CHAPTER 2
Chemical Formulas
and Composition
Stoichiometry
Chapter Goals
1. Chemical Formulas
2. Ions and Ionic
Compounds
3. Names and
Formulas of Some
Ionic Compounds
4. Atomic Weights
5. The Mole
6.
Formula Weights, Molecular
Weights, and Moles
7. Percent Composition and
Formulas of Compounds
8. Derivation of Formulas from
Elemental Composition
9. Determination of Molecular
Formulas
10. Some Other Interpretations of
Chemical Formulas
11. Purity of Samples
2
Chemical Formulas
• Chemical formula shows the chemical
composition of the substance.
– ratio of the elements present in the molecule or
compound
•
•
•
•
He, Au, Na – monatomic elements
O2, H2, Cl2 – diatomic elements
O3, P4, S8 - more complex elements
H2O, C12H22O11 – compounds
Substance consists of two or more elements
3
Chemical Formulas
Compound 1 Molecule Contains
HCl
H2O
NH3
C3H8
1 H atom & 1 Cl atom
2 H atoms & 1 O atom
1 N atom & 3 H atoms
3 C atoms & 8 H atoms
4
Ions and Ionic Compounds
• Ions are atoms or groups of atoms that
possess an electric charge.
• Two basic types of ions:
– Positive ions or cations
• one or more electrons less than neutral
• Na+, Ca2+, Al3+
• NH4+ - polyatomic cation
– Negative ions or anions
• one or more electrons more than neutral
• F-, O2-, N3• SO42-, PO43- - polyatomic anions
5
Ions and Ionic Compounds
• Sodium chloride
– table salt is an ionic compound
6
Systematic Naming
• There are too many compounds to
remember the names of them all.
• Compound is made of two or more
elements.
• Name should tell us how many and what
type of atoms.
Charges on ions
• For most of Group A elements,
location on the Periodic Table can tell
what kind of ion they form
• Elements in the same group have
similar properties.
• Including the charge when they are
ions.
Charge in groups 1A, 2A and
3A is the group number
1+
2+
3+
3- 2- 1-
Naming Monatomic Ions
• 1. Monatomic cations are
– Identified by the element’s name
• 2. Monatomic anions
– Drop the ending of the element name
– Add an “–ide” ending
Name these
lNa1+
3+
lAl
Sodium ion
Calcium ion
Aluminum ion
lLi1+
Lithium ion
2+
lCa
Write Formulas for these
lPotassium
ion
lMagnesium
lCopper(II)
ion
ion
K1+
2+
Mg
2+
Cu
6+
Cr
lChromium(VI) ion
ion
2+
Ba
lMercury(II)
2+
Hg
lBarium
ion
Name these
lCl1-
lN3lBr1lO2lGa3+
Chloride ion
Nitride ion
Bromide ion
Oxide ion
Gallium ion
Write these
lSulfide
lIodide
ion
I1-
ion
lPhosphide
lStrontium
S2ion P3-
ion Sr2+
Binary Ionic Compounds
• Binary Compounds
– Compounds composed of two
different elements
– Binary ionic cmpd, total # of
positive and negative charges
must equal
Writing Formulas for Binary Ionic
Compounds
• Write the symbols for the ions side
by side. ALWAYS write the cation
first!
Al3+ O2-
• Cross over the charges by using the
absolute value of each ion’s charge
as the subscript for the other ion
Al23+ O32-
• Check that the subscripts are in
smallest whole number ratio
Al2O3
Naming Binary Ionic Compounds
Al2O3
–1. Name the cation(don’t change)
Aluminum
–2. Name the anion (drop add-ide)
Oxide
Aluminum Oxide
The Stock System of Nomenclature
• Most transition metals
can have more than one
type of charge.
• Indicate the charge with
Roman numerals in
parenthesis.
– Fe2+Iron(II)
– Fe3+ Iron(III)
• Roman numerals are
never used:
– For anions
– For metals that form
only one ion
Co2+
Cobalt(II) ion
CuCl2
Copper(II) chloride
Naming ions Cont…..
• A few, like silver, zinc and cadmium
only form one kind of ion
• Don’t get roman numerals
• Ag+ silver ion
• Zn2+ zinc ion
• Cd2+ cadmium ion
Compounds Containing Polyatomic
Ions
• Naming a series of similar polyatomic ions
– NO2NO3– Nitrite Nitrate
– Most common ion is given –ate ending, ion
with one less oxygen -ite
• Naming compounds containing polyatomic
ions
– Same as for monatomic ions
Naming Binary Molecular
Compounds
A. Binary Molecular Compounds
– 1. Covalently bonded molecules containing only two elements,
both nonmetals
B. Naming
– 1. Least electronegative element is named first
• Measure of the ability of an atom in chemical compound to
attract e• Tends to increase across periods decrease or remain the
same down
– 2. First element gets a prefix if there is more than 1 atom of that
element
– 3. Second element ALWAYS gets a prefix, and an “-ide” ending
• Examples: N2O3 = dinitrogen trioxide
• CO = carbon monoxide, not monocarbon monoxide
Names and Formulas of
Some Ionic Compounds
• Table 2-1 gives names of several
molecular compounds.
– You must know all of the molecular
compounds from Table 2-2.
• Some examples are:
– H2SO4 - sulfuric acid
– FeBr2 - iron(II) bromide
– C2H5OH - ethanol
22
Names and Formulas of
Some Ionic Compounds
• Table 2-2 displays the formulas, charges,
and names of some common ions
– You must know the names, formulas, and
charges of the common ions in table 2-3.
• Some examples are:
– Anions - Cl1-, OH1-, SO42-, PO43– Cations - Na1+, NH41+, Ca2+, Al3+
23
Acid Nomenclature
Anion
Ending
Acid Name
-ide
hydro-(stem)-ic acid
-ate
(stem)-ic acid
-ite
(stem)-ous acid
Acid Nomenclature
ACIDS
start with 'H'
2 elements
3 elements
hydro- prefix
-ic ending
no hydro- prefix
-ate ending
becomes
-ic
ending
C. Johannesson
-ite ending
becomes
-ous ending
Acid Nomenclature
HBr
2 elements, -ide
H2CO3

