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Introductory Chemistry , 2 nd Edition Nivaldo Tro

Introductory Chemistry , 2 nd Edition Nivaldo Tro. Chapter 6 Chemical Composition. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2006, Prentice Hall. Why is Knowledge of Composition Important?.

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Introductory Chemistry , 2 nd Edition Nivaldo Tro

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  1. Introductory Chemistry, 2nd EditionNivaldo Tro Chapter 6 Chemical Composition Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2006, Prentice Hall

  2. Why is Knowledge of Composition Important? • everything in nature is either chemically or physically combined with other substances • to know the amount of a material in a sample, you need to know what fraction of the sample it is • Some Applications: • the amount of sodium in sodium chloride for diet • the amount of iron in iron ore for steel production • the amount of hydrogen in water for hydrogen fuel • the amount of chlorine in freon to estimate ozone depletion Tro's Introductory Chemistry, Chapter 6

  3. Counting Nails by the Pound • I want to buy a certain number of nails for a project, but the hardware store sells nails by the pound! • How do I know how many nails I am buying when I buy a pound of nails? • Analogy • How many atoms in a given mass of an element? Tro's Introductory Chemistry, Chapter 6

  4. Counting Nails by the Pound A hardware store customer buys 2.60 pounds of nails. A dozen of the nails has a mass of 0.150 pounds. How many nails did the customer buy? Solution map: Tro's Introductory Chemistry, Chapter 6

  5. Counting Nails by the Pound • The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them! Tro's Introductory Chemistry, Chapter 6

  6. Counting Nails by the Pound • What if he bought a different size nail? • Would the mass of a dozen be 0.150 lbs? • Would there be 208 nails in 2.60 lbs? • How would this effect the conversion factors? Tro's Introductory Chemistry, Chapter 6

  7. Counting Atoms by Moles • If we can find the mass of a particular number of atoms, we can use this information to convert the mass of a element sample to the number of atoms in the sample. • The number of atoms we will use is 6.022 x 1023 and we call this a mole • 1 mole = 6.022 x 1023 things • Like 1 dozen = 12 things Tro's Introductory Chemistry, Chapter 6

  8. Chemical Packages - Moles • mole = number of particles equal to the number of atoms in 12 g of C-12 • 1 atom of C-12 weighs exactly 12 amu • 1 mole of C-12 weighs exactly 12 g • The number of particles in 1 mole is called Avogadro’s Number = 6.0221421 x 1023 • 1 mole of C atoms weighs 12.01 g and has 6.022 x 1023 atoms • the average mass of a C atom is 12.01 amu Tro's Introductory Chemistry, Chapter 6

  9. Example 6.1Converting Between Moles and Number of Atoms

  10. Example: A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Tro's Introductory Chemistry, Chapter 6

  11. Write down the given quantity and its units. Given: 1.1 x 1022 Ag atoms Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Tro's Introductory Chemistry, Chapter 6

  12. Write down the quantity to find and/or its units. Find: ? moles Information Given: 1.1 x 1022 Ag atoms Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Tro's Introductory Chemistry, Chapter 6

  13. Collect Needed Conversion Factors: 1 mole Ag atoms = 6.022 x 1023 Ag atoms Information Given: 1.1 x 1022 Ag atoms Find: ? moles Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Tro's Introductory Chemistry, Chapter 6

  14. Write a Solution Map for converting the units : Information Given: 1.1 x 1022 Ag atoms Find: ? moles Conv. Fact.: 1 mole = 6.022 x 1023 Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? atoms Ag moles Ag Tro's Introductory Chemistry, Chapter 6

  15. Apply the Solution Map: Information Given: 1.1 x 1022 Ag atoms Find: ? moles Conv. Fact.: 1 mole = 6.022 x 1023 Sol’n Map: atoms  mole Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? = 1.8266 x 1022 moles Ag • Sig. Figs. & Round: = 1.8 x 1022 moles Ag Tro's Introductory Chemistry, Chapter 6

  16. Check the Solution: Information Given: 1.1 x 1022 Ag atoms Find: ? moles Conv. Fact.: 1 mole = 6.022 x 1023 Sol’n Map: atoms  mole Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? 1.1 x 1022 Ag atoms = 1.8 x 10-2 moles Ag The units of the answer, moles, are correct. The magnitude of the answer makes sense since 1.1 x 1022 is less than 1 mole. Tro's Introductory Chemistry, Chapter 6

