“A” students work (without solutions manual) ~ 10 problems/night. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours W – F 2-3 pm Module #4 Mass and Stoichiometry.
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Transcript “A” students work (without solutions manual) ~ 10 problems/night. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours W – F 2-3 pm Module #4 Mass and Stoichiometry.
“A” students work
(without solutions manual)
~ 10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours W – F 2-3 pm
Module #4
Mass and
Stoichiometry
Chemistry
General
FITCH Rules
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C1. It’s all about charge
C2. Everybody wants to “be like Mike” (grp.18)
C3. Size Matters
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
Particle
Neutron, n
Proton, p
Electron, e
Symbol
1
0
1
1
0
1
n
H
e
Mass
g
1.6749285x10-24
1.6726231x10-24
1.093898x10-28
Amu
1.00867
1.00728
0.00055
These numbers are inconvenient
mass of 1 126 C atom
1amu
Invoke General Rule #5:
12
Chemists Are Lazy
Create some unit (not g) that is more useful (atomic mass unit)
Invoke General Rule #3
19926
.
x10 26 g
There must be some reference state 1amu
12
1amu 166053873
.
x10 24 g
Why a relative reference state of 126C?
Represents a compromise between Physicists and Chemists
Physicists used a reference
Chemists used a reference state
state of 11H to measure
of 168O as an abundant element
Relative velocities of gas
everything (but gold) reacts with.
phase atoms
Measured mass changes
Rock s airg RockOn,s
fire
Chemists must have
Started out chunking
Rocks in fire
12
6C
required
Least adjustments
To put both
Physics and chemistry
On same scale
Properties and Measurements
Property
Size
Volume
Weight
Unit
m
cm3
gram
Temperature
oC,
1.66053873x10-24g
amu
K
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
boiling, freezing of water (specified Pressure
mass of 1C12 atom/12
Expressing masses in terms of amu is more
convenient
Two Examples we Examined already of STABLE isotopes
% Abundance
204
1.36
82 Pb
206
23.6
82 Pb
207
22.6
82 Pb
208
52.1
82 Pb
12
6
C
13
6
C
% Abundance
98.89
1.11
Relative
atomic mass (amu)
203.973
205.9745
206.9759
207.9766
Relative
atomic masses (amu)
12.0000
13.00335
On the periodic table what is the reported atomic mass for lead?
Atomic mass= 207.2
atomic massaverage
% isotope
amuisotope
isotopes 100
On the periodic table what is the reported atomic mass for lead?
Atomic mass= 207.2
Where did this number come from?
atomicmassaverage
% Abundance
1.36
23.6
22.6
52.1
99.98
% isotope
amisotope
isotopes 100
atomic mass (amu)
203.973
205.9745
206.9759
207.9766
Weighted Sum
Average amu
%x(amu)
301.88
4860.9982
4677.65534
10877.17618
20717.70976
207.1770976
207.2
atomicmassPb,average 20717709
.
amu
On the periodic table what is the reported amu for C?
amu=12.01
Where did this number come from?
atomic massaverage
% Abundance
98.89
1.11
100
% isotope
amuisotope
isotopes 100
atomic masses (amu)
12.0000
13.00335
%(amu)/100
11.8668
0.144337185
12.01113719
atomicmassC,average 12.01amu
How many particles are there?
12
1
1amu
6 C atom
12
23 12
6
.
022
x
10
12 g 6 C 12amu 166053873
24
6 C atoms
.
x10 g
General Rule #4: Chemists Are Lazy: make this number unique
N Avogadro mole 6.022x1023 atoms
mole
1
6.022 x10 23 atoms
12
1
1amu
1mole
6 C atom
12
1mole
12 g 6 C 12amu 166053873
.
