2. Honors Chapter 12

3. Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced equation. u We can interpret balanced chemical equations several ways.

4. In terms of Particles u Element- atoms u Molecular compound (non- metals)- molecule u Ionic Compounds (Metal and non-metal) - formula unit

5. 2H 2 + O 2   2H 2 O u Two molecules of hydrogen and one molecule of oxygen form two molecules of water.  2 Al 2 O 3  Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 2Na + 2H 2 O  2NaOH + H 2

6. Look at it differently  2H 2 + O 2   2H 2 O u 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. u 2 x (6.02 x 10 23 ) molecules of hydrogen and 1 x (6.02 x 10 23 ) molecules of oxygen form 2 x (6.02 x 10 23 ) molecules of water. u 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

7. In terms of Moles  2 Al 2 O 3  Al + 3O 2  2Na + 2H 2 O  2NaOH + H 2 u The coefficients tell us how many moles of each kind

8. In terms of mass u The law of conservation of mass applies u We can check using moles  2H 2 + O 2   2H 2 O 2 moles H 2 2.02 g H 2 1 moles H 2 =4.04 g H 2 1 moles O 2 32.00 g O 2 1 moles O 2 =32.00 g H 2 36.04 g H 2

9. In terms of mass  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 2H 2 + O 2   2H 2 O 36.04 g H 2 + O 2 =36.04 g H 2 O

10. Your turn u Show that the following equation follows the Law of conservation of mass.  2 Al 2 O 3  Al + 3O 2

11. Mole to mole conversions  2 Al 2 O 3  Al + 3O 2 u every time we use 2 moles of Al 2 O 3 we 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

12. Mole to Mole conversions u How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose?  2 Al 2 O 3  Al + 3O 2 3.34 moles Al 2 O 3 2 moles Al 2 O 3 3 mole O 2 =5.01 moles O 2

13. Your Turn  2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O u If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? u How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? u If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

14. How do you get good at this?

15. Mass in Chemical Reactions How much do you make? How much do you need?

16. We can’t measure moles!! u What can we do? u We can convert grams to moles. u Periodic Table u Then do the math with the moles. u Balanced equation u Then turn the moles back to grams. u Periodic table

17. For example... u If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?  Fe + CuSO 4  Fe 2 (SO 4 ) 3 + Cu  2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 10.1 g Fe 55.85 g Fe 1 mol Fe = 0.181 mol Fe

18. 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 0.181 mol Fe 2 mol Fe 3 mol Cu = 0.272 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu

19. Could have done it 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu

20. More Examples u To make silicon for computer chips they use this reaction  SiCl 4 + 2Mg  2MgCl 2 + Si u How many grams of Mg are needed to make 9.3 g of Si? u How many grams of SiCl 4 are needed to make 9.3 g of Si? u How many grams of MgCl 2 are produced along with 9.3 g of silicon?

21. For Example u The U. S. Space Shuttle boosters use this reaction  3 Al(s) + 3 NH 4 ClO 4  Al 2 O 3 + AlCl 3 + 3 NO + 6H 2 O u How much Al must be used to react with 652 g of NH 4 ClO 4 ? u How much water is produced? u How much AlCl 3 ?

22. We can also change u Liters of a gas to moles u At STP u 25ºC and 1 atmosphere pressure u At STP 22.4 L of a gas = 1 mole u If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?

23. For Example u If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP?  H 2 O  H 2 + O 2 6.45 g H 2 O 18.02 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O 2 22.4 L O 2

24. Your Turn u How many liters of CO 2 at STP will be produced from the complete combustion of 23.2 g C 4 H 10 ? u What volume of oxygen will be required?

25. Gases and Reactions A few more details

26. Example 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 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

27. Avagadro 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.

28. Example u How many liters of CO 2 at STP are produced by completely burning 17.5 L of CH 4 ?  CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L CH 4 1 L CH 4 2 L CO 2 = 35.0 L CH 4

29. Limiting Reagent u If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make. u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent determines how much product you can make

30. How do you find out? u Do two stoichiometry problems. u The one that makes the least product is the limiting reagent. u For example u Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

31. u If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?  2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent

32. Your turn u If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? u How many grams of solid? u How much excess reagent remains?

