Transcript PowerPoint Presentation - Formulas, Equations, and Moles

Chapter
3
Formulas, Equations,
and Moles
Chemistry, 4th Edition
McMurry/Fay
Dr. Paul Charlesworth
Michigan Technological University
Balancing Chemical Equations
•
A balanced chemical equation represents the
conversion of the reactants to products such that
the number of atoms of each element is conserved.
reactants  products
limestone  quicklime + gas
Calcium carbonate  calcium oxide + carbon dioxide
CaCO3(s)  CaO(s) + CO2(g)
2
Balancing Chemical Equations
CaCO3(s)  CaO(s) + CO2(g)
The letters in parentheses following each substance are
called State Symbols
(g) → gas
(l) → liquid (s) → solid (aq) → aqueous
3
Balancing Chemical Equations
A balanced equation MUST have the same number of
atoms of each element on both sides of the equation.
H2 + O2 → H2O
Not Balanced
H2 + ½O2 → H2O Balanced
2H2 + O2 → 2H2O Balanced
4
Balancing Chemical Equations
The numbers multiplying chemical formulas
in a chemical equation are called:
Stoichiometric Coefficients (S.C.)
2H2 + O2 → 2H2O Balanced
Here 2, 1, and 2 are stoichiometric coefficients.
5
Balancing Chemical Equations
Hints for Balancing Chemical Equations:
1)
Save single element molecules for last.
2)
Try not to change the S.C. of a molecule containing
an element that is already balanced.
3)
If possible, begin with the most complex molecule
that has no elements balanced.
6
Balancing Chemical Equations
Hints for Balancing Chemical Equations:
4) Otherwise, trial and error!!
7
Balancing Chemical Equations
CH4 + O2 → CO2 + H2O
Example 1:
Balance O2 last
C is already balanced
Start by changing S.C. of H2O to balance H
CH4 + O2 → CO2 + 2H2O
8
Balancing Chemical Equations
Example 1:
CH4 + O2 → CO2 + 2H2O
Now C and H are balanced
Balance O by changing the S.C. of O2
CH4 + 2O2 → CO2 + 2H2O
BALANCED!
9
Balancing Chemical Equations
Example 2:
B2H6 + O2 → B2O3 + H2O
Balance O last
B is already balanced
Start by changing S.C. of H2O:
B2H6 + O2 → B2O3 + 3H2O
10
Balancing Chemical Equations
Example 2:
B2H6 + O2 → B2O3 + 3H2O
B and H are balanced
Balance O by changing S.C. of O2
B2H6 + 3O2 → B2O3 + 3H2O
BALANCED!
11
Balancing Chemical Equations
Example 3:
MnO2 + KOH + O2 → K2MnO4 + H2O
Balance O last
Mn is already balanced
Change S.C. of KOH to balance K
MnO2 + 2KOH + O2 → K2MnO4 + H2O
12
Balancing Chemical Equations
Example 3:
MnO2 + 2KOH + O2 → K2MnO4 + H2O
Mn, K, and H are balanced (H was balanced by chance)
Balance O
MnO2 + 2KOH + ½O2 → K2MnO4 + H2O
or
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
13
Balancing Chemical Equations
Example 4:
NaNO2 + H2SO4→
NO + HNO3 + H2O + Na2SO4
Hard one (no single element molecules)
S is balanced
Start with NaNO2 to balance Na
2NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4
14
Balancing Chemical Equations
Example 4:
2NaNO2 + H2SO4→
NO + HNO3 + H2O + Na2SO4
S, Na, and N are balanced
Cannot balance H without changing S.C. for H2SO4!
Boo! Option 1: trial and error
Option 2: Go on to next problem!