x10 24 g 6.022 x1023 atoms
( x amu) g X
x
elementX
1mole atoms
Mole Latin mass=moles of stuff
Amedeo Avogadro
Italian (Turin)
1756-1856
Conte di Quaregna e di Cerreto
Galen, 170
Anders Celsius
1701-1744
1825-1898
Johann Balmer
Marie the Jewess, 300
Amedeo Avogadro
1756-1856
James Maxwell
1831-1879
Jabir ibn
Hawan, 721-815
John Dalton
1766-1844
Johannes
Diderik
Van der Waals
1837-1923
Galileo Galili
1564-1642
Evangelista
Torricelli
1608-1647
Jacques Charles
1778-1850
Johannes Rydberg
1854-1919
TV Mad
scientist
B. P. Emile
Clapeyron
1799-1864
J. J. Thomson
1856-1940
Abbe Jean Picard Daniel Fahrenheit
1620-1682
1686-1737
Germain Henri Hess Thomas Graham
1802-1850
1805-1869
Heinrich R. Hertz,
1857-1894
Max Planck
1858-1947
Blaise Pascal
1623-1662
Justus von Liebig
(1803-1873
Robert Boyle,
1627-1691
James Joule
(1818-1889)
Wolfgang Pauli
1900-1958
Isaac Newton
1643-1727
Rudolph Clausius
1822-1888
Werner Karl
Heisenberg
1901-1976
Charles Augustin
Coulomb 1735-1806
William Thompson
Lord Kelvin, 1824-1907
Linus Pauling
1901-1994
Fitch Rule G3: Science is Referential
Example
Element
H
C
O
Pb
Amu
1.008
12.01
16.00
207.2
Mass (g)
1.008
24.02
48.00
207.2
Moles (n)
1
2
3
1
# atoms
6.022x1023
12.044x1023
18.066x1023
6.022x1023
Properties and Measurements
Property
Size
Volume
Weight
Unit
m
cm3
gram
Temperature
oC,
1.66053873x10-24g
6.022x1023
amu
mole
K
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
boiling, freezing of water (specified Pressure
mass of 1C12 atom/12
atomic mass of an element in grams
Molar Mass
in grams is numerically equal to the sum of the
masses (in amu) of the atoms in the formula
MM
n amu aamu bamu ...... zamu
i
i
i
A
B
Z
Example. Lead carbonate is principle component
of white lead used in all white paints prior to WWII.
Calculate the molar mass of lead carbonate
Element #atoms amu
Lead carbonate = PbCO3
Pb
1
207.2
principle component paint
C
1
12.01
before WWII
O
3
16.00
Molar mass= unknown
Total amu
207.2
12.01
48.00
267.21
267.2
MM
n amu aamu bamu ...... zamu
i
i
i
A
B
Z
Example: Acetylsalicylic acid, C9H 8O4, is the active ingredient of aspirin.
What is the mass in grams of 0.509 mol acetylsalicylic acid?
Acetylsalicylic acid
C9H 8O4
Active ingredient of aspirin
0.509 mol acetylsalicylic acid
Mass?
MM
m
MM
n
MM n m
n amu aamu bamu ...... zamu
i
i
A
i
MM 912.01 81008
. 416.00
g
MM 18015
.
mole
g
.
. g
18015
0.509mole m 917
mole
B
Z
Mass %
atomic weight of element
%element # atoms of that element
100%
molar mass of compound
Calcium carbonate, commonly called is used in many commercial
products to relieve an upset stomach. It has the formula of CaCO3.
Because lead ore bodies form by substitution of lead onto old coral
reefs (calcite or calcium carbonate) some antacid materials have
been tested for their lead composition. What are the mass percents
of Ca, C, and O in calcium carbonate?
CaCO3
commercial product
upset stomach
mass percent Ca
mass percent C
mass percent O
old coral reefs
lead in antacids
Ca
C
O
MM
40.08
12.01
3(16.00)
40.08
12.01
48.00
100.09
n amu
i
i
i
atomic weight of element
%element # atoms of that element
100%
molar mass of compound
Calcium carbonate, commonly called is used in many commercial
products to relieve an upset stomach. It has the formula of CaCO3.
What are the mass percents of Ca, C, and O in calcium carbonate?
Because lead ore bodies form by substitution of lead onto old coral
reefs (calcite or calcium carbonate) some antacid materials have
been tested for their lead composition.
Ca
C
3O
40.08
12.01
3(16.00)
40.08
12.01
48.00
100.09
40.08
%Ca 1
100%
100.09
%Ca 4004
.
1 Ca
12.01
%C 1
100%
100.09
%C 119992%
.
1200%
.
16.00
% O 3
100%
100.09
%O 4795683885
.
4796%
.
Calcium carbonate, commonly called is used in many commercial
products to relieve an upset stomach. It has the formula of CaCO3.
What are the mass percents of Ca, C, and O in calcium carbonate?
Because lead ore bodies form by substitution of lead onto old coral
reefs (calcite or calcium carbonate) some antacid materials have
been tested for their lead composition.
Are We Done?
47.96
12.00
40.04
100%
Check answer
Very Important!
