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Transcript
Simplest (Empirical) Formula
• Simplest integer ratio of the atoms in a compound
• Assume 100 g of compound, %element is then g
element in sample
• g element x 1 mol element = mol element
g element
CHCl3 10.061 g C x 1 mol C = 0.83765 mol C
12.011 g C
Determine simplest integer mole ratio
EXAMPLE: Phosphorus burns in air to
produce a white compound that is 43.7% P
and 56.3% O by mass. What is the empirical
formula of the compound?
Assume 100 g of compound
mass
43.7g P
56.3g O
Relative Number of Atoms
(mass/atomic mass) Divide by Smaller
= 1.41
56.3/15.9994 = 3.52
43.7/30.97
1.41/1.41 = 1.00
3.52/1.41 = 2.50
Empirical Formula  P2O5
Multiply
by Integer
2  1.00  2
2  2.50  5
.25, .33, .5, .67, .75
¼, ⅓, ½, ⅔, ¾
1:1.25 = 4:5
Molecular Formula
• The exact proportions of the elements that
are contained in a molecule
• An integer multiple (X) of the empirical
formula
MF = X  EF
Molecular Formula from
Simplest Formula
empirical formula mass  FM
sum of the atomic weights represented by
the empirical formula
molar mass = MM = X  FM
Molecular Formula from
Simplest Formula
first, knowing MM and FM
X = MM/FM
then
MF = X  EF
EXAMPLE: Our phosphorus compound has a
molar mass of ~285. What is the molecular
formula?
FM = 2 x 30.97 + 5 x 16.00 = 141.94
MM
285
X=
=
=2
FM
141.94
thus MF = 2  EF
P4O10
The empirical formula of a substance is found
to be CH3O and its molecular weight is found
to be roughly 61 g/mol. What is the true
molecular weight of the substance?
1.
2.
3.
4.
5.
130
30.5 g/mol
31.0 g/mol
61.0 g/mol
62.0 g/mol
124.0 g/mol
10
0%
0%
0%
0%
0%
g/
m
ol
12
4.
0
g/
m
ol
62
.0
g/
m
ol
61
.0
g/
m
ol
.0
31
0
30
.5
g/
m
ol
0
Biological Periodic Table
Carbohydrates Cx(H2O)y
Glucose C6H12O6
Sucrose
Glucose + Fructose  Sucrose + Water
Tristearin - Glycerol - Stearic Acid
3H2O+
+3
Chapter 4
Quantities of Reactants
and Products
Balanced Chemical Equation
• Representation of a chemical reaction which
uses stoichiometric coefficients (prefix
numbers) to represent the relative amounts
of reactants and products
• 2 H2 (g) + O2 (g)  2 H2O (l)
• Molecule to Molecule
or Mole to Mole
EXAMPLE How much H2O, in moles
results from burning an excess of H2 in 3.3
moles of O2?
2 H2 + O2  2 H2O
#mol H2O =
(3.3 mol O2) (2 mol H2O)
(1 mol O2)
Mole ratio from
balanced chemical equation
= 6.6 mol H2O
Reaction of H2 and Cl2
H2 (g) + Cl2 (g)  2 HCl (g)
one to one gives two
or
four to four gives eight
Types of Reactions
•
•
•
•
synthesis or combination reactions
decomposition reactions
displacement reactions
exchange reactions
Types of Chemical Reactions
Synthesis or Combination
Reactions
Formation of a compound from simpler compounds
or elements.
Combination Reaction
Decomposition Reactions
Separation into constituents by chemical reaction.
Dynamite
Electrolysis
Displacement Reactions
Reaction of a compound with a more reactive
element to produce a new compound and release
a less reactive element
Displacement Reactions
Exchange Reactions
Reaction where ion partners are exchanged
Pb(NO3)2 (aq) + K2CrO4 (aq)  PbCrO4 (s) + 2 KNO3 (aq)
When Zn(s) is placed in aqueous HCl, hydrogen gas
is evolved and zinc chloride solution is obtained.
Predict the reaction type.
1.
2.
3.
4.
Combination
Combustion
Decomposition
Displacement (single
displacement)
Exchange (double
displacement)
5.
0% 0% 0% 0% 0%
0
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om
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bi
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om
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130
Writing and Balancing
Chemical Equations
• Determine the type of reaction and formulae
of the products
• Write an unbalanced equation with the
correct reactants and products
• Balance the equation by the use of prefixes
(coefficients) to balance the number of each
type of atom on the reactant and product
sides of the equation.
Example
Iron is produced by the reduction of iron(III)
oxide with CO which forms CO2
iron(III) oxide + carbon monoxide  iron + carbon dioxide
Fe2O3 + CO  Fe + CO2
Fe2O3 + CO  2 Fe + CO2
Fe2O3 + 3 CO  2 Fe + 3 CO2
When aluminum reacts with sulfuric
acid to yield aluminum sulfate and
hydrogen what is the SUM of the
coefficients in the balanced equation?
130
10
0%
0%
0%
9
0
0%
7
0%
8
0
6
4
6
7
8
9
4
1.
2.
3.
4.
5.