1. NOTES: Quantifying Chemical Compounds Chapter 11

2. I. Things you should remember From the Moles Unit: –Identify particles as atoms, molecules (mc), and formula units (fun) –1 mole = 6.02 x 10 23 atoms, molecules, or formula units –1 mole atom = mass (in grams) from the periodic table From the Naming & Formulas Unit: –How to write a formula given a chemical name –How to count the number of atoms or ions in a given compound

3. Example 1: Determine the number of atoms/moles and ions in aluminum carbonate Al 3+ CO 3 2- Al 2 (CO 3 ) 3 ( )

4. Example 1: Determine the number of atoms/moles and ions in aluminum carbonate 1 mol ________ = __________ FUN __________ 6.02 x 10 23 Al 2 (CO 3 ) 3

5. CO 3 2- Al 3+ CO 3 2- 1 formula unit of Al 2 (CO 3 ) 3 = 2 atoms Al 3 atoms C 9 atoms O 2 ions Al 3+ 3 ions CO 3 2-

6. CO 3 2- Al 3+ CO 3 2- 1 mole of Al 2 (CO 3 ) 3 = 2 moles Al 3 moles C 9 moles O 2 moles Al 3+ 3 moles CO 3 2-

7. Aluminum Carbonate Al o C o o o C o o o C o o

8. Example 2: Determine the moles of magnesium there are in 1.23 x 10 24 atoms of magnesium 1.23 x 10 24 atoms Mg 1 mole Mg 6.02 x 10 23 atoms Mg = 2.04mol Mg 1 G: 1.23 x 10 24 atoms Mg W: moles Mg R: 1 mole Mg = 6.02 x 10 23 atoms Mg

9. Example 2: Determine the mass of 3.50 mol of copper. 3.50 mol Cu63.546g Cu 1 mol Cu = 222 g Cu 1 G: 3.50 moles Cu W: grams Cu R: 1 mole Cu = 63.546 g Cu

10. Practice: 1.Determine the number of sulfate ions. a. 1 FUN (NH 4 ) 2 SO 4 b. 1 FUN aluminum sulfate 1 ion SO 4 2- 3 ions SO 4 2- (SO 4 ) 2- Al 2 (SO 4 ) 3 Al 3+

11. Practice 2. Determine the number of moles in 11.9 kg of aluminum. 11.9 kg Al1000g Al 1 kg Al = 441 1 mol Al 26.982g Al mol Al 1 G: 11.9 kg Al W: moles Al R: 1 mole Al = 26.982 g Al 1000 g Al = 1 kg Al

12. Extra Practice Determine the number of water molecules in 11.2 moles of water. 11.2 moles H 2 O 6.02 x 10 23 molecules H 2 O = 6.74 x 10 24 1 moles H 2 O molecules H 2 O 1 G: 11.2 moles H 2 O W: molecules H 2 O R: 1 mole H 2 O = 6.02 x 10 23 mc H 2 O

13. II. Quantifying Chemical Compounds

14. We will now do the same thing, but we will have questions that are more detailed about the compounds. First we need to know how to find the molar mass of the whole compound.

15. A. Molar Mass

16. 1. Definition Remember, molar mass is the mass (in grams) of one mole of an element or compound. 1 atom S = 32.066 amu 1 mole S = 6.02 x 10 23 atoms S = 32.066 g S

17. B. Determining the Molar Mass The molar mass of the compound is the sum of the molar masses of each atom in the compound. The units for molar mass are grams per mole (g/mol). grams or g/mol mole

18. To determine the molar mass of a compound: 1. Determine the formula for the compound. 2. Determine the mass of 1 mole of each element in the compound using a periodic table 3. Multiply the molar mass of each element by the total number of atoms (moles) present in the compound. 4. Take the sum for all the elements in the compound. Show your answer using the sig figs.

19. SO 3 Ex 1: Determine the molar mass of sulfur trioxide. + = X 3 (47.997 g) 80.063 g/mol or 80.063 g = 1 mole SO 3

20. To be proper… When calculating molar masses, we are going to use the rules for sig figs! Example: 32.066 +15.999 +15.999 +15.999 80.063 For Addition or Subtraction: You answer is rounded to the same place value as your least precise measurement.

21. Ex: Determine the molar mass of calcium nitrate Ca 2+ (NO 3 ) - Ca(NO 3 ) 2 First you need the right formula right?

22. Ex: Determine the molar mass of calcium nitrate. Ca(NO 3 ) 2 + = X 2+X 6 164.086 g/mol or 164.086 g = 1 mol Ca(NO 3 ) 2

23. Ex: Determine the molar mass of Iron(III) sulfate. Fe 3+ (SO 4 ) 2- Fe 3+ (SO 4 ) 2- Fe 2 (SO 4 ) 3 First you need the right formula right?

