# 4 Mass and Stoichiometry

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Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu

Office Hours W – F 2-3 pm

Module #4
Mass and
Stoichiometry
FITCH Rules
G1: Suzuki is Success
G2. Slow me down
General

G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C2. Everybody wants to “be like Mike” (grp.18)
Chemistry

C3. Size Matters
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
Particle       Symbol      Mass                   Amu
g
1
Neutron, n        n
0        1.6749285x10-24        1.00867
1                                1.00728
Proton, p        1H        1.6726231x10-24
Electron, e         0      1.093898x10-28         0.00055
e
1

These numbers are inconvenient                   mass of 1 12C atom
1amu 
6
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  166053873x10  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  required
6C
To put both
Physics and chemistry
On same scale
Properties and Measurements
Property              Unit         Reference State
Size                  m            size of earth
Volume                cm3          m
Weight                gram         mass of 1 cm3 water at specified Temp
(and Pressure)
Temperature           oC,   K      boiling, freezing of water (specified Pressure

1.66053873x10-24g     amu          mass of 1C12 atom/12

Expressing masses in terms of amu is more
convenient
Two Examples we Examined already of STABLE isotopes
Relative
% Abundance                      atomic mass (amu)
204
82 Pb
1.36                             203.973
206
82 Pb
23.6                             205.9745
207
82 Pb
22.6                             206.9759
208
82 Pb  52.1                             207.9766

Relative
% Abundance                atomic masses (amu)
12
6   C     98.89                      12.0000
13
6   C    1.11                       13.00335

On the periodic table what is the reported atomic mass for lead?
Atomic mass= 207.2

 % isotope 
atomic massaverage                           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?
 % isotope 
atomic massaverage                       amisotope
isotopes  100 

% Abundance    atomic mass (amu)          %x(amu)
1.36                   203.973              301.88
23.6                   205.9745            4860.9982
22.6                   206.9759            4677.65534
52.1                   207.9766           10877.17618
99.98                  Weighted Sum       20717.70976
Average amu        207.1770976
207.2
atomic mass Pb ,average  207.17709amu
On the periodic table what is the reported amu for C?
amu=12.01

Where did this number come from?
 % isotope 
atomic massaverage                        amuisotope
isotopes  100 

% Abundance       atomic masses (amu)        %(amu)/100
98.89                     12.0000            11.8668
1.11                      13.00335           0.144337185
100                                          12.01113719

atomic massC ,average  12.01amu
How many particles are there?
 1 12C atom                    
12 g 6 C  12amu   166053873x10 24 g   6.022 x1023 12C atoms
12          6             1amu
6
.

General Rule #4: Chemists Are Lazy: make this number unique

N Avogadro  mole  6.022 x1023 atoms

mole
1
6.022 x1023 atoms
 1 12C atom                   
12 g 6 C  12amu   166053873x1024 g   6.022 x1023 atoms   1mole
12          6             1amu                    1mole
.

( x  amu) g X
x
elementX     1mole atoms                  Amedeo Avogadro
Italian (Turin)
1756-1856
Mole Latin mass=moles of stuff   Conte di Quaregna e di Cerreto
Galen, 170       Marie the Jewess, 300   Jabir ibn        Galileo Galili   Evangelista      TV Mad              Abbe Jean Picard Daniel Fahrenheit   Blaise Pascal    Robert Boyle,        Isaac Newton          Charles Augustin
Hawan, 721-815   1564-1642        Torricelli       scientist           1620-1682        1686-1737           1623-1662        1627-1691            1643-1727             Coulomb 1735-1806
1608-1647

Anders Celsius     Amedeo Avogadro       John Dalton          Jacques Charles       B. P. Emile         Germain Henri Hess Thomas Graham     Justus von Liebig   James Joule          Rudolph Clausius     William Thompson
1701-1744          1756-1856             1766-1844            1778-1850             Clapeyron           1802-1850          1805-1869         (1803-1873          (1818-1889)          1822-1888            Lord Kelvin, 1824-1907
1799-1864

1825-1898              James Maxwell      Johannes        Johannes Rydberg      J. J. Thomson                                  Max Planck            Wolfgang Pauli         Werner Karl        Linus Pauling
Johann Balmer          1831-1879          Diderik         1854-1919             1856-1940               Heinrich R. Hertz,     1858-1947             1900-1958              Heisenberg         1901-1994
Van der Waals                                                 1857-1894                                                           1901-1976
1837-1923

