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Chapter 3 Mass Relationships in Chemical ReactionsSemester 1/2012
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
1Ref: Raymong Chang/Chemistry/Ninth Edition
Prepared by A. Kyi Kyi Tin
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 amu = times mass of one carbon –12 atom.
By definition:1 atom 12C “weighs” 12 amu
1 amu = x 12 amu
Ex: atomic mass of ‘H’ atom = 8.4% of carbon-12 Atom
= 0.084 x 12.00 amu
= 1.008 amu
12
1
12
1
2
%)91.30(%)09.69( 6529
6329 CuCu
Atomic masses 62.93amu + 64.9278 amu
A.A.M = (0.6909)(62.93amu)+(0.3091x64.9278amu)
= 63.55amu
Ex:3.1 Calculate the average atomic mass of copper.
3
The average atomic mass is the weightedaverage of all of the naturally occurringisotopes of the element.
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)
1 mole = 6.02x1023
Atoms
Molecule
Ions
Molar mass() mass [ in “g” (or) “Kg” ] of 1 mole of units
(atom (or) molecule (or) ion)4
From periodic Table
Element Atomic Mass Molar mass for “Atom” Molecule Molar mass for molecule
H 1.008 amu
O 16.00 amu
Cl 35.5 amu
Na 22.99 amu
C 12.01 amu
5
***For any element
atomic mass (amu) = molar mass (grams)
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 =
gxx
g 2323
10993.11002.6
00.12
1 atom of carbon-12 atom = 12amu
1 amu = gxgx 24
23
10661.112
10933.1
6
7
x 6.022 x 1023 atoms K1 mol K
=
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
0.551 g K 1 mol K
39.10 g Kx
8.49 x 1021 atoms K
3.3 Molecular mass (molecular weight)
Sum of atomic masses (in amu) in the molecule
Ex:
8
For any molecule
molecular mass (amu) = molar mass (grams)
SO2
1S 32.07 amu
2O + 2 x 16.00 amu
SO2 64.07 amu
9
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 H = 6.022 x 1023 atoms H
5.82 x 1024 atoms H
1 mol C3H8O molecules = 8 mol H atoms
72.5 g C3H8O1 mol C3H8O
60 g C3H8Ox
8 mol H atoms
1 mol C3H8Ox
6.022 x 1023 H atoms
1 mol H atomsx =
10
Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound.
1Na 22.99 amu
1Cl + 35.45 amuNaCl 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
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
2 P 2 x 30.97
8 O + 8 x 16.00
310.18 amu
3.5 Percent Composition of the Compounds
Ex: H2O2
1mol of H2O2
2 mol of ‘H’ atom
2 mol of ‘O’ atom
Molar mass of H2O2 = (2x1.008 +32) = 34.016 g / mol
%H =
%O =
%926.5%100016.34
008.12x
g
gx
%06.94%100016.34
162x
g
gx12
13
Percent composition of an element in a compound =
n x molar mass of elementmolar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
C2H6O
%C =2 x (12.01 g)46.07 g x 100% = 52.14%
%H =6 x (1.008 g)46.07 g x 100% = 13.13%
%O =1 x (16.00 g)46.07 g x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
3.6 Empirical Formula
Formula for a compound that contains the smallest whole number ratios for the elements in the compound.
Ex C : H : O
:25.0
500.0 mol:
25.0
50.1 mol25.0
25.0 mol
2 : 6 : 1
i.e C2H6O
Mole ratio 0.500 : 1.50 : 0.25
Smallest whole number ratios
14
15
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 = 0.6330 mol K1 mol K39.10 g K
nMn = 34.77 g Mn x = 0.6329 mol Mn1 mol Mn54.94 g Mn
nO = 40.51 g O x = 2.532 mol O1 mol O16.00 g O
16
Percent Composition and Empirical Formulas
K : ~~ 1.00.63300.6329
Mn : 0.63290.6329
= 1.0
O : ~~ 4.02.5320.6329
nK = 0.6330, nMn = 0.6329, nO = 2.532
KMnO4
Ex:3.11 COMPOUND
Nitrogen 1.52g
Oxygen 3.47g
Mole = :/14
52.1
molg
g217.0:108.0
/16
47.3
molg
g
108.0
108.0
108.0
217.0:
1 : 2
Empirical Formula NO2
Empirical molar mass = 14.01+(16x2) = 46.01g
Smallest whole number ratio
17
201.46
90
..
_
MassMolarEmpirical
massmolar
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 CO2(g)
2 molecules + 1 molecule 2 molecules
2 mol + 1 mol 2 mol18
Molecular Formula= (NO2)2 = N2O4
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 + O2 2NO2INITIAL mole(given) 8 7
Balanced Equation 2mol + 1mol 2 mol
8 mol of “NO” yields…..8 mol of ”NO2” 7 mol of “O2”..yields …14 mol of “NO2”
O2 is ExcessNO is Limiting19
20
Limiting Reagent:
2NO + O2 2NO2
NO is the limiting reagent
O2 is the excess reagent
Reactant used up first in the reaction.
21
In one process, 124 g of Al are reacted with 601 g of Fe2O32Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al mol Al mol Fe2O3 needed g Fe2O3 needed
ORg Fe2O3 mol Fe2O3 mol Al needed g Al needed
124 g Al1 mol Al
27.0 g Alx
1 mol Fe2O3
2 mol Alx
160. g Fe2O3
1 mol Fe2O3
x = 367 g Fe2O3
Start with 124 g Al need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
22
Use limiting reagent (Al) to calculate amount of product thatcan be formed.
g Al mol Al mol Al2O3 g Al2O3
124 g Al1 mol Al
27.0 g Alx
1 mol Al2O3
2 mol Alx
102. g Al2O3
1 mol Al2O3
x = 234 g Al2O3
2Al + Fe2O3 Al2O3 + 2Fe
At this point, all the Al is consumed and Fe2O3 remains in
excess.
3.10 Reaction Yield
%100..
..% x
yeildltheoretica
yeildactualyield
Theoretical yield is the amount of product that wouldresult if all the limiting reagent reacted.
[can be obtained from calculation based on balanced equation.]Actual yield is the amount of product actually obtainedfrom a reaction.
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