1 chapter 7 chemical quantities 7.1 the mole basic chemistry copyright © 2011 pearson education,...
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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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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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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 1023 atoms of an element (Avogadro’s number).
1 mol of Element Number of Atoms
1 mol C = 6.022 x 1023 C atoms
1 mol Na = 6.022 x 1023 Na atoms
1 mol Au = 6.022 x 1023 Au atoms
A Mole of Atoms
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A mole
• of a covalent compound has Avogadro’s number of molecules
1 mol CO2 = 6.022 x 1023 CO2 molecules
1 mol H2O = 6.022 x 1023 H2O molecules
• of an ionic compound contains Avogadro’s number of formula units
1 mol NaCl = 6.022 x 1023 NaCl formula units
1 mol K2SO4 = 6.022 x 1023 K2SO4 formula units
A Mole of A Compound
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Samples of One Mole Quantities
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Avogadro’s number (6.022 x 1023) can be written as an equality and two conversion factors.
Equality:
1 mol = 6.022 x 1023 particles
Conversion Factors:
6.022 x 1023 particles and 1 mol 1 mol 6.022 x 1023
particles
Avogadro’s Number
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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 1023 Cu atoms
1 mol Cu
= 3.0 x 1023 Cu atoms
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Using Avogadro’s Number
Avogadro’s number is used to convert the number of particles of a substance to moles.
How many moles of CO2 are in 2.50 x 1024 molecules of CO2?
2.50 x 1024 molecules CO2
x 1 mol CO2
6.022 x 1023 molecules CO2
= 4.15 mol of CO2
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1. The number of atoms in 2.0 mol of Al is
A. 2.0 Al atoms
B. 3.0 x 1023 Al atoms
C. 1.2 x 1024 Al atoms
2. The number of moles of S in 1.8 x 1024 atoms of S is
A. 1.0 mol of S atoms
B. 3.0 mol of S atoms
C. 1.1 x 1048 mol of S atoms
Learning Check
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1. 2.0 mol Al x 6.022 x 1023 Al atoms
1 mol Al
= 1.2 x 1024 Al atoms (C)
2. 1.8 x 1024 S atoms x 1 mol S
6.022 x 1023 S atoms = 3.0 mol of S atoms (B)
Solution
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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
C6H12O6
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
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Subscripts State Atoms and Moles
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Factors from Subscripts
Subscripts used for conversion factors• relate moles of each element in 1 mol of compound • for aspirin C9H8O4 can be written as
9 mol C 8 mol H 4 mol O
1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4
and
1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4
9 mol C 8 mol H 4 mol O
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Calculating Atoms or Molecules
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Learning Check
How many O atoms are in 0.150 mol of aspirin, C9H8O4?
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Solution
How many O atoms are in 0.150 mol of aspirin, C9H8O4?
STEP 1 Given 0.150 mol of C9H8O4
Need molecules of C9H8O4
STEP 2 Plan
moles of aspirin moles of O atoms of O
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Solution (continued)STEP 3 Equalities/Conversion Factors
1 mol of C9H8O4 = 4 mol of O
1 mol C9H8O4 and 4 mol O
4 mol O 1 mol C9H8O4
1 mol of O = 6.022 x 1023 atoms of O
1 mole O and 6.022 x 1023 atoms O
6.022 x 1023 atoms O 1 mol O
STEP 4 Set Up Problem
0.150 mol C9H8O4 x 4 mol O x 6.022 x 1023 O atoms
1 mol C9H8O4 1 mol O
= 3.61 x 1023 O atoms
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