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Chapter 3: Molecules, Compounds, & Chemical Equations
I) Chemical Bonds
1) Compounds are composed of atoms held together by chemical bonds.
2) The interaction of electrons between elements is what results in chemical
bonding.
3) Two broad types of chemical bonds.
A) Ionic Bonding involves the transfer of electrons from a metal to a
nonmetal.
B) Covalent Bonding involves the sharing of electrons between nonmetal
atoms.
II) Chemical Formulas: Representing Compounds
1) Compounds are most easily represented using chemical formulas for the
individual constituent elements and whole number subscripts to denote how
many atoms of that element are present.
2) Types of Chemical Formulas
A) empirical formula - simplest whole # ratio of atoms in a compound.
B) molecular formula
i) true formula for a compound
ii) actual whole # ratio of atoms in a compound
ii) usually a multiple of the empirical formula
C) structural formula
i) shows connectivity of atoms using dashed lines to denote
covalent bonds
ii) be able to distinguish between single bonds ( ─ ), double bonds
( = ), and triple bonds ( ≡ )
3) Molecular Models
A) Ball-and-Stick models represent atoms as spheres and chemical bonds
as sticks; how the two connect reflects a molecule’s shape.
B) Space-filling models are so named because the atoms fill the space
between them and are our best estimate of how a molecule might
appear to us.
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III) Classifying Matter Revisited: An Atomic Level View
1) Review Figure 3.4 as well as definitions for pure substance, element, and
compound.
2) Atomic vs. Molecular Elements
A) Atomic elements are those that exist in nature with single atoms as
their base units. Most elements fall into this category.
B) Molecular elements exist as molecules in nature and not as atoms.
These are called the diatomic molecules (H2, N2, O2, F2, Cl2, Br2, I2)
and the polyatomic molecules (P4, S8, Se).
3) Covalent Bonds: Molecular Compounds
A) Covalent bonding results when electrons are shared between nonmetal
atoms.
B) Molecule - two or more covalently bonded atoms.
C) Molecular compounds have covalently bonded molecules as their
building blocks
4) Ionic Bonds: Ionic Compounds
A) An ionic bond results from a transfer of electrons from one atom to
another. The transfer, usually from the metal to a nonmetal, results in the
formation of ions.
B) ion - charged atom or group of atoms. There are two types of ions.
i) Cations:
ii) Anions:
lose electrons
gain electrons
(+) ion
(- ) ion
metals
nonmetals
C) The basic unit for an ionic compound is the formula unit which is the
smallest electrically neutral collection of ions.
D) A polyatomic ion is a charged group of covalently bonded atoms.
IV) The Language of Chemistry: Naming Compounds
1) It is imperative to remember that compounds are chemically neutral; that
means that all the positive charges must be equally balanced with all negative
charges that are present.
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2) Binary Ionic Compounds
A) Binary Compounds have two parts.
B) To name a binary ionic compound
i)
ii)
iii)
First identify the positive ion and the negative ion.
The positive ion, usually a metal cation, is named first and
is the same as the name of the element.
The negative ion, usually a nonmetal, is named second.
Here the name of the nonmetal element is used, but the
ending is changed to –ide.
C) Examples
i) KCl
ii) CaF2
iii) BaO
potassium chloride
calcium fluoride
barium oxide
D) Study Tables 3.2 through 3.5 and commit these to memory!
3) Ionic Compounds with a Transition Metal Cation
A) Some metals form more than one type of cation. These are
typically found in the transition metals.
B) To avoid confusion when naming ionic compounds that contain a
transition metal cation, the charge of the metal cation is enclosed in
parenthesis as a Roman numeral. The remaining rules for naming a
binary ionic compound are the same as shown above.
C) An older system uses the endings –ous and –ic to denote the different
charges on a transition metal. The metal cation with the higher charge
takes the –ic ending.
D) Examples
i)
ii)
FeCl2
FeCl3
Iron (II) chloride
Iron (III) chloride
Ferrous chloride
Ferric chloride
4) Naming Binary Molecular (Covalent) Compounds
A) To name a binary covalent compound
i)
Find which nonmetal element is more cationlike from its
position in the periodic table. This nonmetal is named first
in the same fashion as a metallic cation. The other
nonmetal is named as the anion with the –ide ending.
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ii)
iii)
iv)
Use Greek prefixes to account for # atoms in both
nonmetals.
If first element contains only 1 atom, drop mono prefix.
Be familiar with Greek prefixes (see section 3.6).
