Download Prentice Hall Ch 02 Atoms Molecules Ions

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Chapter Two
Atoms, Molecules, and Ions
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Two
2
Laws of Chemical Combination
• Law of Conservation of Mass
– The total mass remains constant during a
chemical reaction.
• Law of Definite Proportions
– All samples of a compound have the same
composition, or …
– All samples have the same proportions, by
mass, of the elements present.
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Chapter Two
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Example 2.1 A Conceptual Example
Jan Baptista van Helmont (1579–1644) first measured
the mass of a young willow tree and, separately, the
mass of a bucket of soil and then planted the tree in the
bucket. After five years, he found that the tree had
gained 75 kg in mass even though the soil had lost only
0.057 kg. He had added only water to the bucket, and
so he concluded that all the mass gained by the tree
had come from the water. Explain and criticize his
conclusion.
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Chapter Two
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Example 2.1 A Conceptual Example
Jan Baptista van Helmont (1579–1644) first measured the mass of a young willow tree
and, separately, the mass of a bucket of soil and then planted the tree in the bucket. After
five years, he found that the tree had gained 75 kg in mass even though the soil had lost
only 0.057 kg. He had added only water to the bucket, and so he concluded that all the
mass gained by the tree had come from the water. Explain and criticize his conclusion.
Analysis and Conclusions
Van Helmont’s explanation anticipated the law of conservation of mass by
dismissing the possibility that the increased mass of the tree could have been
created from nothing. His main failure, however, was in not identifying all the
substances involved. He did not know about the role of carbon dioxide gas in the
growth of plants. In applying the law of conservation of mass, we must focus on all
the substances involved in a chemical reaction.
Exercise 2.1A
A sealed photographic flashbulb containing magnesium and oxygen has a mass of
45.07 g. On firing, a brilliant flash of white light is emitted and a white powder is formed
inside the bulb. What should the mass of the bulb be after firing? Explain.
Exercise 2.1B
Is the mass of the burned match pictured in the photograph the same as, less than, or
more than the mass of the burning match? Explain your answer.
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Chapter Two
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The Law of Definite Proportions
… have the same
composition.
Three different sources of a compound …
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Chapter Two
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Example 2.2
The mass ratio of oxygen to magnesium in the
compound magnesium oxide is 0.6583:1. What mass of
magnesium oxide will form when 2.000 g of magnesium
is completely converted to magnesium oxide by burning
in pure oxygen gas?
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Chapter Two
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Example 2.2
The mass ratio of oxygen to magnesium in the compound magnesium oxide is 0.6583:1.
What mass of magnesium oxide will form when 2.000 g of magnesium is completely
converted to magnesium oxide by burning in pure oxygen gas?
Strategy
We usually write mass ratios in a form such as “the ratio of O to Mg is 0.6583:1.”
The first number represents the mass of the first element named—in this case, a
mass of oxygen, say 0.6583 g oxygen—and the second number represents the mass
of the second element named—here a mass of magnesium. Although written only as
“1,” we assume that the second number is as precisely known as the first, that is,
1.0000 g magnesium. According to the law of constant composition, the mass of
oxygen that combines with the 2.000 g of magnesium must be just the right amount
that makes the mass ratio of oxygen to magnesium in the product 0.6583:1.
According to the law of conservation of mass, the mass of magnesium oxide must
equal the sum of the masses of the magnesium and oxygen that react.
Solution
We can state the mass ratio in the form of a conversion factor and then determine
the required mass of oxygen.
The mass of the sole product, magnesium oxide, equals the total of the masses of
the substances entering into the reaction.
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Chapter Two
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Example 2.2 continued
Assessment
As a simple check, compare the calculated mass with the starting mass of
magnesium. From the law of conservation of mass, the product of the reaction
of magnesium with oxygen must have a mass greater than the starting mass of
magnesium; and 3.317 g magnesium oxide is indeed greater than 2.000 g
magnesium.