  17. Relationship Between Moles and Mass • The mass of one mole of atoms is called the molar mass • The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu Tro's Introductory Chemistry, Chapter 6

  18. 1 mole Sulfur 32.06 g 1 mole Carbon 12.01 g Mole and Mass Relationships Tro's Introductory Chemistry, Chapter 6

  19. Example 6.2Converting Between Grams and Moles of Atoms

  20. Example: Calculate the number of moles of sulfur in 57.8 g of sulfur Tro's Introductory Chemistry, Chapter 6

  21. Write down the given quantity and its units. Given: 57.8 g S Example:Calculate the number of moles of sulfur in 57.8 g of sulfur Tro's Introductory Chemistry, Chapter 6

  22. Write down the quantity to find and/or its units. Find: ? moles S Information Given: 57.8 g S Example:Calculate the number of moles of sulfur in 57.8 g of sulfur Tro's Introductory Chemistry, Chapter 6

  23. Collect Needed Conversion Factors: 1 mole S atoms = 32.06 g Information Given: 57.8 g S Find: ? moles S Example:Calculate the number of moles of sulfur in 57.8 g of sulfur Tro's Introductory Chemistry, Chapter 6

  24. Write a Solution Map for converting the units : Information Given: 57.8 g S Find: ? moles S Conv. Fact.: 1 mole S = 32.06 g Example:Calculate the number of moles of sulfur in 57.8 g of sulfur g S moles S Tro's Introductory Chemistry, Chapter 6

  25. Apply the Solution Map: Information Given: 57.8 g S Find: ? moles S Conv. Fact.: 1 mole S = 32.06 g Sol’n Map: g  moles Example:Calculate the number of moles of sulfur in 57.8 g of sulfur = 1.80287 moles S • Sig. Figs. & Round: = 1.80 moles S Tro's Introductory Chemistry, Chapter 6

  26. Check the Solution: Information Given: 57.8 g S Find: ? moles S Conv. Fact.: 1 mole S = 32.06 g Sol’n Map: g  moles Example:Calculate the number of moles of sulfur in 57.8 g of sulfur 57.8 g sulfur = 1.80 moles sulfur The units of the answer, moles, are correct. The magnitude of the answer makes sense since 57.8 g is more than 1 mole. Tro's Introductory Chemistry, Chapter 6

  27. Example 6.3Converting Between Grams and Number of Atoms

  28. Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Tro's Introductory Chemistry, Chapter 6

  29. Write down the given quantity and its units. Given: 16.2 g Al Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Tro's Introductory Chemistry, Chapter 6

  30. Write down the quantity to find and/or its units. Find: ? atoms Al Information Given: 16.2 g Al Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Tro's Introductory Chemistry, Chapter 6

  31. Collect Needed Conversion Factors: 1 mole Al atoms = 26.98 g Al 1 mole = 6.022 x 1023 Information Given: 16.2 g Al Find: ? atoms Al Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Tro's Introductory Chemistry, Chapter 6

  32. Write a Solution Map for converting the units : Information Given: 16.2 g Al Find: ? atoms Al C F: 1 mol Al = 26.98 g 1 mol = 6.022 x 1023 Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g? g Al mol Al atoms Al Tro's Introductory Chemistry, Chapter 6

  33. Apply the Solution Map: Information Given: 16.2 g Al Find: ? atoms Al CF: 1 mol Al = 26.98 g 1 mol = 6.022 x 1023 SM: g  mol  atoms Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g? = 3.6159 x 1023 atoms Al • Sig. Figs. & Round: = 3.62 x 1023 atoms Al Tro's Introductory Chemistry, Chapter 6

  34. Check the Solution: Information Given: 16.2 g Al Find: ? atoms Al CF: 1 mol Al = 26.98 g 1 mol = 6.022 x 1023 SM: g  mol  atoms Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g? 16.2 g Al = 3.62 x 1023 atoms Al The units of the answer, atoms, are correct. The magnitude of the answer makes sense since 16.2 g is less than the mass of 1 mole, 26.98 g. Tro's Introductory Chemistry, Chapter 6

