1. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole The SI unit for amount is called the mole (mol). A mole is the number of atoms in exactly 12 grams of carbon-12. Scientists use the mole to make counting large numbers of particles easier. The number of particles in a mole is called Avogadro’s Number. Avogadro’s number is 6.02214199  10 23 units/mole. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

2. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole, continued The Mole Is a Counting Unit Section 1 Avogadro’s Number and Molar Conversions The mole is used to count out a given number of particles, whether they are atoms, molecules, formula units, ions, or electrons. The mole is just one kind of counting unit: 1 dozen = 12 objects 1 hour = 3600 seconds 1 mole = 6.022  10 23 particles Chapter 7

3. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts The Mole Chapter 7 Click below to watch the Visual Concept. Visual Concept

4. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole, continued Amount in Moles Can Be Converted to Number of Particles Counting units are used to make conversion factors. The definition of one mole is 6.022  10 23 particles = 1 mol The conversion factor is Section 1 Avogadro’s Number and Molar Conversions Chapter 7

5. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole, continued Choose the Conversion Factor That Cancels the Given Units All conversion factors are equal to 1, so you can use them to convert among different units. You can tell which conversion factor to use, because the needed conversion factor should cancel the units of the given quantity to give you the units of the answer or the unknown quantity. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

6. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Between Amount in Moles and Number of Particles Section 1 Avogadro’s Number and Molar Conversions Chapter 7

7. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Amount in Moles to Number of Particles Sample Problem A Find the number of molecules in 2.5 mol of sulfur dioxide. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

8. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Amount in Moles to Number of Particles Sample Problem A Solution 2.5 mol SO 2  ? = ? molecules SO 2 You are converting from the unit mol to the unit molecules. The conversion factor must have the units of molecules/mol. You use 6.022  10 23 molecules/1 mol. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

9. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Amount in Moles to Number of Particles Sample Problem A Solution, continued 2.5 mol SO 2  ? = ? molecules SO 2 Section 1 Avogadro’s Number and Molar Conversions 2.5 mol SO 2 = 1.5  10 24 molecules SO 2 Chapter 7

10. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Avogadro’s Number and the Mole, continued Number of Particles Can Be Converted to Amount in Moles The reverse calculation is similar to that in Sample Problem A but the conversion factor is inverted to get the correct units in the answer. example: How many moles are 2.54  10 22 iron(III) ions? Section 1 Avogadro’s Number and Molar Conversions Chapter 7

11. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Number of Particles to Amount in Moles Sample Problem B A sample contains 3.01  10 23 molecules of sulfur dioxide, SO 2. Determine the amount in moles. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

12. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem B Solution 3.01  10 23 molecules SO 2  ? = ? mol SO 2 You are converting from the unit molecules to the unit mol. The conversion factor must have the units of mol/molecules. You use 1 mol/6.022  10 23 molecules. Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Amount in Moles

13. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem B Solution, continued 3.01  10 23 molecules SO 2  ? = ? mol SO 2 Section 1 Avogadro’s Number and Molar Conversions 3.01  10 23 molecules SO 2 = 0.500 mol SO 2 Chapter 7 Converting Number of Particles to Amount in Moles

14. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Mass Relates Moles to Grams Amount in Moles Can Be Converted to Mass The molar mass is the mass in grams of one mole of an element or compound. Molar mass is numerically equal to the atomic mass of monatomic elements and the formula mass of compounds and diatomic elements. The units for molar mass are g/mol. Molar mass can be used as a conversion factor in problems converting between mass and amount. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

15. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Chapter 7 Visual Concept Click below to watch the Visual Concept. Molar Mass

16. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Molar Mass as a Conversion Factor Chapter 7 Visual Concept Click below to watch the Visual Concept.

17. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Mass Relates Moles to Grams, continued The Mole Plays a Central Part in Chemical Conversions To convert from number of particles to mass, you must use a two-part process: First, convert number of particles to amount in moles. Second, convert amount in moles to mass in grams. One step common to many problems in chemistry is converting to amount in moles. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

18. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Between Mass, Amount, and Number of Particles Section 1 Avogadro’s Number and Molar Conversions Chapter 7

19. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem C Find the mass in grams of 2.44  10 24 atoms of carbon, whose molar mass is 12.01 g/mol. Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass

20. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Number of Particles to Mass Sample Problem C Solution First part: 2.44  10 24 atoms  ? = ? mol Select the conversion factor that will take you from number of atoms to amount in moles. You use 1 mol/6.022  10 23 atoms. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

21. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem C Solution, continued Second part: ? mol  ? = ? g Select the conversion factor that will take you from amount in moles to mass in grams. You use the molar mass of carbon, 12.01 g C/1 mol. Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass

22. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Avogadro’s Number and Molar Conversions = 48.7 g C Chapter 7 Converting Number of Particles to Mass Sample Problem C Solution, continued

23. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Molar Mass Relates Moles to Grams, continued Mass Can Be Converted to Amount in Moles Converting from mass to number of particles is the reverse of the operation in the previous problem. To convert from mass to number of particles, you must use a two-part process: First, convert mass in grams to amount in moles. Second, convert amount in moles to number of particles. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

24. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Converting Mass to Number of Particles Sample Problem D Find the number of molecules present in 47.5 g of glycerol, C 3 H 8 O 3. The molar mass of glycerol is 92.11 g/mol. Section 1 Avogadro’s Number and Molar Conversions Chapter 7

25. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem D Solution First part: 47.5 g  ? = ? mol Select the conversion factor that will take you from mass in grams to amount in moles. You use the inverse of the molar mass of glycerol: Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles

26. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem D Solution, continued Second part: ? mol  ? = ? molecules Select the conversion factor that will take you from amount in moles to number of particles. You use. Section 1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles

27. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Avogadro’s Number and Molar Conversions = 3.11  10 23 molecules Chapter 7 Sample Problem D Solution, continued Converting Mass to Number of Particles

28. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Remember that an answer must never be given to more significant figures than is appropriate. Round molar masses from the periodic table to two significant figures to the right of the decimal point. Section 1 Avogadro’s Number and Molar Conversions Molar Mass Relates Moles to Grams, continued Remember to Round Consistently Chapter 7

29. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Average Atomic Mass and the Periodic Table Most Elements are a Mixture of Isotopes Isotopes are atoms that have different numbers of neutrons than other atoms of the same element do. Average atomic mass is a weighted average of the atomic mass of an element’s isotopes. If you know the abundance of each isotope, you can calculate the average atomic mass of an element. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

30. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Chapter 7 Visual Concept Average Atomic Mass Click below to watch the Visual Concept.

31. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Average Atomic Mass Sample Problem E The mass of a Cu-63 atom is 62.94 amu, and that of a Cu-65 atom is 64.93 amu. Using the data below, find the average atomic mass of copper. abundance of Cu-63 = 69.17% abundance of Cu-65 = 30.83% Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

32. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem E Solution The contribution of each isotope is equal to its atomic mass multiplied by the fraction of that isotope. contribution of Cu-63:62.94 amu  0.6917 contribution of Cu-65:64.93 amu  0.3083 Average atomic mass is the sum of the individual contributions: (62.94 amu  0.6917) + (64.93 amu  0.3083) = 63.55 amu Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Average Atomic Mass

33. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Formulas and Moles Formulas Express Composition A compound’s chemical formula tells you which elements, as well as how much of each, are present in a compound. Formulas for covalent compounds show the elements and the number of atoms of each element in a molecule. Formulas for ionic compounds show the simplest ratio of cations and anions in any pure sample. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

34. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Any sample of compound has many atoms and ions, and the formula gives a ratio of those atoms or ions. Section 2 Relative Atomic Mass and Chemical Formulas Chemical Formulas and Moles, continued Formulas Express Composition, continued Chapter 7

35. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Formulas and Moles, continued Formulas Give Ratios of Polyatomic Ions Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Formulas for polyatomic ions show the simplest ratio of cations and anions. They also show the elements and the number of atoms of each element in each ion. For example, the formula KNO 3 indicates a ratio of one K + cation to one anion.

36. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Understanding Formulas for Polyatomic Ionic Compounds Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

37. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Formulas and Moles, continued Formulas Are Used to Calculate Molar Masses The molar mass of a molecular compound is the sum of the masses of all the atoms in the formula expressed in g/mol. The molar mass of an ionic compound is the sum of the masses of all the atoms in the formula expressed in g/mol. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

38. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Formula Mass Chapter 7 Visual Concept Click below to watch the Visual Concept.

39. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Mass for Ionic Compounds Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

40. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Molar Mass of Compounds Sample Problem F Find the molar mass of barium nitrate, Ba(NO 3 ) 2. Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7

41. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem F Solution Find the number of moles of each element in 1 mol Ba(NO 3 ) 2 : Each mole has 1 mol Ba, 2 mol N, and 6 mol O. Use the periodic table to find the molar mass of each element in the formula: molar mass of Ba: 137.33 g/mol molar mass of N: 14.01 g/mol molar mass of O: 16.00 g/mol Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds

42. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem F Solution, continued Multiply the molar mass of each element by the number of moles of each element. Add these masses to get the total molar mass of Ba(NO 3 ) 2. mass of 1 mol Ba = 1  137.33 g/mol = 137.33 g/mol mass of 2 mol N = 2  14.01 g/mol = 28.02 g/mol + mass of 6 mol O = 6  16.00 g/mol = 96.00 g/mol Section 2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds molar mass of Ba(NO 3 ) 2 = 261.35 g/mol