hydrobromic acid
3 elements, -ate

carbonic acid

sulfurous acid
H2SO3
3 elements, -ite
Acid Nomenclature
hydrofluoric acid
2 elements
sulfuric acid
 H+ F-
 HF
3 elements, -ic
 H+ SO42-  H2SO4
nitrous acid
3 elements, -ous
 H+ NO2-  HNO2
Names and Formulas of
Some Ionic Compounds
•
•
•
•
•
•
You do it!
What is the formula of nitric acid?
HNO3
What is the formula of sulfur trioxide?
SO3
What is the name of FeBr3?
iron(III) bromide
28
Names and Formulas of
Some Ionic Compounds
•
•
•
•
•
•
You do it!
What is the name of K2SO3?
potassium sulfite
What is charge on sulfite ion?
SO32- is sulfite ion
What is the formula of ammonium sulfide?
(NH4)2S
29
Names and Formulas of
Some Ionic Compounds
•
•
•
•
•
•
You do it!
What is the charge on ammonium ion?
NH41+
What is the formula of aluminum sulfate?
Al2(SO4)3
What are the charges on both ions?
Al3+ and SO4230
31
Atomic Weights
• Weighted average of the
masses of the constituent
isotopes of an element.
– Tells us the atomic masses of
every known element.
– Lower number on periodic
table.
32
The Mole
• A number of atoms, ions, or
molecules that is large
enough to see and handle.
• A mole = number of things
– Just like a dozen = 12 things
– One mole = 6.022 x 1023
things
• Avogadro’s number = 6.022
x 1023
– Symbol for Avogadro’s
number is NA.
33
n
1 mole of hockey pucks would
equal the mass of the moon!
n
1 mole of basketballs would fill a
bag the size of the earth!
• 1 mole of pennies would cover the
Earth 1/4 mile deep!
The Mole
Example 2-1: Calculate the mass of a single Mg
atom, in grams, to 3 significant figures.
35
The Mole
Example 2-1: Calculate the mass of a single Mg
atom, in grams, to 3 significant figures.