Exercise 2.2A
What mass of magnesium oxide is formed when 1.500 g of oxygen combines with
magnesium?
Exercise 2.2B
When a strip of magnesium metal was burned in pure oxygen gas, 1.554 g of oxygen was
consumed and the only product formed was magnesium oxide. What must have been the
masses of magnesium metal burned and magnesium oxide formed?
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Chapter Two
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Law of Multiple Proportions
When two or more different compounds of
the same two elements are compared, the
masses of one element that combine with a
fixed mass of the second element are in the
ratio of small whole numbers.
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Law of Multiple Proportions (cont’d)
Ratio of oxygen-to-carbon in CO2 is
exactly twice the ratio in CO.
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Law of Multiple Proportions (cont’d)
• Four different oxides of nitrogen can be formed by
combining 28 g of nitrogen with:
• 16 g oxygen, forming Compound I
• 48 g oxygen, forming Compound II
• 64 g oxygen, forming Compound III
• 80 g oxygen, forming Compound IV
What is the ratio 16:48:64:80
expressed as small whole numbers?
• Compounds I–IV are N2O, N2O3, N2O4, N2O5
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Chapter Two
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Dalton’s Atomic Theory
Proposed in 1803 to explain the law of conservation
of mass, law of definite proportions, and law of
multiple proportions.
•
•
•
•
•
Matter is composed of atoms: tiny, indivisible particles.
All atoms of a given element are the same.
Atoms of one element differ from atoms of other
elements.
Compounds are formed when atoms of different elements
unite in fixed proportions.
A chemical reaction involves rearrangement of atoms.
No atoms are created, destroyed, or broken apart.
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Chapter Two
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Dalton’s Atomic Theory: Conservation
of Mass and Definite Proportions
Six fluorine atoms and four
hydrogen atoms before reaction …
… six fluorine atoms and four
hydrogen atoms after reaction.
Mass is conserved.
HF always has one H atom
and one F atom; always
has the same proportions
(1:19) by mass.
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Chapter Two
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Subatomic Particles
• Protons and neutrons are located at the center of
an atom (at the nucleus).
• Electrons are dispersed around the nucleus.
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Isotopes
• Atoms that have the same number of protons but
different numbers of neutrons are called isotopes.
• The atomic number (Z) is the number of protons
in the nucleus of a given atom of a given element.
• The mass number (A) is an integral number that
is the sum of the numbers of protons and neutrons
in an atom.
• The number of neutrons = A – Z.
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Isotopes (cont’d)
Atoms can be represented using the element’s
symbol and the mass number (A) and atomic
number (Z):
A
E
Z
35
Cl
17
37
Cl
17
• How many protons are in chlorine-35?
• How many protons are in chlorine-37?
• How many neutrons are in chlorine-37?
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Example 2.3
How many protons, neutrons, and electrons are present
in a 81Br atom?
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Example 2.3
How many protons, neutrons, and electrons are present in a 81Br atom?
Strategy
We can use the identity of the element and several simple relationships between the
subatomic particles described above to determine the numbers of these particles.
Solution
Prentice Hall © 2005
The atomic number of bromine
is not given here, but we can
obtain it from the list of elements
on the inside front cover.
Z = 35 = number of protons
In the bromine atom, the number
of positively charged protons
equals the number of negatively
charged electrons.
Number of protons = number of electrons = 35
From mass number 81 and
atomic number 35, we calculate
the number of neutrons, using
Equation (2.1).
Number of neutrons = A – Z = 81 – 35 = 46
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Chapter Two
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Atomic Mass
• Atoms are very tiny, so a tiny unit is needed to
express the mass of an atom or molecule.
• One atomic mass unit (u) = 1/12 the mass of a C12 atom.
• 1 u = 1.66054 × 10–24 g
• The mass of an atom is not exactly the sum of the
masses of the protons + neutrons + electrons (we
will see why in Chapter 19).
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Chapter Two
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Atomic Mass (cont’d)
• Question: do all isotopes of an element have
the same mass? Why or why not?
