Chapter 3
Molecules, Ions, & their
Compounds
Chapter goals
• Interpret, predict, and write formulas for
ionic and molecular compounds.
• Name compounds.
• Understand some properties of ionic
compounds.
• Calculate and use molar mass.
• Calculate percent composition for a
compound and derive formulas from
experimental data.
Molecule
• an assembly of 2 or more atoms (mostly
of non-metals) bound together in a
particular ratio and a particular manner
• is the smallest identifiable unit into
which a pure substance can be divided
and still retain the composition and
chemical properties of the substance
Elements that exist as
molecules
• Atoms of most of the nonmetals form
discrete molecules, except for the noble
gases.
• Some elements can exist in more than one
form of molecule; the different forms are
called allotropes. Examples are:
• Diamond, graphite, and buckyballs for
carbon.
• O2 and O3 (ozone) for oxygen.
ELEMENTS THAT
EXIST AS MOLECULES
Allotropes of C
ELEMENTS THAT EXIST AS
POLYATOMIC MOLECULES
White P4 and
polymeric red
phosphorus
Sulfur:
crownshaped
rings of S8
molecules
Molecular compounds
models - shapes
Water, H2O
Ammonia, NH3
Methane, CH4
Molecular compounds
• Are made of non-metals
• H2O, carbon dioxide (CO2), ammonia (NH3),
nitric acid (HNO3), ethanol (CH3CH2OH),
sulfuric acid (H2SO4), glucose (C6H12O6),
are examples among thousands
• In the molecules of theses compounds,
atoms share pairs of electrons
Molecular compounds: Formulas
NAME
MOLECULAR
FORMULA
Ethanol
Dimethyl
ether
C2H6O
C2H6O
CONDENSED
FORMULA
CH3CH2OH
CH3OCH3
STRUCTURAL FORMULA
H H
|
|
H─C ─ C─O─H
|
|
H H
H
H
|
|
H─C ─ O─ C─H
|
|
H
H
Ethanol and dimethyl ether are said to be structural isomers.
Ionic Compounds
Ion
• charged particle (atom or group of
atoms)
• cation: + charge
• anion: – charge
Ionic Compounds
Sodium chloride or “table salt” is an
example of an ionic compound.
Ionic compounds
• consist of positive and negative ions,
mostly a metal and a non-metal,
respectively.
• have attractions called ionic bonds
between positively (cations) and
negatively charged ions (anions).
• have high melting and boiling points.
Tm of NaCl = 800 °C = 1472 °F
• are solid at room temperature.
Charge Balance
Group
IA
IIA
VIIA
of periodic table
NaCl, sodium chloride
Periodic Table
Monatomic Cations
Metal atoms of group 1A lose one electron to
produce a mono-positive ion
Monatomic Cations
Metal atoms of group 2A lose two electrons to
produce a di-positive ion
Monatomic Anions
Nonmetals often gain one or more electrons and
form ions having a negative charge equal to
the group number of the element minus 8:
Group
Atom
gained e-
Resulting anion
5A
N
P
3
N3-
6A
O
S
2
O2-
7A
F
Cl, Br, I
1
FCl− , Br− , I−
Charges of Representative Elements
Monatomic Cations
Transition metals (B-group elements) can form
a no easily predictable variety of cations:
Group Atom Electrons loss Resulting cation
7B
Mn
2
Mn2+
8B
Fe
2
Fe2+
8B
Fe
3
Fe3+
1B
Cu
1
Cu+
1B
Cu
2
Cu2+
2B
Zn
2
Zn2+
2B
Cd
2
Cd2+
Transition Metals form Positive Ions
Most transition metals and Group 4A metals form 2 or
more positive ions.
Zn, Ag, and Cd form only one ion.
