Transcript Chapter 3
Chapter 3
Chemical Compounds
3.1 Types of Chemical Compounds
and Their Formulas
Molecular Compounds
• Molecular compounds are
composed of molecules and
contain only nonmetals.
• Electrons are shared
• AKA Covalent Compounds
• Represented by chemical formulas
3
The ratio of the masses of carbon and hydrogen,
C:H in methane is
1. 4:1
2. 1:4
3. 3:1
4. 1:3
5. 1:1
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The ratio of the masses of carbon and hydrogen,
C:H in methane is
1. 4:1
2. 1:4
3. 3:1
4. 1:3
5. 1:1
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The ratio of the masses of carbon and hydrogen,
C:H in ethane is
1. 4:1
2. 1:4
3. 3:1
4. 1:3
5. 1:1
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The ratio of the masses of carbon and hydrogen,
C:H in ethane is
1. 4:1
2. 1:4
3. 3:1
4. 1:3
5. 1:1
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Ions
• When atoms lose or gain electrons, they become
ions.
– Cations are positive and are formed by elements on
the left side of the periodic chart.
– Anions are negative and are formed by elements on
the right side of the periodic chart.
8
Ionic Bonds
Ionic compounds (such as NaCl) are generally formed
between metals and nonmetals.
11
A mole of solid sodium chloride, salt, contains
1. 22.99 g of sodium and 34.45 g of chlorine
A
B
2. 6.02x1023 g of sodium and 6.02x1023 g of
chloride
3. 22.99 u of sodium and 34.45 u of chloride
4. 22.99 g of sodium and 34.45 g of chloride.
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A mole of solid sodium chloride, salt, contains
1. 22.99 g of sodium and 34.45 g of chlorine
A
B
2. 6.02x1023 g of sodium and 6.02x1023 g of
chloride
3. 22.99 u of sodium and 34.45 u of chloride
4. 22.99 g of sodium and 34.45 g of chloride.
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A molecule of solid sodium chloride, salt, contains
1. 22.99 g of Na and 34.45 g of Cl
A
B
2. 22.99 u of Na+ and 34.45 u of Cl3. 22.99 g of Na and 34.45 g of Cl
4. None of the above
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A molecule of solid sodium chloride, salt, contains
1. 22.99 g of sodium and 34.45 g of chlorine
A
B
2. 22.99 u of sodium and 34.45 u of chloride
3. 22.99 g of sodium and 34.45 g of chloride.
4. None of the above; sodium chloride is an
ionic compound, as such, it does not exist
in discrete molecular units as is
characteristic of covalent compounds.
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Would you expect the following to be ionic or
molecular:
• N2O
• Na2O
• CaCl2
• SF4
• CBr4
• FeS
• P4O6
• PbF2
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Types of Formulas
• Empirical formulas give the lowest wholenumber ratio of atoms of each element in a
compound.
• Molecular formulas give the exact number of
atoms of each element in a compound.
• Structural Formulas show the order in which
atoms are bonded
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Types of Formulas
• Structural formulas show the
order in which atoms are
bonded.
• Perspective drawings also show
the three-dimensional array of
atoms in a compound.
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Types of Formulas
• Formula Unit: the smallest neutral collection
of ions (most reduced, for ionic)
For the ball and stick model of naphthalene to the
right, the empirical formula is
1. C10H8
2. C4H5
3. C5H4
4. CH
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For the ball and stick model of naphthalene to the
right, the empirical formula is
1. C10H8
2. C4H5
3. C5H4
4. CH
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For the ball and stick model of pyrimidine to the
right, the molecular formula is:
1. C4N2H4
2. C2N2H2
3. C2NH2
4. (CH)4N2
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For the ball and stick model of pyrimidine to the
right, the molecular formula is:
1. C4N2H4
2. C2N2H2
3. C2NH2
4. (CH)4N2
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3.2 The Mole Concept and
Chemical Compounds
Formula Mass vs. Molar Mass vs.