• The atomic mass given on the periodic table
is the weighted average of the masses of the
naturally occurring isotopes of that element.
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Example 2.4
Use the data cited above to determine the weighted
average atomic mass of carbon.
Example 2.5 An Estimation Example
Indium has two naturally occurring isotopes and a
weighted average atomic mass of 114.82 u. One of the
isotopes has a mass of 112.9043 u. Which is likely to
be the second isotope: 111In, 112In, 114In, or 115In?
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Example 2.4
Use the data cited above to determine the weighted average atomic mass of carbon.
Strategy
The contribution each isotope makes to the weighted average atomic mass is given
by Equation (2.2). The weighted average atomic mass is the sum of the two
contributions.
Solution
The contributions are
Contribution of carbon-12 = 0.98892 x 12.00000 u = 11.867 u
Contribution of carbon-13 = 0.01108 x 13.00335 u = 0.1441 u
The weighted average mass is
Atomic mass of carbon = 11.867 u + 0.1441 u = 12.011 u
Assessment
This is the value listed in a table of atomic masses. As expected, the atomic mass of
carbon is much closer to 12 u than to 13 u.
Exercise 2.4A
There are three naturally occurring isotopes of neon. Their percent abundances and
atomic masses are neon-20, 90.51%, 19.99244 u; neon-21, 0.27%, 20.99395 u; neon-22,
9.22%, 21.99138 u. Calculate the weighted average atomic mass of neon.
Exercise 2.4B
The two naturally occurring isotopes of copper are copper-63, mass 62.9298 u, and
copper-65, mass 64.9278 u. What must be the percent natural abundances of the two
isotopes if the atomic mass of copper listed in a table of atomic masses is 63.546 u?
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Chapter Two
Example 2.5 An Estimation Example
23
Indium has two naturally occurring isotopes and a weighted average atomic mass of
114.82 u. One of the isotopes has a mass of 112.9043 u. Which is likely to be the second
isotope: 111In, 112In, 114In, or 115In?
Analysis and Conclusions
The masses of isotopes differ only slightly from whole numbers, which tells us that
the isotope with mass 112.9043 u is 113In. To account for the observed weighted
average atomic mass of 114.82 u, the second isotope must have a mass number
greater than 114. It can be only 115In.
Exercise 2.5A
The masses of the three naturally occurring isotopes of magnesium are 24Mg,
23.98504 u; 25Mg, 24.98584 u; 26Mg, 25.98259 u. Use the atomic mass given inside the
front cover to determine which of the three is the most abundant. Can you determine
which is the second most abundant? Explain.
Exercise 2.5B
For the three magnesium isotopes described in Exercise 2.5A, (a) could the percent
natural abundance of 24Mg be 60.00%? (b) What is the smallest possible value for the
percent natural abundance of 24Mg?
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Chapter Two
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Mendeleev’s Periodic Table
• Mendeleev arranged the known elements in order
of increasing atomic weight from left to right and
from top to bottom in groups.
• Elements that closely resembled one another were
arranged in the same vertical group.
• Gaps were left where undiscovered elements
should appear.
• From the locations of the gaps, he was able to
predict properties of some of the undiscovered
elements.
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Chapter Two
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Germanium:
Prediction vs. Observation
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The Modern Periodic Table
Except for H,
elements left of
the zigzag line
are metals.
To the right of
the line we find
nonmetals,
including the
noble gases.
Some elements
adjacent to the
line are called
metalloids.
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Chapter Two
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Molecules and Formulas
• A molecule is a group of two or more atoms held
together by covalent bonds.
• A molecular formula gives the number of each kind of
atom in a molecule.
• An empirical formula simply gives the (whole
number) ratio of atoms of elements in a compound.
Compound
Molecular formula Empirical formula
Hydrogen peroxide
H2O2
HO
Octane
C8H18
????
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Structural Formulas and Models
• Structural formulas and models show how atoms
are attached to one another.
