Download Stoichiometry

10/3/09
Stoichiometry
Stoichiometry
Chemistry
Preparatory
Topics
Lecture
5
–
Dr.
Gondran
Composi?on
(Chemical
Formula)
Stoichiometry:
The
propor?ons
of
elements
in
stoichiometric
compounds.
In
H2O,
hydrogen
and
oxygen
is
2:1.
Reac?on
(Chemical
Equa?on)
Stoichiometry:
The
stoichiometric
propor?ons
by
which
species
combine
or
are
produced
in
chemical
equa?ons.
2
H2
+
O2

2
H20
Mass‐to‐Mass
Stoichiometry:
The
mass
stoichiometric
propor?ons
in
which
species
combine
or
are
produced.
(use
formula
masses)
4
grams
of
H2
and
32
grams
of
O2
make
36
grams
of
H2O
Formula
Stoichiometry
Calcula1ng
Molecular
Mass
Formula
or
composi?on
stoichiometry
gives
the
number
of
atoms
of
each
element
that
are
in
a
given
compound
or
molecule
Molar
mass
of
any
compound
or
molecule
is
the
sum
of
the
masses
of
the
cons?tuent
atoms
Example:
Water,
H2O
has
2
atoms
of
H
and
one
atom
of
O
Example:
Glucose,
C6H12O6
has
6
atoms
of
C,
12
atoms
of
H,
and
6
atoms
of
O
Stoichiometry
Composi?on
(Chemical
Formula)
Stoichiometry:
The
propor?ons
of
elements
in
stoichiometric
compounds.
In
H2O,
hydrogen
and
oxygen
is
2:1.
Reac?on
(Chemical
Equa?on)
Stoichiometry:
The
stoichiometric
propor?ons
by
which
species
combine
or
are
produced
in
chemical
equa?ons.
2
H2
+
O2

2
H20
Mass‐to‐Mass
Stoichiometry:
The
mass
stoichiometric
propor?ons
in
which
species
combine
or
are
produced.
(use
formula
masses)
4
grams
of
H2
and
32
grams
of
O2
make
36
grams
of
H2O
Example:
Water,
H2O
has
2
atoms
of
H
and
one
atom
of
O
2
(
1
g/mole)
+
16
g/
mole
=
18
g/mole
Example:
Glucose,
C6H12O6
has
6
atoms
of
C,
12
atoms
of
H,
and
6
atoms
of
O
6
(12
g/mole)
+
12
(1
g/mole)
+
6
(16
g/mole)
=
(72
+
12
+
96)
g/mole
=
180.
g/mole
Chemical
Reac1ons
Chemical
reac?on
occur
when
chemical
bonds
are
broken
and
new
ones
are
formed
to
change
the
molecules
present.
In
the
process
reactants
are
transformed
into
products.
The
number
and
type
of
atoms
does
not
change
Example: 2
H2
+
O2 
2
H2O
reactants
product(s)
1
10/3/09
Types
of
Reac1ons
Combina?on
(or
synthesis)
A
+
X
→
AX
Decomposi?on
AX
→
A
+
X
Single
replacement
A
+
BX
→
AX
+
B
Double
replacement
AX
+
BY
→
AY
+
BX
and
Combus?on
CxHy
+
y/4
O2
→
x
CO2
+
y/2
H2O
Balancing
Equa1ons
Do
not
change
the
iden??es
(formulas)
of
any
of
the
reactants
or
products.
Determine
what
coefficients
are
necessary
to
ensure
that
the
same
number
of
each
type
of
atom
appears
on
both
reactant
and
product
sides.
1) 
2) 
3) 
Balancing
Equa1ons:
Example
1) 
2) 
3) 
Make
a
first
guess
star?ng
with
1
for
the
most
complicated
molecule(s).
Balance
the
number
of
atoms
for
each
element
in
the
star?ng
formula
(compound),
first
those
that
only
occur
in
one
other
compound
then
those
that
occur
several
?me.
Save
atoms
that
occur
in
a
homogeneous
molecule
for
last.
Mul?ply
through
to
reduce
to
eliminate
frac?ons
with
the
lowest
whole
number
coefficients
possible
Make
a
first
guess
star?ng
with
1
for
the
most
complicated
molecule(s).
Balance
the
number
of
atoms
for
each
element
in
the
star?ng
formula
(compound),
first
those
that
only
occur
in
one
other
compound
then
those
that
occur
several
?me.
Save
atoms
that
occur
in
a
homogeneous
molecule
for
last.
Mul?ply
through
to
eliminate
frac?ons
obtaining
the
lowest
whole
number
coefficients
possible
Balancing
Equa1ons:
Example
1) 
2) 
3) 
Make
a
first
guess
star?ng
with
1
for
the
most
complicated
molecule(s).
Balance
the
number
of
atoms
for
each
element
in
the
star?ng
formula
(compound),
first
those
that
only
occur
in
one
other
compound
then
those
that
occur
several
?me.
Save
atoms
that
occur
in
a
homogeneous
molecule
for
last.
Mul?ply
through
to
reduce
to
eliminate
frac?ons
with
the
lowest
whole
number
coefficients
possible
__
(NH4)
2Cr2O7(S)

