Transcript General Chemistry

Molecules, Ions, and Their
Compounds
Chapter 3
Chapter 3
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Molecules and Molecular Compounds
Molecular and Empirical Formulas
Molecular formula – A formula which gives the actual
number and type of atoms in a molecule.
Examples: H2O, CO2, CO, CH4, H2O2, O2, O3, and C2H4.
Empirical formula – A formula which gives the lowest
whole number ratio of atoms in a molecule.
Examples:
Substance
Ethane
Water
Mol. formula
C2H6
H2 O
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Empirical Formula
CH3
H2 O
2
Molecules and Molecular Compounds
Molecular and Empirical Formulas
Condensed formula – A formula which indicates how
atoms are grouped together in a molecule.
Name
Ethane
Diethyl ether
Molecular Formula
C 2 H6
C4H10O
Chapter 3
Condensed formula
CH3CH3
CH3CH2OCH2CH3
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Molecules and Molecular Compounds
Picturing Molecules
Structural Formula – A formula which shows how the
atoms of a molecule are joined.
• Structural formulas do not necessarily show the three
dimensional shape of the molecule.
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Molecules and Molecular Compounds
Molecular Models
These are three-dimensional representations of
molecules.
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Ions and Ionic Compounds
• If an electron is removed or added to a neutral atom
a charged particle or ion is formed.
• A positively charged ion is called a cation.
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Ions and Ionic Compounds
• If an electron is removed or added to a neutral atom
a charged particle or ion is formed.
• A positively charged ion is called a cation.
• A negatively charged ion is called an anion.
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Ions and Ionic Compounds
Predicting Ionic Charge
• Metals tend to form cations
• Non-metals tend to form anions.
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Ions and Ionic Compounds
Molecules can also gain or lose electrons and form
ions, They are called polyatomic ions.
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Ions and Ionic Compounds
Ion Name
Formula
Ion Name
Formula
Peroxide
O22-
Sulfate
SO42-
Triiodide
I3 -
Sulfite
SO32-
Ammonium
NH4+
Phosphate
PO43-
Nitrate
NO3-
Acetate
CH3CO2-
Nitrite
NO2-
Perchlorate
ClO4-
Hydroxide
OH-
Permanganate
MnO4-
Carbonate
CO32-
Dichromate
Cr2O72-
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Ions and Ionic Compounds
Ionic Compounds
Ionic Compound – A compound that contains positively
charged ions and negatively charged ions.
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Ions and Ionic Compounds
Predicting Formulas
Let’s consider a compound containing Mg and N.
• The common charge on Mg is +2 (or Mg2+).
• The common charge on N is –3 (or N3-).
• Since we want to make a neutral (uncharged)
compound, the total charges from the cations and
anions must cancel-out (or sum to zero).
• Therefore, Mg needs to lose 6 electrons (3  2+)
and N gain those 6 electrons (2  3-).
• The resulting formula is: Mg3N2.
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Naming Inorganic Compounds
Names and Formulas of Ionic Compounds
Naming of compounds (nomenclature) is divided into:
• organic compounds (those containing C)
• inorganic compounds (the rest of the periodic
table).
We will consider the naming rules of the Inorganic
compounds.
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Naming Ionic Compounds
Names and Formulas of Ionic Compounds
1. Cations
a) Cations from metal atoms have the same name as
the metal.
b) If the cation can form more than one ion, the
positive charge is indicated by a roman numeral
in parenthesis.
c) Cations of nonmetals end in –ium.
P+3
phosphorium
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Naming Inorganic Compounds
Names and Formulas of Ionic Compounds