The condensed structural formula for acetic acid is
C2H4O2: two C atoms, four H atoms, two O atoms.
CH3COOH.
CH3COOH shows how the atoms are arranged.
Ball-and-stick model
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Space-filling model
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Chapter Two
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Nomenclature
… is the method for naming compounds and writing
formulas for compounds.
• We could have a specific name for each
compound—but we would have to memorize each
one!
– Can you imagine having to memorize the names of half
a million different inorganic compounds? Twenty
million organic compounds??
• Instead we have a systematic method—
conventions and rules—for naming compounds
and writing formulas.
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Nomenclature of
Binary Molecular Compounds
• Binary compounds contain ___ elements.
• Molecular compounds exist as ________.
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Naming Binary Molecular
Compounds
• The name consists of two words.
• First word: name of the element that
appears first in the formula.
• Second word: stem of the name of the
second element, ending with -ide.
• Names are further modified by adding
prefixes to denote the numbers of atoms
of each element in the molecule.
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Which element is named first?
Begin with boron and follow
the line to determine the
order of naming.
Rule of thumb: the element
that is farthest down and to
the left on the periodic table is
usually written first.
In a compound consisting of
arsenic (As) and sulfur (S),
which element is named first?
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Example 2.6
Write the molecular formula and name of a compound
for which each molecule contains six oxygen atoms and
four phosphorus atoms.
Example 2.7
Write (a) the molecular formula of phosphorus
pentachloride and (b) the name of S2F10.
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Example 2.6
Write the molecular formula and name of a compound for which each molecule contains
six oxygen atoms and four phosphorus atoms.
Strategy
We need to write the chemical symbols of the two elements and use the stated
numbers of atoms as subscripts following the symbols. We then determine which
element to place first in the molecular formula.
Solution
We represent the six atoms of oxygen as O6 and the four atoms of phosphorus as P4.
According to the scheme in Figure 2.9, the element O is followed only by F.
Phosphorus therefore comes first in the formula; we write P 4O6.
The name of a compound with four (tetra-) P atoms and six (hexa-) oxygen
atoms in its molecules is tetraphosphorus hexoxide.
Exercise 2.6A
Write the molecular formula and name of a compound for which each molecule contains
four fluorine atoms and two nitrogen atoms.
Exercise 2.6B
Write the molecular formula and name of a compound the molecules of which each
contain one oxygen atom and eight sulfur atoms.
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Example 2.7
Write (a) the molecular formula of phosphorus pentachloride and (b) the name of S2F10.
Solution
(a) Choosing which element symbol goes first. The order of elements shown in the
molecular formula must be the same as the order in the name.
Writing subscripts. The lack of a prefix on phosphorus signifies one P atom per
molecule. The prefix penta- indicates five chlorine atoms. The molecular formula
is PCl5.
(b) The subscripts indicate two (di-) sulfur atoms and ten (deca-) fluorine atoms. The
compound is disulfur decafluoride.
Exercise 2.7A
Write (a) the molecular formula of tetraphosphorus decoxide and (b) the name of S7O2.
Exercise 2.7B
Write a plausible molecular formula for a compound that has one sulfur atom, two
oxygen atoms, and two fluorine atoms in each of its molecules. Comment on any
ambiguity that exists in this case.
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Chapter Two
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Ions and Ionic Compounds
• An atom that either gains or loses electron(s) is
an ion.
• There is no change in the number of protons or
neutrons in the nucleus of the atom.
• Cation – has a positive charge from loss of
electron(s).
• Anion – has a negative charge from gain of
electron(s).
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Chapter Two
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Ions and Ionic Compounds (cont’d)
In an ionic compound,
oppositely charged ions
are attracted to each
other such that the
compound has no net
charge.
There are no
distinct molecules
of sodium chloride.
Sodium chloride simply consists
of sodium ions and chloride ions,
regularly arranged.
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Chapter Two
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Example 2.8
Determine the formula for (a) calcium chloride
and (b) magnesium oxide.