__
N2(g)
+
__
H2O(g)
+
__
Cr2O3(s)
__CaSiO3(aq)
+__HF(aq)
__SiF4(g)
+__CaF2(s)
+
__H2O(l)
1
(NH4)
2Cr2O7(S)

1
N2(g)
+
4
H2O(g)
+
1
Cr2O3(s)
1
CaSiO3(aq)
+
6
HF(aq)

1
SiF4(g)
+
1
CaF2(s)
+
3
H2O(l)
Balancing
Equa1ons:
Example
Moles
to
Moles
1) 
2) 
3) 
Make
a
first
guess
star?ng
with
1
for
the
most
complicated
molecule(s).
Balance
the
number
of
atoms
for
each
element
in
the
star?ng
formula
(compound),
first
those
that
only
occur
in
one
other
compound
then
those
that
occur
several
?me.
Save
atoms
that
occur
in
a
homogeneous
molecule
for
last.
Mul?ply
through
to
reduce
to
eliminate
frac?ons
with
the
lowest
whole
number
coefficients
possible
__
NaHCO3(s)

__
Na2CO3(s)
+
__
H2O(g)
+
__
CO2
(g)
1
NaHCO3(s)

1/2
Na2CO3(s)
+
1/2
H2O(g)
+
1/2
CO2
(g)
Each
balanced
chemical
equa?on
contains
the
informa?on
about
the
ra?os
of
all
of
the
components
CaSiO3(aq)
+
6
HF(aq)

SiF4(g)
+
CaF2(s)
+
3
H2O(l)
1
mole
of
CaSiO3 1
mole
of
CaSiO3
6
moles
of
HF
1
mole
of
SiF4
2
NaHCO3(s)

1
Na2CO3(s)
+
1
H2O(g)
+
1
CO2
(g)
=
1
1
mole
of
CaSiO3 1
mole
of
CaSiO3
1
mole
of
CaF2
3
mole
of
H2O
2
10/3/09
Moles
to
Moles:
Example
SiC
is
made
from
sand,
SiO2
and
elemental
C
with
CO
as
a
by‐product.
How
many
moles
of
CO
are
produced
when
1/3
mole
of
C
is
used?
__
SiO2
+
__
C

__
SiC
+
__
CO
1
SiO2
+
3
C

1
SiC
+
2
CO
1/3
mole
of
C
x
2
moles
of
CO
=
2/9
moles
of
CO
3
moles
of
C
What
Does
a
Mole
Look
Like?
What
is
a
Mole?
The
number
equal
to
the
number
of
carbon
atoms
in
exactly
12
grams
of
pure
12C;
Avogadro’s
number.
One
mole
represents
6.022
×
1023
units.
1)
Molar
mass
and
the
mass
of
the
samples
2)
Molarity
and
volume
of
a
solu?on
3)
Gas
laws
knowing
temperature,
pressure
and
volume
Stoichiometry
Molar
mass
of
any
compound
or
molecule
is
the
sum
of
the
masses
of
the
cons?tuent
atoms
Composi?on
(Chemical
Formula)
Stoichiometry:
The
propor?ons
of
elements
in
stoichiometric
compounds.
In
H2O,
hydrogen
and
oxygen
is
2:1.
Example:
Water,
H2O,
molar
mass
=
18
g/mole
density
1
g/
ml

one
mole
is
18
ml
~
1.2
tbsp
Reac?on
(Chemical
Equa?on)
Stoichiometry:
The
stoichiometric
propor?ons
by
which
species
combine
or
are
produced
in
chemical
equa?ons.
2
H2
+
O2