Converting between g, moles, and number of particles
nN A particles
Mass=m
Molar
grams
Moles=n
Avogadro’s
Moles
Mass
m
MM
n
particles
Formula Units
Number
The Golden Bridge
Stoichiometry or #of moles
Example. Lead carbonate is principle component
of white lead used in all white paints prior to WWII.
Determine the number of moles of lead carbonate in a
sample of 14.8 g lead carbonate
Lead carbonate = PbCO3
principle component paint
before WWII
14.8 g PbCO3
number of moles, n = unknown
m
MM
n
MM
Pb
C
O
207.2
12.01
3(16.00)
207.2
12.01
48.00
267.21
267.2
m
n
MM
n amu aamu bamu ...... zamu
i
i
i
A
B
Z
Example. Lead carbonate is principle component
of white lead used in all white paints prior to WWII.
Determine the number of moles of lead carbonate in a
sample of 14.8 g lead carbonate
Lead carbonate = PbCO3
principle component paint
before WWII
14.8 g PbCO3
number of moles, n = unknown
Pb
C
3O
207.2
12.01
3(16.00)
207.2
12.01
48.00
267.21
267.2
14.8g
n
g
267.2
mole
n 0.055389221mole
n 00554
.
mole
Simplest Formula from Chemical Analysis
(Empirical)
1.
2.
Mass scale is based on atomic number of C
Mass scale is therefore proportional to number of
atoms or moles
3 Convert mass to moles = whole units of atoms
4. If moles are fractional this implies that atoms are
fractional
NOT ALLOWED by chemistry (remember we
chemists do not break up atoms - that is reserved for
physicists)
5. So multiply until we get whole number ratios
Example A 25.00-g sample of an orange compound
contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of
oxygen. Find the simplest formula
(Empirical)
Total =25.00g
K= 6.64g
Cr=8.84 g
O=9.52 g
unknown - simplest formula
orange compound
CHECK THAT THE
COMPOUND IS PURE!!
6.64
8.84
9.52
25.00
n MM ) m
m
n
MM
This means that we have accounted for the total weight of
The compound and we are confident in assigning weight
Ratios.
Example : A 25.00-g sample of an orange compound
contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of
oxygen. Find the simplest formula
(Empirical)
Sig fig
m
n
MM
6.64 gK
n 01698
.
molK 0170
. molK
gK
39.10
1mole
8.84 gCr
n 01700
.
molCr 0170
. molCr
gCr
52.00
mol
9.52 gO
n 0.59500molO 0.595molO
gO
16.00
mol
0.170molK
0.170molCr
0.595mol O
Simplest Formula from Chemical Analysis
(Empirical)
1.
2.
Mass scale is based on atomic number of C
Mass scale is therefore proportional to number of
atoms or moles
3 Convert mass to moles = whole units of atoms
4. If moles are fractional this implies that atoms are
fractional
NOT ALLOWED by chemistry (remember we
chemists do not break up atoms - that is reserved for
physicists)
5. So multiply until we get whole number ratios
Example : A 25.00-g sample of an orange compound
contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of
oxygen. Find the simplest formula
(Empirical)
0.170molK
0.170molCr
0.595mol O
Divide by
Smallest #
0170
. molCr 100
. molCr
0170
. molCr
1molCr
0170
. molK 100
. molK
0170
. molCr
1molCr
0.595molO 350
. molO
0170
. molCr
1molCr
7 O : 2K : 2Cr
2.00molCr
2molCr
2.00molK
2molCr
7.00molO
2molCr
Formula: Cations First
Make this non-fractional, multiply x2
2K : 2Cr : 7 O
K2Cr2O7
Context for the next problem: The Romans weren’t really
Drunkards engaging in orgies
Lead acetate is easily made by boiling wine in a lead pot = sapa, a
sweetner used in many Roman recipes.
Sugar of lead (lead acetate) was used from 0 A.D. to 1750s A.D. to
“sweeten” ethyl alcohol (wine) and to prevent wine from going bad.
One (unproven) theory is that the Roman leaders were poisoned by
excessive lead acetate consumption leading to the fall of Rome.
