INTRONotes - Fundamentals
 CHEM
1211K
 Fall
2010
...

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Unformatted text preview: Fundamentals
 CHEM
1211K
 Fall
2010
 1
 Objec0ves 
 •  Iden1fy
physical
and
chemical
changes
 •  Perform
unit
conversions
 •  Iden1fy
the
subatomic
par1cles
of
an
atom
and
be
 familiar
with
mass
number,
atomic
number,
and
isotopes.
 •  Be
familiar
with
the
organiza1on
of
elements
on
the
 periodic
table
 •  Iden1fy
compounds
and
be
able
to
derive
the
formulas
 of
them
 •  Be
able
to
derive
a
chemical
formula
from
a
name
or
the
 name
from
the
formula
 •  Perform
calcula1ons
using
molar
mass
and
the
mole
 2
 Objec0ves 
 •  Determine
molecular
and
empirical
formulas
 •  Calculate
and
manipulate
solu1on
concentra1ons
 (molarity)
 •  Perform
dilu1on
calcula1ons
 •  Balance
chemical
equa1ons
 •  Predict
products
of
double
displacement
reac1ons
and
 write
total
and
net
ionic
equa1ons
 •  Develop
a
working
knowledge
of
acid
and
bases
 •  Understand
the
fundamentals
of
redox
reac1ons
 •  Carry
out
stoichiometric
calcula1ons
 3
 Why
is
it
important? 
 •  The
topics
covered
in
this
chapter
are
the
 “fundamentals”
of
chemistry.
They
are
 recurrent
topics
throughout
the
two
 semesters
of
Chemical
Principles,
and
they
will
 be
used
in
every
other
chemistry
course
you
 take!
 4
 C:

Compounds
 •  Compounds
are
neutral
substances
consis1ng
 of
two
ore
more
different
elements
bonded
 together
in
a
definite
ra1o.
 –  Binary compounds –  Organic compounds –  Inorganic compounds 5
 E:

Moles
and
Molar
Masses 
 •  One
mole
of
objects
contains
exactly

 



6.022
x
1023
of
those
objects.
 N0
=
6.0221420
X
1023
 Avagradro’s
Number 
 Figure E.3 Mr. Avagadro 6
 E:

Moles
and
Molar
Masses 
 •  Moles
of
different
substances
all
contain
Avagadro’s
 number
of
that
thing,
but
they
all
have
different
 masses.
 Figure E.2 7
 Figure E.4 E:

Moles
and
Molar
Masses 
 •  Since
we
can’t
count
atoms
directly,
we
can
 find
the
amount
of
substance
if
we
know
the
 mass
of
the
sample
and
the
molar mass. –  The
molar
mass
of
an
element
is
the
mass
per
 mole
of
its
atoms.
 –  The
molar
mass
of
a
molecular
compound
is
the
 mass
per
mole
of
its
molecules.
 –  The
molar
mass
of
an
ionic
compound
is
the
mass
 per
mole
of
its
formula
units.
 8
 E:

Moles
and
Molar
Masses 
 •  Example:

How
many
moles
of
CO2
are
present
 in
10.0
g
of
CO2?
 •  Example:

How
many
molecules
of
CO2
are
 present
in
10.0
g
of
CO2?
 9
 E:

Moles
and
Molar
Masses 
 •  The
atomic weights
reported
on
the
periodic
 table
are
weighted
averages
of
the
masses
of
 all
isotopes
of
that
element.
 ∑ [(mass) ( percent) ] + [(mass) n n m ( percent ) m ] + ... + [( mass) z ( percent ) z ] 10
 E:

Moles
and
Molar
Masses 
 •  Example:

What
is
the
weighted
atomic
weight
of
 chlorine
given
the
following
naturally
occurring
 isotopes
and
their
abundances?
 35Cl = 34.9688 amu, 75.77% abundant 37Cl = 36.9659 amu, 24.23% abundant 11
 F:

