Supplemental packet page 117
Supplemental packet page 119
Chem ical Composition & t he M ole
Dr. Gergens - SD Mesa College
I.
Count ing Devices
A. Fam ilar count ing devices for counting physical object s larger t han at oms:
anything
twelve
1 dozen of _______________ = _______________ units of anyt hing
A basket has 18 eggs. How many dozen eggs are cont ained in t he basket?
18 eggs x
1 dozen eggs
________________
=
12 eggs
1.5 dozen eggs
B. In chemist ry, our counting device is t he mole:
23
anything = _______________
6.023 x units
10 of anyt hing
1 mole of _______________
23
6.023 x 10
Gummy
Gummy bears
1 mole of _______________
= _______________
_____________
bears
atoms
6.023 x 10
atoms
1 mole of _______________
= _______________
_____________
23
pennies
23
6.023 x 10
pennies
1 mole of _______________ = _______________
_____________
Called Avogadro’s Number
Supplemental packet page 120
C. In chemist ry, we can't physically count
atoms by inspection, so we like to
count the t otal number of at om s, and m olecules by weighing.
•
Count ing atoms - T he mole has been
given a precise definit ion as the num ber
of at oms cont ained in 12.000000 grams "exactly" of pure carbon-12.
1 mole of
•
12
6
C = 12.000000 grams of
23
12
6
6.023 x 10 at om s of
C = _____________
Molar Mass is t he mass in grams of one mole of any subst ance, and is also
num berically equal to at omic mass unit [amu] expressed in grams.
1Fex55.85 = 55.85
3Nx14.00 = 42.00
9f=x16.00 = 144.00
241.85
12.01
1 mole of carbon = _____________ grams of carbon
55.85
1 mole of iron = _____________ grams of iron
241.85
1 mole of Fe(N O 3 ) 3 = _____________ grams of iron
D.
Molar Mass is num berically equal to at omic mass unit (amu) expressed in
grams.
12
6
C
II.
How can Avogadro's
A. num ber
be used in calculat ing
of atoms for a given num ber
18 moles x
Pb
B. num ber
num ber
t he:
of mol es of an pure element ?
6.023x1023 atoms = 1.0841x1025 atoms
________________
Pb
1 mole atoms
1.1x1025 (correct Sig Figs)
of molecules
for a given num ber
of mol es of a pure subst ance
?
23 molecules
18 moles x 6.023x10
________________
= 1.1x1025 molecules
1 mole molecules
••
of atoms for a given num ber
of gram s of a pure subst ance
?
H O
 H 24
••
C. num ber

molar mass
23 molecule
mole H2O x 6.023x10
3 atoms = 4.8x10 atoms
18 grams x 1__________
_____________
x _______
1 mole
H2O
18.02 g H2O
1 molecule
of H2O
D.
num ber of
molecules
for a given number of
gram s of a pure
subst ance
?
23 molecules
6.023x10
1
mole
H
O
18.0 grams x __________
x __________________= 6.02x1023 molecules
2
H2O
H2O
18.02 g H2O 1 mole
Supplemental packet page 121
Count ing At om s
How many iron atoms are present i n 3.00 moles of i ron metal ?
1 mol Fe = 55.85 g Fe = 6.02 x 10
23
x atoms Fe = 3.00 mol Fe
x mol Fe = 3.00 m ol Fe
at om s Fe
1 mol Fe = 6.02 x 10
x
23
6.02 x 10 23 at om s Fe
1 mol Fe
at om s Fe
=
1.81 x 10
24
at om s Fe
Work out t he following problems (show math set-ups)
x at om s S = 0.174 mol S
0.174 mol S
x
x m ol K = 5.92 x 10
6.02 x 10 23 at om s S
1 mol S
=
5.92 x 10 24 at om s
K
x
24
at om s K
1 mol K
6.02 x 10
How many sulfur atoms are
present in 0.174 moles of S
nonmetal? ANS: 1.05 x 10 m ol S
23
23
=
at om s K
How many moles of K are
present in 5.92 x1024 atoms of
ANS: 9.83 atoms K
K metal?
Let's go over this exam ple together
How many atoms of C are
x atoms C = 27.4 g C
present in 27.4 grams of
carbon nonmetal?
1 mol C = 12.01 g C = 6.02 x 1023 atoms C
Combined we have a grams to particles equivalent statement
27.4 g C
x
1 mol C
12.01 g C
x
6.02 x 10 23 at oms C
1 mol C
(grams) x (mol per grams) Avogradro’s number
moles
x
6.02 x 1023 atoms
1 mole
=
1.37 x 10
24
at oms C
Supplemental packet page 122
Chem ical Compounds
How many atoms are present in a formula unit of sodium sulfate Na2SO4?
Just as a mole of atoms is based on the atomic mass or atomic weight, a mole of a
compound is based upon the formula mass or formula weight.
