Chemical Composition
Chapter 8
Nutrasweet
Aspartic acid
Analysis
Qualitative -- what elements -- C, H, N,
& O.
Quantitative -- how many of each
element --C4H7NO4.
Counting Atoms
Atoms are too small to be seen or counted
individually.
Atoms can only be counted by weighing them.
• all jelly beans are not identical.
• jelly beans have an average mass of 5 g.
• How could 1000 jelly beans be counted?
Jelly Beans & Mints
Mints have an average
mass of 15 g.
How would you count out
1000 mints?
Why do 1000 mints have
a mass greater than
1000 jelly beans?
Atomic Mass Unit
Atoms are so tiny that the gram is much too
large to be practical.
The mass of a single carbon atom is 1.99
x 10-23 g.
The atomic mass unit (amu) is used for atoms
and molecules.
AMU’s and Grams
1 amu = 1.661 x 10 -24 g
Conversion Factors
1.661 x 10-24g/amu
6.022 x 1023amu/g
Calculating Mass Using
AMU’S
1 N atom = 14.01 amu
(23 N atoms)(14.01 amu/1N atom) = 322.2 amu
Calculating Number of Atoms
from Mass
1 O atom = 16.00 amu
(288 amu)(1 O atom/16.00 amu) = 18 atoms O
Atomic Masses
Elements occur in nature as mixtures of
isotopes
Carbon =
98.89% 12C
1.11% 13C
<0.01% 14C
Carbon atomic mass = 12.01 amu
AMU’s & Grams
1 atom C = 12.011 amu = 1.99 x 10-23g
1 mol C = 12.011 g
Use TI-83 or TI-83 Plus to store 6.022 x 1023
to A.
Measurements
dozen = 12
gross = 12 dozen = 144
ream = 500
mole = 6.022 x 1023
The Mole
The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022  1023 units of
that thing
Equal moles of substances have equal
numbers of atoms, molecules, ions,
formula units, etc.
Figure 8.1 (a): All
these sample of
pure elements
contain the same
number (a mole) of
Pb – 207.2g
atoms: 6.022 x 1023
atoms
Cu – 63.55g
Ag – 107.9g
Figure 8.2: Onemole samples of
iron (nails),
iodine crystals,
liquid mercury,
and powdered
sulfur
How many atoms does each substance contain?
The Mole
Substance Average Atomic Mass
(g)
Na
22.99
Cu
63.55
S
32.06
Al
26.98
# Moles
1
1
1
1
# Atoms
6.022 x 1023
6.022 x 1023
6.022 x 1023
6.022 x 1023
Avogadro’s number
equals
23
6.022  10 units
The Mole
• One mole of rice grains is more than the
number of grains of all rice grown since the
beginning of time!
• A mole of marshmallows would cover the
U.S. to a depth of 600 miles!
• A mole of hockey pucks would be equal in
mass to the moon.
Unit Cancellation
How many dozen eggs would 36 eggs be?
(36 eggs)(1 dozen eggs/12 eggs) = 3 dozen eggs
How many eggs in 5 dozen?
(5 dozen eggs)(12 eggs/1 dozen eggs) = 60 eggs
Calculating Moles & Number of Atoms
1 mol Co = 58.93 g
(5.00 x 1020 atoms Co)(1mol/6.022 x 1023 atoms)
= 8.30 x 10-4 mol Co
(8.30 x 10-4 mol)(58.93g/1 mol) = 0.0489 g Co
Moles are the doorway
grams <---> moles <---> atoms
Molar Mass
A substance’s molar mass is the mass in
grams of one mole of the compound.
CO2 = 44.01 grams per mole
1 C = 1 (12.011 g) = 12.011 g
2 O = 2 ( 16.00 g) = 32.00 g
44.01 g
Calculating Mass from Moles
CaCO3
1 Ca = 1 (40.08 g) = 40.08 g
1 C = 1 (12.01 g) = 12.01 g
3 O = 3 (16.00 g) = 48.00 g
100.09 g
(4.86 molCaCO3)(100.09 g/1 mol) = 486 g CaCO3
Calculating Moles from Mass
Juglone
10 C = 10(12.01g) = 120.1 g
6 H = 6(1.008 g) = 6.048 g
3 O = 3(16.00 g) = 48.00 g
174.1 g
(1.56 g juglone)(1 mol/174.1 g) = 0.00896
mol juglone
Percent Composition
Mass percent of an element:
mass of element in compound
mass % 
( 100%)
mass of compound
For iron in iron (III) oxide, (Fe2O3)
11169
. g
mass % Fe 
( 100%)  69.94%
159.69 g
% Composition
CuSO4. 5 H2O
1 Cu = 1 (63.55 g) = 63.55 g
1 S = 1 (32.06 g) = 32.06 g
4 O = 4 (16.00 g) = 64.00 g
5 H2O = 5 (18.02 g) = 90.10 g
249.71 g
% Composition
(Continued)
% Cu = 63.55 g/249.71g (100 %) = 25.45 % Cu
% S = 32.06 g/249.71 g (100 %) = 12.84 % S
% O = 64.00 g/249.71 g (100 %) = 25.63 % O
% H2O = 90.10 g/249.71 g (100 %) = 36.08 % H2O
Check: Total percentages. Should be equal to 100 %
plus or minus 0.01 %.
Formulas
molecular formula = (empirical formula)x
[x = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH
Formulas
Ionic compounds -- empirical formula
NaCl
CaCl2
Covalent compounds -- molecular formula C6H12O6
C2H6
Empirical Formula Determination
1. Base calculation on 100 grams of
compound.
2. Determine moles of each element in 100
grams of compound.
3. Divide each value of moles by the smallest
of the values.
4. Multiply each number by an integer to
obtain all whole numbers (if necessary.)
Calculating Empirical Formulas
63.68 % C, 12.38 % N, 9.80 % H, & 14.14 % O
(63.68 g C)(1 mol/12.01g) = 5.302 mol C/.8837 mol = 6
(12.38 g N)(1 mol/14.01g) = 0.8837 mol N/.8837 mol = 1
(9.80 g H)(1 mol/1.01 g)
= 9.70 mol H/.8837 mol = 11
(14.14 g O)(1 mol/16.00g) = .8838 mol O/.8837 mol = 1
C6NH11O
Calculating Empirical Formulas
4.151 g Al & 3.692 g O
(4.151 g Al)(1 mol/26.98 g) = 0.1539 mol Al/0.1539 mol = 1.000
(3.692 g O)(1 mol/16.00 g) = 0.2308 mol O/0.1539 mol = 1.500
1 Al (2) = 2 Al
1.5 O (2) = 3 O
Al2O3
Molecular Formulas
71.65 % Cl, 24.27 % C, & 4.07 % H
(71.65g Cl)(1 mol/35.45g) = 2.021 mol/2.021 mol = 1
(24.27 g C)(1 mol/12.01g) = 2.021 mol/2.021 mol = 1
(4.07 g H )(1 mol/1.01g) = 4.03 mol/2.021 mol = 2
(EM)x = (MM)
(49.46)x = (98.96)
x = 2
(EF)x
= (MF)
(ClCH2)2 = Cl2C2H4