B) Examples
i) CS2
ii) SO2
iii) PCl5
iv) N2O4
carbon disulfide
sulfur dioxide
phosphorus pentachloride
dinitrogen tetroxide (dinitrogen tetraoxide)
5) Naming Compounds that contain Polyatomic Cations and Anions
A) Compounds that contain polyatomic cations or anions are named
exactly as shown in Table 3.5.
B) Must commit to memory the name, formula, and charges for the
cations and anions in Tables 3.2 to 3.5.
C) Examples
i) NaNO3
ii) (NH4)2SO4
iii) KC2H3O2
sodium nitrate
ammonium sulfate
potassium acetate
6) Naming Hydrates
A) Hydrates are compounds that contain a specific number of water
molecules (called waters of hydration) associated with each formula unit.
B) Waters of hydration can be removed by heating a hydrate, the resulting
product is the anhydrous salt (anhydrous means without water).
C) Examples
i) Na2CrO4 * 4H2O
ii) BaCl2 * 2H2O
sodium chromate tetrahydrate
barium chloride dihydrate
V) Acids & Bases: An Introduction
1) An acid is a substance that produces hydrogen ions (H+ ions) when dissolved
in water.
A) Binary acids named using the form hydro + nonmetal + -ic ending.
B) Examples
i) HCl
ii) HF
hydrochloric acid
hydrofluoric acid
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2) Most acids are oxyacids meaning that they contain hydrogen and an oxyanion
(an anion containing a nonmetal and oxygen).
A) Common oxyanions listed in Table 3.5.
B) Review oxyanions and be familiar with various prefix & suffix rules.
i)
ii)
iii)
The –ate ending is found with the oxoanion that has the
higher number of oxygen atoms.
The –ite ending is found with the oxoanion that has the
lower number of oxygen atoms.
If there are more than two members in a series, then one
must also be familiar with when to use the prefix per- and
when to use the prefix hypo-.
C) Naming oxyacids
i)
If oxyanion base name ends in –ate then the acid name
takes the form base name of oxyanion + -ic acid.
Examples:
ii)
HNO3
H2SO4
nitric acid
sulfuric acid
If oxyanion base name ends in –ite then the acid name
takes the form base name of oxyanion + -ous acid.
Examples:
HNO2
H2SO3
nitrous acid
sulfurous acid
3) A base is a substance that produces hydroxide ions (OH- ions) when dissolved
in water.
A) Bases are often named as metal hydroxides (the word hydroxide
appears in the name).
B) Examples:
i) lithium hydroxide
ii) sodium hydroxide
iii) potassium hydroxide
LiOH
NaOH
KOH
VI) Formula Mass & Mole Concept for Compounds
1) Recall from chapter 2 that the formula mass (FM) represents the sum of
atomic masses of all atoms in a formula unit of any compound (ionic or
molecular).
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2) Other terms that are equivalent to formula mass include molecular mass,
molecular weight, and formula weight. Each of these refer to the same
calculation.
3) To calculate formula mass (FM), one uses the following formula.
FM = Σ (# atoms of each element in formula * atomic mass each element)
4) Example
Formula Mass of Water (H2O):
FM = (2 x 1.01 amu) + (16.00 amu) = 18.02 amu
Formula Mass Carbon Dioxide (CO2):
FM = (12.01 amu) + (2 x 16.00 amu) = 44.01 amu
5) Recall also from chapter 2 that chemists rely on the mole as the basis for
measuring the amounts of substances.
1 mole of X = NA units
(units = atoms, ions, molecules)
NA = Avogadro’s Number = 6.022 x 1023 units/mole
mass of 1 mole of a substance X = molar mass of X in grams.
Molar mass is equal to formula (molecular) mass of X in
grams. Molar mass values are expressed in grams/mole.
VII) Composition of Compounds
1) Ways of Describing Composition of a Compound
A) number of atoms in chemical formula
B) % by mass of elements (called a mass % or weight %)
mass % of element X = (mass of element X in compound / total mass) * 100%
2) Example Calculations
A) Ethyl Alcohol
C2H5OH
? % C, % H, and % O in compound
1 mole ethyl alcohol = 2 mole C = 6 mole H = 1 mole O
MM: [(2 mol x 12.01 g/mol) + (6 mol x 1.008 g/mol) + (1 mol x 16.00g/mol)]
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MM ethyl alcohol = [ 24.02 g + 6.048 g + 16.00 g] = 46.07 g === 1 mole
MM ethyl alcohol = 46.07 g /mole
46.07 g == 1 mole
Mass % = (mass X / total mass) * 100
% C = (24.02 g / 46.07 g) * 100 % = 52.14% C
% H = (6.048 g / 46.07 g) * 100% = 13.13% H
%O = (16.00 g / 46.07 g) * 100 % = 34.73% O
100.00 %
B) Carvone
MM of Carvone
C10H14O
? %, C, H, and O
1 mole Carvone = 10 mol C = 14 mol H = 1 mol O
MM = [(10 mol C x 12.01 g/mol) + (14 mol H x 1.008 g/mol) + (1 mol O x 16.00 g/mol)]
MM Carvone = [ 120.1 g + 14.11 g + 16.00 g] = 150.2 g Carvone === 1 mole
MM Carvone = 150.2 g /mole
150.2 g === 1 mole
% C = (120.1 g / 150.2 g) * 100% = 79.96 % C
% H = (14.11 g / 150.2 g) * 100% = 9.394% H
% O = (16.00 g / 150.2 g) * 100% = 10.65% O
100.00%
3) The mass percent composition of an element in a compound is a conversion
factor between the mass of the element and mass of the compound.
