Download Mass and Stoichiometry

General
FITCH Rules
“A” students work
(without solutions manual)
~ 10 problems/night.
Office Hours W – F 2-3 pm
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
1
amu
=
Invoke General Rule #5:
12
Chemist’s 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
Chemistry
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
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
Why a relative reference state of C12?
Represents a compromise between Physicists and Chemists
Physicists used a reference
Chemists used a reference state
state of H1 to measure
of O16 as an abundant element
Relative velocities of gas
everything reacts with.
phase atoms
Measured precipitate mass
changes
C12 required
Least adjustments
To put both
Physics and chemistry
On same scale
1
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
Two Examples we Examined already of STABLE isotopes
Relative
% Abundance
atomic mass (amu)
204
1.36
203.973
Pb
82
206
25.4
205.9745
Pb
82
207
21.1
206.9759
82 Pb
208
52.1
207.9766
82 Pb
Expressing masses in terms of amu is more
convenient
On the periodic table what is the reported atomic mass for lead?
Atomic mass= 207.2
Where did this number come from?
⎛ % isotope ⎞
⎟ am
atomic massaverage = ∑ ⎜
100 ⎠ isotope
isotopes ⎝
148%
.
( 203.973)
atomic mass Pb ,average =
100
23.6%
( 205.9745)
+
100
22.6%
( 206.9759)
+
100
atomic mass Pb ,average = 207.17709amu
52.3%
( 207.9766)
+
100
12
6
% Abundance
98.89
1.11
C
13
6
C
Relative
atomic masses (amu)
12.0000
13.00335
On the periodic table what is the reported amu for C?
amu=12.01
Where did this number come from?
atomic massaverage =
atomic massC ,average =
⎛ % isotope ⎞
⎜
⎟ amuisotope
100 ⎠
isotopes ⎝
∑
11%
.
98.9%
(13.00006) +
(1199671
)
.
100
100
atomic massC ,average = 12.0077amu
atomic massC ,average = 12.01amu
2
We saw that
Example
⎛ 1 126 C atom ⎞ ⎛
⎞
1amu
23 12
⎟⎜
− 24 ⎟ = 6.022 x10
6 C atoms
12amu ⎠ ⎝ 166053873
.
x10 g ⎠
(12 g C )⎜⎝
12
6
General Rule #4: Chemists Are Lazy: make this number unique
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
N Avogadro = mole = 6.022 x10 23 atoms
1=
⎛
mole
6.022 x10 23 atoms
12
6
⎞⎛
atom
1amu
⎟⎜
(12 g C )⎜⎝ 1 12Camu
.
x10
⎠ ⎝ 166053873
12
6
(( x = amu) g X
− 24
x
elementX
⎞⎛
1mole
⎞
⎟⎜
⎟ = 1mole
g ⎠ ⎝ 6.022 x10 23 atoms ⎠
) = 1mole atoms
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 ) = a(amu ) + b(amu )+ ...... z(amu )
i
i
A
B
Z
i
3
Example: Acetylsalicylic acid, C9H 8O4, is the active ingredient of aspirin.
What is the mass in grams of 0.509 mol acetylsalicylic acid?
Example. Lead carbonate is principle component
of white lead used in all white paints prior to WWII.
Calculate the molar mass of lead carbonate
Lead carbonate = PbCO3
principle component paint
before WWII
Molar mass= unknown
Element #atoms amu
Pb
1
207.2
C
1
12.01
3O
3
16.00
Total amu
207.2
12.01
48.00
267.21
Acetylsalicylic acid
C9H 8O4
Active ingredient of aspirin
0.509 mol acetylsalicylic acid
Mass?
MM =
267.2
MM =
∑ n (amu ) = a(amu ) + b(amu )+ ...... z(amu
i
i
A
B
i
Mass %
⎡ atomic weight of element ⎤
% element = (# atoms of that element )⎢
⎥ (100% )
⎣ molar mass of compound ⎦
Z
( MM ) n = m
∑ n (amu ) = a(amu ) + b(amu )+ ...... z(amu )
i
i
MM =
m
n
g ⎞
⎛
.
. g
⎜ 18015
⎟ 0.509mole = m = 917
⎝
mole ⎠
i
A
B
Z
MM = 9(12.01) + 8(1008
. ) + 4(16.00)
g
MM = 18015
.
mole
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
3O
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 ⎦
4
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 ⎤
100%
% Ca = (1) ⎢
⎣ 100.09 ⎥⎦
% Ca = 40.04
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
⎡ 12.01 ⎤
100%
% C = 1⎢
⎣ 100.09 ⎥⎦
% C = 119992%
.
