CHAPTER 2
CHEMICAL FORMULAS &
COMPOSITION STOICHIOMETRY
1
Chapter Goals
1.
2.
3.
4.
Atoms and Molecules
Chemical Formulas
Ions and Ionic Compounds
Names and Formulas of Some Ionic
Compounds
5. Atomic Weights
6. The Mole
2
Chapter Goals
7. Formula Weights, Molecular Weights, and
Moles
8. Percent Composition and Formulas of
Compounds
9. Derivation of Formulas from Elemental
Composition
10.Determination of Molecular Formulas
11.Some Other Interpretations of Chemical
Formulas
12.Purity of Samples
3
Atoms and Molecules
Dalton’s Atomic Theory - 1808
Five postulates
1.
2.
3.
4.
5.
An element is composed of extremely small,
indivisible particles called atoms.
All atoms of a given element have identical
properties that differ from those of other elements.
Atoms cannot be created, destroyed, or transformed
into atoms of another element.
Compounds are formed when atoms of different
elements combine with one another in small wholenumber ratios.
The relative numbers and kinds of atoms are
constant in a given compound.
Which of these postulates are correct today? 4
Atoms and Molecules
A molecule is the smallest particle of an
element that can have a stable independent
existence.

Usually have 2 or more atoms bonded together
Examples of molecules






H2
O2
S8
H2O
CH4
C2H5OH
5
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, S4, P8 - more complex elements
H2O, C12H22O11 – compounds
 Substance consists of two or more elements
6
Chemical Formulas
Compound 1 Molecule Contains
HCl
H2O
NH3
C3H8
1
2
1
3
H atom & 1 Cl atom
H atoms & 1 O atom
N atom & 3 H atoms
C atoms & 8 H atoms
7
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-, N3SO42-, PO43- - polyatomic anions
8
Ions and Ionic Compounds
Sodium chloride

table salt is an ionic compound
9
Names and Formulas of
Some Ionic Compounds
Table 2-3 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-, PO43Cations - Na1+, NH41+, Ca2+, Al3+
10
Names and Formulas of
Some Ionic Compounds
Formulas of ionic compounds are determined
by the charges of the ions.


Charge on the cations must equal the charge on
the anions.
The compound must be neutral.
NaCl
KOH
CaSO4
Al(OH)3
sodium chloride
(Na1+ & Cl1-)
potassium hydroxide(K1+ & OH1-)
calcium sulfate
(Ca2+ & SO42-)
aluminum hydroxide (Al3+ & 3 OH1-)
11
Names and Formulas of
Some Ionic Compounds
Table 2-2 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
12
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
13
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
14
Names and Formulas of
Some Ionic Compounds
You do it!
What is charge on ammonium ion?
NH41+
What is the formula of aluminum sulfate?
Al2(SO4)3
What is charge on both ions?
Al3+ and SO4215
Atomic Weights
Weighted average of the
masses of the constituent
isotopes if an element.


Tells us the atomic masses of
every known element.
Lower number on periodic
chart.
How do we know what the
values of these numbers are?
16
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.
17
The Mole
How do we know when we have a mole?


count it out
weigh it out
Molar mass - mass in grams numerically
equal to the atomic weight of the element in
grams.
H has an atomic weight of 1.00794 g

1.00794 g of H atoms = 6.022 x 1023 H atoms
Mg has an atomic weight of 24.3050 g

24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms
18
The Mole
Example 2-4: How many moles of Mg atoms
are present in 73.4 g of Mg?
You do it!
19
The Mole
Example 2-4: How many moles of Mg atoms
are present in 73.4 g of Mg?
? mol Mg  73.4 g Mg
20
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 
21
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
22
The Mole
Example 2-1: Calculate the mass of a
single Mg atom in grams to 3 significant
figures.
? g Mg 
23
The Mole
Example 2-1: Calculate the mass of a
single Mg atom in grams to 3 significant
figures.
? g Mg  1 Mg atom
24
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 
25
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

26
The Mole
Example 2-2: Calculate the number of atoms
in one-millionth of a gram of Mg to 3
significant figures.
? Mg atoms 
27
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 
6

? Mg atoms  1.00  10 g Mg
 24.30 g Mg 
28
The Mole
Example 2-2: Calculate the number of
atoms in one-millionth of a gram of Mg
to 3 significant figures.
? Mg atoms  1.00  10
6
 6.022  1023 Mg atoms

1 mol Mg atoms

 1 mol Mg 

g Mg 
 24.30 g Mg 



29
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  1023 Mg atoms 

  2.48  1016 Mg atoms
 1 mol Mg atoms

30
The Mole
Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
? Mg atoms 
31
The Mole
Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
? Mg atoms  1.67 mol Mg
32
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


33
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
34
The Mole
Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
 6.022  1023 Mg atoms 

? Mg atoms  1.67 mol Mg
1 mol Mg


 1.00  1024 Mg atoms
35
Formula Weights, Molecular
Weights, and Moles
How do we calculate the molar mass of
a compound?

add atomic weights of each atom
The molar mass of propane, C3H8, is:
3  C  3  12.01 amu  36.03 amu
8  H  8  1.01 amu  8.08 amu
Molar mass
 44.11 amu
36
Formula Weights, Molecular
Weights, and Moles
The molar mass of calcium nitrate,
Ca(NO3)2 , is:
You do it!
37
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
38
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

