY 2.5 TIMES 2.0 GRAMS, THE MOLAR MASS OF HYDROGEN TO GET 5.0 GRAMS OF H 2(g) NEEDED! THE MOLAR MASS OF OXYGEN IS 32 GRAMS. MULTIPLYING 32 TIMES 1.25 MOLES GIVES 40 GRAMS OF OXYGEN ARE NEEDED.

35. THEN, IN ORDER TO MAKE 45.0 GRAMS OF WATER, USING HYDROGEN GAS AND OXYGEN GAS WE HAVE CALCULATED THAT: THE MOLES OF HYDROGEN NEEDED ARE 2.5 MOLES OR 5.0 GRAMS AND THE MOLES OF OXYGEN NEEDED ARE 1.25 MOLES OR 40.0 GRAMS AND ALL OF THIS HAS BEEN DETERMINED WITHOUT EVER PICKING UP A TEST TUBE BY USING STOICHIOMETRY!!!

36. References Walt Sautter