1 mol Mg atoms
? g Mg  1 Mg atom 
23
6.022

10
Mg atoms

 24.30gMg

 1 mol Mg atoms

 


  4.04  10  23 g Mg

36
The Mole
Example 2-2: Calculate the number of atoms in
one-millionth of a gram of Mg to 3 significant
figures.
37
The Mole
Example 2-2: Calculate the number of atoms in
one-millionth of a gram of Mg to 3 significant
figures.
 1 mol Mg 

? Mg atoms  1.00  10 g Mg 
 24.30 g Mg 
6
 6.022  10 23 Mg atoms

 1 mol Mg atoms

  2.48  1016 Mg atoms

38
The Mole
Example 2-3: How many atoms are contained in
1.67 moles of Mg?
39
The Mole
Example 2-3. How many atoms are contained in
1.67 moles of Mg?
 6.022  10 23 Mg atoms 

? Mg atoms  1.67 mol Mg 
1 mol Mg


 1.00  10 24 Mg atoms
40
The Mole
Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
You do it!
41
The Mole
Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
 1 mol Mg atoms 

? mol Mg  73.4 g Mg 
 24.30 g Mg 
 3.02 mol Mg
IT IS IMPERATIVE THAT YOU KNOW
HOW TO DO THESE PROBLEMS
42
Formula Weights, Molecular
Weights, and Moles
• How do we calculate the formula weight of
a compound?
– sum the atomic weight of each atom
The formula weight of propane, C3H8, is:
3  C  3  12.01 amu
 36.03 amu
8  H  8  1.01 amu
 8.08 amu
Forula Weight
 44.11 amu
43
Formula Weights, Molecular
Weights, and Moles
The formula weight of calcium nitrate, Ca(NO3)2,
is:
You do it!
44
Formula Weights, Molecular
Weights, and Moles
1  Ca  1  40.08 amu  40.08 amu
2  N  2  14.01 amu  28.02 amu
6  O  6  16.00 amu  96.00 amu
Molar mass
 164.10 amu
45
Formula Weights, Molecular
Weights, and Moles
• One Mole of
– Cl2 or 70.90g
Contains
6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
46
Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of C3H8
molecules in 74.6 g of propane.
47
Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of C3H8
molecules in 74.6 g of propane.
? C 3 H 8 molecules  74.6 g C 3 H 8 
 1 mole C 3 H 8  6.022  10 23 C 3 H 8 molecules


1 mole C 3 H 8
 44.11 g C 3 H 8 
24
1.02  10 molecules

 

48
Formula Weights, Molecular
Weights, and Moles
Example 2-6: What is the mass of 10.0 billion
propane molecules?
You do it!
49
Formula Weights, Molecular
Weights, and Moles
Example 2-6. What is the mass of 10.0 billion
propane molecules?
1 mole C3 H 8


? g C3 H 8 molecules  1.00  1010 molecules 

23
 6.022  10 molecules 
 44.11 g C3 H 8

 1 mole C3 H 8

  7.32  10 13 g of C3 H 8

50
Formula Weights, Molecular
Weights, and Moles
Example 2-7: How many (a) moles, (b) molecules,
and (c) oxygen atoms are contained in 60.0 g of
ozone, O3? The layer of ozone in the stratosphere
is very beneficial to life on earth.
You do it!
51
Formula Weights, Molecular
Weights, and Moles
Example 2-7a. How many moles are contained in
60.0 g of ozone, O3
 1 mole 
  1.25 moles
? moles O3  60.0 g O3 
 48.0 g O3 
52
Formula Weights, Molecular
Weights, and Moles
Example 2-7a: How many moles are contained in
60.0 g of ozone, O3?
53
Formula Weights, Molecular
Weights, and Moles
Example 2-7a. How many moles are contained in
60.0 g of ozone, O3
 1 mole 
  1.25 moles
? moles O3  60.0 g O3 
 48.0 g O3 
54
Formula Weights, Molecular
Weights, and Moles
Example 2-7b: How many molecules are contained
in 60.0 g of ozone, O3?
55
Formula Weights, Molecular
Weights, and Moles
Example 2-7b. How many molecules are contained
in 60.0 g of ozone, O3
 6.022 10 23 molecules
? molecules O3  1.25 moles 
1 mole

