AMS Chem Notes

Atoms,
molecules
and
Stoichiometry
 The
rela5ve
atomic
mass,
Ar,
of
an
element,
is
the
 weighted
average
mass
of
an
atom
of
the
element
 compared
to
1/12
the
mass
of
a
12C
atom.


Copied
from
NYJC
chemistry
lecture
 notes,
ACJC
mole
concept
worksheet

and
 ACS(BR)
Chemical
calcula5ons
tutorial




Combus5on
analysis:
Find
empirical
formula
of
CxHy,
CxHyOz


The
rela5ve
isotopic
mass
of
an
isotope
is
the
mass
of
an
atom
 of
the
isotope
compared
to
1/12
the
mass
of
a
a
12C
atom.
 The
rela5ve
molecular
mass,
Mr,
of
an
element
or
compound,
 is
the
weighted
average
mass
of
a
molecule
of
the
element
or
 compound
compared
to
1/12
the
mass
of
a
12C
atom.
 The
rela5ve
formula
mass,
Mr,
of
a
compound,
is
the
 weighted
average
mass
of
a
formula
unit
of
the
 compound

compared
to
1/12
the
mass
of
a
12C
atom.
 E.g.
Chlorine
consists
of
2
isotopes,
35Cl
and
37Cl
in
 the
ra5o
3:1.
Calculate
the
Ar
of
chlorine.
 









(35
x
3)
+
(37
x
1)

 =
35.5
 Ar
=
 




















4
 ‐The
rela5ve
isotopic
mass
should
be
used
instead
if
 available
–
more
accurate.
 E.g.

 Isotope


Rela+ve
isotopic
 mass


Rela+ve
 abundances


35Cl


34.9689


75.77


37Cl


36.9658


24.23


Calculate
the
Ar
of
chlorine
 Ar
of
Cl
=
 (75.77
x
34.9689)
+
(36.9658
x
24.23)
=
35.45
(4s.f.)
 
























100
 One
mole
of
substance
is
the
amount
of
substance
that
 contains
as
many
par5cles
as
there
are
carbon
atoms
in
 exactly
12g
of
12C.
 The
number
of
carbon
atoms
in
12g
of
12C
is
found
to
be
 6.02
x
1023.
(Avogadro’s
constant)
 g


N




m












V
 n
=




=




=
cV
=



 
L





M












Vm
 6.02
x
1023


gmol‐1


Molar
gas
volume
 s.t.p.
:
0oC,
1
atm

Vm=
22.4
dm3
 r.t.p.
:
25oC,
1
atm

Vm
=
24
dm3


The
empirical
formula
of
a
compound
is
the
simplest
 formula
that
shows
the
rela5ve
number
of
atoms
of
 each
element
present
in
a
compound.
 The
molecular
formula
of
a
compound
is
the
formula
 that
shows
the
actual
number
of
atoms
of
each
 element
present
in
one
molecule
of
the
compound.
 E.g.
A
hydrocarbon,
X,
consists
of
83.7%
carbon.
 It
has
Mr
of
86.0.
Calculate
the
empirical
and
 molecular
formula
of
X.
 1.
Use
mass
 ra5o
(Do
not
 simplify
yet)
 2.Find
mol
 ra5o
by
 dividing
by
Ar
 3.
Simplify
 (Divide
by
 smallest
 number)


1.
Mass
ra5o
of
C:H
=
83.7:
16.3
 83.7







16.3
 2.
Mol
ra5o
of
C:
H
=














:
 Ar
of
C




Ar
of
H
 83.7


16.3
 =









:











=
6.975
:
16.3
 12





1


1. Weigh
compound
 2. Burn
compound
completely,
only
products

CO2
+
H2O
 3. H2O
and
CO2
are
drawn
2
u‐tubes,
substance
in
1
u‐tube
 absorbs
H2O.




P4O10(l)
+
6H2O(g)

4H3PO4
(aq)









 Other
U‐tube
contains
substance
that
absorbs
CO2.





 2NaOH
+
CO2

Na2CO3
+H2O
 4. Weigh
each
of
the
tubes
before
and
aeer



























 ‐Increase
in
mass
of
the
1st
tube
is
H2O
absorbed.


































 ‐Increase
in
mass
of
the
2nd
tube
is
CO2
absorbed.

































 Eudiometry:
Steps
 1. Write
out
the
balanced
equa5on
for
combus5on

 e.g.





CxHy
+
(x+
y/4)
O2

xCO2
+
y/2
H2O


2. Use
mole
ra5o
to
determine
x
:

 VCO2











x






 VC Hy

=
 








1

 7
 3.

=
1:


















=
3:
7
 x 
 3 3. Determine
the
volume
of
O2
used:
 Empirical
formula
of
X
is
C3H7.
 



VO2
used
=
VO2
(excess)
–VO2
(Unused)
 Let
the
molecular
formula
be
(C3H7)n,
then
 VO2
used








x
+
y/4
 












Mr
of
X





























86




























 n
=

















































=


























=
2

















































 = 4.
Use
mole
ra5o
to
determine
y:
 V 














1
 Mr
of
empirical
formula











3
x
12
+
7

 CxHy
 Molecular
formula
of
X
is
C6H14.


Limi5ng
reagent
Q
–
2
data
 Normal
Q
–
1
data
 Steps:
 1. Write
equa5on
for
mole
ra5o
 2. 
Convert
to
moles
 3. Use
the
mole
ra5o
to
determine
which
 reactant
produces
the
smallest
amount
of
 product.
This
is
the
limi5ng
reagent.
 E.g.

nH2

if
H2
is
limi5ng
then
H2O
produced
=
 Actual
yield
 Percentage
yield
=































x
100%
 Theore5cal
yield
 






Mass
of
solute
(g)





.







 Mass
conc
=
 Volume
of
solu5on
(dm3)


Back
5tra5on
 ‐ No
definite
end‐point
to
be
detected
 ‐ Add
excess
acid
 ‐ Find
lee‐over
acid
by
5tra5on
 ‐ n(acid
used)
=
n(acid
at
start)
–
n(acid
lee
at
end)
 Double
indicator
5tra5on
 ‐ NaOH
+
HCl
–reac5on
occurs
in
a
single
stage
 ‐ Na2CO3
+
HCl
–
reac5on
occurs
in
2
stages
with
 HCO3‐
as
the
intermediate
product.
 Note:
(2)
and
(3)
 

OH‐
+
H+

H20
‐‐‐‐‐‐
(1)
 require
the
same
 

CO32‐
+
H+

HCO3‐
‐‐‐‐‐‐‐‐‐‐(2)
 

HCO3‐
+
H+

H2O
+
CO2
‐‐‐‐‐‐‐(3)
 amount
of
H+
ions
 With
phenolphthalein
as
the
indicator,
the
end‐point
 is
seen
at
the
comple5on
of
reac5ons
(1)
and
(2)
 With
methyl
orange
as
indicator,
the
end‐point
is
 seen
at
the
comple5on
of
all
3
reac5ons.