24. Ex: Determine the molar mass of Iron(III) sulfate. Fe 2 (SO 4 ) 3 +X 2+X 12 399.88 g/mol or 399.88 g = 1 mol Fe 2 (SO 4 ) 3 X 3

25. Practice: 1. Determine the molar mass of each of the following. a. Li 2 S b. (NH 4 ) 2 CO 3 C. magnesium hydroxide d. copper (II) iodide 45.948 g/mol Li 2 S 96.086 g/mol (NH 4 ) 2 CO 3 58.319 g/mol Mg(OH) 2 317.354 g/mol CuI 2

26. 3. Using Molar Mass in Calculations You would use the molar mass of a compound in dimensional analysis just like you did with elements. Our answers will have the same number of sig figs as our given.

27. Ex: Determine the mass of 48.6 moles of sodium chloride. Step 1: Write the formula for sodium chloride. Na + Cl - NaCl

28. Ex: Determine the mass of 48.6 moles of sodium chloride. Step 2: set up a dimensional analysis problem G: W: R: 48.6 moles NaCl Mass (g) NaCl __g = 1 mol NaCl

29. Ex: Determine the mass of 48.6 moles of sodium chloride. Step 3: Find the molar mass for NaCl because the wanted is MASS Na = 1 X g = 22.990g Cl = 1 X g = 35.453g 58.443g/mol NaCl

30. Ex: Determine the mass of 48.6 moles of sodium chloride. Step 2: set up a dimensional analysis problem G: W: R: 48.6 moles NaCl Mass (g) NaCl 58.443 g = 1 mol NaCl

31. Ex: Determine the mass of 48.6 moles of sodium chloride. 48.6 mol NaCl58.443g NaCl 1 mol NaCl = 2.84 x 10 3 g NaCl = 2840g NaCl G: W: R: 48.6 moles NaCl Mass (g) NaCl 58.443 g = 1 mol NaCl 1

32. Tips for working molar mass problems 1. If mass is involved in your problem, determine molar mass of compound. 1 mole X = ____________ g X 2. If you see the terms: atoms, molecules, or formula units, use Avogadro’s number. 1 mol X = 6.02 X 10 23 ___*___ X * atoms, molecules or formula units 3. If you see 2 different substances in the G and W, you have to think a little harder. Remember, ions make formula units moles of elements make moles of compounds (Ex. 1 mole CO 2 = 2 moles O)

33. Ex: Determine the number of moles in 582 g of magnesium nitrate Mg +2 NO 3 - Mg(NO 3 ) 2

34. Ex: Determine the number of moles in 582 g of magnesium nitrate Step 1: Determine the G & W G: W: R: 582 g Mg(NO 3 ) 2 Moles Mg(NO 3 ) 2

35. Ex: Determine the number of moles in 582 g of magnesium nitrate. Because you saw “grams”, calculate the molar mass for Mg(NO 3 ) 2 Mg = 1 X 24.305 g = 24.305 g N = 2 X 14.007 g = 28.014 g O = 6 X 15.999 g = 95.994 g 148.313 g/mol Mg(NO 3 ) 2 148.313 g = 1 mol Mg(NO 3 ) 2

36. Ex: Determine the number of moles in 582 g of magnesium nitrate G: W: R: 582 g Mg(NO 3 ) 2 Moles Mg(NO 3 ) 2 1 Mol Mg(NO 3 ) 2 = 148.313g Mg(NO 3 ) 2

37. Ex 4: Determine the number of moles in 582 g of magnesium nitrate. Step 3: Set up a dimensional analysis problem. 582g Mg(NO 3 ) 2 148.313g Mg(NO 3 ) 2 1 mol Mg(NO 3 ) 2 = 3.92 mol Mg(NO 3 ) 2 G: W: R: 582 g Mg(NO 3 ) 2 Moles Mg(NO 3 ) 2 1 Mol Mg(NO 3 ) 2 = 148.313g Mg(NO 3 ) 2 1

38. Ex 3: Determine the number of molecules in 47.3 g of sulfuric acid. 47.3g H 2 SO 4 1 mol H 2 SO 4 98.078g H 2 SO 4 6.02 x 10 23 molecules H 2 SO 4 1 mol H 2 SO 4 = 2.90 x 10 23 molecules H 2 SO 4 1 G: W: R: 47.3 g H 2 SO 4 Molecules of H 2 SO 4 1 mol H 2 SO 4 = g H 2 SO 4 1 mol H 2 SO 4 = 6.02 x 10 23 mc H 2 SO 4 98.078