 7.53 10 23 molecules O3



56
Formula Weights, Molecular
Weights, and Moles
Example 2-7c: How many oxygen atoms are
contained in 60.0 g of ozone, O3?
57
Formula Weights, Molecular
Weights, and Moles
Example 2-7c. How many oxygen atoms are
contained in 60.0 g of ozone, O3
 3 O atoms
? O atoms  7.53 10 molecules O3 
 1 O3 molecule
 2.26 10 24 atoms O
23



58
Formula Weights, Molecular
Weights, and Moles
Example 2-8: Calculate the number of O atoms in
26.5 g of Li2CO3.
You do it!
59
Formula Weights, Molecular
Weights, and Moles
Example 2-8. Calculate the number of O atoms in
26.5 g of Li2CO3.
1 mol Li 2 CO 3
? O atoms  26.5 g Li 2 CO 3 

73.8 g Li 2 CO 3
6.022  10 23 form. units Li 2 CO 3
3 O atoms


1 mol Li 2 CO 3
1 formula unit Li 2 CO 3
6.49  10 23 O atoms
60
Formula Weights, Molecular
Weights, and Moles
• Occasionally, we will use millimoles.
– Symbol - mmol
– 1000 mmol = 1 mol
• For example: oxalic acid (COOH)2
– 1 mol = 90.04 g
– 1 mmol = 0.09004 g or 90.04 mg
61
Formula Weights, Molecular
Weights, and Moles
Example 2-9: Calculate the number of mmol in
0.234 g of oxalic acid, (COOH)2.
You do it!
62
Formula Weights, Molecular
Weights, and Moles
Example 2-9: Calculate the number of mmol in
0.234 g of oxalic acid, (COOH)2.
? mmol (COOH) 2  0.234 g (COOH) 2 
 1 mmol (COOH) 2 

  2.60 mmol (COOH) 2
 0.09004 g (COOH) 2 
63
64
Percent Composition and Formulas
of Compounds
• % composition = mass of an individual
element in a compound divided by the total
mass of the compound x 100%
Determine the percent composition of C in C3H8.
65
Percent Composition and Formulas
of Compounds
What is the percent composition of H in C3H8?
You do it!
66
Percent Composition and Formulas
of Compounds
• % composition = mass of an individual
element in a compound divided by the total
mass of the compound x 100%
Determine the percent composition of C in C3H8.
mass C
%C
 100%
mass C 3 H 8
3  12.01 g

 100%
44.11 g
 81.68%
67
Percent Composition and Formulas
of Compounds
What is the percent composition of H in C3H8?
mass H
%H
 100%
mass C 3 H 8
8 H