Example 2.9
What are the names of (a) MgS and (b) CrCl3?
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Example 2.8
Determine the formula for (a) calcium chloride and (b) magnesium oxide.
Solution
(a) First we write the symbols for the ions, with the cation first: Ca 2+ and Cl–. The
simplest combination of these ions that gives an electrically neutral formula unit is
one Ca2+ ion for every two Cl– ions. The formula is CaCl2 .
Ca2+ + 2 Cl– = CaCl2
(b) Figure 2.10 tells us that the ions are Mg2+ and O2–. The simplest ratio for an
electrically neutral formula unit is 1:1. The formula of this binary ionic compound
is MgO.
Mg2+ + O2– = MgO
Exercise 2.8A
Give the formula for each of the following ionic compounds:
(a) potassium sulfide
(b) lithium oxide
(c) aluminum fluoride
Exercise 2.8B
Give the formula for each of the following ionic compounds:
(a) chromium(III) oxide
(b) iron(II) sulfide
(c) lithium nitride
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Example 2.9
What are the names of (a) MgS and (b) CrCl3?
Solution
(a) MgS is made up of Mg2+ and S2– ions. Its name is magnesium sulfide.
(b) From Figure 2.10 we see that there are two simple ions of chromium, Cr3+ and
Cr2+. Because there are three Cl– ions in the formula unit CrCl3, the cation must
have a charge of 3+, that is, it must be Cr3+. Because there are two chromium
cations, there are two chlorides, CrCl2 and CrCl3, and we must assign a different
name to each. Therefore, the name of our compound cannot be simply chromium
chloride; instead, to indicate the 3+ cation, we say the name is chromium(III)
chloride.
Exercise 2.9A
Name the following compounds:
(a) CaBr2
(b) Li2S
(c) FeBr2
(d) CuI
Exercise 2.9B
Write the name and formula for each of the following compounds:
(a) the sulfide of copper(I) (b) the oxide of cobalt(III) (c) the nitride of magnesium
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Chapter Two
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Monatomic Ions
• Group IA metals form ions of 1+ charge.
• Group IIA metals form ions of 2+ charge.
• Aluminum, a group IIIA metal, forms ions with a 3+
charge.
• Nonmetal ions of groups V, VI, and VII usually have
charges of 3–, 2–, and 1–, respectively.
• Group B metal ions (transition metal ions) often have
more than one possible charge. A Roman numeral is used
to indicate the actual charge.
• A few transition elements have only one common ion
(Ag, Zn, Cd), and a Roman numeral is not often used.
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Chapter Two
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Symbols and Periodic Table Locations
of Some Monatomic Ions
Copper forms either
copper(I) or copper(II) ions.
Titanium forms both
titanium(II) and
titanium(IV) ions.
What is the charge on a
zirconium(IV) ion?
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Chapter Two
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Formulas and Names of
Binary Ionic Compounds
• Binary ionic compounds are made up of monatomic
cations and anions.
• These combinations must be electrically neutral.
• The formula unit is the simplest collection of cations
and anions that represents an electrically neutral unit.
• Formula unit is to ion as ________ is to atom.
• To write a formula, combine the proper number of each
ion to form a neutral unit.
• To name a binary ionic compound, name the cation, then
the anion.
• Monatomic anion names end in -ide.
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Chapter Two
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Polyatomic Ions
• A polyatomic ion is a charged group of covalently
bonded atoms.
• There are many more polyatomic anions than
there are polyatomic cations.
• You should (eventually!) commit to memory much
of Table 2.4
• hypo- and per- are sometimes seen as prefixes in
oxygen-containing polyatomic ions (oxoanions).
• -ite and -ate are commonly found as suffixes in
oxygen-containing polyatomic ions.
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Chapter Two
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Chapter Two
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Example 2.10
Write the formula for (a) sodium sulfite and (b)
ammonium sulfate.
Example 2.11
What is the name of (a) NaCN and (b) Mg(ClO4)2?