Determining
Chemical
 Formulas 
 •  There
are
two
types
of
chemical
formulas:
 –  The
empirical formula
represents
the
simplest
 whole‐number
ra1o
of
atoms
in
a
compound.
 –  The
molecular formula
is
the
EXACT
formula
of
 the
molecule.
 12
 F:

Determining
Chemical
 Formulas 
 •  The
mass percent
(aka
weight percent
or
 percent composi:on)
is
a
comparison
of
the
 total
mass
of
each
element
in
one
mole
of
the
 compound
to
the
mass
of
one
mole
of
the
 en1re
compound.
 13
 F:

Determining
Chemical
 Formulas 
 Example:

Tetrodotoxin,
a
potent
poison
found

 in
the
ovaries
and
liver
of
the
globefish,
has
the
 empirical
formula
C11H17N3O8.

Calculate
the
 mass
percentages
of
the
four
element
in
this
 compound.
 14
 F:

Determining
Chemical
 Formulas 
 •  Example:

Maleic
acid
is
an
organic
compound
 composed
of
41.39%
carbon,
3.47%
hydrogen,
and
 the
rest
is
oxygen.
What
is
it’s
empirical
formula?
 •  Strategy:
 –  Assume
that
you
have
100.0
g
of
sample
so
that
all
%
given
 can
be
converted
directly
to
grams.
 –  Use
atomic
weights
to
convert
g
to
moles
of
each
element.
 –  Divide
each
number
of
moles
by
the
smallest
number
of
 moles
present.
 –  The
dividends
are
the
subscripts
in
the
chemical
formula.
 •  You
may
have
to
reduce
with
the
common
denominator
or
 mul1ply
to
get
rid
of
a
frac1on.
 15
 16
 F:

Determining
Chemical
 Formulas 
 •  How
do
you
know
if
the
empirical
formula
is
 the
molecular
formula?
 –  If
0.129
mol
of
maleic
acid
has
a
mass
of
15.0
g,
 what
is
the
molecular
formula?
 17
 G:

Mixtures
and
Solu0ons 
 •  Mixtures
are
not
pure
substances
like
 compounds
or
elements.
 –  Their
composi1ons
may
be
varied.
 –  The
chemical
proper1es
of
the
substances
in
the
 mixture
are
retained.
 18
 G:

Mixtures
and
Solu0ons 
 •  Heterogeneous mixtures
have
component
 par1cles
that
are
so
large
we
can
see
and
 separate
them.
 •  Other
mixtures
have
molecules
so
well
mixed
 that
the
composi1on
is
the
same
throughout,
 no
mager
how
small
the
sample.

These
are
 called
homogeneous
mixtures.
 19
 G:

Mixtures
and
Solu0ons 
 Figure G.1—Heterogeneous granite Figure G.2 Homogeneous mixtures 20
 G:

Mixtures
and
Solu0ons 
 •  Homogeneous
mixtures
are
also
called
 solu%ons.
 –  Solu1ons
are
composed
of
a
solute
and
a
solvent.
 •  Solute Figure G.3 •  Solvent •  Solu1ons
in
which
water
is
the
 


solvent
are
called
aqueous.
 21
 G:

Mixtures
and
Solu0ons 
 •  The
molar concentra:on (or molarity, M)
of
a
 solute
in
a
solu1on
is
defined
as:
 Moles
of
solute 
 L
of
solu0on 
 22
 G:

Mixtures
and
Solu0ons 
 •  Example:

Calculate
the
molarity
of
a
solu1on
 by
dissolving
10.0
g
of
Al(NO3)3
in
enough
 water
to
make
250.0
mL
of
solu1on.
 23
 G:

Mixtures
and
Solu0ons 
 •  Example:

What
volume
of
0.050
M
NaOH
 solu1on
contains
0.025
mol
NaOH?
 24
 G:

Mixtures
and
Solu0ons 
 •  In
a
dilu:on
the
concentra1on
of
a
solu1on
is
 reduced
by
addi1on
of
solvent.
 –  The
number
of
moles of solute does not change.
 M1V1
=
M2V2 25
 H:

