Chemical Formulas and Composition
Stoichiometry: Chapter 2
Chem 101
Fall 2004
Water Molecules
• Ordinary samples contain many
molecules.
• A typical drop of water has mass of
about 0.05 gram.
• How many water molecules in a drop?
Chem 101
Fall 2004
Chapter Outline
• Formula Weights, Molecular Weights, and Moles
• Percent Composition and Formulas of Compounds
• Derivation of Formulas from Elemental
Composition
• Determination of Molecular Formulas
• Purity of Samples
Chem 101
Fall 2004
1
Formula Weights, Molecular
Weights, and Moles
• One Mole of
Contains
• Br2 or 159.80g
6.022 x 1023 Br2 molecules
2 (6.022 x 1023 ) Br atoms
• C6H6O or 94.07 g
6.022 x 1023 C6H6O molecules
6 (6.022 x 1023 ) C atoms
6 (6.022 x 1023 ) H atoms
1 (6.022 x 1023 ) O atoms
Chem 101
Fall 2004
Formula Weights
• How do we calculate the molar mass of a
compound?
• add atomic weights of each atom
• The molar mass of phenol, C6H6O, is:
+
Formula Weight
6 x 12 amu = 72 amu
6 x 1 amu = 6 amu
1 x 16 amu = 16 amu
Carbon
Hydrogen
Oxygen
94 amu
Chem 101
Fall 2004
Formula Weights: Caffeine
• Formula is C8H10N4O2
• Atomic weights:
C=12, H=1, N=14, O = 16
• Molar
8 x 12 = 96
mass
10 x 1 =10
4 x 14 = 56
2 x 16 = 32
194
Chem 101
Fall 2004
2
Empirical and Molecular Formulas
• Empirical Formula: smallest combination
of whole numbers of elements which show
correct ratio
C4H5N2O
• Molecular Formula: True composition
(related to empirical formula by
multiplicative factor)
C8H10N4O2
Chem 101
Fall 2004
Empirical and Molecular Formulas
• In the absence of information about the
molecular (molar) mass the empirical
formula is used.
• Given the molar mass it is straightforward
to find the molecular formula
Chem 101
Fall 2004
Empirical and Molecular Formulas
• Benzene:
CH
C6H6
Empirical
Molecular
• Butane:
C2H5
C4H10
Empirical
Molecular
Chem 101
Fall 2004
3
Water Molecules
H2O molecule = 2 H’s + 1 O
Mass = 2 (1 amu) + 1 (16 amu)
= 18 amu
Chem 101
Fall 2004
Water Molecules
1 amu = 1.66 x 10–27 kg
1 molecule = 18 amu
18 amu
1.66x10 -27 kg 1000 g
×
×
1 molecule
1 amu
1 kg
=
3.0x10 -23 g
molecule
Chem 101
Fall 2004
Water Molecules
1 molecule = 3 x 10-23 g
1 drop = 0.05 g
molecules
1 molecule 0.05 g
×
= 1.7x10 21
- 23
drop
3.0x10 g 1 drop
Chem 101
Fall 2004
4
Derivation of Formulas from
Elemental Composition
Chem 101
Fall 2004
Percent Composition: Caffeine
• Given a chemical formula, it is
straightforward to find mass percentages
• Find molar mass and the mass from each
element
• As an example, calculate mass % of each
element in caffeine
Chem 101
Fall 2004
Nicotine Example
• Nicotine contains 74.0% C, 8.65% H, and
17.35% N. If the molar mass of nicotine is
162, what is the chemical formula of
nicotine?
Atomic weights for C, H, and N are 12, 1,
and 14.
N
Chem 101
Fall 2004
N
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Purity of Samples
• The percent purity of a sample of a
substance is always represented as
mass of pure substance
× 100 %
mass of sample
mass of sample includes impurities
% purity =
Chem 101
Fall 2004
Next Class: Chemical Equations and
Reaction Stoichiometry: Chapter 3
• Continue to work on OWL homework
(Finish Chapter 2)
• work on ‘past due assignments for review’
• Start reading Chapter 3
Chem 101
Fall 2004
6