Names of Some Common Ions
Main group metals
element name only
Nonmetal: change the last
part of name to ide
Naming Cations
• transition metals and In, Sn, Tl, Pb, Bi
new system: element name (charge in
Roman numerals)
eg. Mn2+ manganese(II)
Mn3+ manganese(III)
Cr2+ chromium(II) Cr3+ chromium(III)
Fe2+ iron(II)
Fe3+ iron(III)
Exceptions: when only one cation
Ag+ silver
Zn2+ zinc
Cd2+ cadmium
old system
Latin name-suffix
suffix = -ic for higher charge, -ous for
lower charge
eg. Cu+ copper(I) cuprous ion
Cu2+ copper(II) cupric ion
Co2+ cobalt(II) cobaltous ion
Co3+ cobalt(III) cobaltic ion
Fe2+ iron(II)
ferrous ion
Fe3+ iron(III)
ferric ion
Polyatomic Ions
A polyatomic ion
• is a group of atoms.
• has an overall ionic charge, positive or
negative.
Polyatomic Ions (memorize)
Some examples of polyatomic ions are
NH4+
ammonium
H3O+
hydronium
OH−
hydroxide
N3−
azide
CO32−
carbonate
CN−
cyanide
CH3CO2− acetate
C2O42−
oxalate
NO3−
nitrate
NO2−
nitrite
PO43−
phosphate
PO33−
phosphite
SO42−
sulfate
SO32−
sulfite
CrO42−
chromate
Cr2O72−
dichromate
MnO4−
permanganate
MnO42−
manganate
Hydrogenated Polyatomic Ions
HCO3−
hydrogen carbonate (bicarbonate)
HSO4−
hydrogen sulfate (bisulfate)
HSO3−
hydrogen sulfite (bisulfite)
HPO42−
hydrogen phosphate
H2PO4−
dihydrogen phosphate
HS−
hydrogen sulfide (from sulfide, S2−)
Systematics
• ClO4–
perchlorate
• ClO3–
chlorate
• ClO2–
chlorite
• ClO–
hypochlorite
The same with other halogens (except for F)
• IO4–
periodate
• IO3–
iodate
• IO2–
iodite
• IO–
hypoiodite
1
1A
1 1
1.008
3
2
2A
4
Li
Be
H
Mg
22.99 24.31
19
20
K
Ca
3
3B
21
4
4B
22
5
5B
23
6
6B
24
7
7B
25
8
8B
26
9
8B
27
10
8B
28
11
1B
29
12
2B
30
Sc
Ti
V Cr Cr
Mn
Fe
Co
Ni
Cu
Zn
39.10 40.08 44.96 47.88 50.94
37
38
39
40
41
Rb
Sr
Y
Zr
Nb
85.47 87.62 88.91 91.22 92.91
55
56
72
73
57
Cs
Ba
13
3A
5
14
4A
6
15
5A
7
16
6A
8
17
7A
9
4.003
10
B
C
N
O
F
Ne
He
10.81 12.01 14.01 16.00 19.00 20.18
13
14
15
16
17
18
6.941 9.012
11
12
Na
18
8A
2
La
Hf
Ta
52.00 9854.94 55.85 58.93 58.69 63.55 65.39
42
43
44
45
46
47
48
Mo
Tc
95.94
74
(98)
75
W
Re
Ru
Rh
Pd
Ag
Cd
Al
Si
P
S
Cl
Ar
26.98 28.09 30.97 32.07 35.45 39.95
31
32
33
34
35
36
Ga
Ge
As
Se
Br
Kr
69.72 72.59 74.92 78.96 79.90 83.80
49
50
51
52
54
53
In
Sn
Sb
Te
I
Xe
101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
76
77
79
80
81
82
83
84
85
86
78
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
132.9 137.3 138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0
88
89
104
105
106
107
108
109
87
Fr Ra Ac Unq Unp Unh Uns Uno Une
(223) (226) (227) (257) (260) (263) (262) (265) (266)
Po
At
Rn
(210)
(210)
(222)
58
59
60
61
62
63
64
65
66
67
68
69
70
71
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
(147)
93
150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0
94
95
96
97
98
99
100
101
102
103
140.1 140.9 144.2
92
90
91
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lw
232.0
(231)
238.0
(237)
(242)
(243)
(247)
(247)
(249)
(254)
(253)
(256)
(254)