  35. Molar Mass of Compounds • the relative weights of molecules can be calculated from atomic weights Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu • since 1 mole of H2O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g Tro's Introductory Chemistry, Chapter 6

  36. Example 6.4Converting Between Grams and Moles of Compound

  37. Example: Calculate the mass (in grams) of 1.75 mol of water Tro's Introductory Chemistry, Chapter 6

  38. Write down the given quantity and its units. Given: 1.75 mol H2O Example:Calculate the mass (in grams) of 1.75 mol of water Tro's Introductory Chemistry, Chapter 6

  39. Write down the quantity to find and/or its units. Find: ? g H2O Information Given: 1.75 mol H2O Example:Calculate the mass (in grams) of 1.75 mol of water Tro's Introductory Chemistry, Chapter 6

  40. Collect Needed Conversion Factors: Molar Mass H2O = 2(atomic mass H) + 1(atomic mass O) = 2(1.01) + 1(16.00) = 18.02 g/mol 1 mol H2O = 18.02 g H2O Information Given: 1.75 mol H2O Find: ? g H2O Example:Calculate the mass (in grams) of 1.75 mol of water Tro's Introductory Chemistry, Chapter 6

  41. Write a Solution Map for converting the units : Information Given: 1.75 mol H2O Find: ? g H2O C F: 1 mole H2O = 18.02 g H2O Example:Calculate the mass (in grams) of 1.75 mol of water mol H2O g H2O Tro's Introductory Chemistry, Chapter 6

  42. Apply the Solution Map: Information Given: 1.75 mol H2O Find: ? g H2O C F: 1 mole H2O = 18.02 g H2O Sol’n Map: mol  g Example:Calculate the mass (in grams) of 1.75 mol of water = 31.535 g H2O • Sig. Figs. & Round: = 31.5 g H2O Tro's Introductory Chemistry, Chapter 6

  43. Check the Solution: Information Given: 1.75 mol H2O Find: ? g H2O C F: 1 mole H2O = 18.02 g H2O Sol’n Map: mol  g Example:Calculate the mass (in grams) of 1.75 mol of water 1.75 mol H2O = 31.5 g H2O The units of the answer, g, are correct. The magnitude of the answer makes sense since 31.5 g is more than 1 mole. Tro's Introductory Chemistry, Chapter 6

  44. Example 6.5Converting Between Grams and Number of Molecules

  45. Example: Find the mass of 4.78 x 1024 NO2 molecules? Tro's Introductory Chemistry, Chapter 6

  46. Write down the given quantity and its units. Given: 4.78 x 1024 NO2 molecules Example:Find the mass of 4.78 x 1024 NO2 molecules Tro's Introductory Chemistry, Chapter 6

  47. Write down the quantity to find and/or its units. Find: ? g NO2 Information Given: 4.78 x 1024 molec NO2 Example:Find the mass of 4.78 x 1024 NO2 molecules Tro's Introductory Chemistry, Chapter 6

  48. Collect Needed Conversion Factors: Molar Mass NO2 = 1(atomic mass N) + 2(atomic mass O) = 1(14.01) + 2(16.00) = 46.01 g/mol 1 mole NO2 molec = 46.01 g NO2 1 mole = 6.022 x 1023 Information Given: 4.78 x 1024 molec NO2 Find: ? g NO2 Example:Find the mass of 4.78 x 1024 NO2 molecules Tro's Introductory Chemistry, Chapter 6

  49. Write a Solution Map for converting the units : Information Given: 4.78 x1024 NO2 molec Find: ? g NO2 C F: 1 mol NO2 = 46.01 g NO2 1 mol = 6.022 x 1023 Example:Find the mass of 4.78 x 1024 NO2 molecules molec NO2 mol NO2 g NO2 Tro's Introductory Chemistry, Chapter 6

  50. Apply the Solution Map: Information Given: 4.78 x 1024 molec NO2 Find: ? g NO2 CF: 1 mol NO2 = 46.01 g 1 mol = 6.022 x 1023 SM: molec  mol  g Example:Find the mass of 4.78 x 1024 NO2 molecules = 365.21 g NO2 • Sig. Figs. & Round: = 365 g NO2 Tro's Introductory Chemistry, Chapter 6

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