 100%
C3H 8
8  1.01 g

 100%  18.32%
44.11 g
or
18.32%  100%  81.68%
68
Percent Composition and Formulas
of Compounds
Example 2-10: Calculate the percent composition
of Fe2(SO4)3 to 3 significant figures.
You do it!
69
Percent Composition and Formulas
of Compounds
Example 2-10: Calculate the percent composition
of Fe2(SO4)3 to 3 significant figures.
2  Fe
2  55.8 g
% Fe 
 100% 
 100%  27.9% Fe
Fe 2 (SO 4 ) 3
399.9 g
3 S
3  32.1 g
% S 
 100% 
 100%  24.1% S
Fe 2 (SO 4 ) 3
399.9 g
12  O
12  16.0 g
% O 
 100% 
 100%  48.0% O
Fe 2 (SO 4 ) 3
399.9 g
Total
 100%
70
Derivation of Formulas from
Elemental Composition
• Empirical Formula - smallest whole-number ratio of
atoms present in a compound
– CH2 is the empirical formula for alkenes
– No alkene exists that has 1 C and 2 H’s
• Molecular Formula - actual numbers of atoms of each
element present in a molecule of the compound
– Ethene – C2H4
– Pentene – C5H10
• We determine the empirical and molecular formulas of a
compound from the percent composition of the
compound.
– percent composition is determined experimentally
71
Empirical Formula
1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to
find subscripts.
4. When necessary, multiply
subscripts by 2, 3, or 4 to get
whole #’s.
Derivation of Formulas from
Elemental Composition
Example 2-11: A compound contains 24.74% K,
34.76% Mn, and 40.50% O by mass. What is its
empirical formula?
73
Derivation of Formulas from
Elemental Composition
Example 2-11: A compound contains 24.74% K,
34.76% Mn, and 40.50% O by mass. What is its
empirical formula?
Make the simplifying assumption that we have 100.0 g of
compound.
? mol K
1 mol K
 24.74 g K 
 0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn  34.76 g Mn 
 0.6327 mol Mn
54.94 g Mn
1mol O
? mol O  40.50 g O 
 2.531 mol O
16.00 g O
74
Derivation of Formulas from
Elemental Composition
Example 2-11
? mol K
 24.74 g K 
1 mol K
 0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn  34.76 g Mn 
 0.6327 mol Mn
54.94 g Mn
? mol O  40.50 g O 
1mol O
 2.531 mol O
16.00 g O
obtain smallest whole number ratio
for K 
0.6327
1K
0.6327
for Mn 
0.6327
 1 Mn
0.6327
2.531
4O
0.6327
thus the chemical formula is KMnO 4
for O 
75
Derivation of Formulas from
Elemental Composition
Example 2-12: A sample of a compound contains
6.541g of Co and 2.368g of O. What is the
empirical formula for this compound?
You do it!
76
Derivation of Formulas from
Elemental Composition
Example 2-12: A sample of a compound contains
6.541g of Co and 2.368g of O. What is the
empirical formula for this compound?
? mol Co  6.541 g Co 
? mol O  2.368 g O 
1 mol Co
 0.1110 mol Co
58.93 gCo
1mol O
 0.1480 mol O
16.00 g O
find smallest whole number ratio
0.1110
0.1480
 1 Co for O 
 1.333O
0.1110
0.1110
multipy both by 3 to turn fraction to whole number
for Co 
1 Co  3  3 Co 1.333 O  3  4 O
Thus the compound' s formula is :
Co3O 4
77
Molecular Formula
1.
Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the
empirical mass.
4. Multiply each subscript by the
answer from step 3.
MF mass
n
EF mass
EF n
Determination of Molecular
Formulas
Example 2-13: A compound is found to contain 85.63% C and
14.37% H by mass. In another experiment its molar mass is
found to be 56.1 g/mol. What is its molecular formula?
79
Determination of Molecular
Formulas
Example 2-13: A compound is found to contain 85.63% C and
14.37% H by mass. In another experiment its molar mass is
found to be 56.1 g/mol. What is its molecular formula?
1 mol contains 56.1 g
85.63% is C and 14.37% is H
56.1 g  0.8563  48.0 g of C
56.1 g  0.1437  8.10 g of H
convert masses to moles
48.0 g of C 
1 mol C
 4 mol C
12.0 g C
8.10 g of H 
1 mol H
 8 mol H
1.01 g H
Thus the formula is :
C4 H8
80
Law of Multiple Proportions
• It is possible for two elements, A and B, to
combine to form more than one
compound.
• The ratios of the masses of element B that
combine with a given mass of element A in
each compound can be expressed by
small whole numbers.
81
Law of Multiple Proportions
Example 2-14: Show that the compounds NO2
and N2O5 obey the law of multiple proportions.
82
Law of Multiple Proportions
Example 2-14: Show that the compounds NO2
and N2O5 obey the law of multiple proportions.
For NO2
2 O atoms
2
1 N atoms
For N2O5
5 O atoms
 2.5
2 N atoms
83
Some Other Interpretations of
Chemical Formulas
Example 2-15: What mass of phosphorous is
contained in 45.3 grams of (NH4)3PO4?
84
Some Other Interpretations of
Chemical Formulas
Example 2-15: What mass of phosphorous is
contained in 45.3 grams of (NH4)3PO4?