39. Ex. 4 Determine the number of ions of oxide in 3.6 moles TiO 2. 3.6 mol TiO 2 6.02 x 10 23 FUN TiO 2 1 mol TiO 2 = 4.3 x 10 24 ions O 2- 1 G: W: R: 3.6 moles TiO 2 Ions of oxide 2 ions O = 1 FUN TiO 2 1 mol TiO 2 = 6.02 x 10 23 FUN 2 ions Oxide 1 FUN TiO 2

40. Practice 1.Determine the mass of 9.45 mol of dinitrogen trioxide. 2.Determine the mass (in kg) of 5.83 x 10 23 molecules of HCl. 3.Determine the number of molecules in 782g of N 2 O 3. 0.0353 kg HCl 6.19 x 10 24 molecules N 2 O 3 718 g N 2 O 3

41. 1. Determine the mass of 9.45 mol of dinitrogen trioxide. 9.45 mol N 2 O 3 76.011 g N 2 O 3 1 mol N 2 O 3 = 718 g N 2 O 3 1 G: W: R: 9.45 mol N 2 O 3 g N 2 O 3 1 mol N 2 O 3 = 76.011 g N 2 O 3

42. 2. Determine the mass (in kg) of 5.83 x 10 23 molecules of HCl. 5.83 x 10 23 mc HCl 1 mol HCl 6.02 x 10 23 mc HCl 36.461g HCl 1 mol HCl = 0.0353 kg HCl 1 G: W: R: 5.83 x 10 23 molecules HCl kg of HCl 1 mol HCl = 6.02 x 10 23 mc HCl 1 mol HCl = 36.461 g HCl 1000 g = 1 kg 1 kg 1000 g

43. 3. Determine the number of molecules in 782g of N 2 O 3. 782 g N 2 O 3 1 mol N 2 O 3 76.011 g N 2 O 3 6.02 x 10 23 mc N 2 O 3 1 mol N 2 O 3 = 6.19 x 10 24 molecules N 2 O 3 1 G: W: R: 782 g N 2 O 3 Molecules of N 2 O 3 1 mol N 2 O 3 = 76.011 g N 2 O 3 1 mol N 2 O 3 = 6.02 x 10 23 mc N 2 O 3

44. III. Percent Composition

45. How would we calculate the percent of females in this classroom? % females = females All students X 100

46. III. Percent Composition The relative amounts of each element in a compound are expressed as percentages. The percent by mass of an element in a compound is the total number of grams of the element divided by the molar mass of the compound multiplied by 100%.

47. III. Percent Composition Identify all of the atoms in the compound and solve using the equation below to find the abundance of each element. % composition = Mass element x atoms Molar Mass X 100 (Mass of Compound)

48. Ex: Calculate the % composition of propane (C 3 H 8 ). (In other words, what percent of propane is carbon and what percent is hydrogen?) Carbon 3 x = 36.033g/mol Hydrogen 8 x = 8.064g/mol Molar mass = 44.097 g/mol

49. Ex: Calculate the % composition of propane (C 3 H 8 ) % composition C = 36.033g/mol _C 3 __ 44.097 g/mol C 3 H 8 % composition H = 8.064g/mol __H 8 __ 44.097 g/mol C 3 H 8 X 100 81.713% C 18.29% H X 100

50. Ex: Calculate the % composition of propane (C 3 H 8 ) These calculations tell us that 81.7% of propane is composed of carbon and 18.3% is made up of hydrogen Note: Percentage is part over whole. Mass of one type of element in a compound / mass of the whole compound

51. Ex: Determine the percent nitrogen in zinc nitrate. Zn 2+ NO 3 - Zn(NO 3 ) 2 %N = 28.014 g/mol 189.39 g/mol To Calculate Molar Mass: Zn = 1 x 65.38g/mol = 65.38 g/mo N = 2 x 14.007 g/mol = 28.014 g/mol O = 6 X 15.999g/mol = 95.994 g/mol Molar mass: 189.39 g/mol 14.792% of N in zinc nitrate = X 100 2 x N Zn(NO 3 ) 2

52. You might need to set up a ratio for problems where the mass of an element is being compared the mass of a compound. Element Mass Mass of Compound = Element Mass Mass of Compound

53. EX: Determine the mass of sodium in 450. g of sodium chloride. Na = 22.990 g/mol Cl = 35.453 g/mol 58.443 g/mol 22.990 g/mol Na 58.443 g/mol NaCl Part of a Whole X = 177g Na in 450.g NaCl X g Na = 450. g NaCl