Fitch Rule G3: Science is Referential
Example

Element   Amu     Mass (g)   Moles (n)   # atoms
H         1.008   1.008      1           6.022x1023
C         12.01   24.02      2           12.044x1023
O         16.00   48.00      3           18.066x1023
Pb        207.2   207.2      1           6.022x1023
Properties and Measurements
Property              Unit         Reference State
Size                  m            size of earth
Volume                cm3          m
Weight                gram         mass of 1 cm3 water at specified Temp
(and Pressure)
Temperature           oC,   K      boiling, freezing of water (specified Pressure

1.66053873x10-24g     amu          mass of 1C12 atom/12
6.022x1023            mole         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   aamu   bamu  ...... zamu 
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               Total amu
Lead carbonate = PbCO3       Pb    1     207.2               207.2
principle component paint    C     1     12.01                12.01
before WWII                  O     3     16.00                48.00
Molar mass= unknown                                          267.21

267.2

MM     n amu   aamu   bamu  ...... zamu
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?

m
MM 
Acetylsalicylic acid
C9H 8O4
Active ingredient of aspirin                         n
0.509 mol acetylsalicylic acid
Mass?
 MM  n  m
MM         n amu   aamu   bamu  ...... zamu 
i
i         i                 A                B               Z

MM  912.01  81008  416.00
.
g
MM  18015
.
mole
        g 
 18015
.        0.509mole  m  917 g
.
       mole 
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                   Ca       40.08          40.08
upset stomach                        C        12.01          12.01
mass percent Ca                      O        3(16.00)       48.00
mass percent C                                               100.09
mass percent O
old coral reefs
MM        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.

1 Ca
Ca      40.08          40.08
C       12.01          12.01                   12.01 
3O      3(16.00)       48.00           % C  1         100%
100.09                  100.09 
%C  119992%  12.00%
.
 40.08 
% Ca  1          100%                         16.00 
 100.09                       % O  3         100%
 100.09 
%Ca  4004
.
%O  47.95683885  47.96%
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%
Converting between g, moles, and number of particles

nN A  particles
Mass=m                                      particles
Molar    Moles=n
grams             Moles                   Formula Units
Mass               Number

m                            The Golden Bridge
MM                               Stoichiometry or #of moles
n
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                  Pb     207.2          207.2
principle component paint               C       12.01          12.01
before WWII                             O      3(16.00)        48.00
14.8 g PbCO3                                                 267.21
number of moles, n = unknown
267.2
m                 m
MM                n
n                MM
MM     n amu   aamu   bamu  ...... zamu 
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             Pb     207.2          207.2
principle component paint          C       12.01          12.01
before WWII                        3O     3(16.00)        48.00
14.8 g PbCO3                                            267.21
number of moles, n = unknown
267.2
14.8 g
n                       n  0055389221mole
.
g
267.2
mole                   n  00554mole
.
Simplest Formula from Chemical Analysis
(Empirical)
1. Mass scale is based on atomic number of C
2. 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                         CHECK THAT THE
Cr=8.84 g                        COMPOUND IS PURE!!
O=9.52 g                          6.64         n MM )  m
unknown - simplest formula        8.84             m
orange compound                   9.52         n
MM
25.00

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              6.64 gK
 n  01698molK  0170molK
.           .
MM                 gK
39.10
1mole
8.84 gCr                                      0.170molK
 n  01700molCr  0170molCr
.            .
52.00
gCr                                     0.170molCr
mol
0.595mol O
9.52 gO
 n  0.59500molO  0.595molO
gO
16.00
mol
Simplest Formula from Chemical Analysis
(Empirical)
1. Mass scale is based on atomic number of C
2. 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        Divide by       7 O : 2K : 2Cr
Smallest #
0.595mol O                                  Formula: Cations First
.          .
0170molCr 100molCr           2.00molCr
                                       2K : 2Cr : 7 O
.
0170molCr    1molCr           2molCr
.         .
0170molK 100molK               2.00molK

0170molCr
.          1molCr                                    K2Cr2O7
2molCr
.
0.595molO 350molO              7.00molO

.
0170molCr   1molCr              2molCr
Make this non-fractional, multiply x2
Context for the next problem: The Romans weren’t really
Drunkards engaging in orgies
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
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                                      5.00g ethyl alcohol
Orgies                                      9.55 g CO2
Fall of empire                              5.87 g H2O
intoxicating properties                     simplest formula?
ethyl alcohol                               burned in air
elements C, H, 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        All the C and H in the sample is converted
9.55 g CO2                 In air (contains O2) to CO2 and H2O
5.87 g H2O
simplest formula?          so g C in CO2 represents g C in original
ethyl alcohol
elements C, H, O           And g H in H2O represents g H original
Burned in air

