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Chapter 11 The Mole

Chapter 11 The Mole. A chemists best friend.  Counting units in groups is common.  Dozen  Case  Gross.  Mole - a particular number of atoms, ions, molecules or formula units. It is a large number. 6.02 X 10 23. Just like dozen = 12 Mole = 6.02 x 10 23 particles.

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Chapter 11 The Mole

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  1. Chapter 11 The Mole • A chemists best friend

  2.  Counting units in groups is common  Dozen  Case  Gross  Mole - a particular number of atoms, ions, molecules or formula units It is a large number

  3. 6.02 X 10 23 • Just like dozen = 12 • Mole = 6.02 x 10 23 particles

  4. Use atomic masses to 3 significant digits or to the hundredths' place, whichever gives you more digits 23 Avogadro constant - 6.02  10 A It is an SI standard (symbol = N )

  5. The Mole- 6.02  10 “things” In chemistry “things” are atoms, ions, molecules, or formula units. 23 One mole of particles has a mass in grams equivalent to that of one particle in atomic mass units 23 1 mole of particles contains 6.02  10 particles

  6.  Molar mass- the mass of one mole of molecules, atoms, ions, or formula units Examples: grams to moles molecules to grams grams to atoms

  7. Percent Composition- used to identify unknowns Example: Find the percentage composition of aluminum oxide 2 Al atoms = 53.96 u 3 O atoms = 48.00 u 101.96u Practice prob pg 331 % Al = 53.96  100 = 52.92% 101.96 % O = 48.00 100 = 47.08% 101.96

  8. Empirical Formulas The decomposition of 11.47 g of a compound of copper and oxygen yields 9.16 g of copper. What is the empirical formula? 2.31 g oxygen 9.16 g Cu 1 mole Cu = .144 mole Cu 63.55 g Cu 2.31 g O 1 mole O = .144 mole O 16.00 g O Ratio 1:1 hence CuO

  9. A compound contains carbon 81.7% and Hydrogen 18.3%. Find the empirical formula 81.7 g C 1 mole C = 6.80 moles C 12.01 g C 18.3 g H 1 mole H = 18.1 moles H 1.01 g H Ratio = 6.81 : 18.1 6.81 6.81 = 1:2.66 = 1:2 3 = 1: 8 3:8 C H 2 3 3 3 3 8 Exercises Pg 333

  10.  Molecular Formulas~ Must know molar mass Example: analysis of a compound reveals this composition: 80.0% C 20.0% H Molar mass = 30.0 g What is the molecular formula? C H 80.0 g C 1 mole C = 6.66 mole C 12.01 g C 1 3 20.0 g H 1 mole H = 19.8 mole H 1.01 g H C2H6

  11. Ascorbic acid, vitamin C, is 40.9% carbon, 4.58% hydrogen and 54.5% oxygen. Its molar mass is 176.1 g/mole. What is the molecular formula?

  12. Hydrates ~ water molecules are part of the crystal structure Example: 5.20 g of hydrated BaI2 is heated. The dry sample has a mass of 4.76 g. What is the formula of the hydrate? .44 g H O 2 5.20 g – 4.76 g = .44 g H O 2

  13. 4.76 g BaI 1 mole BaI = .0122 mole BaI 391.13 g BaI 2 2 2 2 .44 g H O 1 mole H O = .024 mole H O 18.02 g H O • 2 2 2 .0122 mole = 1.00 .024 mole = 2.0 .0122 .0122 1:2 Ratio BaI • 2H O 2 2

  14. Atomic mass unit ( u ) a.m.u. H = 1.01 u O = 16.00 u C = 12.00 u Molecular Mass- the sum of all the atomic masses of the atoms in a molecule Formula Mass- used for ionic compounds

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