2
H20
Example:
Glucose,
C6H12O6,
m.
m.=
180
g/mole
density
1.5
g/
ml

one
mole
is
120
ml
Mass‐to‐Mass
Stoichiometry:
The
mass
stoichiometric
propor?ons
in
which
species
combine
or
are
produced.
(use
formula
masses)
4
grams
of
H2
and
32
grams
of
O2
make
36
grams
of
H2O
Mass‐to‐Mass
Stoichiometry
A
balanced
chemical
equa?on
is
a
quan?ta?ve
descrip?on
of
the
reac?on
and
it
tells
you
how
many
moles
of
each
reactant
combine
to
form
how
many
moles
of
each
product.
We
can
not
directly
measure
moles
We
need
to
measure
masses,
volumes,
concentra?ons
Then
convert
to
moles
so
we
can
compare
the
molar
propor?ons
in
the
balances
chemical
equa?on
Finally
we
need
to
convert
back
to
masses,
volumes,
concentra?ons
…
something
we
will
be
able
to
measure
Mass‐to‐Mass
Stoichiometry
1)  Amount
of
reactant

Moles
of
reactant
1)  Molar
mass
2)  Solu?on
molarity
and
volume
3)  Gas
laws
2)  Moles
of
reactant

Moles
of
product
Stoichiometric
coefficients
in
the
balanced
chemical
equa?on
3)  Moles
of
product

Amount
of
product
1)  Molar
mass
2)  Solu?on
molarity
and
volume
3)  Gas
Laws
3
10/3/09
Calcula1ng
Moles
with
Molarity
How
many
moles
of
HCl
are
in
0.25
liters
of
a
0.05
M
solu?on?
moles
=
volume
x
concentra?on
moles
=
0.25
l
x
0.05
moles/
l
=
0.0125
moles
If
the
volume
remained
the
same
when
excess
solid
NaOH
was
added
and
what
would
the
concentra?on
of
NaCl
solu?on?
molarity
(concenta?on)
=
moles
/
volume
molarity
=
0.0125
moles
/
0.25
l
=
0.05
M
Mass‐to‐Mass
Stoichiometry
1)  Amount
of
reactant

Moles
of
reactant
1)  Molar
mass
2)  Solu?on
molarity
and
volume
3)  Gas
laws
2)  Moles
of
reactant

Moles
of
product
Stoichiometric
coefficients
in
the
balanced
chemical
equa?on
3)  Moles
of
product

Amount
of
product
1)  Molar
mass
2)  Solu?on
molarity
and
volume
3)  Gas
Laws
Limi1ng
Reactants
Using
the
stoichiometric
rela?onship
given
in
the
balanced
chemical
equa?on
or
mass‐to‐mass
stoichiometry,
it
is
possible
to
determine
which
(if
any)
reactant
is
present
in
excess
(more
is
present
than
will
react
with
all
of
the
other
component)
and
which
reactant
will
be
consumed
completely
and
thus
limit
how
much
stuff
can
react,
the
“limi?ng
reactant”.
Calcula1ng
Moles
from
Molar
Mass
How
many
moles
are
in
28
gm
of
C3H6O?
3
carbon
atoms
6
hydrogen
atoms
and
1
oxygen
atom
molar
mass
=
3
x
12
+
6
x
1.0
+
1
x
16
=
58
gm/mole
number
of
moles
=
28
gm
x
1
mole
=
0.48
moles
58
gm
How
many
grams
dose
1.45
moles
of
CO2
weigh?
1
carbon
atom
and
2
oxygen
atoms
molar
mass
=
1
x
12
+
2
x
16
=
44
gm/mole
grams
of
CO2
=
1.45
moles
x
44
gm
/mole
=
63.8
grams
Mass‐to‐Mass:
Example
How
much
CO2
is
released
when
you
burn
off
19
grams
of
sugar?
C6H12O6
+
6
O2
→
6
CO2
+
6
H2O
Reactant
‐‐
grams
to
moles
‐‐
moles
to
moles
‐‐
moles
to
grams
19
gm
sugar
x
1
mole
sugar
x
6
moles
CO2
x
44.
gm
CO2
=
180
gm
sugar
1
mole
sugar
1
mole
CO2
28
gm
CO2
Limi1ng
Reactant:
Baking
Analogy
We
are
trying
to
make
as
many
chocolate
chip
cookies
as
possible
using
a
specific
recipe
and
the
ingredients
I
have
on
hand.
The
recipe
calls
for:
recipe 1c
buoer
I
have 2c
buoer
1c
sugar
3c
sugar
2
eggs +
other
stuff