= 12.00%
⎡ 16.00 ⎤
% O = 3⎢
100%
⎣ 100.09 ⎥⎦
% O = 47.95683885 = 47.96%
Are We Done?
47.96
12.00
40.04
100%
Converting between g, moles, and number of particles
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
nN A = particles
Mass=m
Molar
grams
Avogadro’s
Moles
Mass
MM =
Moles=n
m
n
Very Important!
Check answer
particles
Formula Units
Number
The Golden Bridge
Stoichiometry or #of moles
Lead carbonate = PbCO3
principle component paint
before WWII
14.8 g PbCO3
number of moles, n = unknown
m
MNM =
n
MM =
Pb
C
3O
207.2
12.01
3(16.00)
207.2
12.01
48.00
267.21
267.2
m
n=
MM
∑ n (amu ) = a(amu ) + b(amu )+ ...... z(amu )
i
i
A
B
Z
i
5
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
n=
14.8 g
g
267.2
mole
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
n = 0.055389221mole
n = 0.0554mole
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
n=
m
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
n=
m
MM
6.64 gK
.
. molK
molK = 0170
= n = 01698
gK
39.10
(1)mole
8.84 gCr
= n = 01700
.
. molCr
molCr = 0170
gCr
52.00
mol
9.52 gO
= n = 0.59500molO = 0.595molO
gO
16.00
mol
0.170molK
0.170molCr
0.595mol O
6
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
0.595mol O
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
Formula: Cations First
2.00molCr
2molCr
2.00molK
2molCr
7.00molO
2molCr
2K : 2Cr : 7 O
K2Cr2O7
Make this non-fractional, multiply x2
Example : The compound that gives fermented grape juice, malt
liquor, and vodka their intoxicating properties 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
Example : The compound that gives fermented grape juice,
malt liquor, and vodka their intoxicating properties 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
5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
What is the simplest formula of ethyl alcohol?
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
Fermented grape juice
malt liquor
vodka
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
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
5g total ( C + H + O ) ⎯⎯
→ 9.55gCO2 + 587
. gH 2 O
7
Example : The compound that gives fermented grape juice,
malt liquor, and vodka their intoxicating properties 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 COg plus 5.87g H2O
Example : The compound that gives fermented grape juice,
malt liquor, and vodka their intoxicating properties 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 COg plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
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 gCO2 ⎠ ⎝ 1moleCO2 ⎠
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
⎞ ⎛ moleC ⎞
⎛
mole CO2
⎟
⎟⎜
nC = (9.55gCO2 )⎜
⎝ (12.01 + 2(16)) gCO2 ⎠ ⎝ moleCO2 ⎠
elements C, H, O
m
⎛ 1 ⎞
n=
= m⎜
⎟
⎝ MM ⎠
MM
⎛ mole CO2 ⎞ ⎛ moleC ⎞
⎟
⎟⎜
nC = (9.55gCO2 )⎜
⎝ (44.01gCO2 ⎠ ⎝ moleCO2 ⎠
nC = 0.217mole
⎛ moleH 2 O ⎞ ⎛ 2moleH ⎞
⎟
⎟⎜
nh = ( gH 2 O)⎜
⎝ amu gH 2 O ⎠ ⎝ 1moleH 2 O ⎠
⎞ ⎛ 2moleH ⎞
⎛
mole H 2 O
⎟
⎟⎜
n H = (587
. gH 2 O)⎜
. )) gH2 O ⎠ ⎝ moleH2 O ⎠
⎝ (16.00 + 2(1008
⎛ mole H 2 O ⎞ ⎛ 2moleH ⎞
⎟
⎟⎜
n H = (587
. gH 2 O)⎜
⎝ (18.016 gCO2 ⎠ ⎝ moleH 2 O ⎠
nC = 0.6517mole
nC = 0.652mole
Example : The compound that gives fermented grape juice,
malt liquor, and vodka their intoxicating properties 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 COg plus 5.87g H2O
Example : The compound that gives fermented grape juice,
malt liquor, and vodka their intoxicating properties 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 COg plus 5.87g H2O
What is the simplest formula of ethyl alcohol?
What is the simplest formula of ethyl alcohol?
⎛ amugC ⎞
⎜⎜
⎟⎟
⎝ amug CO2 ⎠
5.00g ethyl alcohol How are we going to get it? gc = ( gCO )
9.55 g CO2
gO in original sample?