sodium sulfate, Na
First :
Second:
T hird:
Fourth:
Na2SO4
2
SO4
7 atoms
What is t he mass in am u of one molecule of sodium sulfat e? _________ 142.06 amu
What is t he mass—in grams—of one m ole of sodium sulfat e? _________142.06 g
1.13 x 10–1 mol
How many moles of Na
SO are in 16.0 g Na
SO ?_________
How many atoms are there present per formula unit of Na
2
2
1
4
7
2
2 Na x 22.99 =
1 S x 32.07 =
4 O x 16.00 =
Na
S
O
at om s
x m ol Na 2 SO4 = 16.0 g Na
4
2
SO4 x
2
SO4 ? _________
2
SO4
4
45.98
32.07
64.00
142.06 g Na
1 mol Na 2 SO4
142.06 g Na 2 SO4
=
1.13 x 10
–1
mol Na 2 SO4
Supplemental packet page 122
Molar Mass Calculations; one mole amount of a substance in grams
CH4
CuSO 4 • 5H 2 O
1 C x 12.0 = 12.0
4 H x 1.0 = 4.0
1
1
4
5
This is called a pentahydrate
Cu x 63.6 = 63.6
S x 32.0 = 32.0
O x 16.0 =
64.0
H 2 O x 18.0 = 90.0
ANS: 16.0
C3 H5 Br 2
3 C x 12.0
5 H x 1.0
2 Br x 78.9
aluminum nitrat e Al(NO
= 36.0
= 5.0
= 157.8
)
3 3
1Al x 27.0 = 27.0
3 N x 14.0 = 42.0
9 0 x 16.0 = 144.0
Note you must be able to derive
ANS: 200.9 correct formulas from names ANS: 213.0
C3 H7 OH
3 C x 12.0
8 H x 1.0
1 O x 16.0
ANS: 249.6
calcium dihydrogen phosphate
Ca(H 2 P O4 ) 2
=
=
=
36.0
8.0
16.0
1
4
2
8
ANS: 60.0
Ca x 40.1
H x 1.0
P x 31.0
O x 16.0
= 40.1
= 4.0
= 62.0
= 128.0
ANS: 234.1
Suppl packet page 123
Grams t o Moles and Moles to Grams
How many moles are there in 41.7 g of NaNO3 ?
1 mol NaNO
3
= 85.0 g NaNO
1 Na x 23.0
x 14.0
problem ALWAY13 N
S
O x 16.0
3
=
=
In every calculuation
=
Calculate molar mass; MAKE a Table and Domolar
it. mass
x m
oles grams
= 41.7 g to
NaNO
3
The
moles
x
1 mol
NaNO
conversion
set-up
3
85.0 g NaNO
3
(grams)
(molar mass; g/mol)
moles
=
(molar mass;
g/mol)
Quickly
convert
to moles by
dividing grams by molar mass
Memorize this!!!!
23.0
14.0
48.0
85.0
g/mol
0.491 mol
NaNO3
Calculat e t he number of moles in :
12.6 grams calcium sulfat e
6.18 x 10
1 Ca x 40.1 =
1 S x 32.1 =
4 O x 16.0 =
3
grams ammonium carbonate
1
2
8
3
40.1
32.1
64.0
136.2
C x 12.0 =
N x 14.0 =
H x 1.0 =
O x 16.0 =
12.0
28.0
8.0
48.0
96.0
CaSO
How4 many moles of
(NH
How
many
moles of
4)2CO
3
calcium sulfate atoms are
ammonium carbonate are
Be sure
Most
students
you arefail
able
in to
thewrite
second
correct
half of
formulas.
the semester
Mostbecause
students
3
of
present in 6.18 x 10 grams of
12.6present
g CaSO inx 12.6 1grams
mol CaSO
fail incannot
they
the second
correctly
halfcalculate
of the =semester
a molarbecause
mass they cannot
136.2
g
CaSO
calcium sulfate ionic salt?
ammonium carbonate ionic salt?
correctly
MAKE
write at
a substance’s
TABLE when
formula
calculating molar mass
4
4
4
ANS: 9.25 x 10
–2
m ol CaSO
6.18 x 10 3 g (NH 4 ) 2 CO3
Ca2+
SO42–
CaSO4
4
x
ANS: 64.4mol (NH
1 mol (NH +4 ) 2 CO3
=
96.01 NH
g (NH4 4 ) 2 CO3CO32–
(NH4)2CO3
NH4
+
) CO3
4 2
Calculat e t he number of moles in :
12.6 grams calcium sulfat e
6.18 x 10
1 Ca x 40.1 =
1 S x 32.1 =
4 O x 16.0 =
CaSO
How4 many moles of
calcium sulfate atoms are
12.6
g CaSO
present in 12.6
grams
of 4
calcium sulfate
ionic
salt?
136.2
g/mol
3
grams ammonium carbonate
1
2
8
3
40.1
32.1
64.0
136.2
C x 12.0 =
N x 14.0 =
H x 1.0 =
O x 16.0 =
(NH
How
many
4)2CO
3 moles of
ammonium carbonate are
3 grams
6.18
x103 xg 10
(NH
present
in 6.18
4)2CO3of
ammonium
carbonate
96.0
g/mol ionic salt?
Remember the short
cut for converting
grams to moles!!!!!!!
ANS: 9.25 x 10
–2
m ol CaSO
(grams)
(molar mass; g/mol)
moles
12.0
28.0
8.0
48.0
96.0
4
ANS: 64.4mol (NH
Quickly convert to moles by
dividing grams by molar mass
Memorize this!!!!
) CO3
4 2
Converting Mole Amounts to Grams
Calculat e t he number of grams in (show math set-ups):
4.22 moles of KCl
0.0196 moles barium nitrat e
KCl
1 K x 39.1 =
1 Cl x 35.5 =
4.22 mol KCl
x
74.6 g KCl
1 mol KCl
ANS: 3.15 x 10
K+
39.1
35.5
74.6 g/m ol
2
Ba(NO3)2
=
grams KCl
0.0196 mol Ba(NO
)
3 2
1 Ba x 137.3 =
2 N x 14.0 =
6 O x 16.0 =
x
261.3 g Ba(NO 3 ) 2
1 mol Ba(NO 3 ) 2
ANS: 5.12 gram s Ba(NO
Converting Mole Amounts to Grams, just
2+
–
–
Cl
NO
Ba
3
take moles multiplied by molar
mass
Memorize this!!!!
KCl
137.3
28.0
96.0
261.3 g/mol
Ba(NO3)2
NO3–
=
)
3 2