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Table of Contents

Learn about Avogadro's number, molar conversions, relative atomic mass, chemical formulas, and percentage composition in this comprehensive chapter.

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Table of Contents

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  1. The Mole and Chemical Composition Chapter 7 Table of Contents Section 1Avogadro’s Number and Molar Conversions Section 2Relative Atomic Mass and Chemical Formulas Section 3Formulas and Percentage Composition

  2. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Chapter 7 Bellringer • List as many common counting units as you can. • Determine how many groups of each unit in your list are present in each of the following amounts: • 500 goldfish • 150 unicycles • 50 jet planes

  3. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Objectives • Identify the mole as the unit used to count particles, whether atoms, ions, or molecules. • Use Avogadro’s number to convert between amount in moles and number of particles. • Solve problems converting between mass, amount in moles, and number of particles using Avogadro’s number and molar mass.

  4. Section1 Avogadro’s Number and Molar Conversions Chapter 7 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  1023 units/mole.

  5. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Avogadro’s Number and the Mole, continued The Mole Is a Counting Unit • 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  1023 particles (Scientific Notation)

  6. Section1 Avogadro’s Number and Molar Conversions Chapter 7 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  1023 particles = 1 mol • The conversion factor is

  7. Section1 Avogadro’s Number and Molar Conversions Chapter 7 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.

  8. Converting Between Amount in Moles and Number of Particles Section1 Avogadro’s Number and Molar Conversions Chapter 7

  9. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Amount in Moles to Number of Particles Sample Problem A Find the number of molecules in 2.5 mol of sulfur dioxide.

  10. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Amount in Moles to Number of Particles Sample Problem A Solution 2.5 mol SO2  ? = ? molecules SO2 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  1023 molecules/1 mol.

  11. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Amount in Moles to Number of Particles Sample Problem A Solution, continued 2.5 mol SO2  ? = ? molecules SO2 2.5 mol SO2 = 1.5  1024 molecules SO2

  12. Section1 Avogadro’s Number and Molar Conversions Chapter 7 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  1022 iron(III) ions?

  13. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Amount in Moles Sample Problem B A sample contains 3.01  1023 molecules of sulfur dioxide, SO2. Determine the amount in moles.

  14. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Amount in Moles Sample Problem B Solution 3.01 × 1023 molecules SO2  ? = ? mol SO2 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  1023 molecules.

  15. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Amount in Moles Sample Problem B Solution, continued 3.01  1023 molecules SO2  ? = ? mol SO2 3.01  1023 molecules SO2 = 0.500 mol SO2

  16. Section1 Avogadro’s Number and Molar Conversions Chapter 7 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.

  17. Converting Between Mass, Amount, and Number of Particles Section1 Avogadro’s Number and Molar Conversions Chapter 7

  18. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass Sample Problem C Find the mass in grams of 2.44  1024 atoms of carbon, whose molar mass is 12.01 g/mol.

  19. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass Sample Problem C Solution First part: 2.44  1024 atoms  ? = ? mol Select the conversion factor that will take you from number of atoms to amount in moles. You use 1 mol/6.022  1023 atoms.

  20. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass 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.

  21. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Number of Particles to Mass Sample Problem C Solution, continued = 48.7 g C

  22. Section1 Avogadro’s Number and Molar Conversions Chapter 7 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.

  23. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles Sample Problem D Find the number of molecules present in 47.5 g of glycerol, C3H8O3. The molar mass of glycerol is 92.11 g/mol.

  24. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles 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:

  25. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles 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 .

  26. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Converting Mass to Number of Particles Sample Problem D Solution, continued = 3.11  1023 molecules

  27. Section1 Avogadro’s Number and Molar Conversions Chapter 7 Molar Mass Relates Moles to Grams, continued Remember to Round Consistently • 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.

  28. 7.1 Section Review Questions #1-15. Page 233 Write questions and complete answers in your lab book.

  29. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 Chapter 7 Objectives • Use a periodic table or isotopic composition data to determine the average atomic masses of elements. • Infer information about a compound from its chemical formula. • Determine the molar mass of a compound from its formula.

  30. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 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.

  31. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 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%

  32. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Average Atomic Mass • 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

  33. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 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 compoundsshow the simplest ratio of cations and anions in any pure sample.

  34. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 Chemical Formulas and Moles, continued Formulas Express Composition, continued • Any sample of compound has many atoms and ions, and the formula gives a ratio of those atoms or ions.

  35. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 Chemical Formulas and Moles, continued Formulas Give Ratios of Polyatomic Ions • Formulas for polyatomic ionsshow 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 KNO3 indicates a ratio of one K+ cation to one anion.

  36. Understanding Formulas for Polyatomic Ionic Compounds Section2 Relative Atomic Mass and Chemical Formulas Chapter 7

  37. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 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.

  38. Calculating Molar Mass for Ionic Compounds Section2 Relative Atomic Mass and Chemical Formulas Chapter 7

  39. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds Sample Problem F Find the molar mass of barium nitrate, Ba(NO3)2.

  40. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds • Sample Problem F Solution • Find the number of moles of each element in 1 mol Ba(NO3)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

  41. Section2 Relative Atomic Mass and Chemical Formulas Chapter 7 Calculating Molar Mass of Compounds 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(NO3)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 molar mass of Ba(NO3)2 = 261.35 g/mol

  42. 7.2 Section Review Questions #1-16. Page 240 Write questions and complete answers in your lab book.

  43. Section3 Formulas and Percentage Composition Chapter 7 Objectives • Determine a compound’s empirical formula from its percentage composition. • Determine the molecular formula or formula unit of a compound from its empirical formula and its formula mass. • Calculate percentage composition of a compound from its molecular formula or formula unit.

  44. Section3 Formulas and Percentage Composition Chapter 7 Using Analytical Data • The percentage composition is the percentage by mass of each element in a compound. • Percentage composition helps verify a substance’s identity. • Percentage composition also can be used to compare the ratio of masses contributed by the elements in two different substances.

  45. Percentage Composition of Iron Oxides Chapter 7

  46. Section3 Formulas and Percentage Composition Chapter 7 Using Analytical Data, continued Determining Empirical Formulas • An empirical formula is a chemical formula that shows the simplest ratio for the relative numbers and kinds of atoms in a compound. • An actual formula shows the actual ratio of elements or ions in a single unit of a compound. • For example, the empirical formula for ammonium nitrate is NH2O, while the actual formula is NH4NO2.

  47. Empirical and Actual Formulas Chapter 7

  48. Section3 Formulas and Percentage Composition Chapter 7 Using Analytical Data, continued Determining Empirical Formulas, continued • You can use the percentage composition for a compound to determine its empirical formula. • Convert the percentage of each element to g. • Convert from g to mol using the molar mass of each element as a conversion factor. • Compare these amounts in mol to find the simplest whole-number ratio among the elements.

  49. Section3 Formulas and Percentage Composition Chapter 7 Determining an Empirical Formula from Percentage Composition Sample Problem G Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. Calculate the empirical formula of this substance.

  50. Section3 Formulas and Percentage Composition Chapter 7 Determining an Empirical Formula from Percentage Composition • Sample Problem G Solution • Assume that you have a 100.0 g sample, and convert the percentages to grams. • for C: 60.0%  100.0 g = 60.0 g C • for H: 13.4%  100.0 g = 13.4 g H • for O: 26.6%  100.0 g = 26.6 g O

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