C3H8
You do it!
39
Formula Weights, Molecular
Weights, and Moles
One Mole of

Cl2 or 70.90g

C3H8 or 44.11 g
Contains
6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
6.022 x 1023 C3H8 molecules
3 (6.022 x 1023 ) C atoms
8 (6.022 x 1023 ) H atoms
40
Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3H8 molecules  74.6 g C3H8 
41
Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3 H 8 molecules  74.6 g C3H 8 
 1 mole C3H 8 


 44.11 g C3 H 8 
42
Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3H8 molecules  74.6 g C3H8 
 1 mole C3H8  6.022 10 23 C3H8 molecules


1 mole C3H8
 44.11 g C3H8 

 

43
Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3 H 8 molecules  74.6 g C3 H 8 
 1 mole C3H 8  6.022 10 23 C3H 8 molecules


1 mole C3 H 8
 44.11 g C3 H 8 
24
1.02  10 molecules

 

44
Formula Weights, Molecular
Weights, and Moles
Example 2-6. What is the mass of 10.0
billion propane molecules?
You do it!
45
Formula Weights, Molecular
Weights, and Moles
Example 2-6. What is the mass of 10.0
billion propane molecules?
1 mole C3H8


? g C3H8 molecules  1.00  10 molecules 

23
 6.022 10 molecules 
10
 44.11 g C3H8

 1 mole C3H8

  7.32 10 13 g of C3H8

46
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!
47
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.
(a)
 1 mole 
  1.25 moles
? moles O3  60.0 g O3 
 48.0 g O3 
48
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.
(b)
 6.022 10 23 molecules 

? molecules O 3  1.25 moles 
1 mole


 7.53 10 23 molecules O 3
49
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.
(b)
 3 O atoms 
23

? O atoms  7.53 10 molecules O3 
 1 O3 molecule 
 2.26 10 24 atoms O
50
Formula Weights, Molecular
Weights, and Moles
Example 2-8. Calculate the number of O
atoms in 26.5 g of Li2CO3.
You do it!
51
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 CO3
? O atoms  26.5 g Li 2 CO3 

73.8 g Li 2 CO3
6.022 10 23 form. units Li 2 CO3
3 O atoms


1 mol Li 2 CO3
1 formula unit Li 2 CO3
6.49 10 23 O atoms
52
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
53
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!
54
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 
55
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 C3 H 8
3 12.01 g

100%
44.11 g
56
 81.68%
Percent Composition and
Formulas of Compounds
What is the percent composition of H in C3H8?
You do it!
57
Percent Composition and
Formulas of Compounds
What is the percent composition of H in C3H8?
mass H
%H
 100%
mass C3 H 8
8 H

 100%
C3H 8
8  1.01 g

 100%  18.32%
44.11 g
or
18.32%  100%  81.68%
58
Percent Composition and
Formulas of Compounds
Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 significant
figures.
You do it!
59
Percent Composition and
Formulas of Compounds
Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 sig. fig.
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%
60
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
61
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.
In 100.0 g of compound there are:



24.74 g of K
34.76 g of Mn
40.50 g of O
62
Derivation of Formulas from
Elemental Composition
1 mol K
? mol K  24.74 g K 
 0.6327 mol K
39.10 g K
63
Derivation of Formulas from
Elemental Composition
? 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
64
Derivation of Formulas from
Elemental Composition
1 mol K
? 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
obtain smallest whole number ratio
65
Derivation of Formulas from
Elemental Composition
1 mol K
? 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
obtain smallest whole number ratio
0.6327
for K 
1K
0.6327
0.6327
for Mn 
 1 Mn
0.6327
66
Derivation of Formulas from
Elemental Composition
? 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
1mol O
? mol O  40.50 g O 
 2.531 mol O
16.00 g O
obtain smallest whole number ratio
0.6327
for K 
1K
0.6327
0.6327
for Mn 
 1 Mn
0.6327
2.531
for O 
4O
0.6327
thus the chemical formula is KMnO 4
67
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!
68
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?
1 mol Co
? mol Co  6.541 g Co 
 0.1110 mol Co
58.93 gCo
1mol O
? mol O  2.368 g O 
 0.1480 mol O
16.00 g O
find smallest whole number ratio
69
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?
0.1110
0.1480
for Co 
 1 Co for O 
 1.333O
0.1110
0.1110
multipy both by 3 to turn fraction to whole number
1 Co  3  3 Co 1.333 O  3  4 O
Thus the compound' s formula is :
Co3O 4
70
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?

short cut method
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
71
Determination of Molecular
Formulas
convert masses to moles
1 mol C
48.0 g of C 
 4 mol C
12.0 g C
1 mol H
8.10 g of H 
 8 mol H
1.01 g H
Thus the formula is :
C4H8
72
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
73
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
74
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
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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
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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 3 PO 4
unit factor
100.0 g sample
77
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 3 PO 4
unit factor
100.0 g sample
98.3 g Na 3PO 4
? g Na 3PO 4  250.0 g sample 
100.0 g sample
= 246 g Na 3 PO 4
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