Stoichiometry
Sec%on
7.2
pg.
286‐293

Stoichiometry
  A
method
of
problem‐solving
using
known
quan%%es
of
one
en%ty
in
a
chemical
reac%on
to
find
out
unknown
quan%%es
of
another
en%ty
in
the
same
reac%on
  Always
requires
a
balanced
chemical
equa%on
  Series
of
steps
for
measurement
calcula%ons
  Always
involves
a
mole
ra%o
to
“switch
substances”
  Wri%ng
chemical
reac%ons,
net
ionic
equa%ons
and
dissocia%on
equa%ons
is
essen%al
  Conver%ng
grams
to
moles
(and
vice
versa)
is
also
essen%al
Gravimetric
Stoichiometry
  Gravimetric
Stoichiometry
–
method
used
to
calculate
the
masses
of
reactants
or
products
in
a
chemical
reac%on
  Gravimetric
=
mass
measurement
  Using
gravimetric
stoichiometry,
we
can
apply
our
knowledge
of
balanced
chemical
reac6ons
and
mass
to
mol
conversions
to:
  Predict
the
mass
of
product
we
will
get
from
a
reac%on
  Es%mate
the
amount
of
reactant
we
need
for
a
reac%on
to
produce
a
certain
mass
of
product
Gravimetric
Stoichiometry
Steps:
  Step
1:
Write
a
balanced
chemical
reac%on
equa%on
  List
the
measured
mass
(given),
the
unknown
quan%ty
(required)
and
conversion
factors
(M
=
molar
mass)
  Step
2:
Convert
the
mass
of
the
measured
substance
to
moles
  Step
3:
Calculate
the
moles
of
the
unknown
substance
using
the
mole
ra%o
required
given
  Step
4:
Convert
from
moles
of
the
required
substance
to
its
mass.
Stoichiometry
Calculations
(Measured
quan6ty)
solids/liquids
m

n
mole
ra6o
solids/liquids
m

n
(Required
quan6ty)
How
many
grams
of
oxygen
are
required
to
completely
burn
10.0
g
of
propane
(C3H8(g))?
  The
balanced
chemical
equa%on
is:
C3H8(g)
+
5
O2(g)

3
CO2(g)
+
4
H2O(g)
  The
number
of
moles
of
propane
reac%ng
is:
(MASS
to
MOLES)
10.0
g
x
1
mol
=
0.227
mol
44.11
g
  The
number
of
moles
of
oxygen
produced
is:
(MOLE
RATIO)
0.227
mol
x
5
mol
=
1.13
mol
1
mol
  The
mass
of
oxygen
produced
is:
(MOLES
TO
MASS)
1.1335
mol
x
32.00
g
=
36.3
g
1
mol
Practice
#2
  20.0
g
of
butane
(C4H10(g))
are
completely
burned
in
a
lighter.
How
many
grams
of
CO2(g)
are
produced?
  Start
with
a
balanced
chemical
equa%on:
2
C4H10(g)
+
13
O2(g)

8
CO2(g)
+
10
H2O
(g)
m
=
20.0g
m
=
?
M
=
58.84
g/mol
M
=
44.01
g/mol
Practice
#2
(Team
Unit
Analysis)
2
C4H10(g)
+
13
O2(g)

8
CO2(g)
+
10
H2O
(g)
m
=
20.0g
m
=
?
M
=
58.14
g/mol M
=
44.01
g/mol
2)
Mass
to
moles,
mole
ra%o,
moles
to
mass
20.0
g
x
1
mol
x
.
8
mol
x
44.01
g
=
60.6
g
58.14
g
2
mol
1
mol
Practice
#3
(Unit
Analysis)
What
mass
of
iron
(III)
oxide
is
required
to
produce
100.0
g
of
iron?
Fe2O3(s)
+
3
CO(g)

