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Introductory Chemistry , 3 rd Edition Nivaldo Tro

Introductory Chemistry , 3 rd Edition Nivaldo Tro. Chapter 6 Chemical Composition. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2009, Prentice Hall. Why Is Knowledge of Composition Important?.

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Introductory Chemistry , 3 rd Edition Nivaldo Tro

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  1. Introductory Chemistry, 3rd EditionNivaldo Tro Chapter 6 Chemical Composition Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2009, Prentice Hall

  2. Why Is Knowledge of Composition Important? • All matter is either chemically or physically combined into substances. • Knowing the fraction of material you have can tell you: • 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. How much seed do you plant? • In a garden you count the seeds by hand. How many seeds would you know to plant in a field?

  4. Counting by Weighing • Building a house requires a lot of nails. • If you know that a single nail weighs .0122 g, than 100 nails weigh 1.22 g, a 1000 nails weigh 12.2 g and so on. • Analogy: • You want to make 100 lbs of Al2O3, how much aluminum do you use Tro's "Introductory Chemistry", Chapter 6

  5. Counting Nails by the Pound, Continued A hardware store customer buys 2.60 pounds of nails. A dozen nails has a mass of 0.150 pounds. How many nails did the customer buy? 1 dozen nails = 0.150 lbs. 12 nails = 1 dozen nails Solution map: Tro's "Introductory Chemistry", Chapter 6

  6. Counting Nails by the Pound, Continued • 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

  7. Counting Nails by the Pound, Continued • What if he bought a different size nail? • Would the mass of a dozen be 0.150 lbs? • Would there still be 12 nails in a dozen? • Would there be 208 nails in 2.60 lbs? • How would this effect the conversion factors? Tro's "Introductory Chemistry", Chapter 6

  8. 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 an 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. • Avogadro’s number. • Like a kilo = 1000 or a Google = 1×10100 Tro's "Introductory Chemistry", Chapter 6

  9. Chemical Packages—Moles • Mole = Number of carbon atoms “in” 12 g of C-12. • 1 mole protons or 1 mole of neutrons = 1 amu • C-12 exactly 6 protons and 6 neutrons • since 1 mole × 1 amu = 1 g. • 1 mole of C-12 (which is 12 amu) weighs exactly 12 g. • In 12 g of C-12 there are 6.022 x1023 C-12 atoms. Tro's "Introductory Chemistry", Chapter 6

  10. Example 6.1: 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 Conversion Factor: 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 Conversion Factor: 1 mole = 6.022 x 1023 Solution 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 10-2 moles Ag • Significant figures and round: = 1.8 x 10-2 moles Ag Tro's "Introductory Chemistry", Chapter 6

  16. Check the solution: Information: Given: 1.1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6.022 x 1023 Solution 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. Practice—Calculate the Number of Atoms in 2.45 Mol of Copper. Tro's "Introductory Chemistry", Chapter 6

  18. mol Cu atoms Cu Practice—Calculate the Number of Atoms in 2.45 Mol of Copper, Continued. Given: Find: 2.45 mol Cu atoms Cu Solution Map: Relationships: 1 mol = 6.022 x 1023 atoms Solution: Check: Since atoms are small, the large number of atoms makes sense. Tro's "Introductory Chemistry", Chapter 6

  19. 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. • The lighter the atom, the less a mole weighs. • The lighter the atom, the more atoms there are in 1 g. Tro's "Introductory Chemistry", Chapter 6

  20. Mole and Mass Relationships 1 mole carbon 12.01 g 1 mole sulfur 32.06 g Tro's "Introductory Chemistry", Chapter 6

  21. g S mol S Example 6.2—Calculate the Moles of Sulfur in 57.8 g of Sulfur. Given: Find: 57.8 g S mol S Solution Map: Relationships: 1 mol S = 32.07 g Solution: Check: Since the given amount is much less than 1 mol S, the number makes sense. Tro's "Introductory Chemistry", Chapter 6

  22. Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead. Tro's "Introductory Chemistry", Chapter 6

  23. g C mol C Practice—Calculate the Moles of Carbon in 0.0265 g of Pencil Lead, Continued. Given: Find: 0.0265 g C mol C Solution Map: Relationships: 1 mol C = 12.01 g Solution: Check: Since the given amount is much less than 1 mol C, the number makes sense. Tro's "Introductory Chemistry", Chapter 6

  24. Example 6.3: How many aluminum atoms are in an aluminum can with a mass of 16.2 g? Tro's "Introductory Chemistry", Chapter 6

  25. g Al mol Al atoms Al Example 6.3—How Many Aluminum Atoms Are in a Can Weighing 16.2 g? Given: Find: 16.2 g Al atoms Al Solution Map: Relationships: 1 mol Al = 26.98 g, 1 mol = 6.022 x 1023 Solution: Check: Since the given amount is much less than 1 mol Cu, the number makes sense. Tro's "Introductory Chemistry", Chapter 6

  26. 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

  27. mol H2O g H2O Example 6.4—Calculate the Mass of 1.75 Mol of H2O. Given: Find: 1.75 mol H2O g H2O Solution Map: Relationships: 1 mol H2O = 18.02 g Solution: Check: Since the given amount is more than 1 mol, the mass being > 18 g makes sense. Tro's "Introductory Chemistry", Chapter 6

  28. Practice—How Many Moles Are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00) Tro's "Introductory Chemistry", Chapter 6

