Lecture 7 Use of the Mole

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Lecture 7 - Use of the Mole
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Do some reading...

• Chapter 3 should be review for you
• Read sections 3.1-3.6, 3.11-3.13
• Problems from chapter 3

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Review

• orbital filling order
• outer shell (valence) electrons determine
  chemistry
• periodic trends in radius, ionization energy

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Anomalous Electronic Configurations

• Sometimes AUFBAU gets it wrong
• filled and half-filled subshells are anomalously
  stable

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For example, Mo

• AUFBAU predicts Kr 5s24d4
• but Mo is actually Kr 5s14d5
• (two half-filled subshells)

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For example, Ag

• AUFBAU predicts Kr 5s24d9
• but Ag is actually Kr 5s14d10
• (trades one filled for one filled plus one
  half-filled)

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The Mole

• The Mole (Avogadros Number)
• 6.022 x 1023

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for example,

• atomic dimensions 10-10 m
• limits of human vision 10-5 m

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1 amu is also rather small...

• 1 amu 1.66 10-27 kg
• atomic mass (amu) NA
• molar atomic mass (grams)

atomic weight
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for example,

• 12C 12.000 amu (per atom)
• 12.000 g (per mole of atoms)

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for example,

• C 12.011 amu (per atom)
• 12.011 g (per mole of C atoms)
• H 1.0079 amu (per atom)
• 1.0079 g (per mole of H atoms)

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Same thing applies to compounds

• H2O
• Molecular
• Weight 2 1.0079 15.9994 amu
• 18.0152 amu molecule-1
• 18.0152 g mol-1

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e.g. cholesterol, C27H46O

• MW (27 12.01) (46 1.01) 16.00
• 386.7 g/mol

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Percent Composition

• e.g. what is the mass percent of each element in
  ammonium nitrate?
• (1) Find molecular weight of NH4NO3
• (2 14.01) (4 1.01) (3 16.00)
• 80.06 g/mol

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2. Calculate Mass Ratios
28.02 g N
x 100 35.0
N
80.06 g NH4NO3
4.04 g H
x 100 5.0
H
80.06 g NH4NO3
48.00 g N
x 100 60.0
O
80.06 g NH4NO3
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e.g. fluoride in Colgate toothpaste

• 0.76 by weight Na2PO3F. Find mass of F in a 150
  g tube.
• MW (2 x 22.99) 30.97 (3 x 16.00) 19.00
• 144.0 g/mol

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e.g. fluoride in Colgate toothpaste
0.76 g Na2PO3F
150 g toothpaste x
100 g toothpaste
19.0 g F
x
0.150 g F
144.0 g Na2PO3F
i.e. toothpaste is 0.1 F
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Empirical Formulas

• O2 in sample Mg(ClO4)2 NaOH

(absorbs H2O) (absorbs CO2)
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example

• given 0.3629 g THC, when burned produces 1.067
  g CO2 and 0.3120 g H2O
• find (a) mass percent composition
• (b) empirical formula

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• g CO2 mol CO2 mol C g C
• C in sample
• and
• g H2O mol H2O mol H g H
• H in sample

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(a) mass percent composition
1 mol CO2
gC 1.067 g CO2 x
44.01 g CO2
1 mol C
12.01 g C
x
x
1 mol CO2
1 mol C
0.2912 g C
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(a) mass percent composition
0.2912 g C
x 100
mass C
0.3629 g sample
80.24 C
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(a) mass percent composition
1 mol H2O
gH 0.3120 g H2O x
18.01 g H2O
2 mol H
1.008 g H
x
x
1 mol H2O
1 mol H
0.0349 g H
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(a) mass percent composition
0.0349 g H
x 100
mass H
0.3629 g sample
9.62 H
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(a) mass percent composition

• O 100.00 - C - H
• 100.00 - 80.24 - 9.62
• 10.14

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(b) empirical formula

• formula is CxHyOz
• but what are x, y and z?

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(b) empirical formula
assume a 100 g sample
1 mol C
80.24 g C x
6.68 mol C
12.01 g C
1 mol H
9.62 g H x
9.54 mol H
1.01 g H
1 mol O
10.14 g O x
0.634 mol O
16.00 g O
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then divide each by the smallest...
C
H
O
6.68
9.54
0.634
0.634
0.634
0.634
C10.5H15O
C21H30O2
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Balancing Reactions

• e.g. burning ethane
• C2H6 7/2 O2 2 CO2 3 H2O
• or
• 2 C2H6 7 O2 4 CO2 6 H2O

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Assume we burn 10 g ethane...

• how much oxygen is required?
• how much CO2 and H2O are produced?

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method
mass reactant
MW of reactant
moles reactant
reaction stoichiometry
moles product
MW of product
mass product
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oxygen required
1 mol C2H6
10.00 g C2H6 x
30.1 g C2H6
7 mol O2
32.0 g O2
x
x
2 mol C2H6
1 mol O2
37.3 g O2
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CO2 produced
10.00 g C2H6
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CO2 produced
1 mol C2H6
10.00 g C2H6 x
30.1 g C2H6
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CO2 produced
1 mol C2H6
10.00 g C2H6 x
30.1 g C2H6
4 mol CO2
x
2 mol C2H6
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CO2 produced
1 mol C2H6
10.00 g C2H6 x
30.1 g C2H6
4 mol CO2
44.0 g CO2
x
x
2 mol C2H6
1 mol CO2
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CO2 produced
1 mol C2H6
10.00 g C2H6 x
30.1 g C2H6
4 mol CO2
44.0 g CO2
x
x
2 mol C2H6
1 mol CO2
29.3 g CO2
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H2O produced
1 mol C2H6
10.00 g C2H6 x
30.1 g C2H6
6 mol H2O
18.1 g H2O
x
x
2 mol C2H6
1 mol H2O
18.0 g H2O
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mass balance

• Reactants 10.00 g C2H6 37.3 g O2
• 47.3 g
• Products 29.3 g CO2 18.0 g H2O
• 47.3 g