Chapter 6
Chemical Composition
2006, Prentice Hall
CHAPTER OUTLINE
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The Mole Concept
Molecular & Molar Mass
Calculations Using the Mole
Percent Composition
Chemical Formulas
Calculating Empirical Formulas
Molecular Formulas
2
Why is Knowledge of
Composition Important?
• everything in nature is either chemically or
physically combined with other substances
• to know the amount of a material in a
sample, you need to know what fraction of
the sample it is
• Some Applications:
–
–
–
–
the amount of sodium in sodium chloride for diet
the amount of iron in iron ore for steel production
the amount of hydrogen in water for hydrogen fuel
the amount of chlorine in freon to estimate ozone
depletion
Tro's Introductory Chemistry,
Chapter 6
THE MOLE
CONCEPT
 Chemists find it more convenient to use
mass relationships in the laboratory, while
chemical reactions depend on the number
of atoms present.
 In order to relate the mass
and number of atoms,
chemists use the SI unit
mole (abbreviated mol).
4
THE MOLE
CONCEPT
 The number of particles in a mole is called
Avogadro’s number and is 6.02x1023.
1 mole
equals to
6.02 x 1023
Avogadro’s
number (NA)
5
THE MOLE
CONCEPT
A mole is a very large quantity
6.02x1023
If 10,000 people started to count
Avogadro’s number and counted at
the rate of 100 numbers per minute
each minute of the day, it would
take over 1 trillion years to count
the total number.
6
THE MOLE
CONCEPT
1 mole H atoms = 6.02x1023 H atoms
1 mole H2 molecules = 6.02x1023 H2 molecules
= 2 x (6.02x1023) H atoms
1 mole H2O molecules = 6.02x1023 H2O molecules
= 2 x (6.02x1023) H atoms
= 6.02x1023 O atoms
1 mole Na+ ions = 6.02x1023 Na+ ions
7
THE MOLE
CONCEPT
 The atomic mass of one atom expressed
in amu is numerically the same as the
mass of 1 mole of atoms of the element
expressed in grams.
Mass
Mass
Massof
of
of111H
Mg
Clatom
atom
atom
= =1.008
= 35.45
24.31
amu
amu
amu
Mass
Mass
Massof
of
of111mol
mol
molHMg
Cl
atoms
atoms
atoms
==
1.008
= 35.45
24.31
grams
gg
8
MOLE
DAY
 Chemists and chemistry students
celebrate two days in the year in honor of
the Mole and call them Mole Days.
October 23rd
Jun 2nd
6:02 a.m.
10:23 a.m.
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MOLECULAR MASS
 The sum of atomic masses of all the atoms
in one molecule of a substance is called
molecular mass, and is measured in amu.
Mass of one molecule of H2O
2 H atoms = 2 (1.008 amu) = 2.016
amu
1 O atom = 1 (16.00 amu) = 16.00 amu
Molecular mass
18.02 amu
10
Relationship Between
Moles and Mass
• The mass of one mole of atoms is called
the molar mass
• The molar mass of an element, in grams,
is numerically equal to the element’s
atomic mass, in amu
MOLAR MASS
 The mass of one mole of a substance is
called molar mass, and is measured in
grams.
Mass of one mole of H2O
2 mol H atoms = 2 (1.008 g) = 2.016 g
1 mol O atom = 1 (16.00 g) = 16.00 g
Molar mass
18.02 g
12
CALCULATIONS
USING THE MOLE
 Conversions between mass, mole and
particles can be done using molar mass
and Avogadro’s number.
Avogadro’s
Molar
number
mass
Mass of a
substance
MM
Moles of a
substance
NA
Particles of
a
substance
13
Example 1:
What is the mass of 5.00 mol of water?
18.02 g
5.00 m ol H 2 O x
= 90.1 g H 2 O
1 m ol
3 significant figures Molar mass
14
Example 2:
How many Mg atoms are present in 5.00 g of Mg?
mass  mol  atoms
5.00 g M g x
Molar mass
1
m ol
24.3
g
x
6.02 x 1023
1
atom s
=
m ol
1.24x1023 atoms Mg Avogadro’s
3 significant figures
number
15
Example 3:
How many molecules of HCl are present in 25.0 g
of HCl?
mass  mol  molecules
25.0 g H C l x
1 m ol
36.45 g
x
6.02 x 1023 m olecu les
1
=
m ol
4.13 x 1023 molecules HCl
3 significant figures
16
PERCENT
COMPOSITION
 The percent composition of a compound
is the mass percent of each element in the
compound.
M ass % X =
(n o. of X in form u la) x (m olar m ass of X )
x 100
m olar m ass of com p ou n d
Total mass of element
Total mass of compound
17
Example 1:
Calculate the percent composition of sodium
chloride (NaCl).
Step 1: determine molar mass of NaCl
1 mol Na atoms 1 (22.99 g) = 22.99
= 1
g (35.45 g) = 35.45 g
1 mol Cl atom
=
Molar
58.44
mass
g/mol
18
Example 1:
Step 2: calculate the mass % of each element
22.99 g
% Na =
58.44 g
35.45 g
% Cl =
x 100 = 39.34%
x 100 = 60.66%
58.44 g
Sum = 100%
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Example 2:
1.63 g of zinc combines with 0.40 g of oxygen to form
zinc oxide. Determine the % composition of the
compound formed.