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Naming Inorganic Compounds
Names and Formulas of Ionic Compounds
2. Anions
a) Monoatomic anions have names formed by
dropping the ending of the name of the element
and adding –ide.
b) Polyatomic anions containing oxygen have names
ending in –ate or –ite.
c) Anions derived by adding H+ to an oxyanion are
named by adding as a prefix the word hydrogen- or
dihydrogen-.
HSO4H2PO4-
Hydrogensulfate
Dihydrogenphsophate
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Naming Inorganic Compounds
Names and Formulas of Ionic Compounds
2. Anions
d) Oxyanions
There are rules for these, but they are confusing.
Ion
ClO4ClO3ClO2ClO-
Name
perchlorate ion
chlorate ion
chlorite ion
hypochlorite ion
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Naming Inorganic Compounds
Names and Formulas of Ionic Compounds
3. Ionic Compounds
Name the compound by naming the cation
followed by the anion.
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Naming Inorganic Compounds
Naming Binary Molecular Compounds
Binary molecular compounds have two elements.
1. The name of the left-most element is written first.
2. If the elements are in the same group the lower
element is written first.
3. The name of the second element ends in –ide.
4. Prefixes are used to indicate the number of atoms of
each element.
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Naming Inorganic Compounds
Naming Binary Molecular Compounds
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Naming Inorganic Compounds
Naming Binary Molecular Compounds
Binary molecular compounds have two elements.
1. The name of the left-most element is written first.
2. If the elements are in the same group the lower
element is written first.
3. The name of the second element ends in –ide.
4. Prefixes are used to indicate the number of atoms of
each element.
• Mono is never used in the first element.
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Naming Inorganic Compounds
N2O
•
•
•
•
This is a molecular compound.
The first element (N), just takes its name,
Nitrogen.
The second compound takes its name, ending in
-ide, Oxide.
Now we must consider how to show that there
are two nitrogen atoms, use di- as a prefix.
Dinitrogen Oxide
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Naming Inorganic Compounds
N2O5
•
•
•
•
•
This is a molecular compound.
The first element (N), just takes its name,
Nitrogen.
The second compound takes its name, ending in
-ide, Oxide.
Now we must consider how to show that there
are two nitrogen atoms, use di- as a prefix.
Finally, we must consider how to show that there
are five oxygen atoms, use penta- as a prefix.
Dinitrogen Pentoxide
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Formulas, Compounds, and the Mole
Formula and Molecular Weights
Molecular weight
The sum of the atomic weights of each atom in the
molecular formula.
• Formula weight is the general term, molecule weight
refers specifically to molecular compounds.
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Formulas, Compounds, and the Mole
Formula and Molecular Weights
Formula weight (FW)
The sum of the atomic weights of each atom in the
chemical formula.
Example: CO2
Formula Weight = 1(AW, carbon) + 2(AW, oxygen)
Formula Weight = 1(12.011amu) + 2(16.0amu)
Formula Weight = 44.0 amu
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Formulas, Compounds, and the Mole
Formula and Molecular Weights
Chemistry “trick”
• The masses of the atoms are on a “gram equivalent
scale”.
1 atom (average)
1 mole
C
12.01 amu
12.01 g
H
1.008 amu
1.008 g
• So, the mass of a single atom or a mole is numerically
equvalent.
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Formulas, Compounds, and the Mole
Converting Between Mass, Moles, Molecules and
Atoms
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The Mole
Moles  Numbers of Particles
 6.021023 particles