Molecular Mass
• Formula Mass: the mass of a formula unit (amu)
• Molecular Mass: the mass of a molecule (amu)
• Molar Mass: the mass of one mole of a
compound (grams)
H2O: Molecule Mass 18.0153 amu (1 molecule)
Molar Mass 18.0153 grams (1 mole)
Molecular Formulas
Diatomic molecules: H2 O2 N2 F2 Cl2 Br2 I2
Molecules: P4 (White Phosphorus)
S8 (Sulfur)
Distinguish between molecule and atom!
Example 3-1A
How many grams of MgCl2 would you need to
obtain 5.0 x 10-23 Cl- ions?
Example 3-2A
Gold has a density of 19.32 g/cm3. A piece of
gold foil is 2.50 cm on each side and 0.100mm
thick. How many atoms of gold are in this piece
of gold foil?
Example 3-3A
Halothane: C2HBrClF3
How many grams of Br are contained in 25.00
mL of halothane (d=1.871g/mL)
3.3 Composition of Chemical
Compounds
Percent Composition
One can find the percentage of the mass of a
compound that comes from each of the
elements in the compound by using this
equation:
% element =
(number of atoms)(MM of element)
(MM of the compound)
x 100
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Percent Composition
So the percentage of carbon in ethane is…
%C =
(2)(12.0 g/mol)
(30.0 g/mol)
24.0 g/mol
x 100
=
30.0 g/mol
= 80.0%
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Example 3-4A
• What are the mass percent composition in
C10H16N5P3O13?
Calculating Empirical Formulas
One can calculate the empirical formula from the
percent composition
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Establishing Formulas From % Comp
• Assume 100g, % grams
• Grams Moles
• Divide by smallest moles
– Get whole numbers in most reduced form
Ration= MM Molecular/ MM Empirical
Multiply Ratio by Empirical to get Molecular Formula
Example 3-5A
Sorbitol is a sweetener that has a molecular
mass of 182 u and percent composition of
39.56% C
7.74% H
52.70 % O
What are the empirical and molecular formulas?
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen
it listed as PABA on your bottle of sunscreen) is composed of
carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and
oxygen (23.33%). Find the empirical formula of PABA.
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Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
61.31 g x
H:
5.14 g x
N:
10.21 g x
O:
23.33 g x
1 mol
12.01 g
1 mol
1.01 g
1 mol
14.01 g
1 mol
16.00 g
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
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Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of
moles:
C:
5.105 mol
0.7288 mol
= 7.005 7
H:
5.09 mol
0.7288 mol
= 6.984 7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001 2
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Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
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Combustion Analysis
• Compounds containing C, H and O are routinely analyzed
through combustion in a chamber like this
– C is determined from the mass of CO2 produced
– H is determined from the mass of H2O produced
– O is determined by difference after the C and H have been
determined
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Example
The combustion of 5.00 grams of an alcohol
produces 9.55 g of CO2 and 5.87 g of H2O. Find
the empirical formula.
Solution: First you need to find the individual
masses of the elements.
The general equation would look like
this:
CxHyOz + O2 CO2 + H2O
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The mass of the carbon in the CO2 all
came from the alcohol so…
9.55 gCO2 X 12.01 g C
2.61 g C
44.01 gCO2
The mass of the hydrogen in H2O all
came from the alcohol so…
5.87 g H2O X 2.02 g H
0.658 g H
18.02 g H2O
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Because the mass of the oxygen in both
CO2 and H2O is derived from both the
alcohol and the oxygen from combustion
you need to find the mass of oxygen only
in the alcohol….
By subtracting the mass of the carbon and
hydrogen from the total mass of the compound.
5.00 g – (2.61 g + 0.658 g) = 1.73 g O
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Now you can determine the number of moles of
each element in the formula…..
Moles of C = 2.61 g X 1.00 mole C
0.217 mol C
12.01 g C
Moles of H = 0.658 g X 1.00 mole H
0.651 mol H
1.01 g H
Moles of O = 1.73 g O X 1.00 mole O
0.108 mol O
16.00 g O
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From the number of moles we find
the mole ratios,….