Step 1: determine total mass of sample
mass of sample = 1.63 g + 0.40 g = 2.03 g
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Example 2:
Step 2: calculate the mass % of each element
% Zn =
%O=
1.63 g
x 100 = 80.3%
2.03 g
0.40 g
x 100 = 19.7%
2.03 g
Sum = 100%
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Example 3:
Calculate the percent composition of sodium
hydroxide (NaOH).
Step 1: determine molar mass of NaOH
1 mol Na atoms = 1 (23.0 g) = 23.0 g
1 mol O atom = 1 (16.0 g) = 16.0 g
1 mol H atoms = 1 (1.01 g) = 1.01 g
Molar mass
40.0 g/mol
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Example 3:
Calculate the percent composition of sodium
hydroxide (NaOH).
Step 1: determine molar mass of NaOH
1 mol Na atoms = 1 (23.0 g) = 23.0 g
1 mol O atom = 1 (16.0 g) = 16.0 g
1 mol H atoms = 1 (1.01 g) = 1.01 g
Molar mass
40.0 g/mol
23
CHEMICAL
FORMULAS
Molecular
formula
Shows
Can be the
written
actual
for
number
molecular
of atoms
compounds
in a compound
only
Empirical
formula
Shows
simplest
Can bethe
written
for
ratio
of atoms
in a
molecular
and
compound
ionic
compounds
24
CHEMICAL
FORMULAS
Molecular
Empirical
Multiplier
H2O2
HO
2
H2O
H2O
1
C6H12O6
CH2O
6
C6H6
CH
6
C2H2
CH
2
C6H12
CH2
6
25
CHEMICAL
FORMULAS
 Several compounds may possess the same
percent composition and empirical formula,
but different molecular formulas.
Formula
CH
Composition
%C
%H
92.3
7.7
Molar mass
(g/mol)
13.02
C2H2
92.3
7.7
26.04 (2x13.02)
C6H6
92.3
7.7
78.12 (6x13.02)
26
Finding an Empirical Formula
1) convert the percentages to grams
a) skip if already grams
2) convert grams to moles
a) use molar mass of each element
3) write a pseudoformula using moles as
subscripts
4) divide all by smallest number of moles
5) multiply all mole ratios by number to make all
whole numbers
a) if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67,
multiply all by 3, etc.
b) skip if already whole numbers
If, after dividing by the smallest number of moles, the
subscripts are not whole numbers, multiply all the
subscripts by a small whole number to arrive at wholenumber subscripts.
CALCULATING
EMPIRICAL FORMULAS
Arsenic (As) reacts with oxygen (O) to form a
compound that is 75.7% As and 24.3% oxygen
by mass. What is the empirical formula for this
compound?
Step 1: Percent to mass
Assume 100 g
75.7% As
75.7 g As
24.3% O
24.3 g O
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CALCULATING
EMPIRICAL FORMULAS
Step 2: Mass to mole
75.7 g As x
24.3 g O x
1 m ol
= 1.01 m ol A s
74.9 g
1 m ol
= 1.52 m ol O
16.0 g
Use atomic mass
of oxygen
30
CALCULATING
EMPIRICAL FORMULAS
Step 3: Divide by small
As =
1.01 m ol
= 1.00
1.01 m ol
O=
1.52 m ol
= 1.50
1.01 m ol
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CALCULATING
EMPIRICAL FORMULAS
Step 4: Multiply till Whole
AsAs
1.002O3
1.50
x2=2
x2=3
32
Example 1:
Determine the empirical formula for a compound
containing 11.2% H and 88.8% O.
mol H =
mol O =
11.2 g x
1 m ol
= 11.1 m ol H
1.01 g
88.8 g x
1 m ol
11.1
= 2.0
5.55
= 5.55 m ol O
16.0 g
5.55
= 1.0
5.55
H2O
33
Example 2:
Determine the empirical formula for a compound
with the following percent
52.14% C,
C2Hcomposition:
O
6
13.12% H, 34.73% O.
mol C =
52.14 g x
mol H = 13.12
mol O =
g x
34.73 g x
1 m ol
= 4.345 m ol C
4.345
12.0 g
2.17
1 m ol
13.0
= 13.0 m ol H
1.01 g
2.17
1 m ol
2.17
16.0 g
= 2.17 m ol O
= 2.0
= 6.0
= 1.0
2.17
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MOLECULAR
FORMULAS
 Molecular formula can be calculated from
empirical formula if molar mass is known.
Molecular formula = (empirical formula) n
n (m ultiplier) =
M olar m ass
M ass of em pirical form ula
35
Example 1:
A compound of N and O with a molar mass of 92.0
g, has the empirical formula of NO2. What is its
molecular formula?
Mass of empirical formula = 14.0 + 2(16.0) = 46.0
n =
92.0 g
= 2
46.0 g
Molecular formula = 2 x (NO2) = N2O4
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Example 2:
Calculate the empirical and molecular formulas of
a compound that contains 80.0% C and 20.0% H,
and has a molar mass of 30.00 g.
mol C =
mol H =
80.0 g x
20.0 g x
Empirical
formula
1 m ol
= 6.667 m ol C
6.667
12.0 g
6.667
1 m ol
19.8
= 19.8 m ol H
1.01 g
= 1.0
= 3.0
6.667
CH3
37
Example 2:
Empirical formula = CH3
Mass of empirical formula = 12.0 + 3(1.01) = 15.0
n =
30.0 g
= 2
15.0 g
Molecular formula = 2 x (CH3) = C2H6
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THE END
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