2. Chapter 6 Chemical Quantities

3. How you measure how much? You can measure mass, or volume, or you can count pieces. We measure mass in grams. We measure volume in liters. We count pieces in MOLES.

4. Moles Defined as the number of carbon atoms in exactly 12 grams of carbon-12. 1 mole is 6.02 x 10 23 particles. Treat it like a very large dozen 6.02 x 10 23 is called Avagadro’s number.

5. Representative particles The smallest pieces of a substance. For a molecular compound it is a molecule. For an ionic compound it is a formula unit. For an element it is an atom.

6. Types of questions How many oxygen atoms in the following? –CaCO 3 –Al 2 (SO 4 ) 3 How many ions in the following? –CaCl 2 –NaOH –Al 2 (SO 4 ) 3

7. Types of questions How many molecules of CO 2 are the in 4.56 moles of CO 2 ? How many moles of water is 5.87 x 10 22 molecules? How many atoms of carbon are there in 1.23 moles of C 6 H 12 O 6 ? How many moles is 7.78 x 10 24 formula units of MgCl 2 ?

8. Measuring Moles Remember relative atomic mass? The amu was one twelfth the mass of a carbon 12 atom. Since the mole is the number of atoms in 12 grams of carbon-12, the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

9. Gram Atomic Mass The mass of 1 mole of an element in grams. 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron. We can right this as 12.01 g C = 1 mole We can count things by weighing them.

10. Examples How much would 2.34 moles of carbon weigh? How many moles of magnesium in 24.31 g of Mg? How many atoms of lithium in 1.00 g of Li? How much would 3.45 x 10 22 atoms of U weigh?

11. What about compounds? in 1 mole of H 2 O molecules there are two moles of H atoms and 1 mole of O atoms To find the mass of one mole of a compound –determine the moles of the elements they have –Find out how much they would weigh –add them up

12. What about compounds? What is the mass of one mole of CH 4 ? 1 mole of C = 12.01 g 4 mole of H x 1.01 g = 4.04g 1 mole CH 4 = 12.01 + 4.04 = 16.05g The Gram Molecular mass of CH 4 is 16.05g The mass of one mole of a molecular compound.

13. Gram Formula Mass The mass of one mole of an ionic compound. Calculated the same way. What is the GFM of Fe 2 O 3 ? 2 moles of Fe x 55.85 g = 111.70 g 3 moles of O x 16.00 g = 48.00 g The GFM = 111.70 g + 48.00 g = 159.70g

14. Molar Mass The generic term for the mass of one mole. The same as gram molecular mass, gram formula mass, and gram atomic mass.

15. Examples Calculate the molar mass of the following and tell me what type it is. Na 2 S N 2 O 4 C Ca(NO 3 ) 2 C 6 H 12 O 6 (NH 4 ) 3 PO 4

16. Using Molar Mass Finding moles of compounds Counting pieces by weighing

17. Molar Mass The number of grams of 1 mole of atoms, ions, or molecules. We can make conversion factors from these. To change grams of a compound to moles of a compound.

18. For example How many moles is 5.69 g of NaOH?

19. For example How many moles is 5.69 g of NaOH?

20. For example How many moles is 5.69 g of NaOH? l need to change grams to moles

21. For example How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH

22. For example How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g

23. For example How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

24. For example How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

25. For example How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

26. Examples How many moles is 4.56 g of CO 2 ? How many grams is 9.87 moles of H 2 O? How many molecules in 6.8 g of CH 4 ? 49 molecules of C 6 H 12 O 6 weighs how much?

27. Gases and the Mole

28. Gases Many of the chemicals we deal with are gases. They are difficult to weigh. Need to know how many moles of gas we have. Two things effect the volume of a gas Temperature and pressure Compare at the same temp. and pressure.

29. Standard Temperature and Pressure 0ºC and 1 atm pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume Avagadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

30. Examples What is the volume of 4.59 mole of CO 2 gas at STP? How many moles is 5.67 L of O 2 at STP? What is the volume of 8.8g of CH 4 gas at STP?

31. Density of a gas D = m /V for a gas the units will be g / L We can determine the density of any gas at STP if we know its formula. To find the density we need the mass and the volume. If you assume you have 1 mole than the mass is the molar mass (PT) At STP the volume is 22.4 L.

32. Examples Find the density of CO 2 at STP. Find the density of CH 4 at STP.

33. The other way Given the density, we can find the molar mass of the gas. Again, pretend you have a mole at STP, so V = 22.4 L. m = D x V m is the mass of 1 mole, since you have 22.4 L of the stuff. What is the molar mass of a gas with a density of 1.964 g/L? 2.86 g/L?