48
cookies
3
eggs If
I
divided
the
amounts
I
have
by
the
amounts
needed
per
batch
I
have
enough
buoer sugar eggs
to
make
a
double
triple
1.5
batch
If
I
try
to
make
a
triple
batch
it
won’t
come
out
right!
The
most
I
can
make
using
the
correct
propor?ons
is
a
1.5
batch.
Use
1.5
c
buoer 1.5
c
sugar 3
eggs to
make
72
cookies
Leaving
0.5
c
buoer 1.5
c
sugar 0
eggs
4
10/3/09
Limi1ng
Reactant:
by
Moles
1
CaSiO3(aq)+
6
HF(aq)1
SiF4(g)+
1
CaF2(s)+
3
H2O(l)
How
much
CaF2
is
formed
if
you
start
with
2
moles
of
CaSiO3
and
3
moles
of
HF?
2
moles
CaSiO3
x
1
mole
CaF2
=
2
moles
of
CaF2
1
mole
CaSiO3
3
moles
HF
x
1
mole
CaF2
=
½
mole
of
CaF2
6
mole
HF
Limi1ng
Reactant
Limi1ng
Reactant:
by
Mass
Mass‐to‐Mass
Stoichiometry
1)  Amount
of
reactant

Moles
of
reactant
1)  Molar
mass
2)  Solu?on
molarity
and
volume
3)  Gas
laws
2)  Moles
of
reactant

Moles
of
product
Stoichiometric
coefficients
in
the
balanced
chemical
equa?on
3)  Moles
of
product

Amount
of
product
1)  Molar
mass
2)  Solu?on
molarity
and
volume
3)  Gas
Laws
Theore1cal,
Actual
and
Percent
Yield
How
many
grams
of
CO2
are
produced
when
45
gm
of
C2H6
are
burned
in
600
gms
of
O2?
2
C2H6
+
7
O2

4
CO2
+
6
H20
The
amount
of
the
limi?ng
reactant
will
determine
the
maximum
theore?cal
yield
for
the
reac?on
45
gm
C2H6
x
1
mole
C2H6
x
4
moles
CO2
=
3
moles
CO2
30
gm
C2H6
2
moles
C2H6
600
gm
O2
x
1
mole
O2
x
4
moles
CO2
=
11
moles
CO2
32
gm
O2
7
moles
O2
The
amount
of
CO2
is
limited
by
the
amount
of
C2H6
grams
of
CO2
=
3
moles
CO2
x
44
gm/mole
=
132
gm
percent
yield
=
100%
x
(actual
yield
/
theore?cal
yield)
Percent
Yield
Problem
Summary
51
gm
SiC
are
produced
from
100
gm
of
sand,
SiO2
and
excess
elemental
C.
What
is
the
percent
yield
for
the
process?
SiO2
+
3
C

SiC+
2
CO
How
much
SiC
should
be
produced
from
100
g
of
sand?
100
gm
SiO2
x
1
mole
SiO2
x
1
mole
SiC
x
40
g
SiC
=
67
gm
60
gms
1
mole
SiO2
1
mole
SiC
Percent
yield
=
actual
yield
x
100%
=
51
gm
=
76
%
theore?cal
yield
67
gm
In
prac?ce,
the
actual
yield
will
be
some
what
lower.
Example:
If
we
expected
to
produce
17.64
g
of
ZnCl
and
we
only
produced
or
only
collected
16.05
g
of
ZnCl
The
percent
yield
=
100%
x
(16.05
g
/
17.64
g)
=
91.17%
Defined
stoichiometry
Composi?on
stoichiometry
‐
Molar
mass
Reac?on
stoichiometry
‐
Balancing
equa?ons
Mass‐to‐mass
stoichiometry
Moles
from
mass
or
molarity
Limi?ng
reactants
Theore?cal,
actual
and
percent
yields
5
10/3/09
One
More
…
?
The
Haber
process
for
making
ammonia
from
nitrogen
in
air
is
given
by
the
equa?on:
N2
+
3
H2
‐>
2
NH3
What
the
mass
of
hydrogen
must
be
supplied
to
make
500.
kg
of
ammonia
in
a
system
that
has
an
88.8%
yield?
Solu1on
N2
+
3
H2
‐>
2
NH3
What
the
mass
of
H2
must
be
supplied
to
make
500.
kg
of
NH3
in
a
system
that
has
an
88.8%
yield?
Percent
yield
=
100%
x
actual
yield/theore?cal
yield
Theore?cal
yield
=
100%
x
actual
yield/percent
yield
=
100%
x
500.
kg
of
NH3
/
88.8%
=
563
kg
of
NH3
563,000
gm
NH3
x
1
mole
NH3
x
3
moles
H2
x
2.0
gm
H2
17
gm
NH3
2
moles
NH3
mole
H2
=
99,000
gm
H2
=
99
kg
H2
6