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Example 2.10
Write the formula for (a) sodium sulfite and (b) ammonium sulfate.
Solution
(a) It is possible to identify the sulfite ion without memorizing all the ions in Table 2.4.
If you remember the name and formula of one of the sulfur–oxygen polyatomic
anions, you should be able to deduce the names of others. Suppose you remember
that sulfate is SO42–. The -ite anion has one fewer oxygen atom, 3 instead of 4, and
so it is SO32–. The charges of the two ions in a formula unit must balance, which
means the formula unit of sodium sulfite must have Na + and SO32– in the ratio 2:1.
The formula is therefore Na2SO3.
(b) The ammonium ion is NH4+, and the sulfate ion is SO42–. A formula unit of
ammonium sulfate has two NH4+ ions and one SO42– ion. To represent the two
NH4+ ions, we place parentheses around the NH4, followed by a subscript 2,
(NH4)2, and thus arrive at the formula (NH4)2SO4.
Exercise 2.10A
What is the formula for (a) ammonium carbonate, (b) calcium hypochlorite, and
(c) chromium(III) sulfate?
Exercise 2.10B
Write a plausible formula for
(a) potassium aluminum sulfate
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(b) magnesium ammonium phosphate
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Chapter Two
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Example 2.11
What is the name of (a) NaCN and (b) Mg(ClO4)2?
Solution
(a) The ions in this compound are Na+, sodium ion, and CN–, cyanide ion
(see Table 2.4). The name of the compound is sodium cyanide.
(b) The ions present are Mg2+, magnesium ion, and ClO4–, perchlorate ion. The name
of the compound is magnesium perchlorate.
Exercise 2.11A
Name each of the following compounds:
(a) KHCO3
(b) FePO4
(c) Mg(H2PO4)2
Exercise 2.11B
Give a plausible name for the following:
(a) Na2SeO4
(b) FeAs
(c) Na2HPO3
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Chapter Two
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Hydrates
• A hydrate is an ionic compound in which the formula unit
includes a fixed number of water molecules associated
with cations and anions.
• To name a hydrate, the compound name is followed by
“___hydrate” where the blank is a prefix to indicate the
number of water molecules.
• The number of water molecules associated with each
formula unit is written as an appendage to the formula unit
name separated by a dot.
• Examples: BaCl2 . 2 H2O; CuSO4 . 5 H2O
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Chapter Two
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Hydrates (cont’d)
How many atoms are in one
formula unit of copper(II)
sulfate pentahydrate?
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Chapter Two
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Acids …
• Taste sour, if diluted with enough water to be
tasted safely.
• May produce a pricking or stinging sensation on
the skin.
• Turn the color of litmus or indicator paper from
blue to red.
• React with many metals to produce ionic
compounds and hydrogen gas.
• Also react with bases, thus losing their acidic
properties.
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Chapter Two
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Bases …
• Taste bitter, if diluted with enough water to be
tasted safely.
• Feel slippery or soapy on the skin.
• Turn the color of litmus or indicator paper from
red to blue.
• React with acids, thus losing their basic properties.
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Chapter Two
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Acids and Bases:
The Arrhenius Concept
• There are several definitions which may be used to
describe acids and bases.
• An Arrhenius acid is a compound that ionizes in
water to form a solution of H+ ions and anions.
• An Arrhenius base is a compound that ionizes in
water to form solutions of OH– and cations.
• Neutralization is the process of an acid reacting
with a base to form water and a salt.
• A salt is the combination of the cation from a base
and the anion from an acid.
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Chapter Two
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Chapter Two
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Acid Nomenclature
• Notice that the acid name is related to the
anion name.
–
–
–
–
–
Hydrochloric acid, chloride ion
Hydrosulfuric acid, sulfide ion
Phosphoric acid, phosphate ion
Nitric acid, nitrate ion
Nitrous acid, nitrite ion
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Chapter Two
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Organic Compounds
• Organic chemistry is the study of carbon and its
compounds.