34. All the things we can change

35. We have learned how to change moles to grams moles to atoms moles to formula units moles to molecules moles to liters molecules to atoms formula units to atoms formula units to ions

36. Moles Mass

37. Moles Mass PT

38. Moles Mass Volume PT

39. Moles Mass Volume PT 22.4 L

40. Moles Mass Volume Representative Particles PT 22.4 L

41. 6.02 x 10 23 Moles Mass Volume Representative Particles PT 22.4 L

42. Moles Mass Volume Representative Particles 6.02 x 10 23 PT Atoms 22.4 L

43. Moles Mass Volume Representative Particles 6.02 x 10 23 PT Atoms Ions 22.4 L

44. Percent Composition Like all percents Part x 100 % whole Find the mass of each component, divide by the total mass.

45. Example Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.

46. Getting it from the formula If we know the formula, assume you have 1 mole. Then you know the pieces and the whole.

47. Examples Calculate the percent composittion of C 2 H 4 ? Aluminum carbonate.

48. Empirical Formula From percentage to formula

49. The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula the actual ration of elements in a compound. The two can be the same. CH 2 empirical formula C 2 H 4 molecular formula C 3 H 6 molecular formula H 2 O both

50. Calculating Empirical Just find the lowest whole number ratio C 6 H 12 O 6 CH 4 N It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen. In one molecule of CO 2 there is 1 atom of C and 2 atoms of O.

51. Calculating Empirical Means we can get ratio from percent composition. Assume you have a 100 g. The percentages become grams. Can turn grams to moles. Find lowest whole number ratio by dividing by the smallest.

52. Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 1.01 gH 45.11 g N x 1mol N = 3.219 mole N 14.01 gN

53. Example The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N C 1 H 5 N 1 A compound is 43.64 % P and 56.36 % O. What is the empirical formula? Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

54. Empirical to molecular Since the empirical formula is the lowest ratio the actual molecule would weigh more. By a whole number multiple. Divide the actual molar mass by the the mass of one mole of the empirical formula. Caffeine has a molar mass of 194 g. what is its molecular mass?

55. Example A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be 98.96 g. What is its molecular formula?

56. Chapter 8

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

58. In terms of Particles Atom - Element Molecule –Molecular compound (non- metals) –or diatomic (O 2 etc.) Formula unit –Ionic Compounds (Metal and non-metal)

59. 2H 2 + O 2   2H 2 O 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

61. Look at it differently 2H 2 + O 2   2H 2 O 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. 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. 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

62. In terms of Moles 2 Al 2 O 3  Al + 3O 2 The coefficients tell us how many moles of each kind

63. 36.04 g reactants In terms of mass The law of conservation of mass applies 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 O 2

64. 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

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

66. Mole to mole conversions 2 Al 2 O 3  Al + 3O 2 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

67. Mole to Mole conversions 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

68. Your Turn 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? How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

69. Mole to Mole Conversions 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

70. We can’t measure moles!! What can we do? We can convert grams to moles. Periodic Table Then use moles to change chemicals Balanced equation Then turn the moles back to grams. Periodic table

71. Periodic Table Moles A Moles B Mass g B Periodic Table Balanced Equation Mass g A Decide where to start based on the units you are given and stop based on what unit you are asked for

72. Conversions 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O How many moles of C 2 H 2 are needed to produce 8.95 g of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many g of CO 2 are formed?

73. For example... 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 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu

74. 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

75. 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

76. More Examples To make silicon for computer chips they use this reaction SiCl 4 + 2Mg  2MgCl 2 + Si How many moles of Mg are needed to make 9.3 g of Si? 3.74 mol of Mg would make how many moles of Si? How many grams of MgCl 2 are produced along with 9.3 g of silicon?

77. For Example 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 How much Al must be used to react with 652 g of NH 4 ClO 4 ? How much water is produced? How much AlCl 3 ?

78. How do you get good at this?

79. Gases and Reactions

80. We can also change Liters of a gas to moles At STP 0ºC and 1 atmosphere pressure At STP 22.4 L of a gas = 1 mole If 6.45 moles of water are decomposed, how many liters of oxygen will be produced at STP?

81. For Example 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 2H 2 O  2H 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

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

83. Example 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

84. Avagadro told us Equal volumes of gas, at the same temperature and pressure contain the same number of particles. Moles are numbers of particles You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.

85. Example 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 1 L CO 2 = 17.5 L CO 2

86. Particles We can also change between particles and moles. 6.02 x 10 23 –Molecules –Atoms –Formula units

87. Limiting Reagent 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? The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make

88. How do you find out? Do two stoichiometry problems. The one that makes the least product is the limiting reagent. For example 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?

89. 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

90. How much excess reagent? Use the limiting reagent to find out how much excess reagent you used Subtract that from the amount of excess you started with

91. Your turn Mg(s) +2 HCl(g)  MgCl 2 (s) +H 2 (g) If 10.1 mol of magnesium and 4.87 mol of HCl gas are reacted, how many moles of gas will be produced? How much excess reagent remains?

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

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

95. Example 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 What is the actual yield? What is the theoretical yield? What is the percent yield? If you had started with 9.73 g of Al, how much copper would you expect?

96. Details Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. How would you get good at this?