5gtotal (C H  O) 9.55gCO2  587 gH2 O
O2
.
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                 moleCO2   1moleC 
nC   gCO2                      
9.55 g CO2                          amu gCO2   1moleCO2 
5.87 g H2O
                        moleC 
nC  9.55gCO2 
moleCO2
simplest formula?                                                            
 (12.01  2(16)) gCO2   moleCO2 

elements C, H, O                                   mole CO2   moleC 
nC  9.55gCO2                       
m        1                                    (44.01gCO2   moleCO2 
n       m      
MM        MM 
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                 moleH2 O   2moleH 
nh   gH2 O                       
9.55 g CO2                          amu gH2 O   1moleH2 O 
5.87 g H2O
                           2moleH 
nH  587 gH2 O
mole H2 O
simplest formula?                .                                             
 (16.00  2(1008)) gH2 O   moleH2 O 
.

elements C, H, O                                    mole H2 O   2moleH 
n H  587 gH2 O
.                                   
m        1                                     (18.016 gH2 O   moleH2 O 
n       m      
MM        MM                 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 
2

9.55 g CO2          gO in original sample?
2

 12.01gC 
5.87 g H2O                                    g c  9.55g CO 
 44.01g   2.61g C

simplest formula? g total  g C  g H  g O                           CO 
2
2
3 sig fig
 amugH 
elements C, H, O 500gtotal  2.61gC  0.657 g H  gO
.                                           gH  gH O              
 amug    
2
     H2 O 

Now know:          gO  500  2.61  0.657
.                            3 sig fig 
2(1008) gC 
g c  587 g CO 
.
.                     
               0.657 g C
mol C 0.217    gO  1733
.                                   18.016 g CO 
2
2

mol H 0.652    gO  173
.
Need to know mol 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          500gtotal  2.61gC  0.657 g H  gO
.
9.55 g CO2                                              gO  500  2.61  0.657
.
5.87 g H2O                                                           gO  173
.
 moleO 
simplest formula?                nO   gO        
 amu gO 
elements C, H, O                               1mole O 
nH  173gO
.               
m         1                            (16.00gO 
n        m     
MM         MM 
nO  0108125mole
.

nO  0108mole
.
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
molO=0.108           .
0108molO
9.55 g CO2
molH=0.652                   1
5.87 g H2O                                     .
0108molO
molC=0.217
simplest formula?                             0.652mol H
 6.037  6
.
0108molO
elements C, H, O
0.217molC
 2.009  2
.
0108molO
C2H6O
Mass Relations and Stoichiometry

Chemical Formulas

Write reactions

Relate to Mass
Predict                        Ratios and
Change in amounts              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                        Right Hand Side

Reactants                             Products
number atoms A as reactants =   number atoms A as products
number atoms Bas reactants =    number atoms B as products
number atoms C as reactants =   number atoms C as products
.
.
.
Number atoms Z as reactants =   number atoms Z as products
Steps to Balance Reaction Equations
1.   Write a “skeleton” equation with molecular formulas of
reactants on left, products on right
2    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
3.   Chose an element that appears in only one molecular formula
on each side of the equation
4.   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
5.   Continue for the other elements
6.   The best answer is the one which is simplest whole-number
coefficients
“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.
galena                            minium
PbS Burned in O2 To make Pb3O4 and SO2
PbS  O2  Pb3O4  SO2
PbS

Pb3O4
1S                 1S

PbS  O2  Pb3O4  SO2

1Pb          3Pb
3S                1S      Subscript
Number in
indicating number
3 PbS   O2  Pb3O4 
Front indicates
Number of                                   SO2 Pb within a single
molecules                                       molecule
3Pb          3Pb
3S                 3S
Count up total O
3PbS  O2  Pb3O4  3SO2

2O                  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

3 PbS ( s )  5O2 ( g )  Pb3 O4 s   3SO2 ( g )

3            5           1              3    STOICHIOMETRY

Sets up proportionalities         BUT How doe we make use
Of those proportionalities?
Mass Relations and Stoichiometry

Chemical Formulas

Write reactions

Relate to Mass
Predict                        Ratios and
Change in amounts              Stoichiometry

Consider what happens when one part necessary for the
Reaction (recipe) is limiting
Mass Relations and Stoichiometry

Chemical Formulas

Write reactions

Relate to Mass
Predict                        Ratios and
Change in amounts              Stoichiometry

Consider what happens when one part necessary for the
Reaction (recipe) is limiting
2 MAIN Strategies to solving these problems

Step wise                                  String wise

1. Easy conceptually               1. Not easy conceptually
= series of proportionalities      when you are starting out

1. Longer time to compute           1. Shorter time to compute
2. Easy to lose track of units      2. Easy to keep track of units