(257)
•
•
•
•
PO53–
PO43–
PO33–
PO23–
perphosphate
phosphate
phosphite
hypophosphite
•
•
•
•
AsO53–
AsO43–
AsO33–
AsO23–
perarsenate
arsenate
arsenite
hypoarsenite
Oxyanions that end in ate
Group
III A
BO33–
borate
Group
IV A
Group
VA
CO32–
carbonate
NO3–
nitrate
SiO32–
silicate
PO43–
phosphate
AsO43–
arsenate
Group
VI A
SO42–
sulfate
Group
VII A
ClO3–
chlorate
SeO42–
BrO3–
selenate bromate
TeO42–
tellurate
IO3–
iodate
Other oxyanions:
Once ions ending in ate are memorized, those
with different number of O atoms are named with
following prefixes and suffixes
2 oxygen less 1 oxygen less
than the “ate” than the
“ate” ion is
ion is
“ate” ion is
____ate
hypo__ite
___ite
BrO–
BrO2–
BrO3–
BrO4–
hypobromite
bromite
bromate
perbromate
1 oxygen
more than
“ate” ion is
per__ate
Formula Unit
• combination of ions in simplest whole
number ratio to be electrically neutral (the
simplest unit of an ionic compound)
• consists of positively and negatively charged
ions.
• is neutral.
• has charge balance.
total positive charge = total negative charge
The symbol of the metal is written first
followed by the symbol of the nonmetal.
Naming Ionic Compounds with Two
Elements (binary compound)
To name a compound
that contains two
elements,
• identify the cation
and anion.
• name the cation
first followed by the
name of the anion.
Examples, name each below
• KBr
K+
potassium
Br–
bromide
potassium bromide
• AlF3
Al3+
aluminum
F–
fluoride
aluminum fluoride
• Sr3P2
Sr
Sr2+
3(+2) = +6
strontium
P3–
phosphide
strontium phosphide
P
2(−3) = −6
• CuCl2
Cu
Cl–
1(+2) = +2
chloride
Cu2+
copper(II)
cupric
copper(II) chloride
cupric chloride
CuCl: copper(I) chloride
Cl
2(−1) = −2
• WF6
W
F
F–
w + 6x(−1) = 0
fluoride
w = 6+
W6+
tungsten(VI)
tungsten(VI) fluoride
Naming Compounds with
Polyatomic Ions
The positive ion is named first followed by the
name of the polyatomic ion.
NaNO3
sodium nitrate
K2SO4
potassium sulfate
Fe(HCO3)3
iron(III) bicarbonate
or iron(III) hydrogen carbonate
fe + 3x(−1) = 0
(NH4)3PO3
fe = 3+
iron(III)
ammonium phosphite
Writing Formulas with Polyatomic Ions
The formula of an ionic compound
• containing a polyatomic ion must have a
charge balance that equals zero (0).
Na+ and NO3−  NaNO3
• with two or more polyatomic ions has the
polyatomic ions in parentheses.
Mg2+ and 2NO3−  Mg(NO3)2
subscript 2 for charge balance
• Aluminum sulfate
2Al3+ and 3SO42−  Al2(SO4)3
Learning Check
Match each formula with the correct name.
A. MgS
MgSO3
MgSO4
1) magnesium sulfite
2) magnesium sulfate
3) magnesium sulfide
B. Ca(ClO3)2
CaCl2
Ca(ClO2)2
1) calcium chlorate
2) calcium chlorite
3) calcium chloride
Name each of the following compounds:
A. Mg(NO3)2
B. Cu(ClO3)2
C. PbO2
D. Fe2(SO4)3
E. Ba3(PO3)2
magnesium nitrate
copper(II) chlorate
lead(IV) oxide
iron(III) sulfate
barium phosphite
Mg2+ and 2 NO3−
Cu2+ and 2 ClO3−
Pb4+ and 2 O2−
2 Fe3+ and 3 SO42 −
3 Ba2+ and 2 PO33 −
Learning Check
Select the correct formula for each.
