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Chapter 8

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Chapter 8

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  1. Chapter 8 Chemical Composition

  2. Students will learn… • Mole as a counting unit • amu = atomic mass unit • molar mass • % composition of compounds • Formulas of compounds • empirical formulas • molecular formulas

  3. Mole (mol) A) Is a counting unit B) 1 mole = 6.02 x 1023 particles (ex) 1 mol of pencils = 6.02 x 1023 pencils 1 mol of iron = 6.02 x 1023 Fe atoms 1 mol of CO2 = 6.02 x 1023 CO2molecules 1 mol of Ca2+ ions = 6.02 x 1023 Ca2+ ions C) Also called Avogadro’s number

  4. Know the Particle Types! • molecules: particles of covalent compounds (nonmetal + nonmetal) • formula units: particles of ionic compounds (metal + nonmetal) • ions: positively or negatively charged particles (cation & anion) • atoms: particles of element

  5. Conversion Factors 1) always in fraction form and equal to 1 (Ex) 2) Two conversion factors are written from one equal relationship. (Ex) 3) Also called as unit or dimensional analysis

  6. 4) Always follow the same set-up. Given quantity Conversion Wanted quantity factor (Ex 1) How many oxygen atoms are in 4.5×108 formula units of CuSO4?

  7. Ex2) Find the number of C atoms in 2.8 mol of C.

  8. amu (atomic mass unit), u • a mass unit for a very small mass • defined by assigning 12.0 amu to the mass of a carbon-12 atom • 1 amu= mass of 1/12 of a C-12 atom =1.66053886 × 10-27kg • 1 proton ≈ 1 amu 1 neutron ≈ 1 amu 1 electron ≈ 0 amu

  9. amu continues… • The amu of other element is determined by comparing its weight to the weight of a carbon-12 atom. • The decimal number in the periodic table is the average amu of the isotopes of each element • The word, average, will be left out. • Isotopes = atoms with the same atomic number but different atomic mass number

  10. Molar Mass • The mass of 1 mole of a pure substance • The mass of 6.02x1023particles • Numerically the same as the average atomic mass unit (amu) (Ex) Mass of 1 C-12 atom = 12.0 amu Mass of 1 mol C-12 = 12.0 g

  11. Example The (average) mass of one Cu atom =63.546 amu The(average) atomic mass of Cu = 63.546 g

  12. Molar Mass of Compound • mass of H2O? 1 mol H2O = 2 mol H + 1 mol O • Formula weight of Ca(NO3)2? 1 mol Ca(NO3)2 = 1 mol Ca + 2 mol N + 6 mol O

  13. Use the correct terms • For elements, molar mass = atomic mass • For compounds, molar mass = formula mass ** In chemistry, mass = weight

  14. Practice Find the molar mass of: 1) C6H5CH3 (toluene) 2) CaHPO4 3) Ca3(PO4)2 4) CrSO4∙3H2O

  15. Answers 1) 92.15 g 2) 136.1 g 3) 310.2 g 4) 202.1 g

  16. Learned So Far mol Avogadro’s # Molar mass # of particles (atoms, molecules, formula units, ions) mass

  17. Problem Solving Strategy • Know what are you solving for. • Know what is the given quantity to convert. • Know what is the conversion factor. • 1 mol = 6.02x1023 particles • 1 mol = mass of 1 mol of a substance • Use the same set-up format *Called the “box” or “railroad track”

  18. Examples Mol  grams 1.A certain sample of calcium phosphate contains 4.86 mol. What is the mass in grams of this sample?

  19. Grams  mol 2.The molar mass of sodium sulfate, Na2SO4. How many moles of sodium sulfate are in 300.0 g?

  20. Grams  # of particles 3.Isopentyl acetate, C7H14O2 (MW=130.2 g) the compound responsible for the scent of bananas, can be produced commercially. How many molecules of isopentyl acetate are in 1 µg?

  21. % Composition of Compound • Mass percent of an element: • (Ex) For iron in iron(III) oxide, (Fe2O3):

  22. Chemical Formulas • Empirical formula ≈ simplest formula • shows the lowest whole-number ratio of the atoms in a compound (Ex) a benzene molecule = C6H6 Empirical formula of benzene = CH • Molecular Formula • shows the actual composition of molecules (Ex) Molecular formula of benzene = C6H6

  23. EF & MF Relationship • MF = n·EF • n = a whole number ≥ 1

  24. From mass % to MF • Get the mass % of each element • Get the mass of each element in 100 g sample of compound • Convert the mass of each element to moles • Reduce the moles to the lowest whole numbers • Do not apply the “5 and up” rule.

  25. Why “5 and up” can’t be applied? Compounds made of 2 elements, C and H: • 1 mol C: 0.5 mol H  1 mol C:1 mol H • 1 mol C: 1.2 mol H  1 mol C:1 mol H • 1 mol C: 1.33 mol H  1 mol C:1 mol H *Do you see a problem?

  26. 1 mol C: 0.5 mol H  10 mol C: 5 mol H  2 mol C: 1 mol H 1 mol C: 1.2 mol H  10 mol C:12 mol H 5 mol C: 6 mol H 1 mol C: 1.33 mol H  100 mol C:133 mol H  3 mol C: 4 mol H ** Handy to remember: 0.25 = ¼; 0.2 = 1/5; 0.33 = 1/3

  27. How to round correctly? • Multiply the moles by a factor of 10 until all moles become whole numbers • Reduce the whole-number moles to the lowest whole numbers

  28. Examples • mass % to EF The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). What is the empirical formula? C3H5O2

  29. The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. The empirical formula is C3H5O2. What is the molecular formula? C6H10O4 • Mass % to MF