C = 0.217/0.108 = 2.01
H = 0.651/ 0.108 = 6.03
O = 0.108/ 0.108 = 1.00
Thus the formula will be C2H6O or C2H5OH
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Practice Problem
1.540 g of an organic acid burns completely to
produce 2.257 g CO2 and 0.9241 g H2O. Find
the empirical formula. If the molecular mass
is 90.0 grams what is the molecular formula?
C 2.257 gCO2 X 12.01/44.01 = 0.6159g
H 0.9241g H2O X 2.02/18.02 = 0.1036g
O 1.540 –(0.6159 + 0.1036) = 0.8205g
C 0.6159/12.01= 0.0513
Empirical formula= CH2O
H 0.1026/ 1.01 = 0.1026
Molecular formula = (CH2O)x
O 0.8205/16.00= 0.05128
X= molecular mass/empirical
mass
X= 90/30= 3
C3H6O3
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Example 3-6B
• Combustion of a 1.505 g sample of thiophene,
a carbon-hydrogen-sulfur compound yields
3.149 g CO2, 0.645 g H2O and 1.146 g SO2 as
the only combustion products. What is the
empirical formula?
3.4 Oxidation States
Tell the number of electrons gained or lost when
forming compounds
• Handout
Common Oxidation States
Rules for Assigning Oxidation States (OS)
1. The OS of a free element is 0
2. The total OS of all atoms in a compound is 0
3. The total OS of all atoms in a ion is equal to the
charge of the ion
4. Group 1 = +1, Group 2 = +2
5. Fluorine = -1
6. Hydrogen usually +1
7. Oxygen usually -2
8. In Binary Ionic : Group 17 = -1, Group 16 = -2,
Group 15 = -3
Exceptions
• Rule 6: H bonded to metals; LiH, NaH, CaH2
• Rule 7: O-F Bonds OF2, H2O2, and KO2
Example
What are the oxidation states of:
S8
Cr2O72MgCl2
Al2O3
MnO4-
The oxidation number for N in nitric acid,
HNO3, is
1. 1
2. 2
3. 3
4. 4
5. 5
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The oxidation number for N in nitric acid,
HNO3, is
1. 1
2. 2
3. 3
4. 4
5. 5
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3.5 Naming Compounds: Organic
and Inorganic Compounds
Cations
1. Cations from metal atoms have the
same name as the metal
Na+ sodium ion
Ba2+ barium ion
2. Use Roman numerals to indicate the
charge on a metal ion that can form more
than one positive charge
Fe2+ Iron (II) ion Fe3+ Iron(III) ion
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Cations
Latin names for some metals still are used
and the ending –ous and –ic are used to
indicate the charge
Fe2+ ferrous ion
Sn2+ stannous ion
Fe3+ ferric
Sn4+ stannic ion
Other latin root names include:
Plumbous
Auric
Cuprous
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Cations
Cations from nonmetal atoms end in-ium
NH4+ ammonium ion
H3O+ hydronium ion
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Anions
Monatomic anions are named are named
by adding –ide at the end
H- Hydride S2- sulfide
P3- phosphide
Some polyatomics have –ide endings
OH- hydroxide
CN- cyanide
O22- peroxide
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Anions
Polyatomic anions containing oxygen
end in –ite or -ate
The –ate ending is used for the most common
oxyanion of the element . The –ite ending is
used for the oxyanion that has the same charge
but one O atom less.
NO3- nitrate
SO42- sulfate
NO2- nitrite
SO32- sulfite
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Anions
Prefixes are used when there is a series of four
oxyanions. (usually the halogens)
Per- is used to indicate one more O than the –ate
ending and hypo- is used for one less O than the
–ite ending.
ClO4- perchlorate
ClO3- chlorate
ClO2- chlorite
ClO- hypochlorite
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Anions
Anions with H+ are named by adding the
prefix hydrogen or dihydrogen
HCO3- hydrogen carbonate (bicarbonate)
HSO4- dihydrogen sulfate (bisulfate)
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Writing Formulas
• Because compounds are electrically neutral, one
can determine the formula of a compound this
way:
– The charge on the cation becomes the subscript on the
anion.
– The charge on the anion becomes the subscript on the
cation.
– If these subscripts are not in the lowest whole-number
ratio, divide them by the greatest common factor.
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Potassium dichromate is used in breathalyzers.
