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1 Chapter 7 Chemical Quantities 7.1 The Mole Basic Chemistry Copyright © 2011 Pearson Education, Inc. Collections of items include dozen, gross, and mole.

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2 Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans Basic Chemistry Copyright © 2011 Pearson Education, Inc. Collections of items include dozen, gross, and mole.

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3 A mole (mol) is a collection that contains the same number of particles as there are carbon atoms in 12.01 g of carbon. 6.022 x 10 23 atoms of an element (Avogadro’s number). 1 mol of Element Number of Atoms 1 mol C = 6.022 x 10 23 C atoms 1 mol Na = 6.022 x 10 23 Na atoms 1 mol Au = 6.022 x 10 23 Au atoms A Mole of Atoms Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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4 A mole of a covalent compound has Avogadro’s number of molecules 1 mol CO 2 = 6.022 x 10 23 CO 2 molecules 1 mol H 2 O = 6.022 x 10 23 H 2 O molecules of an ionic compound contains Avogadro’s number of formula units 1 mol NaCl = 6.022 x 10 23 NaCl formula units 1 mol K 2 SO 4 = 6.022 x 10 23 K 2 SO 4 formula units A Mole of A Compound Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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5 Samples of One Mole Quantities Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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6 Avogadro’s number (6.022 x 10 23 ) can be written as an equality and two conversion factors. Equality: 1 mol = 6.022 x 10 23 particles Conversion Factors: 6.022 x 10 23 particles and 1 mol 1 mol 6.022 x 10 23 particles Avogadro’s Number Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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7 Using Avogadro’s Number Avogadro’s number converts moles of a substance to the number of particles. How many Cu atoms are in 0.50 mol of Cu? 0.50 mol Cu x 6.022 x 10 23 Cu atoms 1 mol Cu = 3.0 x 10 23 Cu atoms Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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8 Using Avogadro’s Number Avogadro’s number is used to convert the number of particles of a substance to moles. How many moles of CO 2 are in 2.50 x 10 24 molecules of CO 2 ? 2.50 x 10 24 molecules CO 2 x 1 mol CO 2 6.022 x 10 23 molecules CO 2 = 4.15 mol of CO 2 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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9 1. The number of atoms in 2.0 mol of Al is A. 2.0 Al atoms B. 3.0 x 10 23 Al atoms C. 1.2 x 10 24 Al atoms 2. The number of moles of S in 1.8 x 10 24 atoms of S is A. 1.0 mol of S atoms B. 3.0 mol of S atoms C. 1.1 x 10 48 mol of S atoms Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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10 1.2.0 mol Al x 6.022 x 10 23 Al atoms 1 mol Al = 1.2 x 10 24 Al atoms (C) 2.1.8 x 10 24 S atoms x 1 mol S 6.022 x 10 23 S atoms = 3.0 mol of S atoms (B) Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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11 Subscripts and Moles The subscripts in a formula state the relationship of atoms in the formula the moles of each element in 1 mol of compound Glucose C 6 H 12 O 6 1 molecule: 6 atoms of C 12 atoms of H 6 atoms of O 1 mol: 6 mol of C 12 mol of H 6 mol of O Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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12 Subscripts State Atoms and Moles Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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13 Factors from Subscripts Subscripts used for conversion factors relate moles of each element in 1 mol of compound for aspirin C 9 H 8 O 4 can be written as 9 mol C 8 mol H 4 mol O 1 mol C 9 H 8 O 4 1 mol C 9 H 8 O 4 1 mol C 9 H 8 O 4 and 1 mol C 9 H 8 O 4 1 mol C 9 H 8 O 4 1 mol C 9 H 8 O 4 9 mol C 8 mol H 4 mol O Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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Calculating Atoms or Molecules 14 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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15 Learning Check How many O atoms are in 0.150 mol of aspirin, C 9 H 8 O 4 ? Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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16 Solution How many O atoms are in 0.150 mol of aspirin, C 9 H 8 O 4 ? STEP 1 Given 0.150 mol of C 9 H 8 O 4 Need molecules of C 9 H 8 O 4 STEP 2 Plan moles of aspirin moles of O atoms of O Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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17 Solution (continued) STEP 3 Equalities/Conversion Factors 1 mol of C 9 H 8 O 4 = 4 mol of O 1 mol C 9 H 8 O 4 and 4 mol O 4 mol O 1 mol C 9 H 8 O 4 1 mol of O = 6.022 x 10 23 atoms of O 1 mole O and 6.022 x 10 23 atoms O 6.022 x 10 23 atoms O 1 mol O STEP 4 Set Up Problem 0.150 mol C 9 H 8 O 4 x 4 mol O x 6.022 x 10 23 O atoms 1 mol C 9 H 8 O 4 1 mol O = 3.61 x 10 23 O atoms Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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