Chemical
Equa0ons 
 •  Chemical reac:ons are
processes
by
which
 one
or
more
substances
are
converted
into
 other
substances.
 –  The
star1ng
materials
are
the
reactants.
 –  The
substances
formed
are
the
products. 
3
Br2
(l)
+
2
Al
(s)

→

Al2Br6
(s) 


Reactants














Products





 26
 H:

Chemical
Equa0ons 
 •  To
obey
the
law of conserva:on of mass,
 chemical
reac1ons
must
be
balanced.
 •  The
number
mul1plying
an
en%re
chemical
 formula
in
a
chemical
equa1on
is
called
the
 stoichiometric coefficient.
 •  Chemical
reac1ons
also
show
state symbols:
 
(s):

solid 

(l):

liquid




(g):
gas 
(aq):
aqueous
 27
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 •  Soluble
substances
are
those
that
dissolve
to
a
 significant
extent
in
a
specified
solvent.
 –  Substances
are
considered
“insoluble”
if
they
do
 not
dissolve
to
more
than
about
0.1
M.
 Figure I.1 28
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 •  Solutes
may
be
present
as
ions
or
as
 molecules.


 –  How
do
we
know
which
one?
 29
 Figure I.4 I:

Aqueous
Solu0ons
and
 Precipita0on 
 •  An
electrolyte
is
a
substance
that
is
present
as
 ions
in
solu1on.


 –  Strong electrolytes –  Weak electrolytes
 •  A
nonelectrolyte
does
not
form
ions
in
 solu1on.
 30
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 Figure I.2 31
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 •  In
a
precipita:on reac:on,
an
insoluble
solid
 product
forms
when
we
mix
two
electrolyte
 solu1ons.
 –  The
solid
formed
is
called
a
precipitate.
 AgNO3(aq)
+
NaCl(aq)


AgCl(s)
+
NaNO3(aq) 
 32
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 •  Complete ionic equa:ons
show
all
the
species
 as
they
actually
exist
in
solu1on—dissolved
 ionic
compounds
exist
as
separate
aqueous
 ions,
so
the
ions
are
shown
separately.
 AgNO3(aq)
+
NaCl(aq)


AgCl(s)
+
NaNO3(aq) 
 33
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 •  Because
the
Na+
and
NO3‐
ions
appear
as
both
 reactants
and
products,
they
play
no
direct
 role
in
the
reac1on.

They
are
spectator ions.
 Ag+(aq)
+
NO3‐(aq)
+
Na+(aq)
+
Cl‐(aq)


 
 
 
 
 
 
 
 
AgCl(s)
+

Na+(aq)

+

NO3‐(aq) 
 34
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 •  Spectator
ions
can
be
“cancelled”
to
give
the
 total ionic equa:on,
the
chemical
equa1on
 that
displays
only
the
net
change
taking
place
 aoer
the
reac1on.
 35
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 Figure I.5 36
 I:

Aqueous
Solu0ons
and
 Precipita0on 
 •  Example:
Write
the
net‐ionic
equa1on
for
the
 reac1on
of
potassium
hydroxide
and
copper
 (II)
percholorate.
 37
 L:

Reac0on
Stoichiometry 
 •  The
quan1ta1ve
aspect
of
chemical
reac1ons
 is
called
reac:on stoichiometry.
 •  The
key
to
reac1on
stoichiometry
is
the
 balanced
chemical
equa1on.
It
is
used
to
set
 up
the
mole ra:o,
a
factor
that
is
used
to
 convert
the
amount
of
one
substance
into
the
 amount
of
another.
 38
 L:

Reac0on
Stoichiometry 
 •  Example:

How
many
moles
of
O2(g)
are
 required
to
produced
5
moles
of
H2O(l)?
 2H2(g)
+
O2(g)


2H2O(l) 
 39
 L:

Reac0on
Stoichiometry 
 •  Example:

How
many
grams
of
Pb(SO4)
can
be
 produced
from
1.
45
g
of
Pb
in
the
presence
of
 excess
PbO2
and
H2SO4?
 PbO2 + Pb + 2H2SO4 → 2PbSO4 + 2 H2O 40
 L:

Reac0on
Stoichiometry 
 • Reac1on
stoichiometry
can
also
be
used
for
 reac1ons
that
occur
in
the
aqueous
phase.
 –  Molarity
and
volume
can
be
used
to
convert
to
 moles,
and
then
mole
ra1os
can
be
used.
 •  A
common
laboratory
technique
to
 determining
the
concentra1on
of
a
solute
is
 :tra:on.
 –  Acid‐base
and
redox
1tra1ons
are
commonly
 used.
 41
 L:

Reac0on
Stoichiometry 
 •  In
a
1tra1on,
the
volume
of
one
solu1on
is
 known,
and
we
measure
the
volume
of
the
 other
solu1on
require
for
complete
reac1on.
 –  The
solu1on
being
analyzed
is
called
the
analyte,
 and
a
known
volume
is
transferred
into
a
flask.
 –  A
solu1on
with
known
concentra1on
of
reactant
is
 called
a
:trant.
 42
 L:

Reac0on
Stoichiometry 
 Buret containing the titrant Contains the analyte 43
 L:

Reac0on
Stoichiometry 
 •  In
an
acid‐base
1tra1on,
the
analyte
is
a
 solu1on
of
a
base
and
the
1trant
is
a
solu1on
 of
an
acid
or
vice
versa.
 •  An
indicator
(a
water
soluble
dye)
is
used
to
 detect
the
stoichiometric point,
the
stage
at
 which
the
volume
of
1trant
added
is
exactly
 that
required
by
the
stoichiometric
 rela1onship
between
the
1trant
and
analyte.
 44
 L:

Reac0on
Stoichiometry 
 •  Example:

What
is
the
concentra1on
of
a
25.0
 mL
sample
of

HNO3
if
52.4
mL
of
0.10
M
 NaOH
was
required
to
reach
the
 stoichiometric
point?
 45
 M:

Limi0ng
Reactants 
 •  Stoichiometric
calcula1ons
of
the
amount
of
 product
formed
in
a
reac1on
suppose,
for
 instance,
that
all
the
reactants
react
exactly
as
 described
in
the
chemical
equa1on.
 •  In
prac1ce,
that
might
not
be
so.
 –  Compe1ng
reac1ons
 –  Reac1on
may
not
be
complete
at
1me
of
 measurement
 –  The
reac1on
may
not
go
to
comple1on
 46
 M:

Limi0ng
Reactants 
 •  The
theore:cal yield
from
a
chemical
reac1on
 is
the
yield
(amount
of
product
formed)
 calculated
by
assuming
that
the
reac1on
goes
 to
comple1on.
 •  The
amount
of
pure
product
that
you
actually
 obtain
from
a
given
reac1on
is
the
actual yield.
 47
 M:

Limi0ng
Reactants 
 •  The
percent yield
of
a
reac1on
is
used
to
 indicate
how
much
of
a
product
is
obtained
 rela1ve
to
the
amount
it
is
possible
to
 produce.
 48
 M:

Limi0ng
Reactants 
 •  In
all
of
the
examples
we’ve
looked
at,
we
had
 exactly
the
right
amount
of
both
reactants.
 –  We
completely
“used”
or
consumed
all
of
the
 reactants
to
make
a
certain
amount
of
product.
 –  We
did
the
same
with
the
food.
 •  But
what
happens
when
you
don’t
have
 exactly
the
right
amounts
of
both
reactants?
 49
 M:

Limi0ng
Reactants 
 •  There
are
26
slices
of
bread
in
the
loaf.

There
 are
20
slices
of
ham
in
the
package,
and
I
have
 10
slices
of
cheese.
 •  According
to
the
equa1on,
how
many
 sandwiches
can
I
make?
 2
bread
+
4
ham
+
1
cheese
→
1
sandwich 
 50
 M:

Limi0ng
Reactants 
 •  2
slices
of
bread
=
1
sandwich
 


26
slices
of
bread
=
13
sandwiches
 •  4
slices
of
ham
=
1
sandwich
 


20
slices
of
ham
=
5
sandwiches
 •  1
slice
of
cheese
=
1
sandwich
 


10
slices
of
cheese
=
10
sandwiches
 51
 M:

Limi0ng
Reactants 
 •  In
this
case,
the
amount
of
ham
limits
the
number
of
 sandwiches
that
we
can
actually
make.