A. aluminum nitrate
1) AlNO3
2) Al(NO)3
3) Al(NO3)3
B. copper(II) nitrate
1) CuNO3
2) Cu(NO3)2
3) Cu2(NO3)
C. iron(III) hydroxide
1) FeOH
2) Fe3OH
3) Fe(OH)3
D. tin(IV) hydroxide
1) Sn(OH)4 2) Sn(OH)2
3) Sn4(OH)
• CaSO4
Ca2+
1(+2) = +2
calcium
SO42–
sulfate
calcium sulfate
1(−2) = −2
• (NH4)2S
NH4+
ammonium
S2–
sulfide
ammonium sulfide
• (NH4)3PO4
NH4+
ammonium
PO43–
phosphate
ammonium phosphate
• Mo(BrO)6
BrO–
mo + 6x(−1) = 0 mo = 6+
hypobromite
Mo6+
molybdenum(VI)
molybdenum(VI) hypobromite
Write formula for
• rubidium bromide
Rb
Rb+
Br
Br–
RbBr
• calcium phosphide
Ca
Ca2+
P
For neutrality we need
P3–
3 Ca2+ and 2 P3–
Ca3P2
that is, +6 – 6 = 0
• niobium(IV) sulfite
Nb4+
In order to have the same
SO32–
total + and – charge, we need
one Nb4+ ion and two SO32–, that is
1(+4) + 2 (–2) = 4 – 4 =0 (neutrality)
Nb(SO3)2
• barium phosphite: Ba2+ and PO33–
for the compound to be neutral, we need
3 Ba2+ and 2 PO33–, that is, +6 – 6 = 0
Ba3(PO3)2
Electrostatic Forces
The oppositely charged ions in ionic
compounds are attracted to one another by
ELECTROSTATIC FORCES.
These forces are governed by COULOMB’S
LAW.
Electrostatic Forces
COULOMB’S LAW
As ion charge increases, the attractive force
increases.
As the distance between ions (d) increases, the
attractive force decreases.
This idea is important and will come up many times in
future discussions!
Q cQ a
F 
d2
Product of charges
NaCl
Na+
Cl–
1
CaCl2
Ca2+
Cl–
2
CaS
Ca2+
S–2
4
Al2S3
Al3+
S2–
6 strongest
force
Q cQ a
F 
d2
d
Na+ F–
d
Na+
Cl–
NaF has the strongest force and NaI the weakest
d
d
Na+
I–
Na+
Br–
Acids
• H+ is only cation in acids
Binary Acids
hydrogen bonded to one other
element
hydro-anion-ic acid
Examples
• HCl in aqueous solution (water is the solvent)
H+ and Cl–
anion is chloride
hydro-chlor-ic acid
Same for F, Br, and I
hydrochloric acid
• H2S
hydrosulfuric acid
• H3P
hydrophosphoric acid
• H2Te
hydrotelluric acid
Oxoacids
hydrogen, oxygen, and another
element
anion-suffix acid
anion
suffix: -ate  -ic, -ite  -ous
acid
Examples, name
• H3PO4
phosphoric acid
PO43–
phosphate
H2SO4
sulfuric acid
SO42–
sulfate
• H3PO3
phosphorous acid
PO33–
phosphite
H2SO3
sulfurous acid
SO32–
sulfite
If more than two oxoacids, that with less O
than the –ous will be hypo– –ous; that with
more O than the –ic will be per– –ic.
• HBrO
hypobromous acid
BrO–
hypobromite
HBrO2
bromous acid
BrO2–
bromite
• HBrO3
HBrO4
bromic acid
perbromic acid
BrO3–
BrO4–
bromate
perbromate
bromic acid
perbromic acid
The same applies to Cl and I, not to F
Molecular Compounds
• contain only nonmetals bound by covalent
bonds
methane, CH4
Covalent bonds form
• when atoms share electrons to complete
octets. There are no ions.
