Download - Chapters 8-10
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Chapters 8-10Chapters 8-10
The mole, equations and stoichiometry
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The Structure of AtomsThe Structure of Atoms
Atomic Mass Unit
1 amu = 1/12 of the mass of one atom of Carbon-12
1 amu = 1.6605 x 10-24 g
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Mass: proton = 1.00728 amuneutron = 1.0086 amuelectron = 0.0005486 amu12C atom = 12.00000 amu13C atom = 13.00335 amu
Atomic and Molecular MassAtomic and Molecular Mass
1 amu mass of carbon 12 atom12
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Atomic and Molecular MassAtomic 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
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Atomic and Molecular MassAtomic 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
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Avogadro and the MoleAvogadro 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.
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Avogadro and the MoleAvogadro and the Mole
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Avogadro and the MoleAvogadro 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 MoleAvogadro and the Mole
• The Mole: Allows us to
make comparisons
between substances
that have
different
masses.
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Counting atomsCounting atoms
Amount in moles can be converted to number of atoms:
the conversion is similar to changing 20 dozen eggs into the equivalent number of individual eggs:
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Convert 2.66 mol of an element to its equivalent number of atoms:
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How many moles in 2.54 x 1024 Fe atoms? 2.54 x 1024 atoms x
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The moleThe mole
# particles# particles
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Avogadro’s number X /
The moleThe mole
# Particles# Particles
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Practice problems p 278Practice problems p 278
1) How many atoms are present in 3.7 mol of sodium?
2) How many atoms are present in 155 mol of arsenic?
3) How many moles of xenon is 5.66 x 1026 atoms?4) How many moles of silver is 2.888 x 1015 atoms?
5) Recall that 1 mol = 6.02 x 1023 atoms
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Balancing Chemical EquationsBalancing 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)
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Molar mass relates moles to gramsMolar mass relates moles to grams
Molar mass- is numerically equal to the element’s atomic mass and has the unit grams per mole (g/mol)
- molar masses are used as conversion factors in the relationship between moles and the mass in grams of a substance.
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Avogadro’s number X / (x 1mol/6.02EE23)6.02EE23
x mass x 1 mol 1 mol mass
The MoleThe Mole
# of Particles # of Particles
Mass (g)Mass (g)
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Balancing Chemical EquationsBalancing 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
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Balancing Chemical EquationsBalancing 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
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Balancing Chemical EquationsBalancing 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.
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Balancing Chemical EquationsBalancing 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.
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Balancing Chemical EquationsBalancing Chemical Equations
Hints for Balancing Chemical Equations:
4) Otherwise, trial and error!!
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Balancing Chemical EquationsBalancing Chemical Equations
Example 1: CH4 + O2 → CO2 + H2O
Balance O2 last
C is already balanced
Start by changing S.C. of H2O to balance H
CH4 + O2 → CO2 + 2H2O
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Balancing Chemical EquationsBalancing 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!
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Balancing Chemical EquationsBalancing 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
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Balancing Chemical EquationsBalancing 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!
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Balancing Chemical EquationsBalancing 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
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Balancing Chemical EquationsBalancing 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
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Balancing Chemical EquationsBalancing 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
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Balancing Chemical EquationsBalancing 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!
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Balancing Chemical EquationsBalancing Chemical Equations
• Balance the following equations:
C6H12O6 → C2H6O + CO2
Fe + O2 → Fe2O3
NH3 + Cl2 → N2H4 + NH4Cl
KClO3 + C12H22O11 → KCl + CO2 + H2O
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Balancing Chemical EquationsBalancing 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 EquationsBalancing Chemical Equations
• Balance the following equations:
C6H12O6 → 2C2H6O + 2CO2
4Fe + 3O2 → 2Fe2O3 (balance O first)
NH3 + Cl2 → N2H4 + NH4Cl
KClO3 + C12H22O11 → KCl + CO2 + H2O
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Balancing Chemical EquationsBalancing Chemical Equations
• Balance the following equations:
C6H12O6 → 2C2H6O + 2CO2
4Fe + 3O2 → 2Fe2O3 (balance O first)
NH3 + Cl2 → N2H4 + NH4Cl
N:H is 1:3 on left, must get 1:3 on right!
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Balancing Chemical EquationsBalancing Chemical Equations
NH3 + Cl2 → N2H4 + NH4Cl
N:H is 1:3 on left, must get 1:3 on right!
4NH3 + Cl2 → N2H4 + 2NH4Cl
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Balancing Chemical EquationsBalancing Chemical Equations
• Balance the following equations:
C6H12O6 → 2C2H6O + 2CO2
4Fe + 3O2 → 2Fe2O3
4NH3 + Cl2 → N2H4 + 2NH4Cl
KClO3 + C12H22O11 → KCl + CO2 + H2O (tough!)
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Balancing Chemical EquationsBalancing 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
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Balancing Chemical EquationsBalancing Chemical Equations
• Balance the following equations:
8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O
balance K (and hope Cl is balanced)
8KClO3 + C12H22O11 → 8KCl + 12CO2 + 11H2O
Balanced!
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Balancing Chemical EquationsBalancing Chemical Equations
• Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below:
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Avogadro and the MoleAvogadro and the Mole
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StoichiometryStoichiometry
• Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities.
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StoichiometryStoichiometry
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?
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StoichiometryStoichiometry
2 NaOH + Cl2 → NaOCl + NaCl + H2O
25.0 g Cl2 reacts with ? g NaOH
22
22 353.0
90.70
10.25 Clmoles
Clg
ClmoleClg
NaOHgNaOHmole
NaOHg
Clmole
NaOHmoles
Clg
ClmoleClg 2.28
1
0.40
1
2
90.70
10.25
22
22
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Avogadro and the MoleAvogadro 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)
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Avogadro and the MoleAvogadro and the Mole
Fe2O3 + CO → Fe + CO2
Balance (not a simple one)Save Fe for last
C is balanced, but can’t balance OIn 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
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Avogadro and the MoleAvogadro and the Mole
Fe2O3 + 3CO → 2Fe + 3CO2
COmolesOFemole
COmoleOFemoles 50.1
1
3500.0
3232
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StoichiometryStoichiometry
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StoichiometryStoichiometry
• 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?
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StoichiometryStoichiometry
Salicylic acid + Acetic anhydride →
Aspirin + acetic acid
C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H
C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2
Balanced!
Equal # moles for all
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StoichiometryStoichiometry
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
..0326.0..12.138
..1..50.4 ASmoles
ASg
ASmoleASg
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StoichiometryStoichiometry
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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StoichiometryStoichiometry
• 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.
100% yieldproduct lTheoretica
yieldproduct Actual yield%
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StoichiometryStoichiometry
• 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%?
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StoichiometryStoichiometry
CH4 + Cl2 → CH2Cl2 + HCl
Balance
CH4 + 2Cl2 → CH2Cl2 + 2HCl
1.85 kg CH4 = ? moles CH4
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StoichiometryStoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
1.85 kg CH4 = ? moles CH4
MW CH4 = 1x12.01 + 4x1.008 = 16.04 g/mole
44
44 115
4.16
1100085.1 CHmoles
CHg
CHmole
kg
gCHkg
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StoichiometryStoichiometry
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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StoichiometryStoichiometry
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
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StoichiometryStoichiometry
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
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StoichiometryStoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
Suppose the reaction went to completion
(100% yield)
Is mass conserved?
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StoichiometryStoichiometry
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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StoichiometryStoichiometry
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