1. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-1 Chapter 4: Chemical Composition

2. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-2 Questions for Consideration 1. How can we describe the mass composition of elements in a compound? 2. How can we determine the number of atoms in a given mass of a material? The number of molecules? The number of formula units? 3. How can we use the masses of elements in a compound to determine its chemical formula? 4. How can we express the composition of a solution?

3. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-3 Chapter 4 Topics: 1. Percent Composition 2. Mole Quantities 3. Determining Empirical and Molecular Formulas 4. Chemical Composition of Solutions

4. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-4 4.1 Mole Quantities This math toolbox gives a concise overview of the mole-quantity calculations that are introduced in more detail throughout this chapter. This math toolbox gives a concise overview of the mole-quantity calculations that are introduced in more detail throughout this chapter. Chapter 4 Math Toolbox:

5. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-5 4.1 Percent Composition We have used chemical formulas to express the composition of compounds, but where do these formulas come from? We have used chemical formulas to express the composition of compounds, but where do these formulas come from? We cannot directly determine a chemical formula. However, we can measure the mass of each element in a sample of a compound by using appropriate analytical techniques. We cannot directly determine a chemical formula. However, we can measure the mass of each element in a sample of a compound by using appropriate analytical techniques. Because the mass of each element in a compound will vary from one sample of the compound to another, we must have some expression of composition that is the same for all samples. Because the mass of each element in a compound will vary from one sample of the compound to another, we must have some expression of composition that is the same for all samples.

6. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-6 Percent Composition A convenient method for expressing composition is percent composition by mass. A convenient method for expressing composition is percent composition by mass. For any element, E, in a compound, the percent composition by mass is given by the following equation: For any element, E, in a compound, the percent composition by mass is given by the following equation:

7. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-7 Moles, Masses, and Particles How can we describe the composition of a compound if we know the mass of the elements in the compound? How can we describe the composition of a compound if we know the mass of the elements in the compound? A 3.67-g sample of the mineral chalcopyrite was determined to contain 1.27 g Cu, 1.12 g Fe, and 1.28 g S. A 3.67-g sample of the mineral chalcopyrite was determined to contain 1.27 g Cu, 1.12 g Fe, and 1.28 g S. What is the mass percent of each element in this compound? What is the mass percent of each element in this compound?

8. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-8 Percent Composition The percent composition for all the elements present in a compound must add up to 100%. The percent composition for all the elements present in a compound must add up to 100%. Figure 4.5

9. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-9 Composition of Chalcopyrite Would this mass % differ for a different sample of this mineral? Would this mass % differ for a different sample of this mineral? If you had a 100-gram sample, what mass of copper would it contain? If you had a 100-gram sample, what mass of copper would it contain? Figure 4.5

10. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-10 Activity: Percent Composition A sample of a copper compound weighs 1.63 g and contains 1.30 g of Cu. The other element in the compound is oxygen. What is the percent composition of this compound? A sample of a copper compound weighs 1.63 g and contains 1.30 g of Cu. The other element in the compound is oxygen. What is the percent composition of this compound? By difference, the sample contains 0.33 g O. Applying the formula for percent composition gives: By difference, the sample contains 0.33 g O. Applying the formula for percent composition gives:

11. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-11 Activity: Percent Composition What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S? What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S?

12. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-12 Activity Solution: Percent Composition 1. What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S?

13. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-13 Activity: Percent Composition A 4.55-g sample of limestone (CaCO 3 ) contains 1.82 g of calcium. What is the percent Ca in limestone? A 4.55-g sample of limestone (CaCO 3 ) contains 1.82 g of calcium. What is the percent Ca in limestone?

14. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-14 Activity Solutions: Percent Composition A 4.55-g sample of limestone (CaCO 3 ) contains 1.82 g of calcium. What is the percent Ca in limestone? A 4.55-g sample of limestone (CaCO 3 ) contains 1.82 g of calcium. What is the percent Ca in limestone?

15. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-15 4.2 Mole Quantities When working with amounts of a substance on a macroscopic scale, we cannot simply count atoms or molecules. There are too many. Instead, we use the moles of atoms, which are related by Avogadro’s number: When working with amounts of a substance on a macroscopic scale, we cannot simply count atoms or molecules. There are too many. Instead, we use the moles of atoms, which are related by Avogadro’s number: 1 mole = 6.022 × 10 23 particles 1 mole C = 6.022 × 10 23 carbon atoms 1 mole C = 6.022 × 10 23 carbon atoms 1 mole H 2 S = 6.022 × 10 23 H 2 S molecules 1 mole H 2 S = 6.022 × 10 23 H 2 S molecules 1 mol Cu 2 O = 6.022 × 10 23 Cu 2 O formula units 1 mol Cu 2 O = 6.022 × 10 23 Cu 2 O formula units

16. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-16 The Mole The mole unit acts as a bridge between the microscopic world and the macroscopic world. The mole unit acts as a bridge between the microscopic world and the macroscopic world. One mole of substance contains as many basic particles (atoms, molecules, or formula units) as there are atoms in exactly 12 g of carbon-12. One mole of substance contains as many basic particles (atoms, molecules, or formula units) as there are atoms in exactly 12 g of carbon-12. One mole of a substance contains 6.022 × 10 23 particles (molecules, atoms, ions, formula units, etc.) One mole of a substance contains 6.022 × 10 23 particles (molecules, atoms, ions, formula units, etc.) This number is called Avogadro’s number. This number is called Avogadro’s number.

17. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-17 Moles of Various Elements and Compounds Figure 4.8

18. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-18 Activity: Atoms in H 2 S How many sulfur atoms are in 1 mol of H 2 S? How many sulfur atoms are in 1 mol of H 2 S? How many hydrogen atoms are in 1 mol of H 2 S? How many hydrogen atoms are in 1 mol of H 2 S? Figure 4.6

19. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-19 Molar Mass The Mass of 1 Mole Avogadro’s number has been defined so that the mass of 1 mol of C-12 has a mass of exactly 12 grams. Avogadro’s number has been defined so that the mass of 1 mol of C-12 has a mass of exactly 12 grams. This means that the average mass of an atom of any substance in amus is the same numerical value as the mass of 1 mole of that substance in grams (molar mass). This means that the average mass of an atom of any substance in amus is the same numerical value as the mass of 1 mole of that substance in grams (molar mass). The molar mass of carbon is 12.01 g/mol. The molar mass of carbon is 12.01 g/mol. The molar mass of oxygen is 16.00 g/mol The molar mass of oxygen is 16.00 g/mol The molar mass of CO 2 is: The molar mass of CO 2 is: 12.01 + 2(16.00) = 44.01 g/mol We’ll use 4 sig. figs

20. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-20 Activity: Molar Mass Complete the table. Complete the table. Element or Compound Atomic Mass Molar Mass H O Na Ca(ClO 3 ) 2 H2OH2OH2OH2O NaCl

21. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-21 Activity Solutions: Molar Mass Element or Compound Atomic Mass Molar Mass H 1.008 amu 1.008 g/mol O 16.00 amu 16.00 g/mol Na 22.99 amu 22.99 g/mol Ca(ClO 3 ) 2 206.98 amu 206.98 g/mol H2OH2OH2OH2O 18.02 amu 18.02 g/mol NaCl 58.44 amu 58.44 g/mol

22. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-22 Molar Mass Which contains the greatest number of atoms? Which contains the greatest number of atoms? 1 mole of copper or 1 mole of gold? 1 mole of copper or 1 mole of gold? 1 gram of copper or 1 gram of gold? 1 gram of copper or 1 gram of gold? 6.022 × 10 23 Answer: Because 1 mole is a fixed quantity of particles (6.022 × 10 23 ), 1 mole of copper contains the same number of atoms as 1 mole of gold. Answer: Because each copper atom has less mass than each gold atom, 1 gram of copper contains more atoms than 1 gram of gold.

23. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-23 Converting Between Grams and Moles 1. Convert 10.0 grams of CO 2 to moles. 2. Convert 0.50 mol CO 2 to grams. Figure from p. 137

24. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-24 Activity: Converting Between Grams and Moles Convert 10.0 g O 2 to moles. Convert 10.0 g O 2 to moles. Figure from p. 137

25. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-25 Grams  Moles  Particles Once we know the number of moles of a substance, we can use Avogadro’s number (6.022×10 23 ) to determine the number of particles in that sample of the substance. Once we know the number of moles of a substance, we can use Avogadro’s number (6.022×10 23 ) to determine the number of particles in that sample of the substance. 1 mole = 6.022 × 10 23 particles Figure from p. 139

26. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-26 Activity: Determining Number of Particles How many CO 2 molecules are in a 100-g sample of CO 2 ? How many CO 2 molecules are in a 100-g sample of CO 2 ? 1.37 × 10 24 CO 2 molecules 1.37 × 10 24 CO 2 molecules How many carbon atoms are in a 100-g sample of CO 2 ? How many carbon atoms are in a 100-g sample of CO 2 ? 1.37 × 10 24 C atoms 1.37 × 10 24 C atoms How many oxygen atoms are in a 100-g sample of CO 2 ? How many oxygen atoms are in a 100-g sample of CO 2 ? 2.74 × 10 24 O atoms 2.74 × 10 24 O atoms

27. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-27 Activity: Number of Particles What mass of MgCl 2 contains 6.022 × 10 23 Cl  ions? What mass of MgCl 2 contains 6.022 × 10 23 Cl  ions? The molar mass of MgCl 2 is 95.21 g/mol. The molar mass of MgCl 2 is 95.21 g/mol. Figure from p. 139

28. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-28 Extra Activity: Conversions with Molar Mass How many moles of aspartame (C 14 H 18 N 2 O 5 ) are found in 40.0 mg of aspartame? How many molecules of aspartame are found in this mass? How many moles of aspartame (C 14 H 18 N 2 O 5 ) are found in 40.0 mg of aspartame? How many molecules of aspartame are found in this mass?

29. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-29 Extra Activity Solution: Conversions with Molar Mass How many moles of aspartame (C 14 H 18 N 2 O 5 ) are found in 40.0 mg of aspartame? How many molecules of aspartame are found in this mass? How many moles of aspartame (C 14 H 18 N 2 O 5 ) are found in 40.0 mg of aspartame? How many molecules of aspartame are found in this mass? We first need to convert from mg to g: Next, we need to find the molar mass of C 14 H 18 N 2 O 5 :

30. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-30 Extra Activity Solution: Conversions with Molar Mass How many moles of aspartame (C 14 H 18 N 2 O 5 ) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass? How many moles of aspartame (C 14 H 18 N 2 O 5 ) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass? Next, convert from grams to moles: Finally, we convert from moles to molecules:

31. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-31 Extra Activity: Conversions with Molar Mass If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C 9 H 8 O 4 ), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets? If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C 9 H 8 O 4 ), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets?

32. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-32 Extra Activity Solution: Conversions with Molar Mass If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C 9 H 8 O 4 ), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets? If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C 9 H 8 O 4 ), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets?

33. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-33 4.3 Determining Empirical and Molecular Formulas The formula for a substance also tells us about the composition of a compound: The formula for a substance also tells us about the composition of a compound: A formula unit for an ionic compound tells us the ratio of ions of different elements in the compound. (MgCl 2 has a 1:2 ratio of Mg 2+ to Cl .) A formula unit for an ionic compound tells us the ratio of ions of different elements in the compound. (MgCl 2 has a 1:2 ratio of Mg 2+ to Cl .) A molecular formula tells the number of atoms of each element in a molecule and the atom ratio. (CO 2 has a 1:2 ratio of C to O atoms.) A molecular formula tells the number of atoms of each element in a molecule and the atom ratio. (CO 2 has a 1:2 ratio of C to O atoms.)

34. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-34 Empirical and Molecular Formulas Empirical formula Empirical formula Expresses the simplest ratios of atoms in a compound Expresses the simplest ratios of atoms in a compound Written with the smallest whole-number subscripts Written with the smallest whole-number subscripts Molecular formula Molecular formula Expresses the actual number of atoms in a compound Expresses the actual number of atoms in a compound Can have the same subscripts as the empirical formula or some multiple of them Can have the same subscripts as the empirical formula or some multiple of them

35. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-35 Empirical Formulas What is the same about these two compounds? What is the same about these two compounds? Figure 4.13

36. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-36 Empirical Formulas The empirical formula of a substance is the ratio of atoms of different elements, in terms of the smallest whole numbers. The empirical formula of a substance is the ratio of atoms of different elements, in terms of the smallest whole numbers. What are the empirical formulas of these two compounds? What are the empirical formulas of these two compounds? What are the empirical formulas of H 2 O 2 and H 2 O? Are they different compounds? What are the empirical formulas of H 2 O 2 and H 2 O? Are they different compounds? Figure 4.13

37. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-37 What is the empirical formula for copper(II) oxide? Figure 4.12

38. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-38 Empirical or Molecular Formulas

39. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-39 Activity: Empirical Formula Figure 4.14 Determine the empirical formulas of these substances. For which substances is the empirical formula the same as the molecular formula?

40. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-40 Activity Solutions: Empirical Formula Figure 4.14 Determine the empirical formulas of these substances. For which substances is the empirical formula the same as the molecular formula?

41. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-41 Determining Empirical Formulas from % Composition If we know the masses of the elements in a compound, or its percent composition, we can determine its mole ratio, and therefore the compound’s empirical formula. If we know the masses of the elements in a compound, or its percent composition, we can determine its mole ratio, and therefore the compound’s empirical formula. Consider the compound commonly called chalcopyrite. Any sample will have the following % composition: Consider the compound commonly called chalcopyrite. Any sample will have the following % composition: Figure 4.5

42. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-42 Determining Empirical Formulas 1. Since the percent composition does not change from sample to sample, assume any size sample. The most convenient is 100 grams so % value = mass value. 2. Convert grams to moles for each element using its molar mass as a conversion factor. 3. Without changing the relative amounts, change moles to whole numbers. Do this by dividing all by the same smallest value. If all do not convert to whole numbers, multiply to get whole numbers.

43. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-43 Determining Empirical Formulas 1. 100 grams chalcopyrite contains: 30.5 g Fe 30.5 g Fe 34.6 g Cu 34.6 g Cu 34.9 g S 34.9 g S 2. mol Fe = (30.5g Fe)(1mol/55.85 g) = 0.5461 mol Fe mol Cu = (34.6g Cu)(1mol/63.55 g) = 0.5444 mol Cu mol S = (34.9g S)(1mol/32.07 g) = 1.088 mol S 3.(0.5461 mol Fe)/(0. 5444) = 1.003 mol Fe (0.5444 mol Cu)/(0. 5444) = 1.000 mol Cu (1.088 mol S)/(0. 5444) = 1.999 mol S FeCuS 2 Figure 4.5

44. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-44 Steps for Determining Empirical Formula Figure from p. 143

45. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-45 Activity: Empirical Formula from Percent Composition A compound was determined to have the following percent composition: A compound was determined to have the following percent composition: 50.0% sulfur 50.0% sulfur 50.0% oxygen 50.0% oxygen What is the empirical formula for the compound? What is the empirical formula for the compound? SO 2

46. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-46 Activity: Determining Empirical Formulas Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S. Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S.

47. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-47 Activity Solution: Determining Empirical Formulas Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S. Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S. First, reassign the percentages to units of grams: 66.5 g Cu and 33.5 g S. Then, convert to moles and divide both numbers by the lowest number. The whole numbers then become our subscripts. The empirical formula is therefore: CuS

48. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-48 Activity: Determining Empirical Formulas Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite. Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite.

49. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-49 Activity Solutions: Determining Empirical Formulas Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite. Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite. Therefore, the empirical formula is: Cu 5 Si 4 O 14 H 2

50. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-50 Molecular Formulas from Empirical Formulas Benzene and acetylene have the same empirical formulas but different molecular formulas. Benzene and acetylene have the same empirical formulas but different molecular formulas. How much greater in mass is benzene than acetylene? How much greater in mass is benzene than acetylene? How much greater in mass is each of these than the empirical formula? How much greater in mass is each of these than the empirical formula? Figure 4.13

51. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-51 Empirical and Molecular Formulas What are the ratios of masses for the molecular and empirical formulas? What are the ratios of masses for the molecular and empirical formulas?

52. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-52 Activity: Molecular Formulas from Empirical Formulas A compound was determined to have an empirical formula of CH 2. Its molar mass was determined to be 42.12 g/mol. What is the molecular formula for this compound? A compound was determined to have an empirical formula of CH 2. Its molar mass was determined to be 42.12 g/mol. What is the molecular formula for this compound? The mass of the empirical formula is 14.03 g/mol, so the ratio of masses is 3.00. The ratio of the formulas therefore is 3, giving a molecular formula of (CH 2 ) 3 or C 3 H 6.

53. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-53 Molecular Formulas To determine a molecular formula, the problem must give a piece of experimental data, such as a molar mass, MM. To determine a molecular formula, the problem must give a piece of experimental data, such as a molar mass, MM. To find the molecular formula: To find the molecular formula: 1. Find the empirical formula first. 2. Divide the empirical formula’s molar mass into the experimental molar mass (which is given). Figure 4.15

54. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-54 Activity: Molecular Formulas Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate? Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate?

55. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-55 Activity Solution: Molecular Formulas Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate? Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate? First, find the empirical formula: Thus, the empirical formula is KSO 4.

56. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-56 Activity Solution: Molecular Formulas Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate? Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate? To find the molecular formula: Calculate the MM of KSO 4 = (1 mol K × 39.10 g/mol K) + (1 mol S × 32.07 g/mol S) + (4 mol O × 16.00 g/mol O) = 135.17 g/mol KSO 4 Thus, the molecular formula is K 2 S 2 O 8.

57. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-57 Determining Percent Composition from Empirical or Molecular Formula If you know the formula for a compound, you can determine the mass percent composition. If you know the formula for a compound, you can determine the mass percent composition. Assume you have 1 mole of compound, and convert moles of the element and compound to grams. Assume you have 1 mole of compound, and convert moles of the element and compound to grams. Mass Percent Composition Mass Ratio Empirical or Molecular Formula Atom or Mole Ratio

58. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-58 Determining Percent Composition Assume 1 mole of compound, and remember the mass of 1 mole is the molar mass. Assume 1 mole of compound, and remember the mass of 1 mole is the molar mass. Figure from p. 149

59. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-59 Activity: Determining Percent Composition What is the percent sodium in Na 2 O? What is the percent sodium in Na 2 O?

60. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-60 Activity: Percent Composition Which of these compounds has the greater percentage of Cu? Which of these compounds has the greater percentage of Cu? Cuprite, Cu 2 O Chalcocite, Cu 2 S Figure 4.16 Answer: cuprite

61. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-61 4.4 Chemical Composition of Solutions A solution is a homogeneous mixture. A solution is a homogeneous mixture. This is a solution being prepared by adding the CuSO 4 (solute) to water (solvent). This is a solution being prepared by adding the CuSO 4 (solute) to water (solvent). The composition of the solution depends on the relative amounts of the solute and solvent. The composition of the solution depends on the relative amounts of the solute and solvent. Figure 4.17

62. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-62Concentration Which aqueous CuSO 4 solution has the greatest concentration (most concentrated)? Which is most dilute? Which aqueous CuSO 4 solution has the greatest concentration (most concentrated)? Which is most dilute? Figure 4.18

63. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-63 Percent by Mass of Solute Solution concentration is often expressed as the mass percent of solute: Solution concentration is often expressed as the mass percent of solute:

64. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-64 Activity: Percent by Mass of Solute What is the mass percent of NaCl in a solution that is prepared by adding 10.0 g NaCl to 50.0 g water? What is the mass percent of NaCl in a solution that is prepared by adding 10.0 g NaCl to 50.0 g water?

65. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-65 Molarity (M) Another common way to express the concentration of a solution is in molarity units: Another common way to express the concentration of a solution is in molarity units:

66. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-66 Activity: Preparing a CuSO 4 Solution 6.25 grams (0.0250 mol) of CuSO 4  5H 2 O is added to a 250-mL volumetric flask. 6.25 grams (0.0250 mol) of CuSO 4  5H 2 O is added to a 250-mL volumetric flask. Water is added to the mark so that the total volume is 250.0 mL. Water is added to the mark so that the total volume is 250.0 mL. What is the molarity of this solution? What is the molarity of this solution? Figure 4.19

67. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-67 Activity: Molarity How many moles of NaCl are in 1.85 L of a 0.25 M NaCl solution? How many moles of NaCl are in 1.85 L of a 0.25 M NaCl solution? moles NaCl = 1.85 L × 0.25 mol/L = 0.46 mol

68. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-68 Activity: Solution Concentration Bluestone is copper(II) sulfate pentahydrate, CuSO 4 5H 2 O, with a molar mass of 249.7 g/mol. A sample of pond water was found to have a concentration of 6.2  10 5 M copper(II) sulfate. If the pond has a volume of 1.8  10 7 L, then what mass of bluestone did the farmer add to the pond as an algicide? Bluestone is copper(II) sulfate pentahydrate, CuSO 4 5H 2 O, with a molar mass of 249.7 g/mol. A sample of pond water was found to have a concentration of 6.2  10 5 M copper(II) sulfate. If the pond has a volume of 1.8  10 7 L, then what mass of bluestone did the farmer add to the pond as an algicide?

69. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-69 Activity Solution: Solution Concentration Bluestone is copper(II) sulfate pentahydrate, CuSO 4 5H 2 O, with a molar mass of 249.7 g/mol. A sample of pond water was found to have a concentration of 6.2  10 5 M copper(II) sulfate. If the pond has a volume of 1.8  x 10 7 L, then what mass of bluestone did the farmer add to the pond as an algicide? Bluestone is copper(II) sulfate pentahydrate, CuSO 4 5H 2 O, with a molar mass of 249.7 g/mol. A sample of pond water was found to have a concentration of 6.2  10 5 M copper(II) sulfate. If the pond has a volume of 1.8  x 10 7 L, then what mass of bluestone did the farmer add to the pond as an algicide?

70. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-70Dilution Suppose you want to dilute a 0.25 M solution to a concentration of 0.025 M. What are some ways to do this? Suppose you want to dilute a 0.25 M solution to a concentration of 0.025 M. What are some ways to do this?

71. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-71Dilution Figure 4.21

72. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-72 Describe this process Figure from p. 168

73. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-73Dilution Moles initial = Moles final Moles initial = Moles final Moles = Molarity × Volume Moles = Molarity × Volume Moles = mol/L × L Moles = mol/L × L M initial V initial = M final V final M initial V initial = M final V final

74. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-74 Activity: Dilution M initial V initial = M final V final M initial V initial = M final V final What is the concentration of a solution prepared by adding water to 25.0 mL of 6.00 M NaOH to a total volume of 500.0 mL? What is the concentration of a solution prepared by adding water to 25.0 mL of 6.00 M NaOH to a total volume of 500.0 mL?

75. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-75 Activity: Dilution If 42.8 mL of 3.02 M H 2 SO 4 solution is diluted to a final volume 500.0 mL, what is the molarity of the diluted solution of H 2 SO 4 ? If 42.8 mL of 3.02 M H 2 SO 4 solution is diluted to a final volume 500.0 mL, what is the molarity of the diluted solution of H 2 SO 4 ?

76. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-76 Activity Solution: Dilution If 42.8 mL of 3.02 M H 2 SO 4 solution is diluted to a final volume 500.0 mL, what is the molarity of the diluted solution of H 2 SO 4 ? If 42.8 mL of 3.02 M H 2 SO 4 solution is diluted to a final volume 500.0 mL, what is the molarity of the diluted solution of H 2 SO 4 ? M con V con = M dil V dil 3.02 M × 42.8 mL = M dil × 500.0 mL

77. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-77 Activity: Dilution What is the concentration of a solution prepared by diluting 35.0 mL of 0.150 M KBr to 250.0 mL? What is the concentration of a solution prepared by diluting 35.0 mL of 0.150 M KBr to 250.0 mL?

78. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 4-78 Activity Solution: Dilution What is the concentration of a solution prepared by diluting 35.0 mL of 0.150 M KBr to 250.0 mL? What is the concentration of a solution prepared by diluting 35.0 mL of 0.150 M KBr to 250.0 mL? M con V con = M dil V dil 0.150 M × 35.0 mL = M dil × 250.0 mL