Naming Molecular (Covalent) Compounds
To name covalent
compounds
• STEP 1: Name the first
nonmetal as an
element.
• STEP 2: Name the
second nonmetal with
an ide ending.
• STEP 3: Use prefixes
to indicate the number
of atoms (subscript) of
each element.
Prefixes Used in Naming
Covalent Compounds
# of Atoms
Prefix
1
Mono
2
Di
3
Tri
4
Tetra
5
Penta
6
Hexa
7
Hepta
8
Octa
9
Nona
10
Deca
Exception
• prefix mono- omitted from name of
first element
• eg. CO2
carbon dioxide, not monocarbon
dioxide
Examples, name
What is the name of SO3?
1. The first nonmetal is S sulfur.
2. The second nonmetal is O named oxide.
3. The subscript 3 of O is shown as the
prefix tri.
SO3  sulfur trioxide
The subscript 1 (for S) or mono is
understood
Examples, name
• P2O5
phosphorus
Diphosphorus (two P atoms)
oxygen
oxide
Pentoxide(5 O atoms) (not pentaoxide: a
of penta- is dropped)
diphosphorus pentoxide
• P3Br6
phosphorus
triphosphorus
bromine
bromide
hexabromide
triphosphorus hexabromide
Learning Check
Select the correct name for each compound.
A.
SiCl4
1) silicon chloride
2) tetrasilicon chloride
3) silicon tetrachloride
B.
P2O5
1) phosphorus oxide
2) phosphorus pentoxide
3) diphosphorus pentoxide
C.
Cl2O7
1) dichlorine heptoxide
2) dichlorine oxide
3) chlorine heptoxide
Learning Check
Write the name of each molecular (covalent)
compound.
CO
_____________________
CO2
_____________________
PCl3
_____________________
CCl4
_____________________
N2O
_____________________
Learning Check
Write the correct formula for each of the
following.
A. phosphorus pentachloride
B. dinitrogen trioxide
C. sulfur hexafluoride
Common Names
• H 2O
water
• PH3
phosphine
• CaCO3
limestone
• CaO
lime
• Ca(OH)2
slaked lime
• CH4
methane
• NaCl
table salt
• N 2O
laughing gas
• NaHCO3
baking soda
• Na2CO3•10H2O
washing soda
Sodium carbonate decahydrate
• MgSO4•7H2O
epsom salt
• Mg(OH)2
milk of magnesia
• Ca(SO4)•2H2O
gypsum
Nomenclature Details
•
acid (aqueous solution)
HBr(g) hydrogen bromide (gas phase)
HBr(aq) hydrobromic acid (aqueous
solution)
Same for HF, HCl, and HI.
Hydrogen
• belongs in group all its own
• generally considered a nonmetal
• forms both H+ and H– in ionic compounds
HCl (in aqueous solution)
H+ and Cl–
NaH (sodium hydride, same with Li, K,
Rb, Ca, Ba…)
Na+ and H–
• mostly forms molecular compounds
e.g.. CH4, NH3, C2H6SO, ...
Chemical Formulas
• empirical formula
– indicates the elements present and
their simplest, whole-number ratio in
a compound
• molecular formula
– indicates elements present and the
exact number of atoms of each in a
unit of a compound
• structural formula
– a molecular formula that includes
structural information
Benzene
structural formula
H
C
H C
H C
C H
C H
C
molecular formula C6H6
H
dividing by smallest subscript (6)
empirical formula CH
Dimethylether
Dimethyl
ether
C2H6O
CH3OCH3
H
H
|
|
H─C ─ O─ C─H
|
|
H
H
• condensed formula H3COCH3
• molecular formula C2H6O (cannot be
divided to get whole numbers; by diving
by 2 we get O0.5)
• empirical formula
C2H6O
Calcium Chloride (ionic)
• structural formula
none (it is ionic)
• molecular formula
none (it is ionic)
• formula unit
CaCl2
• empirical formula
CaCl2
Other examples
NAME
butanoic acid
Molecular
Empirical
formula
formula
C 4H 8O 2 C 4H 8O 2
C 2H 4O
2
diboron
hexahydride
B 2H 6
2 2
B 2H 6
2
sodium ditionate
Na2S2O4
hexane
C6H14
BH3
2
NaSO2
CH2.333
C3H7
Molar Mass of a Compound
is the mass (g) of one mole (6.022 x 1023) of
• formula units for ionic compounds
• molecules for molecular compounds
The molar mass of a compound is the sum of
the molar masses of the elements in the
Formula (times respective coefficients.)