⎛ 12.01gC ⎞
5.87 g H2O
⎟⎟
g c = (9.55g CO )⎜⎜
⎝ 44.01g CO ⎠
simplest formula? g total = g C + g H + g O
3 sig fig
2
2
= 2.61g C
2
⎛ amugH ⎞
⎟⎟
g H = ( g H O )⎜⎜
elements C, H, O 5.00 g total = 2.61g C + 0.657 g H + g O
⎝ amug H O ⎠
3
sig
fig
Now know:
gO = 5.00 − 2.61 − 0.657
⎛ 2(1008
. ) gC ⎞
⎟⎟ = 0.657 g C
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
2
5.00g ethyl alcohol
9.55 g CO2
5.87 g H2O
simplest formula?
elements C, H, O
m
⎛ 1 ⎞
n=
= m⎜
⎟
⎝ MM ⎠
MM
5.00 g total = 2.61g C + 0.657 g H + g O
gO = 5.00 − 2.61 − 0.657
⎛ moleO ⎞
⎟
nO = ( gO)⎜
⎝ amu gO ⎠
gO = 173
.
⎛ 1mole O ⎞
n H = (173
. gO)⎜
⎟
⎝ (16.00 gO ⎠
nO = 0108125
.
mole
nO = 0108
. mole
8
Example : The compound that gives fermented grape juice,
malt liquor, and vodka their intoxicating properties 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 COg 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?
molO=0.108
molH=0.652
molC=0.217
elements C, H, O
0108
. molO
=1
0108
. molO
0.652mol H
= 6.037 = 6
0108
. molO
0.217molC
= 2.009 = 2
. molO
0108
C2H6O
Mass Relations and Stoichiometry
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.
Mass Relations In Reactions
Writing and Balancing Chemical Reactions
Chemical Formulas
Lead Hand Side
Right Hand Side
Write reactions
Predict
Change in amounts
Relate to Mass
Ratios and
Stoichiometry
Reactants
number atoms A as reactants =
number atoms Bas reactants =
number atoms C as reactants =
.
.
.
Number atoms Z as reactants =
Products
number atoms A as products
number atoms B as products
number atoms C as products
number atoms Z as products
Consider what happens when one part necessary for the
Reaction (recipe) is limiting
9
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
PbS Burned in O2 To make Pb3 O4 and SO2
PbS + O2 → Pb3 O4 + SO2
1S
PbS
1S
PbS + O2 → Pb3 O4 + SO2
1Pb
3Pb
3S
1S
3( PbS ) + O2 → Pb3 O4 +
Pb3O4
3Pb
3S
Subscript
indicating number
SO2 Pb within a single
molecule
3Pb
3S
3 PbS + O2 → Pb3 O4 + 3SO2
2O
4O+(3x2)O=10 O
10
3 PbS + O2 → Pb3 O4 + 3SO2
From previous slide
Mass Relations and Stoichiometry
4O+3x2O=10O
3 PbS + 5O2 → Pb3 O4 + 3SO2
Chemical Formulas
4O+3x2O=10O
5x2O=10O
Write reactions
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
Sets up proportionalities
Predict
Change in amounts
STOICHIOMETRY
BUT How doe we make use
Of those proportionalities?
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
Mass Relations and Stoichiometry
Chemical Formulas
Write reactions
Step wise
1. Easy conceptually
= series of proportionalities
Predict
Change in amounts
Relate to Mass
Ratios and
Stoichiometry
Consider what happens when one part necessary for the
Reaction (recipe) is limiting
1. Longer time to compute
2. Easy to lose track of units
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!!!
11
3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g )
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.
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 )
⎢⎣ 5molO2 ( g ) ⎥⎦
(
)
⎡ gSO2 ⎤
gSO2 = 0.804molSO2 ⎢
⎥
⎣ molSO2 ⎦
⎡ Samu + 2(Oamu) ⎤
gSO2 = 0.804molSO2 ⎢
⎥
molSO2
⎣
⎦
Alternative Strategy is to fold all calculations together
m = nMM
⎡ (32.07 + 2(16.00)) gSO2
gSO2 = 0.804molSO2 ⎢
molSO2
⎢⎣
⎡ 64.07 gSO2 ⎤
gSO2 = 0.804molSO2 ⎢
⎥
⎣ molSO2 ⎦
gSO2 = 5151228
.