 •  Some1mes
this
happens
in
chemical
reac1ons,
too.


 •  We
call
the
reactant
that
we
run
out
of
first
the
 limi:ng reactant or limi:ng reagent
because
it
 limits
the
amount
of
product
that
can
actually
be
 produced.

 •  The
other
reactant
is
called
the
excess reactant or reagent.
 52
 M:

Limi0ng
Reactants 
 •  Consider
the
balanced
chemical
equa1on:
 3
Br2
(l)
+
2
Al
(s)
→

Al2Br6
(s) •  How
many
moles
of
Al2Br6
can
be
produced
 from
8
moles
of
Br2
and
4
moles
Al?
 53
 M:

Limi0ng
Reactants 
 •  How
many
grams
of
Al2Br6
can
be
produced
 from
1278.4
g
Br2
and
107.9
g
Al?
 54
 M:

Limi0ng
Reactants 
 •  The
image
below
represents
a
reac1on
vessel
prior
to
 chemical
reac1on.

According
to
the
following
equa1on,
how
 much
CO2
and
SO2
can
be
formed?


 O 2 + CS2 → CO 2 + SO 2 Balance : 3O 2 + CS2 → CO 2 + 2SO 2 Count : 4 CS2 1 CO 2 = 4 CO 2 1 1 CS2 9 O 2 1 CO 2 = 3 CO 2 1 3 O 2 Limiting for one product = limiting for both! 9 O 2 2 SO 2 = 6 SO 2 1 3 O 2 55
 M:

Limi0ng
Reactants 
 •  Some1mes
it
is
also
important
to
determine
 how much of the excess reagent remains.


 56
 M:

Limi0ng
Reactants 
 •  There
are
several
op1ons
for
this
calcula1on:
 –  Use
the
ini1al
amount
of
the
limi1ng
reagent
 •  Convert
to
mols
 •  Use
the
mol
ra1o
to
get
to
mols
of
excess
reagent
 •  Convert
to
grams
(this
is
the
amount
actually
used)
 •  Subtract
from
the
ini1al
amount
of
excess
reagent
 –  Use
the
amount
of
product
actually
formed
 •  Convert
to
mols
 •  Use
the
mol
ra1o
to
get
to
mols
of
excess
reagent
 •  Convert
to
grams
(this
is
the
amount
actually
used)
 •  Subtract
from
the
ini1al
amount
of
excess
reagent
 57
 M:

Limi0ng
Reactants 
 •  What
is
the
maximum
mass
of
sulfur
dioxide
that
can
 be
produced
by
the
reac1on
of
95.6
g
of
carbon
 disulfide
with
110.
g
of
oxygen?

How
much
of
the
 excess
reagent
remains?
 58
 M:

Limi0ng
Reactants 
 110 g O 2 1 mol O 2 1 mol CS2 76.2 g CS2 = 87.3 g CS2 used 32.00 g O 2 3 mol O 2 1 mol CS2 1 OR 147 g SO 2 1 mol SO 2 1 mol CS2 76.2 g CS2 = 87.4 g CS2 used 64.1 g O 2 2 mol SO 2 1 mol CS2 1 95.6 g CS2 initial - 87.3 g CS2 used 8.3 g CS2 remaining 59
 € M:

Limi0ng
Reactants 
 Suppose
that
1.00
g
of
sodium
and
1.00
g
of
chlorine
react
to
 form
sodium
chloride
(NaCl).

Which
of
these
is
in
excess,
and
 what
mass
(in
grams)
of
it
remains
when
all
of
the
limi1ng
 reactant
is
consumed?
 2
Na
+
Cl2
→
2
NaCl 
 60
 ...
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