BUT will eventually
Will model this one
Only use this method!!!
To begin with
3 PbS ( s )  5O2 ( g )  Pb3 O4 s   3SO2 ( g )

Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
1.       Find moles of SO2 by Stoichiometry
2.       Find grams of SO2 from those moles
x       3molSO2 ( g )
                      STOICHIOMETRY
134molO2
.          5molO2 ( g )                                            m  nMM
 32.07  2(16.00) gSO2
 3molSO2 ( g ) 
x               
 5molO2 ( g ) 
 134molO2  0804molSO2 ( g )
.         .
gSO2  0.804molSO2 


molSO2
               
 64.07 gSO2 
 gSO2                         gSO2  0.804molSO2             
gSO2  0.804molSO2                                                   molSO2 
 molSO2 
gSO2  5151228
.
 Samu  2(Oamu) 
gSO2  0.804molSO2                 
    molSO2                    gSO2  515gSO2
.
3 PbS ( s )  5O2 ( g )  Pb3 O4 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


 3molSO2 ( g )   g  amus  2 g  amuO2    

x  134molO2
.                            
 5molO2 ( g )           molSO2              
                                            

 3molSO2 ( g )   gSO 

x  134molO2
.                

      2

5molO2 ( g )   molSO2 
               

 3molSO2 ( g )   64.07 gSO 

x  134molO2
.                              
 5molO2 ( g )   molSO2 
2
  5151228g  515g
.          .
               
3 PbS ( s )  5O2 ( g )  Pb3 O4 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
m  nMM             n
MM                                                           103 g 
 103 g                                   1kgPb3 O4          
1kgPb3 O4                                                      kg 
molPb3 O4 formed 
 kg                                        685.6 gPb3 O4 
molPb3 O4 formed                                                                        
 gPb3O4                                            molPb3 O4 
            
 molPb3O4 
molPb3O4 formed  1458576429  1459
.            .
 10 3 g 
1kgPb3O4          
 kg              molPb3O4 formed  1459
molPb3 O4 formed                                                        .
 3(207.2)  4(16.00) gPb3 O4 
                             
         molPb3O4            
3 PbS ( s )  5O2 ( g )  Pb3 O4 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
x         5molO2

1459molPb3 O4 molPb3 O4
.
m  n MM 
x  14595molO2   7.295molO2
.                                                                      gO2 
x  7.295molO2        
 molO2 

 32.00gO2 
x  7.295molO2             233.4 g
 molO2 
3 PbS ( s )  5O2 ( g )  Pb3 O4 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


x  1kg Pb3O4
x  1kg Pb3O
x  1kg Pb3OO4     10 33g mol Pb3OO4  5molO22   g O2
 10 3 g mol Pb3 4
 kg  g                     


 kg  gPb3OO4  mol Pb33O44   molO2
Pb3 4 4
 Pb3 4                          

 10 3 g   mol Pb3O4       5molO2        32.00 g O2   

x  1kg Pb3O4         kg   685.6 g
        

  mol Pb3O4

  molO2


       Pb3O4                                
x  23337 g
.
3 PbS ( s )  5O2 ( g )  Pb3 O4 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  nMM          m
n                                            x   01125molPbS  M PbS
.
MM
6.00gO2                                          207.2amu Pb  32.07amuS 
x                   01875molO2
.              x  01125molPbS 
 32.00gO2                           .                                    
        molPbS           
          
  molO2                               239.27 gPbS 
x  01125molPbS 
.                          26.917875g
 molPbS 
x       3molPbS
                                                     x  26.9 g
.
01875molO2    5molO2 x  01125mol
.
3 PbS ( s )  5O2 ( g )  Pb3 O4 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

x.00gO  molO22 3molPbS   239.27 gPbS 
x  6.  600gO 2 molO
.00gO                  3molPbS
x 6          22                   
32.00gO22 5molO22   molPbS 
32.00gO       5molO              

x  26.917 g
x  26.9 g
Chemical Formulas

Write reactions

Relate to Mass
Ratios and
Predict
Stoichiometry
Change in amounts

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

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
a. What is the limiting reagent?

1pkgchips  2eggs  3cflour  1csugar  1tbspvanilla  2tbsppowder heat  36cookies
           ?  2 pkg               72cookies
Eggs 2                         12 eggs        2 pkg     1 pkg                1pkg   
           ?  12eggs              192cookies
Sugar 1c                       >1cup         12eggs     2eggs                2eggs  
vanilla 1tbsp                  >1tbsp         Smallest yield = 72 cookies
Powder 2tbsp                   >2tbsp         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 
 72  4 
% yield cookies            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
afitch@luc.edu

Office Hours W-F 2-3 pm

```
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