When it comes in contact with alcohol vapor it
turns from orange to green. It is an ionic
compound where the polyatomic anion has the
formula Cr2O72-. What is the chemical formula
for potassium dichromate?
1. KCr2O7
2. K(Cr2O7)2
3. K2Cr2O7
4. K2(Cr2O7)3
5. K3(Cr2O7)2
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Potassium dichromate is used in breathalyzers.
When it comes in contact with alcohol vapor it
turns from orange to green. It is an ionic
compound where the polyatomic anion has the
formula Cr2O72-. What is the chemical formula
for potassium dichromate?
1. KCr2O7
2. K(Cr2O7)2
3. K2Cr2O7
4. K2(Cr2O7)3
5. K3(Cr2O7)2
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Sodium phosphate is an active component in
some constipation medicines and enemas. The
chemical formula for sodium phosphate is
1. NaPO4
2. Na2PO3
3. Na2PO4
4. Na3PO4
5. Na3(PO4)2
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Sodium phosphate is an active component in
some constipation medicines and enemas. The
chemical formula for sodium phosphate is
1. NaPO4
2. Na2PO3
3. Na2PO4
4. Na3PO4
5. Na3(PO4)2
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Marble consists of primarily of calcium carbonate.
Acids react with and dissolve marble as evident
by this marble statue eroded by acid rain. The
chemical formula for calcium carbonate is
1. CaCO3
2. Ca2CO3
3. CaCO4
4. Ca3(CO3)2
5. Ca(CO3)2
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Marble consists of primarily of calcium carbonate.
Acids react with and dissolve marble as evident
by this marble statue eroded by acid rain. The
chemical formula for calcium carbonate is
1. CaCO3
2. Ca2CO3
3. CaCO4
4. Ca3(CO3)2
5. Ca(CO3)2
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Common Cations
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Common Anions
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Inorganic Nomenclature
• Write the name of the cation.
• If the anion is an element, change its
ending to -ide; if the anion is a polyatomic
ion, simply write the name of the
polyatomic ion.
• If the cation can have more than one
possible charge, write the charge as a
Roman numeral in parentheses.
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Name the following
compounds:
(a)K2SO4
(b) Ba(OH)2
(c) FeCl3
Write the chemical formulas for the
following compounds:
(a)potassium sulfide,
(b) calcium hydrogen carbonate,
(c) nickel(II) perchlorate.
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Acid Nomenclature
• If the anion in the acid ends in -ide,
change the ending to -ic acid and add the
prefix hydro- :
– HCl: hydrochloric acid
– HBr: hydrobromic acid
– HI: hydroiodic acid
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Acid Nomenclature
• If the anion in the acid ends in -ite, change the
ending to -ous acid:
– HClO: hypochlorous acid
– HClO2: chlorous acid
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Acid Nomenclature
• If the anion in the acid ends in -ate, change
the ending to -ic acid:
– HClO3: chloric acid
– HClO4: perchloric acid
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Nomenclature of Binary Covalent
Compounds
• The less electronegative
atom is usually listed first.
• A prefix is used to denote
the number of atoms of
each element in the
compound (mono- is not
used on the first element
listed, however.)
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Nomenclature of Binary Compounds
• The ending on the more
electronegative element is
changed to -ide.
– CO2: carbon dioxide
– CCl4: carbon tetrachloride
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Nomenclature of Binary Compounds
If the prefix ends with a or
o and the name of the
element begins with a
vowel, the two successive
vowels are often elided
into one:
N2O5: dinitrogen pentoxide
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3.6 Names and Formulas of
Inorganic Compounds
Organic compounds
• Alkanes-Carbon
atoms bonded to
four other atoms
• Methane CH4
• Ethane C2H6
• Propane C3H8
• Butane C4H10
• Alkanes with 5 or more
carbons use the
following prefixes
• Penta
• Hexa
• Hepta
• Octa
• Nona
• Deca
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Organic functional groups
• An alcohol is an
alkane with an –OH
group and is named
by adding an –ol
ending
• Methanol
• In naming the
carbon with the
functional group is
identified by a
number
• 2-Propanol
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