num bersof particles m oles
1 m ol


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The Mole
Mass  Moles
 gramsof substance 

moles 
 formulaweight 
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The Mole
Moles  Mass
mass  formulaweight  molesof sample
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The Mole
A sample of hormone, estradiol, C18H24O2, contains
3.0 x 1020 atoms of hydrogen. How many atoms of
carbon does it contain?
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The Mole
A sample of hormone, estradiol, C18H24O2, contains
3.0 x 1020 atoms of hydrogen. How many atoms of
carbon does it contain?
ratioof C to H
18C : 24H
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The Mole
A sample of hormone, estradiol, C18H24O2, contains
3.0 x 1020 atoms of hydrogen. How many atoms of
carbon does it contain?
ratioof C to H, 18C : 24H
or
18C
24H
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The Mole
A sample hormone, estradiol, C18H24O2, contains 3.0 x
1020 atoms of hydrogen. How many atoms of
carbon does it contain?
ratioof C to H, 18C : 24H
or
18C
x

24H 3.0  1020 atom sH
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The Mole
A sample hormone, estradiol, C18H24O2, contains 3.0 x
1020 atoms of hydrogen. How many atoms of
carbon does it contain?
ratioof C to H, 18C : 24H
or
18C
x

24H 3.0  1020 atom sH
x  2.3  1020 atom sC
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The Mole
A sample hormone, estradiol, C18H24O2, contains 3.0 x
1020 atoms of hydrogen. How many molecules of
estradiol does it contain?
ratioof moleculesto H
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The Mole
A sample hormone, estradiol, C18H24O2, contains 3.0 x
1020 atoms of hydrogen. How many molecules of
estradiol does it contain?
ratioof moleculesto H
1 molecule: 24H
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The Mole
A sample hormone, estradiol, C18H24O2, contains 3.0 x
1020 atoms of hydrogen. How many molecules of
estradiol does it contain?
ratioof moleculest o H
1 molecule: 24H
or
1 m olecule
24H
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The Mole
A sample hormone, estradiol, C18H24O2, contains 3.0 x
1020 atoms of hydrogen. How many molecules of
estradiol does it contain?
ratioof moleculesto H
1 molecule: 24H
or
1 m olecule
x

24H
3.0  1020 atom sH
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The Mole
A sample hormone, estradiol, C18H24O2, contains 3.0 x
1020 atoms of hydrogen. How many molecules of
estradiol does it contain?
ratioof moleculest o H
1 molecule: 24H
or
1 m olecule
x

24H
3.0  1020 atom sH
x  1.3  1019 m olecules
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Describing Compound Formulas
Percentage Composition from Formulas


m assof elem ent
100
% elem ent 
 form ulaweightof m olecule
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Describing Compound Formulas
Percentage Composition from Formulas
Example:
Calculate the percent oxygen in CH3CH2OH.
Formula weight of ethanol:
2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amu
Mass of oxygen:
1(16.00amu) = 16.00amu
% oxygen
 16.00am u

100
 46.08am u
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Describing Compound Formulas
Percentage Composition from Formulas
Example:
Calculate the percent oxygen in CH3CH2OH.
Formula weight of ethanol:
2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amu
Mass of oxygen:
1(16.00amu) = 16.00amu
% oxygen
 16.00am u 