4


1 mole ( NH 4)3 PO 4 
? grams P  45.3 g ( NH 4)3 PO

149.09 g ( NH 4)3 PO 4 

 30.9738 g P 
1 mole P



 1 mole ( NH 4)3 PO 4  1 mole P 
 9.41 g P
85
Some Other Interpretations of
Chemical Formulas
Example 2-16: What mass of ammonium phosphate,
(NH4)3PO4, would contain 15.0 g of N?
86
Some Other Interpretations of
Chemical Formulas
Example 2-16: What mass of ammonium phosphate,
(NH4)3PO4, would contain 15.0 g of N?
molar mass of (NH 4 ) 3 PO 4  149.0 g/mol
1 mol N
? mol N  15.0 g of N 
 1.07 mol N
14.0 g N
1 mol (NH 4 ) 3 PO 4
1.07 mol N 
 0.357 mol (NH 4 ) 3 PO 4
3 mol N
149.0 g (NH 4 ) 3 PO 4
0.357 mol (NH 4 ) 3 PO 4 
 53.2 g (NH 4 ) 3 PO 4
1 mol (NH 4 ) 3 PO 4
87
Purity of Samples
• The percent purity of a sample of a
substance is always represented as
mass of pure substance
% purity =
 100%
mass of sample
- mass of sample includes impurities
88
Purity of Samples
Example 2-18: A bottle of sodium phosphate,
Na3PO4, is 98.3% pure Na3PO4. What are the
masses of Na3PO4 and impurities in 250.0 g of this
sample of Na3PO4?
89
Purity of Samples
Example 2-18: A bottle of sodium phosphate,
Na3PO4, is 98.3% pure Na3PO4. What are the
masses of Na3PO4 and impurities in 250.0 g of this
sample of Na3PO4?
98.3 g Na PO
unit factor
100.0 g sample
98.3 g Na PO
? g Na PO  250.0 g sample 
100.0 g sample
= 246 g Na PO
? g impurities = 250.0 g sample - 246 g Na PO
90
= 4 g impurities
3
4
3
3
4
4
3
4
3
4
Synthesis Problem
In 1986, Bednorz and Muller succeeded in making
the first of a series of chemical compounds that
were superconducting at relatively high
temperatures. This first compound was La2CuO4
which superconducts at 35K. In their initial
experiments, Bednorz and Muller made only a few
mg of this material. How many La atoms are
present in 3.56 mg of La2CuO4?
91
Synthesis Problem
How many La atoms are present in 3.56 mg of
La2CuO4?
92
Synthesis Problem
How many La atoms are present in 3.56 mg of
La2CuO4?
molar mass of La 2 CuO 4 = 405.3 g/mol
 1g 
 
3.56 mg La 2CuO 4 
 1000 mg 
 1 mol La 2 CuO 4 

  8.78  10 6 mol La 2 CuO 4
 405.3 g La 2 CuO 4 
23

6
.
022

10
molecules La 2 CuO 4 
6

 
(8.78  10 mol La 2 CuO 4 )
1 mol La 2 CuO 4




2 La atoms

  1.06  1019 La atoms
 molecule La 2 CuO 4 
93
Group Activity
• Within a year after Bednorz and Muller’s
initial discovery of high temperature
superconductors, Wu and Chu had
discovered a new compound, YBa2Cu3O7,
that began to superconduct at 100 K. If
we wished to make 1.00 pound of
YBa2Cu3O7, how many grams of yttrium
must we buy?
94