• molecular or formula weight =
amu/molecule
• molar mass = g/mol of molecules
= g/mol of formula units (for
ionic compounds)
Molar Mass of a Compound
Example: Calculate the molar mass of CaCl2 (ionic)
Element Number of Moles Atomic Mass Total Mass
Ca
1
40.1 g/mole
40.1 g
Cl
2
35.5 g/mole
71.0 g
CaCl2
111.1 g/mole
For glucose: C6H12O6 (molecular, covalent)
C
6
12.0 g/mole
72.0 g
H
12
1.0 g/mole
12.0 g
O
6
16.0 g/mole
96.0 g
C6H12O6
180.0 g/mol
Molar Mass (M), Avogadro’s
number, and atoms of elements
1 mole = 6.022 x 1023 particles = molar mass(g)
molecules or formula units for compounds
From the molecular formula of glucose, C6H12O6,
1 glucose molecule contains 6 C, 12 H, and 6 O atoms
1 mol glucose contains 6 mol C,12 mol H, and 6mol
O
x 6.022 x 1023 particles (molecules and atoms)
M (g/mol) = 6x AW C + 12 AW H + 6 AW O
(from periodic table)
Molar Mass and Avogadro’s number
Avogadro’s number (6.022 x 1023) can be written
as equalities and conversion factors.
Equality: (atoms, molecules, ions, electrons, protons)
1 mole = 6.022 x 1023 particles = molar mass (g)
Particles here are molecules or Formula Units (ionic)
6.022 x 1023 particles
1 mole
and
6.022 x 1023 particles
molar mass (g)
and
1 mole
molar mass (g)
and
1 mole
6.022 x 1023 particles
molar mass (g)
6.022 x 1023 particles
molar mass (g)
1 mole
The mass (in grams) of a single
molecule
The molar mass of water, H2O, is
2 x 1.008 + 16.00 = 18.02 g/mol (about 18 mL)
Does a water molecule have a mass of 18.02 g?
No, it doesn’t.
18.02 g (the molar mass)
1 mol H2O
——————————
6.022 x 1023 H2O molecules
by diving, we get 2.99x10−23 g per H2O molecule
• What is the mass of 12.0 million benzene
(C6H6) molecules?
Molar mass = (6 mol C)(12.01 g/mol C)
+ (6 mol H)(1.008 g/mol H) = 78.11 g/mol C6H6
12.0 million= 12,000,000 = 1.20 x 107 molecules
1.20 x 107 molec.x
1 mol
x 78.11 g
6.022 x 1023 molecules 1 mol
78.11 g
= 1.99 x 10–17 mol C6H6  ──────=
mol C6H6
1.56 x 10–15 g C6H6
Consider 4.49 g Ca3(PO4)2 . Its formula weight
FW = 3x40.08 + 2x(30.97 + 4x16.00)=310.18 g/mol
• How many mol of Ca3(PO4)2 is this?
1 mol
4.49 g x ───── = 0.0145 moles Ca3(PO4)2
310.18 g
• How many mol of Ca2+ ions does this contain?
3 moles Ca2+
0.0145 mol Ca3(PO4)2 x─────── = 0.0435 mol Ca2+
1 mol Ca3(PO4)2
• How many Ca2+ ions are there?