A recipe of
Apicius which
Uses ½ cup
sapa
Example : The compound that the Romans supposedly drank in
excess is ethyl alcohol, which contains the elements carbon, hydrogen,
and oxygen. When a sample of ethyl alcohol is burned in air, it is
found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
Romans
Orgies
Fall of empire
intoxicating properties
ethyl alcohol
elements C, H, O
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
burned in air
Example : The compound that the Romans supposedly drank in
excess is ethyl alcohol, which contains the elements carbon, hydrogen,
and oxygen. When a sample of ethyl alcohol is burned in air, it is
found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
elements C, H, O
Burned in air
All the C and H in the sample is converted
In air (contains O2) to CO2 and H2O
so g C in CO2 represents g C in original
ethyl alcohol
And g H in H2O represents g H original
O2
5gtotal (C H O)
9.55gCO2 587
. gH2 O
Example : The compound that the Romans supposedly drank in
excess is ethyl alcohol, which contains the elements carbon, hydrogen,
and oxygen. When a sample of ethyl alcohol is burned in air, it is
found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
elements C, H, O
m
1
n
m
MM
MM
moleCO2 1moleC
nC gCO2
amu
gCO
1
moleCO
2
2
moleC
moleCO2
nC 9.55gCO2
(
12
.
01
2
(
16
))
gCO
moleCO
2
2
mole CO2 moleC
nC 9.55gCO2
(44.01gCO2 moleCO2
nC 0.217mole
Example : The compound that the Romans supposedly drank in
excess is ethyl alcohol, which contains the elements carbon, hydrogen,
and oxygen. When a sample of ethyl alcohol is burned in air, it is
found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
elements C, H, O
m
1
n
m
MM
MM
moleH2 O 2moleH
nh gH2 O
amu
gH
O
1
moleH
O
2
2
2moleH
mole H2 O
n H 587
. gH2 O
(
16
.
00
2
(
1008
.
))
gH
O
moleH
O
2
2
mole H2 O 2moleH
n H 587
. gH2 O
(18.016 gH2 O moleH2 O
nH 0.6517mole
nH 0.652mole
Example : The compound that the Romans supposedly drank in
excess is ethyl alcohol, which contains the elements carbon, hydrogen,
and oxygen. When a sample of ethyl alcohol is burned in air, it is
found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
5.00g ethyl alcohol How are we going to get it? gc gCO amugC
amug CO
9.55 g CO2
gO in original sample?
12.01gC
5.87 g H2O
2.61g C
g c 9.55gCO
44.01gCO
simplest formula? gtotal gC g H gO
3 sig fig
2
2
2
2
amugH
amug
H2 O
. gtotal 2.61gC 0.657 g H gO
gH gH O
elements C, H, O 500
3 sig fig
Now know:
gO 5.00 2.61 0.657
2(1008
. ) gC
g c 587
. g CO
mol C 0.217
gO 1733
.
18.016 g CO
mol H 0.652
gO 173
.
Need to know mol O.
2
2
2
0.657 g C
Example : The compound that the Romans supposedly drank in
excess is ethyl alcohol, which contains the elements carbon, hydrogen,
and oxygen. When a sample of ethyl alcohol is burned in air, it is
found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
elements C, H, O
m
1
n
m
MM
MM
500
. gtotal 2.61gC 0.657 g H gO
gO 5.00 2.61 0.657
moleO
nO gO
amu gO
1moleO
nH 173
. gO
(16.00gO
nO 0108125
.
mole
nO 0108
. mole
gO 173
.
Example : The compound that the Romans supposedly drank in
excess is ethyl alcohol, which contains the elements carbon, hydrogen,
and oxygen. When a sample of ethyl alcohol is burned in air, it is
found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
Divide by smallest
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
elements C, H, O
molO=0.108
molH=0.652
molC=0.217
0108
. molO
1
0108
. molO
0.652mol H
6.037 6
0108
. molO
0.217molC
2.009 2
0108
. molO
C2H6O
Mass Relations and Stoichiometry
Chemical Formulas
Write reactions
Predict
Change in amounts
Relate to Mass
Ratios and
Stoichiometry
Consider what happens when one part necessary for the
Reaction (recipe) is limiting
Atoms are Conserved: no alchemy
Mass Relations In Reactions
Writing and Balancing Chemical Reactions
Lead Hand Side
Reactants
number atoms A as reactants =
number atoms Bas reactants =
number atoms C as reactants =
.
.
.
Number atoms Z as reactants =
Right Hand Side
Products
number atoms A as products
number atoms B as products
number atoms C as products
number atoms Z as products
Steps to Balance Reaction Equations
1.
2
3.
4.
5.
6.
Write a “skeleton” equation with molecular formulas of
reactants on left, products on right
Indicate the physical state of the reactants and products
a.
(g) for a gas
b.
