Transcript Document

Chapter 3 Mass Relationships in Chemical Reactions
Semester 2/2012
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: Raymond Chang/Chemistry/Tenth 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.
By definition:1 atom 12C “weighs” 12 amu
1 amu = 1 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
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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 
The average atomic mass is the weighted
average of all of the naturally occurring
isotopes of the element.
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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
Chemist Measure Atoms and molecules in a unit called “moles”
( A unit to count numbers of particles)
Atoms
1 mole = 6.02x1023
Molecule
Ions
Molar mass()  mass [ in “g” (or) “Kg” ] of 1 mole of units
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(atom (or) molecule (or) ion)
From periodic Table
Element Atomic Mass
H
1.008 amu
O
16.00 amu
Cl
35.5 amu
Na
22.99 amu
C
12.01 amu
Molar mass for
“Atom”(g/mol)
Molecule
Molar mass for molecule
(g/mol)
***For any element
atomic mass (amu) = molar mass (grams)
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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
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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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Example:How many atoms are in 0.551 g of potassium (K) ?
1 mol “K” = 39.10 g “K”
1 mol “K” = 6.022 x 1023 atoms “K”
1 mol K
0.551 g K x
x
39.10 g K
6.022 x 1023 atoms K
=
1 mol K
8.49 x 1021 atoms K
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3.3
Molecular mass (molecular weight)
Sum of atomic masses (in amu) in the molecule
Ex:
1S
2O
SO2
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
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How many “H” atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
8 mol H atoms
6.022 x 1023 H atoms
1 mol C3H8O
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
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Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
NaCl
1Na
22.99 amu
1Cl
NaCl
+ 35.45 amu
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
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What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08
2P
2 x 30.97
8O
+ 8 x 16.00
310.18 amu
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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%
34.016g
%O =
2 x16g
x100%  94.06%
34.016g
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Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole of
the compound
C2H6O
2 x (12.01 g)
%C =
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
52.14% + 13.13% + 34.73% = 100.0%
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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
0.500 mol
:
0.25
2
H
:
:
0.25
0.25 mol
0.25
1.50 mol
:
0.25
:
O
6
:
i.e
C2H6O
1
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Percent Composition and Empirical Formulas
Determine the empirical formula of a
compound that has the following
percent composition by mass: K 24.75,
Mn 34.77, O 40.51 percent.
nK = 24.75 g K x 1 mol K = 0.6330 mol K
39.10 g K
nMn = 34.77 g Mn x 1 mol Mn= 0.6329 mol Mn
54.94 g Mn
nO = 40.51 g O x 1 mol O = 2.532 mol O
16.00 g O
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Percent Composition and Empirical Formulas
nK = 0.6330, nMn = 0.6329,
nO = 2.532
KMnO4
0.6330 ~
K:
1.0
~
0.6329
0.6329
= 1.0
Mn :
0.6329
2.532 ~
O:
4.0
~
0.6329
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
Ex:3.11
COMPOUND
Nitrogen
1.52g
Mole =
Smallest whole
number ratio
1.52g
3.47g
:
 0.108: 0.217
14g / m ol 16g / m ol
0.1 0 8
0.1 0 8
1

Oxygen
3.47g
:
:
0.217
0.108
2
Empirical Formula NO2
Empirical molar mass = 14.01+(16x2) = 46.01g
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m olar_ m ass
90

2
Em pirical.Molar.Mass 46.01
 Molecular Formula= (NO2)2 = N2O4
Molecular Mass(or) Molar Mass = 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
18
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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Limiting Reagent:
Reactant used up first in the
reaction.
2NO + O2
2NO2
NO is the limiting reagent
O2 is the excess reagent
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In one process, 124 g of Al are reacted with 601 g of
Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
g Fe2O3
mol Fe2O3 needed
OR
mol Al needed
mol Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
g Fe2O3 needed
1 mol Fe2O3
2 mol Al
Start with 124 g Al
g Al needed
160. g Fe2O3
=
x
1 mol Fe2O3
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
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Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
g Al2O3
Al2O3 + 2Fe
1 mol Al2O3
2 mol Al
102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
At this point, all the Al is
consumed and Fe2O3 remains in
excess.
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3.10 Reaction Yield
actual..yeild
% yield 
x100%
theoretical..yeild
Theoretical yield is the amount of product that would
result if all the limiting reagent reacted.
[can be obtained from calculation based on balanced
equation.]
Actual yield is the amount of product actually obtained
from a reaction.
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