4) Always keep in mind that percent means per hundred, so we know for example
that in 100.00 g of Carvone, there are 79.96 g of carbon.
100.00 g Carvone : 79.96 g carbon
5) Since chemical formulas contain within them inherent relationships between
atoms (or moles of atoms) and molecules (or moles of molecules), we can use
these to find mass. The general approach often uses the following concept map.
Mass compound → moles compound → moles element → mass element
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VIII) Determining Formula of a Compound
1) Terminology
A) empirical formula - simplest whole # ratio of atoms in a compound.
B) molecular formula
i) true formula for a compound.
ii) usually a multiple of the empirical formula
iii) must be given MM to figure out the multiple factor.
2) Methods for Determining Formulas from % Composition
A) Example Problem
Given & Unknowns:
71.65 % Cl
24.27 % C
4.07% H
Solutions:
MM of compound = 98.96 g/mol
? empirical formula
? molecular formula
Two Methodologies
Method 1: Traditional Approach
(Works Every Single Time)
Step 1: Assume 100.00 g sample of compound (Why?)
[% s = mass of element (g)]
71.65 g Cl x ( 1 mol Cl / 35.45 g Cl) = 2.021 mol Cl
24.27 g C x ( 1 mol C / 12.01 g C) = 2.021 mol C
4.07 g H ( 1 mol H / 1.008 g H) = 4.04 mol H
Note: Although one expresses the intermediate # moles to the proper # sig. figs., always
remember the DNRUD principle!!!
Step 2: Divide # moles obtained from step 1 by smallest # moles present. This gives the
number of atoms present.
2.021 mol Cl / 2.021 mol = 1.000 Cl atoms
2.021 mol C / 2.021 mol = 1.000 C atoms
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4.04 mol H / 2.021 mol = 2.00 H atoms
Note: If one obtains a value that is very close to the next whole # (e.g., 1.98 or 2.03),
then one may round those values to the next whole number. DO NOT round
values like 2.5, 3.67, 1.33, etc. to the nearest whole number....
Step 3: Check to see that # atoms from step 2 is a whole # for each entry. If so, then one
has the empirical formula directly. If the values are not whole #s, one must now
introduce a factor to convert the values to whole #s.
Since # atoms above are all whole #s, the empirical formula is CH2Cl
Step 4: Using the empirical formula obtained in step 3, find empirical MM (molecular
mass corresponding to the given empirical formula). From that value, one finds
molecular formula by using the molecular MM (must be given in problem) as
shown in the following equation
empirical MM (x) = molecular MM
where x is the factor that must be used to go from empirical formula to molecular
formula.
Empirical MM = [ (1 x 12.01 g/mol) + (2 x 1.008 g/mol) + (1 x 35.45 g/mol)]
Empirical MM = (12.01 g/mol + 2.016 g/mol + 35.45 g/mol) = 49.48 g/mol
49.48 g/mol (x) = 98.96 g/mol
x = 2.00
Step 5: Once the factor (x) is known, simply plug into equation shown below and write
out the molecular formula
Molecular formula = (Empirical formula) * x.
Molecular Formula = (CH2Cl) * 2 = C2H4Cl2 = Answer
_____________________________________________________________________
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Method 2:
Combustion Analysis involves burning a sample in the presence of pure
oxygen and isolating the products CO2 and H2O. The mass of each product is obtained
and from this data the empirical formula is evaluated.
Step 1:
Write as given the masses of each combustion product (CO2 and H2O)
along with mass of sample if it is provided.
Step 2:
Convert mass of CO2 and H2O to moles using molar mass values for CO2
and H2O.
Step 3:
Convert moles of CO2 and H2O from step 2 to moles of C and H using
chemical formulas for CO2 and H2O.
