3. Atomic Mass l Atoms are so small, it is difficult to discuss how much they weigh in grams. l Use atomic mass units. l an atomic mass unit (amu) is one twelth the mass of a carbon-12 atom. l This gives us a basis for comparison. l The decimal numbers on the table are atomic masses in amu.

4. They are not whole numbers l Because they are based on averages of atoms and of isotopes. l can figure out the average atomic mass from the mass of the isotopes and their relative abundance. l add up the percent as decimals times the masses of the isotopes.

5. Examples l There are two isotopes of carbon 12 C with a mass of 12.00000 amu(98.892%), and 13 C with a mass of 13.00335 amu (1.108%). l There are two isotopes of nitrogen, one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundance of each?

6. The Mole l The mole is a number. l A very large number, but still, just a number. l 6.022 x 10 23 of anything is a mole l A large dozen. l The number of atoms in exactly 12 grams of carbon-12.

7. The Mole l Makes the numbers on the table the mass of the average atom.

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

9. More Stoichiometry

10. Molar mass l Mass of 1 mole of a substance. l Often called molecular weight. l To determine the molar mass of an element, look on the table. l To determine the molar mass of a compound, add up the molar masses of the elements that make it up.

11. Find the molar mass of l CH 4 l Mg 3 P 2 l Ca(NO 3 ) 3 l Al 2 (Cr 2 O 7 ) 3 l CaSO 4 · 2H 2 O

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

13. © 2009, Prentice-Hall, Inc. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (FW of the compound) x 100

14. © 2009, Prentice-Hall, Inc. Percent Composition So the percentage of carbon in ethane is… %C = (2)(12.0 amu) (30.0 amu) 24.0 amu 30.0 amu =x 100 = 80.0%

15. Copyright © Cengage Learning. All rights reserved l Mass percent of an element: l For iron in iron(III) oxide, (Fe 2 O 3 ):

16. Percent Composition l Percent of each element a compound is composed of. l Find the mass of each element, divide by the total mass, multiply by a 100. l Easiest if you use a mole of the compound. l Find the percent composition of CH 4 l Al 2 (Cr 2 O 7 ) 3 l CaSO 4 · 2H 2 O

17. © 2009, Prentice-Hall, Inc. Finding Empirical Formulas

18. © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

19. Chapter 3: Stoichiometry18 Determining Empirical Formulas EOS A compound is comprised of 40.01% carbon, 6.72% hydrogen, and 53.27% oxygen. Calculate the empirical formula of the compound. 1) Given the percent composition, assume a mass of sample - use 100.00 g for convenience 6.72% H 40.01% C 53.27% O 6.72 g H 40.01 g C 53.27 g O

20. © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas The compound para -aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

21. © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas Assuming 100.00 g of para -aminobenzoic acid, C:61.31 g x = 5.105 mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= 0.7288 mol N O:23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g

22. © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= 7.005  7 H:= 6.984  7 N:= 1.000 O:= 2.001  2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol

23. © 2009, Prentice-Hall, Inc. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

24. Working backwards l From percent composition, you can determine the empirical formula. l Empirical Formula the lowest ratio of atoms in a molecule. l Based on mole ratios. l A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O what is its empirical formula.

25. Pure O 2 in CO 2 is absorbed H 2 O is absorbed Sample is burned completely to form CO 2 and H 2 O

26. l A 0.2000 gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O 2. 0.2998 g of CO 2 and 0.0819 g of H 2 O are produced. What is the empirical formula?

27. Empirical To Molecular Formulas l Empirical is lowest ratio. l Molecular is actual molecule. l Need Molar mass. l Ratio of empirical to molar mass will tell you the molecular formula. l Must be a whole number because...

28. Example l A compound is made of only sulfur and oxygen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its formula?

29. Copyright © Cengage Learning. All rights reserved 28 Exercise The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol.  What is the empirical formula? C3H5O2C3H5O2  What is the molecular formula? C 6 H 10 O 4

30. Chemical Equations l Are sentences. l Describe what happens in a chemical reaction. Reactants  Products l Equations should be balanced. l Have the same number of each kind of atoms on both sides because...

31. Meaning l A balanced equation can be used to describe a reaction in molecules and atoms. l Not grams. l Chemical reactions happen molecules at a time l or dozens of molecules at a time l or moles of molecules.

32. Stoichiometry l Given an amount of either starting material or product, determining the other quantities. l use conversion factors from –molar mass (g - mole) –balanced equation (mole - mole) l keep track.

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

34. Examples l One way of producing O 2 ( g ) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O 2 (g) are produced? l How many grams of potassium chloride? l How many grams of oxygen?

35. Examples l A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess HCl(aq). How many grams of H 2 ( g ) are produced? How many grams of each reactant are needed to produce 15 grams of iron form the following reaction? Fe 2 O 3 ( s ) + Al( s )  Fe( s ) + Al 2 O 3 ( s )

36. Examples K 2 PtCl 4 ( aq ) + NH 3 ( aq )  Pt(NH 3 ) 2 Cl 2 ( s )+ KCl( aq ) l what mass of Pt(NH 3 ) 2 Cl 2 can be produced from 65 g of K 2 PtCl 4 ? l How much KCl will be produced? l How much from 65 grams of NH 3 ?

37. Gases and the Mole

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

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

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

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

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

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

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

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

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

47. Mole to Mole conversions l 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

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

49. 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

50. For example... l 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

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

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

53. Gases and Reactions

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

55. For Example l 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

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

57. Yield How much you get from an chemical reaction

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

59. Limiting Reagent l Reactant that determines the amount of product formed. l The one you run out of first. l Makes the least product. l Book shows you a ratio method. l It works. l So does mine

60. Limiting reagent l To determine the limiting reagent requires that you do two stoichiometry problems. l Figure out how much product each reactant makes. l The one that makes the least is the limiting reagent.

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

62. l 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

63. Example Ammonia is produced by the following reaction N 2 + H 2  NH 3 What mass of ammonia can be produced from a mixture of 100. g N 2 and 500. g H 2 ? l How much unreacted material remains?

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

65. Excess Reagent l The reactant you don’t run out of. l The amount of stuff you make is the yield. l The theoretical yield is the amount you would make if everything went perfect. l The actual yield is what you make in the lab.

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

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

69. Percent Yield l % yield = Actual x 100% Theoretical l % yield = what you got x 100% what you could have got

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

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

72. Examples l Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.

73. Examples Years of experience have proven that the percent yield for the following reaction is 74.3% Hg + Br 2  HgBr 2 If 10.0 g of Hg and 9.00 g of Br 2 are reacted, how much HgBr 2 will be produced? l If the reaction did go to completion, how much excess reagent would be left?

74. Examples Commercial brass is an alloy of Cu and Zn. It reacts with HCl by the following reaction Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) Cu does not react. When 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnCl 2 are eventually isolated. What is the composition of the brass?