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Chapter 3 Mass Relationships in Chemical Reactions
Semester 1/2011
3.1 Atomic Mass
3.2 Avogadro’s Number and the Molar Mass of an element
3.3 Molecular Mass
3.5 Percent Composition of Compounds
3.6 Experimental Determination of Empirical Formulas
3.7 Chemical Reactions and Chemical Equations
3.8 Amounts of Reactants and Products
3.9 Limiting Reagents
3.10 Reaction Yield
Ref: Raymong Chang/Chemistry/Ninth Edition
Prepared by A. Kyi Kyi Tin
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3.1 Atomic Mass(Atomic weight)
Mass of the atom in atomic mass units (amu), which is based on the
carbon-12 isotope scale.
amu = atomic mass unit
Define: 1amu
1
 1 amu = 12 times mass of one carbon –12 atom.
Mass of one carbon-12 atom = 12 amu
1
1 amu =
x 12 amu
12
Ex:
atomic mass of ‘H’ atom
=
8.4% of carbon-12 Atom
=
0.084 x 12.00 amu
=
1.008 amu
2
Ex:3.1 Calculate the average atomic mass of copper.
Cu (69.09%)  Cu (30.91%)
63
29
65
29
Atomic masses 62.93amu
+
64.9278 amu
A.A.M = (0.6909)(62.93amu)+(0.3091x64.9278amu)
= 63.55amu 
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3.2 Avogadro’s Number and the Molar Mass of an Element
(Italian scientist..Amedeo Avogadro)
Amedeo Avogadro’s number  NA
Pair  2 items , Dozen = 12 items , Gross = 144 items
Chemist Measure Atoms and molecules in a unit called “moles”
Atoms
1 mole = 6.02x1023
Molecule
Ions
Molar mass()  mass [ in “g” (or) “Kg” ] of 1 mole of units
(atom (or) molecule (or) ion)
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From periodic Table
Element Atomic Mass
Molar mass for “Atom”
Molecule
Molar mass for molecule
H
1.008 amu
1.008 g / mol
H2
1.008x2 = 2.016g/mol
O
16.00 amu
16.00 g / mol
O2
16.00x2 = 32.00g/mol
Cl
35.5 amu
35.5 g / mol
Cl2
HCl
35.5x2 = 71.00 g / mol
(1.008+35.5)
=36.5.08g / mol
Na
22.99 amu
22.99 g / mol
Na
22.99 g / mol
C
12.01 amu
12.01 g / mol
CO
(12.01+16.00)
= 28.01 g / mol
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1 mol of ‘H’ atom = 1.008 g = 6.02x1023 atoms of ‘H’ atom
1 mol of ‘H2’ moleule = (1.008x2) g = 6.02 x1023 molecules of ‘H2’ molecule
1 mol of ‘Na’ atom = 22.99 g = 6.02x1023 atoms of ‘Na’ atom
1 mol of ‘O’ atom = 16.00 g = 6.02x1023 atoms of ‘O’ atom
1 mol of ‘O2’ moleule = (16.00x2)g = 6.02x1023 molecule of ‘O2’ molecule
1 mol of carbon-12 atom = 12g = 6.02x1023 atoms of carbon-12 atom
6.02x1023 atoms of carbon-12 atom = 12 g
1 atom of carbon-12 atom =
12.00 g
 1.993 x10  23 g
23
6.02 x10
1 atom of carbon-12 atom = 12amu
1.933x10 23 g
 1.661x10 24 g
 1 amu =
12
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Solve this problem in two ways: ‘He’
Ex: 3.2
1st method. 1 mol of ‘He’ atom = 4.003g = 6.02x1023 atoms of ‘He’ atom
i.e 4.003g 1 mol of ‘He’ atom
6.46g  ?
(6.46 g ) x(1mol)
of ' He ' atom
4.003g
=1.61 mol of ‘He’ atom
2nd. Method
Apply Conversion factor
1mol ( He )
4.003g
....(or )........
4.003g
1mol ( He )
mass
mol 
molar.mass
m
n
M
1mol( He)
x6.46 g  1.61mol Of ‘He’ atom
4.003g
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3.3
Molecular mass (molecular weight)
Sum of atomic masses (in amu) in the molecule
Ex:
H2O
2(atomic mass of H)+1 atomic mass of O
2(1.008 amu) + 16.00amu = 18.02amu
Note: For Ionic compounds like NaCl and MgO
WE USE THE TERM “Formula Mass”
Formula mass of NaCl = 22.99 amu
+ 35.45 amu = 58.44 amu
Atomic mass + Atomic mass
of “Na”
of “Cl”
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3.5 Percent Composition of the Compounds
Ex:
H2O2
2 mol of ‘H’ atom
1mol of H2O2
2 mol of ‘O’ atom
Molar mass of H2O2 = (2x1.008 +32) = 34.016 g / mol
%H = 2 x1.008g x100%  5.926%
34.016 g
%O =
2 x16 g
x100%  94.06%
34.016 g
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3.6
Empirical Formula
Formula for a compound that contains the smallest whole number
ratios for the elements in the compound.
Ex
Mole ratio
Smallest
whole
number
ratios
C
0.500
:
: 1.50
0.500mol
:
0.25
2
H
:
:
0.25
0.25mol
0.25
1.50mol
:
0.25
:
O
6
:
i.e
C2H6O
1
10

Ex:3.11
COMPOUND
Nitrogen
1.52g
Mole =
Smallest whole
number ratio
1.52 g
3.47 g
:
 0.108 : 0.217
14 g / mol 16 g / mol
0.108
0.108
1

Oxygen
3.47g
:
:
0.217
0.108
2
Empirical Formula NO2
Empirical molar mass = 14.01+(16x2) = 46.01g
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molar _ mass
90

2
Empirical .Molar .Mass 46.01
(NO2)2 = N2O4 = 28.02+64 = 92.02g/mol
3.8 Amounts of Reactants and Products
Stoichiometry is the quantitative study of reactants and
products in a balanced chemical reaction.
2 CO (g)
+
O2 (g)
2 molecules + 1 molecule
2 CO2(g)
2 molecules
12
2 mol
+ 1 mol
2 mol
3.9 Limiting Reagents (L.R)
Limiting Reagent….. The reactant used up first in a
reaction.
Excess Reagent.. The reactants present in quantities
greater than necessary to react with the quantity of the
limiting reagent.
Ex:
2NO
INITIAL mole(given) 8
Balanced Equation 2mol +
+
O2

2NO2
7
1mol

2 mol
8 mol of “NO” yields…..8 mol of ”NO2”
7 mol of “O2”..yields …14 mol of “NO2”
NO is Limiting
O2 is Excess
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3.10 Reaction Yield
actual.. yeild
% yield 
x100%
theoretical.. yeild
Theoretical yield can be obtained from calculation based
on balanced equation.
Actual yield can be obtained from the given problem.
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