Step 4:
If the compound contains an element other than C and H, find the mass of
the other element by subtracting the sum of the masses of C and H
(obtained in step 3) from the mass of the sample. Finally convert the mass
of the other element to moles.
Step 5:
Divide # moles obtained for each element by smallest # moles present.
This gives the number of atoms present.
Step 6:
Check to see that # atoms from step 5 is a whole # for each entry. If so,
then one has the empirical formula directly. If the values are not whole #s,
one must now introduce a factor to convert the values to whole #s.
Review Example Problem 3.19 & 3.20
IX) Chemical Reactions:
1. Components of a Chemical Equation
A) Reactants (left side)
B) Arrow
(produces or yields)
C) Products (right side)
D) Physical States
(s) = solid, (l) = liquid, (g)=gas, (aq) = aqueous
E) Other information (catalysts, temperature, pressure, etc.)
2) Example Reaction
A) Combustion of Methane CH4 (g) + O2 (g) ------> CO2 (g) + H2O (g)
3) Characteristics Shared by All Reactions
A) Bonds are broken; atoms rearrange; new bonds are formed
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B) Conservation of Matter must hold true.
i) All reactions must be balanced!
ii) Balancing a chemical reaction involves finding out how many
formula units of each different substance takes part in the
reaction. A formula unit is one unit, (atom, molecule, or ion),
corresponding to a given formula.
Most simple chemical reactions can be balanced by inspection.
Ex:
CH4 (g) + 2 O2 (g) ------> CO2 (g) + 2 H2O (g)
1C
4H
4O
1C
4H
4O
iii) Use coefficients (written in front of reactant or product) to
balance the reaction.
iv) Only coefficients can be changed when balancing an
equation; the formulas themselves can not be changed.
v) Any species that lacks a coefficient is ALWAYS understood to
be 1. The coefficient 1 is NEVER written in a chemical reaction...
vi) Get into habit of checking your reactions to see that all
atoms are balanced.
4) Reaction Classification
(General Scheme)
A) 4 Main Classes of Reactions
(A + B  AB)
(AB  A + B)
i) Synthesis (Combination)
ii) Decomposition
-
heat required for decomposition reactions.
iii) Single Replacement
iv) Double Replacement
(A + BC ------> AC + B)
(AB + CD ------> AD + CB)
- also termed metathesis reaction.
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5) Methodology for Balancing Chemical Reactions
A) Write out reaction correctly, including all necessary information about
the reaction. Determine what is occurring.
B) Construct the table listing the individual elements and how many of
each species is initially present.
C) Balance the reaction using whole number coefficients. Keep track of
how many atoms are present once the coefficients are changed
(coefficients affect the entire reactant or product species).
D) As a general rule, start with most complicated molecule or an item that
appears by itself on either side of the reaction.
E) Examples
C2H5OH (l) + O2 (g) -------> CO2 (g) + H2O (g)
2C
6H
7O
i) Balance C first
ii) Balance H next
iii) Balance O next
2C
6H
7O
(need 2 in front of C product)
(need 3 in front of H product)
(need 3 in front of oxygen reactant)
C2H5OH (l) + 3 O2 (g) -------> 2 CO2 (g) + 3 H2O (g)
iv) check to make sure it is balanced
2C
6H
7O
2C
6H
7O
Example 2
heat
(NH4)2Cr2O7 (s)  Cr2O3 (s) + N2 (g) + H2O (g)
2N
2 Cr
7O
8H
2N
2 Cr
4O
2H
i) Cr and N are already balanced, so leave them alone.
ii) Balance H next (place 4 in front of water)
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iii) O is balanced once coefficient is placed in front of water.
heat
(NH4)2Cr2O7 (s)  Cr2O3 (s) + N2 (g) + 4 H2O (g)
iv) Check your answer!
2N
2 Cr
7O
8H
2N
2 Cr
7O
8H
X) Introduction to Organic Compounds
1) Combustion reactions involve an organic compound reacting with oxygen to
form carbon dioxide and water.
2) Organic chemistry investigates the chemistry of carbon containing
compounds.
A) Carbon is essential to all life.
B) Carbon is unique because it has the ability to bond with itself forming
chain, branched, and ring structures.
C) Carbon always forms four bonds.
3) Organic compounds that contain only carbon and hydrogen are called
hydrocarbons.
A) Alkanes contain only C-C single bonds.
B) Alkenes contain one C=C double bond.
C) Alkynes contain one C ≡ C triple bond.
4) Functionalized hydrocarbons contain a functional group which is a
characteristic atom or group of atoms that are incorporated into the
hydrocarbon.
5) The functional group gives rise to the unique chemistry found in each organic
family.
6) Know Table 3.8 and be able to recognize organic functional groups.