gSO2 = 515
. gSO2
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
2. Use Stoichiometry to find moles of O2
3. Find grams of O2 from those moles
m = nMM
n=
m
MM
⎡ 10 3 g ⎤
1kgPb3 O4 ⎢
⎥
⎣ kg ⎦
molPb3 O4 formed =
⎡ gPb3 O4 ⎤
⎢
⎥
⎣ molPb3 O4 ⎦
⎡ 10 3 g ⎤
1kgPb3O4 ⎢
⎥
⎣ kg ⎦
molPb3 O4 formed =
⎡ 3(207.2) + 4(16.00) gPb3 O4 ⎤
⎢
⎥
molPb3O4
⎣
⎦
Determine:
a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts
(
⎡ 3molSO2 ( g ) ⎤ ⎡ amus + 2 amuO2
⎥⎢
x = 134
. molO2 ⎢
molSO2
⎢⎣ 5molO2 ( g ) ⎥⎦ ⎢
⎣
(
)
) ⎤⎥
⎥
⎦
⎡ 3molSO2 ( g ) ⎤ ⎡ gSO ⎤
2
⎥⎢
x = 134
. molO2 ⎢
⎥
⎢⎣ 5molO2 ( g ) ⎥⎦ ⎣ molSO2 ⎦
(
)
⎡ 3molSO2 ( g ) ⎤ ⎡ 64.07 gSO ⎤
2
⎥⎢
x = 134
. molO2 ⎢
.
g = 515
. g
⎥ = 5151228
⎢⎣ 5molO2 ( g ) ⎥⎦ ⎣ molSO2 ⎦
(
)
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
⎡ 10 3 g ⎤
1kgPb3 O4 ⎢
⎥
⎣ kg ⎦
molPb3 O4 formed =
⎡ 685.6 gPb3 O4 ⎤
⎢
⎥
⎣ molPb3 O4 ⎦
molPb3 O4 formed = 1458576429
.
= 1459
.
molPb3 O4 formed = 1459
.
5molO2
x
=
1459
. molPb3 O4 molPb3 O4
x = (1459
. )(5molO2 ) = 7.295molO2
m = n( MM )
⎡ gO2 ⎤
x = 7.295molO2 ⎢
⎥
⎣ molO2 ⎦
⎡ 32.00 gO2 ⎤
x = 7.295molO2 ⎢
⎥ = 233.4 g
⎣ molO2 ⎦
12
3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g )
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
(((
)))
(
⎡ 10 3 g ⎤ ⎡ mol Pb3O4
⎢ kg ⎥ ⎢ 685.6 g
⎣
⎦ ⎢⎣
Pb3O4
molPbPb3O3O44 ⎤⎤⎡ 5molO2 ⎤ ⎡ g O2 ⎤
10 gg⎤⎤⎡⎡mol
x = 1kg Pb3O4 ⎡⎡10
kgPbPb
xx == 11kg
⎥
⎢
⎢
O
⎢
⎢
⎥
3
4
Pb3O
O
3 44
kg ⎦⎦⎢⎢ ggPbPbOO ⎥⎥⎥⎥⎢⎢ mol Pb O ⎥⎥ ⎢⎢ molO ⎥⎥
⎣⎣ kg
3 3 4 4 ⎦⎦⎣
3 4 ⎦⎣
2 ⎦
⎣⎣
x = 1kg Pb3O4
)
333
⎤ ⎡ 5molO2
⎥⎢
⎥⎦ ⎢⎣ mol Pb3O4
⎤ ⎡ 32.00 g O2
⎥⎢
⎥⎦ ⎢⎣ molO2
⎤
⎥
⎥⎦
x = 233.37 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=
6.00 gO2
⎡ 32.00 gO2 ⎤
⎢
⎥
⎣ molO2 ⎦
molO2
.
= 01875
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
molO2
01875
.
5molO2 x = 01125
x = 26.9 g
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 = (6.00 gO2 )⎡ molO2 ⎤ ⎡ 3molPbS ⎤ ⎡ 239.27 gPbS ⎤
x = (6.00 gO2 )⎢
⎥⎢
⎥⎢
⎥
⎣ 32.00 gO2 ⎦ ⎣ 5molO2 ⎦ ⎣ molPbS ⎦
Chemical Formulas
Write reactions
x = 26.917 g
x = 26.9 g
Predict
Change in amounts
Relate to Mass
Ratios and
Stoichiometry
Consider what happens when one part necessary for the
Reaction (recipe) is limiting
13
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 it 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.
Limiting Reagents
Liebig’s Law of the Limiting (1810-1820 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?
1 pkgchips + 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⎢
⎥ = 192cookies
12eggs
2eggs
⎣ 2eggs ⎦
Smallest yield = 72 cookies
Limiting reagent = chips
While I wasn’t looking, my son snitched 4 cookies.
⎡ yield exp erimental ⎤
% yield = ⎢
⎥ 100%
⎣ yield theoretical ⎦
⎡ 72 − 4 ⎤
% yield cookies = ⎢
100% = 94.44444 = 94%
⎣ 72 ⎥⎦
“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
Actual Yield=Theoretical
Yield
(most of the time)
14