(l) for a liquid
c.
(s) for a solid
d.
(aq) for an ion or molecule in water (aqueous) solution
Chose an element that appears in only one molecular formula
on each side of the equation
Balance the equation for mass of that element
a.
placing coefficients in front of the molecular formula
NOT by changing subscripts in the molecular formula
Continue for the other elements
The best answer is the one which is simplest whole-number
coefficients
Example: Lead ore, Lead sulfide (with the common name
“galena”), was “calcined” by early metallurgists to form a lead
oxide used to purify silver from other metals. Calcined means to
burn in the presence of oxygen. At low temperatures the
yellow lead oxide PbO, litharge, is formed. At higher temperatures
the a red lead oxide, minium, is formed, Pb3O4. The sulfur is
converted in the reaction to the gas SO2. Minium was
the predominate source of red paints and pigments for illuminated
bibles from which we derive the word miniature. Write a balanced
chemical equation for the reaction of lead ore, galena with oxygen gas
(O2) to form minium and SO2 gas.
minium
galena
PbS
Burned in O2 To make Pb3 O4 and SO2
PbS O2 Pb3O4 SO2
PbS
Pb3O4
1S
1S
PbS O2 Pb3O4 SO2
1Pb
3Pb
3S
Number in
Front indicates
Number of
molecules
1S
3 PbS O2 Pb3O4
3Pb
3S
Subscript
indicating number
SO2 Pb within a single
molecule
3Pb
3S
3PbS O2 Pb3O4 3SO2
2O
Count up total O
4O+(3x2)O=10 O
3PbS O2 Pb3O4 3SO2
From previous slide
4O+3x2O=10O
3PbS 5O2 Pb3O4 3SO2
4O+3x2O=10O
5x2O=10O
Be sure to indicate the physical state of the reactants and products
3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g )
3
5
1
Sets up proportionalities
3
STOICHIOMETRY
BUT How doe we make use
Of those proportionalities?
Mass Relations and Stoichiometry
Chemical Formulas
Write reactions
Predict
Change in amounts
Relate to Mass
Ratios and
Stoichiometry
Consider what happens when one part necessary for the
Reaction (recipe) is limiting
Mass Relations and Stoichiometry
Chemical Formulas
Write reactions
Predict
Change in amounts
Relate to Mass
Ratios and
Stoichiometry
Consider what happens when one part necessary for the
Reaction (recipe) is limiting
2 MAIN Strategies to solving these problems
Step wise
1. Easy conceptually
= series of proportionalities
1. Longer time to compute
2. Easy to lose track of units
3. Leads to rounding errors
Will model this one
To begin with
String wise
1. Not easy conceptually
when you are starting out
1. Shorter time to compute
2. Easy to keep track of units
BUT will eventually
Only use this method!!!
3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g )
Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
1.
2.
Find moles of SO2 by Stoichiometry
Find grams of SO2 from those moles
3molSO2 ( g )
x
134
. molO2
5molO2 ( g )
STOICHIOMETRY
3molSO2 ( g )
134
x
. molO2 0.804molSO2 ( g )
5
molO
2 ( g)
gSO2
gSO2 0.804molSO2
molSO
2
Samu 2(Oamu)
gSO2 0.804molSO2
molSO
2
m nMM
32.07 2(16.00) gSO2
gSO2 0804
. molSO2
molSO2
64.07 gSO2
gSO2 0.804molSO2
molSO
2
gSO2 5151228
.
gSO2 515
. gSO2
3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g )
Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
Alternative Strategy is to fold all calculations together
x 134
. molO2
3molSO2 ( g ) g amus 2 g amuO2
molSO2
5molO2 ( g )
3molSO2 ( g ) gSO
2
5
molO
molSO
2 ( g)
2
3molSO2 ( g ) 64.07 gSO
2
.
g 515
. g
5151228
5molO2 ( g ) molSO2
x 134
. molO2
x 134
. molO2
3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g )
Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
b) The mass in grams of O2 required to form 1.000 kg of Pb3O4
Stepwise calculations
1. Find moles of Pb3O4 formed
m nMM
m
n
MM
10 3 g
1kgPb3 O4
kg
molPb3 O4 formed
gPb3 O4
molPb
O
3 4
10 3 g
1kgPb3 O4
kg
molPb3 O4 formed
3(207.2) 4(16.00) gPb3 O4
molPb3 O4
10 3 g
1kgPb3 O4
kg
molPb3 O4 formed
685.6 gPb3 O4
molPb
O
3 4
molPb3O4 formed 1458576429
.