15
Balancing Chemical Equations
•
Balance the following equations:
C6H12O6 → C2H6O + CO2
Fe + O2 →
Fe2O3
NH3 + Cl2 →
N2H4 + NH4Cl
KClO3 + C12H22O11 → KCl + CO2 + H2O
16
17
Balancing Chemical Equations
•
Balance the following equations:
C6H12O6 → 2C2H6O + 2CO2
Fe + O2 →
Fe2O3
NH3 + Cl2 →
N2H4 + NH4Cl
KClO3 + C12H22O11 → KCl + CO2 + H2O
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Balancing Chemical Equations
•
Balance the following equations:
C6H12O6 → 2C2H6O + 2CO2
4Fe + 3O2 →
NH3 + Cl2 →
2Fe2O3 (balance O first)
N2H4 + NH4Cl
KClO3 + C12H22O11 → KCl + CO2 + H2O
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Balancing Chemical Equations
•
Balance the following equations:
C6H12O6 → 2C2H6O + 2CO2
4Fe + 3O2 →
NH3 + Cl2 →
2Fe2O3 (balance O first)
N2H4 + NH4Cl
N:H is 1:3 on left, must get 1:3 on right!
20
Balancing Chemical Equations
NH3 + Cl2 →
N2H4 + NH4Cl
N:H is 1:3 on left, must get 1:3 on right!
4NH3 + Cl2 →
N2H4 + 2NH4Cl
21
Balancing Chemical Equations
•
Balance the following equations:
C6H12O6 → 2C2H6O + 2CO2
4Fe + 3O2 →
4NH3 + Cl2 →
2Fe2O3
N2H4 + 2NH4Cl
KClO3 + C12H22O11 → KCl + CO2 + H2O (tough!)
22
Balancing Chemical Equations
•
Balance the following equations:
KClO3 + C12H22O11 → KCl + CO2 + H2O
balance C
KClO3 + C12H22O11 → KCl + 12CO2 + H2O
balance H
KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O
balance O
8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O
23
Balancing Chemical Equations
•
Balance the following equations:
8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O
balance K (and hope Cl is balanced)
8KClO3 + C12H22O11 → 8KCl + 12CO2 + 11H2O
Balanced!
24
Balancing Chemical Equations
•
Write a balanced equation for the reaction of element
A (red spheres) with element B (green spheres) as
represented below:
25
Atomic and Molecular Mass
mass of carbon12 atom
1 amu 
12
Mass:
proton = 1.00728 amu
neutron = 1.0086 amu
electron = 0.0005486
12C atom = 12.00000 amu
13C atom = 13.00335 amu
26
Atomic and Molecular Mass
•
The atomic masses as tabulated in the periodic table
are the averages of the naturally occurring isotopes.
•
Mass of C = average of 12C and 13C
= 0.9889 x 12 amu + 0.0111 x 13.0034 amu
= 12.011 amu
27
Atomic and Molecular Mass
The mass of a molecule is just the sum of the
masses of the atoms making up the molecule.
m(C2H4O2) = 2·mC + 4·mH + 2·mO
= 2·(12.01) + 4·(1.01) + 2·(16.00)
= 60.06 amu
28
Avogadro and the Mole
•
One mole of a substance is the gram mass value
equal to the amu mass of the substance.
•
One mole of any substance contains 6.02 x 1023
units of that substance.
•
Avogadro’s Number (NA, 6.022 x 1023) is the
numerical value assigned to the unit, 1 mole.
29
Avogadro and the Mole
30
Avogadro and the Mole
•
Methionine, an amino acid used by organisms to
make proteins, is represented below. Write the
formula for methionine and calculate its molar mass.
(red = O; gray = C; blue = N; yellow = S; ivory = H)
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Avogadro and the Mole
•
The Mole: Allows us to
make comparisons
between substances
that have
different
masses.
32
Avogadro and the Mole
33
Stoichiometry
•
Stoichiometry: Relates the moles of products and
reactants to each other and to measurable
quantities.
34
Stoichiometry
Aqueous solutions of NaOCl (household bleach)
are prepared by the reaction of NaOH with Cl2:
2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react
with 25.0 g of Cl2?
35
Stoichiometry
2 NaOH + Cl2 → NaOCl + NaCl + H2O
25.0 g Cl2 reacts with ? g NaOH
25.0 g Cl2 
1 m oleCl2
 0.353m olesCl2
70.90 g Cl2
1m oleCl2 2 m olesNaOH 40.0 g NaOH
25.0 g Cl2 


 28.2 g NaOH
70.90 g Cl2
1m oleCl2
1m oleNaOH
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Avogadro and the Mole
•
Calculate the molar mass of the following:
Fe2O3 (Rust)
C6H8O7 (Citric acid)
C16H18N2O4 (Penicillin G)
•
Balance the following, and determine how many
moles of CO will react with 0.500 moles of Fe2O3.