6.022 x1023 Ca2+
0.0435 mol Ca2+ x ───────────= 2.61x1022 ions
1 mol Ca2+
Consider 4.49 g Ca3(PO4)2
FW = 310.18 g/mol
• How many mol of P does that amount of Ca3(PO4)2
contain?
2 moles P
0.0145 mol Ca3(PO4)2 x─────── = 0.0290 mol P
1 mol Ca3(PO4)2
• How many mol of O does that amount contain?
8 moles O
0.0145 mol Ca3(PO4)2 x─────── = 0.116 mol O
1 mol Ca3(PO4)2
• How many grams of O are in that amount of salt?
16.0 g O
0.116 mol O x─────── = 1.86 g O
1 mol O
Consider 4.49 g Ca3(PO4)2
FW = 310.18 g/mol
How many phosphate ions does that amount of
Ca3(PO4)2 contain?
First we calculate how many moles of phosphate
2 moles PO43−
0.0145 mol Ca3(PO4)2 x───────── = 0.0290 mol PO43−
1 mol Ca3(PO4)2
Second, using Avogadro’s number,
6.02 x 1023 PO43−
0.0290 mol PO43− x ─────────── = 1.75 x 1022 PO43−
1 mol PO43−
ions
What mass of C4H8O is required to
supply 1.43 x 1016 C atoms?
• Molar mass = 72.10 g/mol
1 mol C atoms
1 mol C4H8O
1.43 x 1016 C atoms x──────────── x ───────
6.022x1023 C atoms 4 moles C
72.10 g C4H8O
x──────────── = 4.28 x 10−7 g C4H8O
1 mol C4H8O
Empirical Formula from Analysis
It will be shown by example.
We will use The Percent Composition, i.e,
the amount (grams) of each element
forming one compound in 100 g of that
compound.
(An example of calculating %s)
For the compound AxByCz (x, y, z are
unknown)
% A + % B + %C = 100
• A compound was found to be 40.92 % C,
4.58 % H, and 54.50 % O by mass.
• Determine its empirical formula.
CxHyOz
x, y, and z are the simplest whole number
ratios of atoms in compound.
Assume exactly 100 g of compound
then 40.92 g C, 4.58 g H, and 54.50 g O
are contained in those 100 g
Firstly, calculate the number of moles of C,
H, and O in that amount of compound… next
1 mole C
40.92 g C x──────── = 3.407 moles C
12.01 g C
1 mole H
4.58 g H x ──────── = 4.54 moles H
1.008 g H
1 mole O
54.50 g O x──────── = 3.406 moles O
16.00 g O
C3.407H4.54O3.406
Now, divide all # by the smallest
C3.407H4.54O3.406
3.406
The result is C1.0H1.33O1.0
3.406 3.406
Finally, 1.33 x 3 = 4
And the same for 1.0 x 3
Empirical formula is
C3H4O3
Determine the empirical formula of a compound
that is 36.4% Mn, 21.2% S, and 42.4% O by mass.
assuming exactly 100 g compound
1 mole Mn
36.4 g Mn x──────── = 0.663 moles Mn
54.9 g Mn
1 mole S
21.2 g S x ──────── = 0.660 moles S
32.1 g S
1 mole O
42.4 g O x──────── = 2.65 moles O
16.0 g O
Mn0.663S0.660O2.65
0.660
0.660
The result is Mn1.0S1.0O4.02
0.660
empirical formula: MnSO4
Homework: Determine the empirical
formula of a compound that is
27.6% Mn, 24.2% S, & 48.2% O
answer
Mn2S3O12
molecular formula = n x (empirical formula)
• n = whole number
molar mass (or formula weight)
• n = ─────────────────────────
weight of the empirical formula
Example: benzene, molecular formula: C6H6
empirical formula: CH
n=6
6 x (CH) = C6H6
Example: A 4.99 g sample of a compound was
found to contain 1.52 g N and oxygen.