100 34.72%
 46.08am u
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Empirical Formulas from Analyses
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Empirical Formulas from Analyses
Analysis
Hg  73.9%
Cl  26.1%
-assume 100g sample
Hg  73.9 g
Cl  26.1g
-convert grams to moles
Hg  73.9g / 200.59g/mol =
Cl  26.1g/ 35.45g/mol =
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Empirical Formulas from Analyses
Analysis
Hg  73.9%
Cl  26.1%
-assume 100g sample
Hg  73.9 g
Cl  26.1g
-convert grams to moles
Hg  73.9g / 200.59g/mol = 0.368 mol
Cl  26.1g/ 35.45g/mol = 0.736 mol
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Empirical Formulas from Analyses
Analysis
Hg  73.9%
Cl  26.1%
-assume 100g sample
Hg  73.9 g
Cl  26.1g
-convert grams to moles
Hg  73.9g / 200.59g/mol = 0.368 mol
Cl  26.1g/ 35.45g/mol = 0.736 mol
-determine the empirical formula by using the moles of the
elements to get the smallest whole number ratio of the elements.
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Empirical Formulas from Analyses
Analysis
Hg  73.9%
Cl  26.1%
-assume 100g sample
Hg  73.9 g
Cl  26.1g
-convert grams to moles
Hg  73.9g / 200.59g/mol = 0.367 mol
Cl  26.1g/ 35.45g/mol = 0.736 mol
-determine the empirical formula by using the moles of the
elements to get the smallest whole number ratio of the elements.
Hg0.367 Cl0.736
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Empirical Formulas from Analyses
Analysis
Hg  73.9%
Cl  26.1%
-assume 100g sample
Hg  73.9 g
Cl  26.1g
-convert grams to moles
Hg  73.9g / 200.59g/mol = 0.368 mol
Cl  26.1g/ 35.45g/mol = 0.736 mol
-determine the empirical formula by using the moles of the
elements to get the smallest whole number ratio of the elements.
Hg0.368 Cl 0.736
0.368
0.368
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Empirical Formulas from Analyses
Analysis
Hg  73.9%
Cl  26.1%
-assume 100g sample
Hg  73.9 g
Cl  26.1g
-convert grams to moles
Hg  73.9g / 200.59g/mol = 0.368 mol
Cl  26.1g/ 35.45g/mol = 0.736 mol
-determine the empirical formula by using the moles of the
elements to get the smallest whole number ratio of the elements.
Hg1Cl2
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Empirical Formulas from Analyses
Determine the empirical formula of the compound with the following
compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.
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Empirical Formulas from Analyses
Determine the empirical formula of the compound with the following
compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.
Analysis:
C  10.4%
S  27.8%
Cl  61.7%
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Empirical Formulas from Analyses
Determine the empirical formula of the compound with the following
compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.
Assume 100g sample
C  10.4g
S  27.8g
Cl  61.7g
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Empirical Formulas from Analyses
Determine the empirical formula of the compound with the following
compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.
Moles of each element
C  10.4g/12.011g/mol = 0.866 mol
S  27.8g/32.066g/mol = 0.867 mol
Cl  61.7g/35.453g/mol = 1.74 mol
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Empirical Formulas from Analyses
Determine the empirical formula of the compound with the following
compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.
Moles of each element
C  10.4g/12.011g/mol = 0.866 mol
S  27.8g/32.066g/mol = 0.867 mol
Cl  61.7g/35.453g/mol = 1.74 mol
C0.866 S0.867Cl1.74
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Empirical Formulas from Analyses
Determine the empirical formula of the compound with the following
compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.
Moles of each element
C  10.4g/12.011g/mol = 0.866 mol
S  27.8g/32.066g/mol = 0.867 mol
Cl  61.7g/35.453g/mol = 1.74 mol
C0.866 S 0.867 Cl 1.74
0.866
0.866
Chapter 3
0.866
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Empirical Formulas from Analyses
Determine the empirical formula of the compound with the following
compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%.
Moles of each element
C  10.4g/12.011g/mol = 0.866 mol
S  27.8g/32.066g/mol = 0.867 mol
Cl  61.7g/35.453g/mol = 1.74 mol
C1S1Cl2
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Empirical Formulas from Analyses
Molecular Formula from Empirical Formula
To determine the molecular formula from an empirical
formula, you must have the molecular weight of the
substance.
m olecularweight
num berof form ulaunits
em pirical form ulaweight
molecularformula numberof formulaunitsempirical formula
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Empirical Formulas from Analyses
Molecular Formula from Empirical Formula
Empirical formula: CH
Empirical formula weight: 13.019 g/mol
Molecular weight: 78.114g/mol
78.114g / m ol
form ulaunits
13.019g / m ol
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Empirical Formulas from Analyses
Molecular Formula from Empirical Formula
Empirical formula: CH
Empirical formula weight: 13.019 g/mol
Molecular weight: 78.114g/mol
78.114g / m ol
form ulaunits
6
13.019g / m ol
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Empirical Formulas from Analyses
Molecular Formula from Empirical Formula
Empirical formula: CH
Empirical formula weight: 13.019 g/mol
Molecular weight: 78.114g/mol
78.114g / m ol
form ulaunits
6
13.019g / m ol
m olecularform ula 6C1 H1 
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Empirical Formulas from Analyses
Molecular Formula from Empirical Formula
Empirical formula: CH
Empirical formula weight: 13.019 g/mol
Molecular weight: 78.114g/mol
78.114g / m ol
form ulaunits
6
13.019g / m ol
m olecularform ula 6C1 H 1 
 C6 H 6
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End of Chapter Problems
6, 8, 12, 14, 20, 22, 28, 32, 36, 42, 48, 54, 68, 86
Chapter 3
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