54. Another way to solve the same problem EX: Determine the mass of sodium in 450. g of sodium chloride. 450. g NaCl 1 mol NaCl 58.443 g NaCl 1 mol Na 1 mol NaCl = 177 g Na 1 G: W: R: 450. g NaCl g Na 1 mol NaCl = 58.443 g NaCl 1 mol Na = 22.990 g Na 1 mol NaCl = 1 mol Na 22.990 g Na 1 mol Na

55. Practice 1. Calculate the percent composition of a. calcium chloride b. potassium nitrate 2. Determine the percent oxygen in calcium carbonate 3. Calculate the mass of hydrogen in a. 350 g of C 2 H 6 b. 2.14 g NH 4 Cl 36.1 % Ca, 63.9 % Cl 38.7 % K, 13.9 % N, 47.5 %O 48.0 % O 70.4 g H 0.161 g H CaCl 2 KNO 3 CaCO 3

56. 3a. Calculate the mass of hydrogen in 350 g of C 2 H 6 using proportions X = (6 x 1.008) 30.07 g C 2 H 6 70 g H = 350 g

57. 3a. Calculate the mass of hydrogen in 350 g of C 2 H 6 using dimensional analysis 350 g C 2 H 6 1 mol C 2 H 6 30.07 g C 2 H 6 6 moles H 1 mol C 2 H 6 = 70 g H 1 G: W: R: 350 g C 2 H 6 g H 1 mol C 2 H 6 = 28.054 g C 2 H 6 1 mol H = 1.008 g H 6 moles H = 1 mol C 2 H 6 1.008 g H 1 mol H

58. 3b. Calculate the mass of hydrogen in 2.14 g NH 4 Cl using proportions X = (4 x 1.008) 53.492 g NH 4 Cl 0.161 g H = 2.14 g

59. 3b. Calculate the mass of hydrogen in 2.14 g NH 4 Cl 2.14 g NH 4 Cl 1 mol NH 4 Cl 53.492 g NH 4 Cl 4 moles H 1 mol NH 4 Cl = 0.161 g H 1 G: W: R: 2.14 g NH 4 Cl g H 1 mol NH 4 Cl= 53.492 g NH 4 Cl 1 mol H = 1.008 g H 4 moles H = 1 mol NH 4 Cl 1.008 g H 1 mol H

60. IV. Empirical Formulas

61. How are they related? Molecular Formulas C 2 H 4 = ethene C 3 H 6 = cyclopropane C 4 H 8 = butene Empirical Formula CH 2

62. Empirical Formula C 6 H 11 O 6 What is the empirical formula for C 12 H 22 O 12 All are divisible by 2 __ __ __ 2 2 2

63. Empirical Formula SO 3 What is the empirical formula for S 2 O 6 Both divisible by 2 __ 2 2

64. Empirical Formula C3H6O2C3H6O2 What is the empirical formula for C 9 H 18 O 6 All are divisible by 3 __ __ __ 3 3 3

65. Empirical Formulas The formula that represents the smallest whole-number ratio of the elements in a compound is called the empirical formula. The empirical formula may or may not be the same as the actual molecular formula

66. Empirical Formulas  To determine the empirical formula for an unknown compound, the compound is analyzed to find the amount of each element in the compound. The amount can be expressed in moles, mass, or percent.

67. To determine an empirical formula: 1)Change the % composition to grams. (Remember, % composition is based on 100 g of the compound.) 2)Determine the number of moles of each element (using molar mass). 3)Determine the lowest whole number ratio of moles by dividing the smallest number of moles into the other numbers. (Keep in mind: 1.04 ≈ 1; 1.9 ≈ 2) 4)Write subscripts using these ratios. 5)If the numbers aren’t whole numbers, multiply by a factor you know will make them whole. Ex: 0.25 x 4 = 1; 0.33 x 3 = 1; 0.50 x 2 = 1

68. Ex.An iron oxide is composed of 78% iron 78% Fe 100% - 78% Fe = 22% O =78g Fe =22g O How to translate a % into a mass

69. Ex: Determine the empirical formula for a sulfur oxide that is 60% oxygen? 40% S 60% O 40g S 60g O

70. Ex: Determine the empirical formula for a sulfur oxide that is 60% oxygen? 60% oxygen = 60g O 60g O x 1 mol O = 15.999g O 40% sulfur = 40g S 40g S x 1 mol S = 32.066g S 3.75 mol O 1.25 mol S (100% - 60%) S 1.25 O 3.75