1459
.
molPb3O4 formed 1459
.
3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g )
Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
b) The mass in grams of O2 required to form 1kg of Pb3O4
1. Find moles of Pb3O4 formed
2. Use Stoichiometry to find moles of O2
3. Find grams of O2 from those moles
5molO2
x
1459
. molPb3O4 molPb3O4
x 1459
. 5molO2 7.295molO2
m n MM
gO2
x 7.295molO2
molO
2
32.00gO2
x 7.295molO2
233.4 g
molO2
3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g )
Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
b) The mass in grams of O2 required to form 1kg of Pb3O4
Alternative string strategy
10 3 g mol Pb3O4
kg 685.6 g
Pb3O4
x 1kg Pb3O4
kgPbPb
xx 11kg
PbO
O4
3O
33
44
x 1kg Pb3O4
x 233.37 g
333 g mol
10
10 g molPbPb3O3O4 4 5molO2 g O2
kg
gg
kg
PbPb3O3O4 4 mol Pb3O4 molO2
5molO2
mol Pb3O4
32.00 g O2
molO2
3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g )
Stepwise method
Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
b) The mass in grams of O2 required to form 1kg of Pb3O4
c) The mass in grams of PbS required to react with 6.00 g of O2
1. Get moles of O2
2. Use Stoichiometry to get moles of PbS
3. Get grams of PbS
m
m nMM
n
MM
x
6.00gO2
32.00gO2
molO
2
01875
.
molO2
x 01125
.
molPbS M PbS
207.2amu Pb 32.07amuS
x 01125
.
molPbS
molPbS
239.27 gPbS
x 01125
.
molPbS
26.917875g
molPbS
x
3molPbS
.
mol
01875
.
molO2
5molO2 x 01125
x 26.9 g
3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g ) Stringwise
Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
b) The mass in grams of O2 required to form 1kg of Pb3O4
c) The mass in grams of PbS required to react with 6.00 g of O2
molO
molO22 33molPbS
molPbS 239.27 gPbS
x
6
.
00
gO
xx 66.00
.00gO
gO222
32
32
.
00
.
00
gO
gO
5
5
molO
molO
molPbS
22
22
x 26.917 g
x 26.9 g
Chemical Formulas
Write reactions
Predict
Change in amounts
Relate to Mass
Ratios and
Stoichiometry
Consider what happens when one part necessary for the
Reaction (recipe) is limiting
Rules for Limiting Reagent
1. Calculate the amount of product that would be formed if the
first reactant were completely consumed
2. Repeat for the second reactant
3. Choose the smaller of the two amounts. This is the
theoretical yield of the product.
4. The reactant that produces the smaller amount of the
product is the limiting reagent.
5. The other reagent is in excess, only part of it is consumed.
Expelled from grammar school for detonating an explosive
Limiting Reagents
Made from chemicals obtained from his father’s business
Justus von Liebig’s Law of the Limiting (1803-1873 Germany)
I have 1 dozen eggs, 2 packages of chocolate chips, and
An entire carton of flour, an entire carton of sugar, a new bottle
Of vanilla, and a new box of baking powder. The chocolate chip recipe
calls for 2 eggs, 3 cups flour, 1 cup Sugar, 1 package chocolate chips, 1
tbsp vanilla, and 2 Tbsp baking powder. The recipe results in 36
Chocolate chip cookies.
a. What is the limiting reagent?
1pkgchips 2eggs 3cflour 1csugar 1tbspvanilla 2tbsppowder heat
36cookies
Chips 1pkg
Eggs 2
Flour 3c
Sugar 1c
vanilla 1tbsp
Powder 2tbsp
Yield 36 cookies
2pkg
12 eggs
>3cups
>1cup
>1tbsp
>2tbsp
Yield??
?
36cookies
36cookies
? 2 pkg
72cookies
2 pkg
1 pkg
1
pkg
?
36cookies
36cookies
? 12eggs
192cookie
12eggs
2eggs
2
eggs
Smallest yield = 72 cookies
Limiting reagent = chips
While I wasn’t looking, my son snitched 4 cookies and ran off to Panama.
yield exp erimental
% yield
100%
yield theoretical
% yield cookies
72 4
100% 94.44444 94%
72
Free at
last
Actual Yield=Theoretical
Yield
(most of the time)
“A” students work
(without solutions manual)
~ 10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours W-F 2-3 pm