33. Your Turn II u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? u How much excess reagent will remain?

35. Yield u The amount of product made in a chemical reaction. u There are three types u Actual yield- what you get in the lab when the chemicals are mixed u Theoretical yield- what the balanced equation tells you you should make. u Percent yield u Percent yield = Actual x 100 % Theoretical

36. Example u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the actual yield? u What is the theoretical yield? u What is the percent yield?

37. Details u Percent yield tells us how “efficient” a reaction is. u Percent yield can not be bigger than 100 %.

38. Energy in Chemical Reactions How Much? In or Out?

39. 38 Energy u Energy is measured in Joules or calories u Every reaction has an energy change associated with it u Exothermic reactions release energy, usually in the form of heat. u Endothermic reactions absorb energy u Energy is stored in bonds between atoms

40. 39 C + O 2  CO 2 Energy ReactantsProducts  C + O 2 395kJ + 395 kJ

41. 40 In terms of bonds C O O C O O Breaking this bond will require energy C O O O O C Making these bonds gives you energy In this case making the bonds gives you more energy than breaking them

42. 41 Exothermic u The products are lower in energy than the reactants u Releases energy

43. 42 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ CaCO 3 + 176 kJ  CaO + CO 2

44. 43 Endothermic u The products are higher in energy than the reactants u Absorbs energy

45. 44 Chemistry Happens in u MOLES u An equation that includes energy is called a thermochemical equation  CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u 1 mole of CH 4 makes 802.2 kJ of energy. u When you make 802.2 kJ you make 2 moles of water

46. 45 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u If 10. 3 grams of CH 4 are burned completely, how much heat will be produced? 10. 3 g CH 4 16.05 g CH 4 1 mol CH 4 802.2 kJ =514 kJ

47. 46 CH 4 + 2 O 2  CO 2 + 2 H 2 O + 802.2 kJ u How many liters of O 2 at STP would be required to produce 23 kJ of heat? u How many grams of water would be produced with 506 kJ of heat?

48. Heats of Reaction

49. 48 Enthalpy u The heat content a substance has at a given temperature and pressure u Can’t be measured directly because there is no set starting point u The reactants start with a heat content u The products end up with a heat content u So we can measure how much enthalpy changes

50. 49 Enthalpy u Symbol is H  Change in enthalpy is  H u delta H u If heat is released the heat content of the products is lower   H is negative (exothermic) u If heat is absorbed the heat content of the products is higher   H is negative (endothermic)

51. 50 Energy ReactantsProducts  Change is down  H is <0

52. 51 Energy ReactantsProducts  Change is up  H is > 0

53. 52 Heat of Reaction u The heat that is released or absorbed in a chemical reaction  Equivalent to  H  C + O 2 (g)  CO 2 (g) +393.5 kJ  C + O 2 (g)  CO 2 (g)  H = -393.5 kJ u In thermochemical equation it is important to say what state  H 2 (g) + 1/2O 2 (g)  H 2 O(g)  H = -241.8 kJ  H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -285.8 kJ

54. 53 Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance  C 2 H 4 + 3 O 2  2 CO 2 + 2 H 2 O  C 2 H 6 + O 2  CO 2 + H 2 O  2 C 2 H 6 + 5 O 2  2 CO 2 + 6 H 2 O  C 2 H 6 + (5/2) O 2  CO 2 + 3 H 2 O

55. 54 Standard Heat of Formation  The  H for a reaction that produces 1 mol of a compound from its elements at standard conditions u Standard conditions 25°C and 1 atm. u Symbol is u The standard heat of formation of an element is 0 u This includes the diatomics

56. 55 What good are they? u There are tables (pg. 190) of heats of formations u The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products

57. 56 Examples  CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = -74.86 kJO 2 (g) = 0 kJCO 2 (g) = -393.5 kJH 2 O(g) = -241.8 kJ   H= [-393.5 + 2(-241.8)]-[-74.68 +2 (0)]   H= 802.4 kJ

58. 57 Examples  2 SO 3 (g)  2SO 2 (g) + O 2 (g)

59. 58 Why Does It Work?  If H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H=-285.5 kJ  then H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ u If you turn an equation around, you change the sign  2 H 2 O(g)  H 2 (g) + O 2 (g)  H =+571.0 kJ u If you multiply the equation by a number, you multiply the heat by that number.

60. 59 Why does it work? u You make the products, so you need there heats of formation u You “unmake” the products so you have to subtract their heats. u How do you get good at this