71. Empirical Formula SO 3 S 1.25 O 3.75 3.75 mol O 1.25 mol S = 3 mol O = 1 mol S

72. In case you didn’t know The order elements will be in for Molecular compound formulas will depend on its position on the periodic table. –Left-most elements go first –If they are in the same group, the lower elements are listed first –Again, carbon likes to be the center of attention, so you will see it in the front of formulas very often. (Ex. C x H y O z )

73. Ex. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% carbon, 8.16% hydrogen, and 43.20% oxygen. 48.64% = 48.64g C 8.16% = 8.16g H 43.20% = 43.20g O 12.011g C = 4.05 mol C 1 mol C 1.008g H = 8.10 mol H 1 mol H 15.999g O = 2.70 mol O 1 mol O

74. Empirical Formula C 1.5 H 3 O 1 Determining the subscripts for the empirical formula 4.05 mol C 8.10 mol H 2.70 mol O 2.70 mol = 1.5 mol C = 3.00 mol H = 1 mol O X 2 = 3 = 6 = 2 C3C3 H6H6 O2O2 Empirical Formula C3H6O2C3H6O2

76. V. Molecular Formulas

77. Remember this? Molecular Formulas C 2 H 4 = ethene C 3 H 6 = cyclopropane C 4 H 8 = butene Empirical Formula CH 2

78. Molecular Formulas A molecular formula is the actual chemical formula for a compound A molecular formula can be determined after determining the empirical formula of a compound The molar mass of a substance must be given to determine the molecular formula The formula mass will refer to the mass of the empirical formula

79. To go from the molecular formula to an empirical formula, divide by the greatest common factor. Empirical Formula C 6 H 11 O 6 What is the empirical formula for C 12 H 22 O 12 All are divisible by 2 __ __ __ 2 2 2

80. Empirical Formula SO 3 What is the empirical formula for S 2 O 6 Both divisible by 2 __ 2 2

81. Empirical Formula is the same as the molecular formula C3H6O2C3H6O2 What is the empirical formula for C 3 H 6 O 2 GCF is 1

82. Determining Molecular Formulas 1.Determine the empirical formula 2.Determine the formula mass of the empirical formula 3.Molar Mass = multiplying factor formula mass (empirical formula’s mass) 4.Multiply each of the subscripts by the multiplying factor to create the molecular formula

83. Ex. If the molar mass of succinic acid is 118.1 g and its empirical formula of C 2 H 3 O 2, what is its molecular formula? 59.044 g/mol C 2 H 3 O 2 = ( 2 x 12.011)+(3 x 1.008)+(2 x 15.999) Formula mass C 2 H 3 O 2 = 59.044 g/mol 118.1 g/mol = 2 = multiplying factor (Mass of empirical formula) (molecular mass)

84. Using the multiplying factor C2H3O2C2H3O2 2 X = C 4 H 6 O 4

85. Ex. A compound composed of hydrogen and oxygen is analyzed and a sample of the compound yields 0.590 g of hydrogen and 9.40 g of oxygen. The molecular mass of this compound is 34.0 g/mol. Determine the empirical formula and the molecular formula for this compound H?O?H?O? H + O 0.59g9.40g

86. “A compound composed of hydrogen and oxygen “ x 1 mol H = 1.008 g H x 1 mol O = 15.999g O 0.585 mol H 0.588 mol O Empirical Formula HO 0.590g H 9.40g O 0.585 0.585 = 1 = 1

87. Determining the molecular formula “The molecular mass of this compound is 34.0 g/mol” Formula = 17.007g/mol Mass for HO 17.007 g/mol = 1.999 ~ 2 = multiplying factor HO 34.0 g/mol (Mass of empirical formula) (molecular mass)

88. Using the multiplying factor H1O1H1O1 2 X = H 2 O 2 Hydrogen peroxide

89. Determining the multiplying factor Formula mass (empirical formula) = multiplying factor Molar mass

90. Practice: 1. Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH 4 N. 30.05 g/mol 60.0 g/mol = 1.997 ~ 2 = multiplying factor CH 4 N has a formula mass of 30.05. CH 4 N 2 X = C 2 H 8 N 2

91. Practice: 2. A colorless liquid composed of 46.68% nitrogen and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula? 46.68% = 46.68 g N 53.32% = 53.32 g O 14.007 g N = 3.333 mol N 1 mol N 15.999 g O = 3.333 mol O 1 mol O

92. Determining the subscripts for the empirical formula 3.333 mol N 3.333 mol O 3.333 mol N 3.333 mol O = 1 mol N = 1 mol O Empirical Formula NO NO has a formula mass of 30.006 “ has a molar mass of 60.01 g/mol. What is the molecular formula?” 60.01 g/mol 30.006g/mol = 1.999 ~ 2 NO= 2 X molecular formula N 2 O 2 (formula mass)