Fe2O3(s) + CO(g)
Fe(s) + CO2(g)
37
Avogadro and the Mole
Fe2O3 + CO → Fe + CO2
Balance (not a simple one)
Save Fe for last
C is balanced, but can’t balance O
In the products the ratio C:O is 1:2 and can’t change
Make the ratio C:O in reactants 1:2
Fe2O3 + 3CO → 2Fe + 3CO2
38
Avogadro and the Mole
Fe2O3 + 3CO → 2Fe + 3CO2
3 m oleCO
0.500 m oles Fe2O3 
 1.50m olesCO
1m ole Fe2O3
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Stoichiometry
40
Stoichiometry
•
Aspirin is prepared by reaction of salicylic acid
(C7H6O3) with acetic anhydride (C4H6O3) to form aspirin
(C9H8O4) and acetic acid (CH3CO2H). Use this
information to determine the mass of acetic anhydride
required to react with 4.50 g of salicylic acid. How many
grams of aspirin will result? How many grams of acetic
acid will be produced as a by-product?
41
Stoichiometry
Salicylic acid + Acetic anhydride →
Aspirin + acetic acid
C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H
C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2
Balanced!
Equal # moles for all
42
Stoichiometry
4.50 g Salicylic acid (C7H6O3) = ? moles
MW C7H6O3 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00
= 138.12 g/mole
1m ole S. A.
4.50 g S. A. 
 0.0326m oles S. A.
138.12 g S. A.
43
Stoichiometry
Since all compounds have the same S.C., there must be
0.0326 moles of all 4 of them involved in the reaction.
g Aspirin (C9H8O4) = 0.0326 moles x MW Aspirin
= .0326 x [9x12.01 + 8x1.008 + 4x16.00]
=.0326 mole x 180.15 g/mole
5.87 g Aspirin
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Stoichiometry
•
Yields of Chemical Reactions: If the actual amount
of product formed in a reaction is less than the
theoretical amount, we can calculate a percentage
yield.
Actual product yield
% yield 
 100%
Theoretical product yield
45
Stoichiometry
•
Dichloromethane (CH2Cl2) is prepared by reaction
of methane (CH4) with chlorine (Cl2) giving
hydrogen chloride as a by-product. How many
grams of dichloromethane result from the reaction
of 1.85 kg of methane if the yield is 43.1%?
46
Stoichiometry
CH4 + Cl2 → CH2Cl2 + HCl
Balance
CH4 + 2Cl2 → CH2Cl2 + 2HCl
1.85 kg CH4 = ? moles CH4
47
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
1.85 kg CH4 = ? moles CH4
MW CH4 = 1x12.01 + 4x1.008 = 16.04 g/mole
1000g 1 m oleCH 4
1.85 kg CH 4 

 115m olesCH 4
kg
16.4 g CH 4
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Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
115 moles CH4
in theory we should produce:
115 moles of CH2Cl2 and 230 moles of HCl
And use up 230 moles of Cl2
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Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
115 moles of CH2Cl2 = ? g
MW CH2Cl2 = 12.01 + 2x1.008 + 2x35.45 = 84.93
115 moles x (84.03 g/mole) = 9770 g
50
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
Expect 9770 g CH2Cl2
but the yield is 43.1%
So we produced just 0.431 x 9770 g
4.21 kg CH2Cl2
51
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
Suppose the reaction went to completion
(100% yield)
Is mass conserved?
52
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
Start with 115 moles CH4 and 230 moles Cl2
total mass = 115x16.04 + 230x70.90
= 1850 + 16300 = 18150
only 3 sig. figs. → 18.2 kg
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Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
End with 115 moles CH2Cl2 and 230 moles HCl
total mass = 115x84.93 + 230x36.46
= 9770 + 8390 = 18160
only 3 sig. figs → 18.2 kg
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