  29. g PbO2 mol PbO2 Practice—How Many Moles Are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00), Continued Given: Find: 50.0 g mol PbO2 moles PbO2 Solution Map: Relationships: 1 mol PbO2 = 239.2 g Solution: Check: Since the given amount is less than 239.2 g, the moles being < 1 makes sense. Tro's "Introductory Chemistry", Chapter 6

  30. molecules mol NO2 g NO2 Example 6.5—What Is the Mass of 4.78 x 1024 NO2 Molecules? Given: Find: 4.78 x 1024 NO2 molecules g NO2 Solution Map: Relationships: 1 mol NO2 = 46.01 g, 1 mol = 6.022 x 1023 Solution: Check: Since the given amount is more than Avogadro’s number, the mass > 46 g makes sense. Tro's "Introductory Chemistry", Chapter 6

  31. Counting and ratio’s It takes me .2 gal of gas to get to IVC. It is a very simple ratio: = What if I only had .1 gal 200 2X4’s, 3 sinks, 2 showers, you can make a house with 3 bathrooms and 3 bedrooms. What if you had 12 sinks…how many houses could you make. 200 3 2 = 1 3 3

  32. Chemical Formulas as Conversion Factors • 1 spider  8 legs. • 1 chair  4 legs. • 1 H2O molecule  2 H atoms  1 O atom. Tro's "Introductory Chemistry", Chapter 6

  33. Mole Relationships inChemical Formulas • Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound. Tro's "Introductory Chemistry", Chapter 6

  34. mol CaCO3 mol O Example 6.6—Calculate the Moles of Oxygen in 1.7 Moles of CaCO3. Given: Find: 1.7 mol CaCO3 mol O Solution Map: Relationships: 1 mol CaCO3 = 3 mol O Solution: Check: Since the given amount is much less than 1 mol S, the number makes sense. Tro's "Introductory Chemistry", Chapter 6

  35. Example 6.7: Carvone (C10H14O) is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liqueurs, soaps, and perfumes. Find the mass of carbon in 55.4 g of carvone. Tro's "Introductory Chemistry", Chapter 6

  36. Write down the given quantity and its units. Given: 55.4 g C10H14O Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). Tro's "Introductory Chemistry", Chapter 6

  37. Write down the quantity to find and/or its units. Find: ? g C Information: Given: 55.4 g C10H14O Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). Tro's "Introductory Chemistry", Chapter 6

  38. Collect needed conversion factors: Molar mass C10H14O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O) = 10(12.01) + 14(1.01) + (16.00) = 150.2 g/mol 1 mole C10H14O = 150.2 g C10H14O 1 mole C10H14O  10 mol C 1 mole C = 12.01 g C Information: Given: 55.4 g C10H14O Find: g C Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). Tro's "Introductory Chemistry", Chapter 6

  39. Write a solution map for converting the units: Information: Given: 55.4 g C10H14O Find: g C Conversion Factors: 1 mol C10H14O = 150.2 g 1 mol C10H14O  10 mol C 1 mol C = 12.01 g Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). g C10H14O mol C10H14O mol C g C Tro's "Introductory Chemistry", Chapter 6

  40. Apply the solution map: Information: Given: 55.4 g C10H14O Find: g C Conversion Factors: 1 mol C10H14O = 150.2 g 1 mol C10H14O  10 mol C 1 mol C = 12.01 g Solution Map: g C10H14O  mol C10H14O  mol C  g C Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). = 44.2979 g C • Significant figures and round: = 44.3 g C Tro's "Introductory Chemistry", Chapter 6

  41. Percent Composition • Percentage of each element in a compound. • By mass. • Can be determined from: • The formula of the compound. • The experimental mass analysis of the compound. • The percentages may not always total to 100% due to rounding. Tro's "Introductory Chemistry", Chapter 6

  42. Example 6.9—Find the Mass Percent of Cl in C2Cl4F2. Given: Find: C2Cl4F2 % Cl by mass Solution Map: Relationships: Solution: Check: Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense.

  43. Practice—Determine the Mass Percent Composition of the Following: CaCl2 (Ca = 40.08, Cl = 35.45) Tro's "Introductory Chemistry", Chapter 6

  44. Practice—Determine the Percent Composition of the Following, Continued: CaCl2 Tro's "Introductory Chemistry", Chapter 6

  45. Mass Percent as a Conversion Factor • The mass percent tells you the mass of a constituent element in 100 g of the compound. • The fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na. • This can be used as a conversion factor. • 100 g NaCl  39 g Na Tro's "Introductory Chemistry", Chapter 6

  46. Empirical Formulas • The simplest, whole-number ratio of atoms in a molecule is called the empirical formula. • Can be determined from percent composition or combining masses. • The molecular formula is a multiple of the empirical formula. Tro's "Introductory Chemistry", Chapter 6

  47. Empirical (CH2O)-vs- Molecular Formula molecular formula Formaldehyde CH2O 1 30.03 g/mol Acetic Acid C2H4O2 2 60.06 g/mol Lactic Acid C3H6O3 3 90.09 g/mol Erythrose C4H8O4 4 120.12 g/mol Ribose C5H10O5 5 150.15 g/mol Glucose C6H12O6 6 180.18 g/mol

  48. Empirical Formulas, Continued Hydrogen Peroxide Molecular formula = H2O2 Empirical formula = HO Benzene Molecular formula = C6H6 Empirical formula = CH Glucose Molecular formula = C6H12O6 Empirical formula = CH2O

  49. Example 6.11—Finding an Empirical Formulafrom Experimental Data Tro's "Introductory Chemistry", Chapter 6

  50. Example: A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53% Tro's "Introductory Chemistry", Chapter 6

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