• Carbon compounds can have an almost unlimited
diversity, because carbon atoms can bond to one
another, and to other atoms, to form chains and
rings.
• Carbon compounds containing one or more of the
elements H, O, N, or S are especially common.
• Many organic compounds have common names as
well as systematic names.
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Chapter Two
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Alkanes
• Hydrocarbons are molecules that contain only hydrogen
and carbon atoms.
• Alkanes are saturated (have the maximum number of
hydrogen atoms possible for the number of carbon atoms).
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Chapter Two
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Alkanes
Isomers are compounds
with the same molecular
formula but different
structural formulas.
Alkane molecules with ring
structures are named with the
prefix cyclo- and are called
cycloalkanes.
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Chapter Two
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Propane, used in gas grills,
is an alkane with three
carbon atoms
Butyric acid, which gives
rancid butter its “fragrance,”
contains four carbon atoms.
Octane, a component of
gasoline, is a(n) ______ which
contains _____ carbon atoms.
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Chapter Two
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Types of Organic Compounds
• Many organic compounds contain a functional
group.
• A functional group is an atom or group of atoms
attached to the hydrocarbon chain, which confers
particular physical and/or chemical properties
upon the compound.
• Compounds with the same functional group often
undergo similar reactions.
• A list of common functional groups is found in
Table D.1.
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Chapter Two
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Types of Organic Compounds (cont’d)
For alcohols, the
functional group is a
hydroxyl group attached
to the carbon chain.
Carboxylic acids have a carboxyl group
(–COOH) attached to the carbon chain;
they are acidic (of course! Why else
would they be called carboxylic acids??).
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Chapter Two
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Cumulative Example
Show that the following experiment is consistent with the
law of conservation of mass (within the limits of
experimental error): A 10.00-g sample of calcium
carbonate was dissolved in 100.0 mL of hydrochloric
acid solution (d = 1.148 g/mL). The products were
120.40 g of solution (a mixture of hydrochloric acid and
calcium chloride) and 2.22 L of carbon dioxide gas (d =
0.0019769 g/mL).
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Chapter Two
Cumulative Example
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Show that the following experiment is consistent with the law of conservation of mass
(within the limits of experimental error): A 10.00-g sample of calcium carbonate was
dissolved in 100.0 mL of hydrochloric acid solution (d = 1.148 g/mL). The products were
120.40 g of solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L of
carbon dioxide gas (d = 0.0019769 g/mL).
Strategy
This problem may give the initial impression of being quite formidable because
several data are given. Upon examination, it is much less challenging than you
might think. We must show that mass is conserved in the experiment. That means
that we must compare the mass of the starting materials to the mass of the end
products of the chemical change. If mass is conserved, the two masses should be
identical, within the limits of experimental error. Our job then is to find the masses
of the starting materials and of the end products.
Solution
Let’s begin by identifying the starting materials and end products. The context of
the problem makes it clear that calcium carbonate reacts with a hydrochloric acid
solution, and so calcium carbonate and the HCl solution are the starting materials.
The end products are another solution and carbon dioxide gas.
Starting mass: The mass of calcium carbonate is given. We can use the density and
the volume of the HCl solution to find its mass.
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Chapter Two
Cumulative Example continued
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Solution continued
Then we can add the masses of the two starting materials.
Mass of products: This time, the mass of the solution is given. We must use volume
and density of carbon dioxide gas to find its mass. However, we must first convert
the volume in liters to milliliters because the density is given in grams per milliliter.
Then we can add the masses of the two products.
Assessment
We note that the masses of reactants and products are not exactly the same.
However, the 124.8 g of reactants is reported to four significant figures, and so it is
precise only to 0.1 g. The difference between the masses of the starting materials
and end products is less than 0.1 g. Clearly, the difference in masses is smaller than
the uncertainty in the mass of the starting materials. We can therefore conclude that
this experiment is consistent with the law of conservation of mass.
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Chapter Two