2
Fe(s)
+
3
CO2(g)
m
=
?
m
=
100g
M
=
159.70g/mol
M
=
55.85
g/mol
m
Fe2O3(s):
100.0
g
x
1
mol
x
1
mol
x
159.70
g
=
143.0
g
Fe2O3
55.85
g
2
mol
1
mol
  Remember
to
keep
the
unrounded
values
in
your
calculator
for
further
calcula%on
un%l
the
final
answer
is
reported.
  Pg.
290
#8‐14
  (Answers
pg.
785)
Percent
Yield
for
Reactions
  We
can
use
stoichiometry
to
test
experimental
designs,
technological
skills,
purity
of
chemicals,
etc.
We
evaluate
these
by
calcula%ng
a
percent
yield.
  This
is
the
ra%o
of
the
actual
(experimental)
quan1ty
of
product
obtained
and
the
theore1cal
(predicted)
quan1ty
of
product
obtained
from
a
stoichiometry
calcula%on
  Percent
yield
=
actual
yield
x
100
predicted
yield
  Some
forms
of
experimental
uncertain%es:
  All
measurements
(limita%ons
of
equipment)
  Purity
of
chemical
used
(80‐99.9%
purity)
  Washing
a
precipitate
(some
mass
is
lost
through
filter
paper)
  Es%ma%on
of
reac%on
comple%on
(qualita%ve
judgements
i.e.
color)
Percent
Yield
Example
#1
  Example:
In
a
chemical
analysis,
3.00
g
of
silver
nitrate
in
solu%on
was
reacted
with
excess
sodium
chromate
to
produced
2.81
g
of
filtered,
dried
precipitate.
  Predicted
value:
2AgNO3(aq)
+
Na2CrO7(aq)
Ag2CrO7(s)
+2NaNO3(aq)
m
=
3.00
g
m
=
M
=
169.88g/mol
M
=
331.74
g/mol
3.00g
x
1
mol
x
1
mol
x
331.
74
g
=
2.
93
g
169.88g
2
mol
1
mol
  Percent
yield
=
actual
yield
x
100%
=
2.81g
x
100%
=
95.9%
predicted
yield
2.93g
Testing
the
Stoichiometric
Method
  Stoichiometry
is
used
to
predict
the
mass
of
precipitate
actually
produced
in
a
reac%on.
  Filtra%on
is
used
to
separate
the
mass
of
precipitate
actually
produced
in
a
reac%on
Testing
the
Stoichiometric
Method
  What
mass
of
lead
is
produced
by
the
reac%on
of
2.13
g
of
zinc
with
an
excess
of
lead(II)
nitrate
solu%on?
  Design:
A
known
mass
of
zinc
is
place
in
a
beaker
with
an
excess
of
lead(II)
nitrate
solu%on.
The
lead
is
produced
in
the
reac%on
is
separated
by
filtra%on
and
dried.
The
mass
of
the
lead
is
determined
  Predic6on:
Zn(s)
+
Pb(NO3)2(aq)

Zn(NO3)2(aq)
+
Pb(s)
m
=
2.13
g m
=
?
M
=
65.41
g/mol
M
=
207.2
g/mol
2.13
g
x
1
mol
x
1
x
207.2
g
=
6.75
g
65.41
g
1
1
mol
Testing
the
Stoichiometric
Method
  Predic6on:
6.75
g
of
lead
will
be
produced
(Stoichiometric
calcula6on)
  Evidence:
In
the
beaker,
crystals
of
a
shiny
black
solid
were
produced,
and
all
the
zinc
disappeared.
Mass
of
filter
paper
=
0.92
g Mass
of
dried
filter
paper
plus
lead
=
7.60g
  Analysis:
mass
of
lead
product
=
7.60g
–
0.92g
=
6.68g
According
to
the
evidence
collected,
6.68
g
of
lead
precipitated.
  Evalua6on:
%
difference
=
(experimental
–
predicted)
x
100 %
predicted
6.68
–
6.75
x
100%
=
1%
6.75
Given
such
a
small
difference,
which
can
readily
be
accounted
for
by
normal
sources
of
error,
the
predic1on
is
clearly
verified,
and
so
the
stoichiometric
method
for
predic1ng
the
mass
of
lead
formed
in
this
experiment
is
judged
to
be
acceptable.
Testing
the
Stoichiometric
Method
  Predic6on:
6.75
g
of
lead
will
be
produced
(Stoichiometric
calcula6on)
  Analysis:
mass
of
lead
product
=
7.60g
–
0.92g
=
6.68g
According
to
the
evidence
collected,
6.68
g
of
lead
precipitated.
  Evalua6on:
 
Could
you
also
calculate
the
%
yield
?
Percent
yield
=
actual
yield
x
100%
=
6.68
g
x
100%
=
99.0%
predicted
yield
6.75g
Homework
  Inves%ga%on
7.2
–
Guided
Lab
Report
  Pg.
293
#1‐2,
6‐10