Determine its empirical and molecular formula
if its MW is 92.04
• determine empirical formula
g O = 4.99
−1.52
1 mol N
= 3.47 g
1.52 g N─────── = 0.108 mol N
14.01 g N
1 mol O
3.47 g O ─────── = = 0.217 mol O
16.00 g O
N0.108O0.217
0.108
0.108
NO2
• determine molecular formula
empirical formula NO2
empirical weight = 14.01+ 216.00 = 46.01
molecular weight = 92.04 (given in
previous slide)
92.04
n = ───── = 2
46.01
Molecular formula = 2 x (emp. form.)
= 2(NO2)
molecular formula = N2O4
A compound was found to contain only B
and H. Analysis of an 8.247 g sample
indicated that it contained 1.803 g H and
had a MW of ~30. Determine its molecular
formula and its MW to 4 SF.
• determine empirical formula
total mass = g B + g H
8.247 g = g B + 1.803 g H
g B = 8.247 g compound −1.803 g H =6.444 g
1 mol B
6.444 g B ───────= 0.5961 mol of boron
10.811 g B
1 mol H
1.803 g H ─────── = 1.789 mol H
1.008 g H
B0.5961 H1.789
0.5961
BH3 is the Empirical formula
0.5961
• determine molecular formula
MW
30
n = ───────  ──── = 2
emp. wt.
13.83
mol. form. = 2(BH3) = B2H6
B 2H 6
MW = 210.811 + 6 1.008 = 27.67 g/mol
Hydrates
compounds that contain intact H2O
eg. CaSO4•2H2O
Two H2O moles per mole of CaSO4
compound name prefix-hydrate
(prefix indicates number of H2O molecules)
calcium sulfate dihydrate
• LiCl•8H2O
lithium chloride octahydrate
• tungsten(VI) sulfate heptahydrate
W6+
SO42–
W(SO4)3•7H2O
• Hydrated nickel(II) chloride is a beautiful green color,
crystalline compound. When heated strongly, the
compound is dehydrated. If 0.235 g of NiCl2•xH2O gives
0.128 g of NiCl2 on heating, what is the value of x?
g of NiCl2•xH2O = g NiCl2 + g H2O
Then,
hydrated dehydrated
g H2O = 0.235 g – 0.128 g = 0.107 g
1 mol NiCl2
0.128 g NiCl2 x ──────── = 0.000987 moles NiCl2
129.7 g NiCl2
1 mol H2O
0.107 g H2O x ──────── =
18.0 g H2O
NiCl2 0.000987 H2O0.00594
0.000987
0.000987
0.00594 moles H2O
NiCl2•6H2O
x = 6 moles of water
per mole of NiCl2
Percent Composition
The percent of the total mass of a
substance represented by each element
within that substance
In other words, how many grams of every
element are in 100 g of the compound
Determine the % composition of aluminum
bromite
•
•
•
•
Al3+
BrO2–
Al(BrO2)3 Al
Br
O
FW = 26.982 + 3 x(79.904 + 2 x 16.000)
or = 26.982 + 3x79.904 + 6x16.000
= 362.68 g/mol
g Al
% Al = ──────── x 100
g compound
26.982 g
% Al = ──────── x 100 = 7.440 %
362.68 g
g Br
% Br = ──────── x 100
g compound
3 x 79.904 g
% Br = ──────── x 100 = 66.095 %
362.68 g
% O = 100% – 7.440% – 66.095 = 26.465%
3.69 Silver chloride, often used in silver plating,
contains 75.27% Ag. What mass of silver chloride is
required to plate 155 mg of pure silver? AgCl
The 75.27% means there is 75.27 g of Ag in 100 g AgCl.
That can be used as a conversion factor (ratio)
110−3 g Ag
155 mg Ag  ──────── = 0.155 g Ag
1 mg
100 g AgCl
0.155 g Ag  ──────── = 0.206 g AgCl
75.27 g Ag
or, with mg,
100 mg AgCl
155 mg Ag  ──────── = 206 mg AgCl (with mg)
75.27 mg Ag