S
The Mole
6.02 X 1023
IT IS JUST A STANDARD NUMBER OF ATOMS/MOLECULES
Relative Masses
To understand relative scales, let’
let’s ignore
electrons and compare atoms by total
number of nuclear particles.
Hydrogen, H
1 particle
Helium, He
4 particles
Lithium, Li
7 particles
Carbon, C
12 particles
Oxygen, O
16 particles
1
Atomic Mass
The atomic mass of an element is the
mass of an atom of the element compared
with the mass of an atom of carbon taken
as 12 atomic mass units (u).
Atomic masses of a few
elements:
Hydrogen, H = 1.0 u
Carbon,
Helium,
He = 4.0 u
Nitrogen, N = 14.0 u
Lithium,
Li = 7.0 u
Beryllium, Be = 9.0 u
Neon,
C = 12.0 u
Ne = 20.0 u
Copper, Cu = 64.0 u
Molecular Mass
The molecular mass M is the sum of the
atomic masses of all the atoms making
up the molecule.
Consider Carbon Dioxide (CO2)
The molecule
has one
carbon atom
and two
oxygen atoms
1 C = 1 x 12 u = 12 u
2 O = 2 x 16 u = 32 u
CO2 = 44 u
2
Definition of a Mole
One mole is that quantity of any
substance that contains the same
number of particles as there are in 12 g
of carboncarbon-12.
(6.022 x 1023 particles)
1 mole of Carbon has a mass of 12 g
1 mole of Helium has a mass of 4 g
1 mole of Neon has a mass of 20 g
1 mole of Hydrogen (H2) = 1 + 1 = 2 g
1 mole of Oxygen (O2) is 16 + 16 = 32 g
A Mole of Particles
Contains 6.02 x 1023 particles
1 mole C
= 6.02 x 1023 C atoms
1 mole H2O = 6.02 x 1023 H2O molecules
1 mole NaCl = 6.02 x 1023 NaCl “molecules”
(technically, ionics are compounds not
molecules so they are called formula units)
6.02 x 1023 Na+ ions and
6.02 x 1023 Cl– ions
3
Just How Big is a Mole?
• Enough soft drink cans to cover the
surface of the earth to a depth of
over 200 miles.
• If you had Avogadro's number of
unpopped popcorn kernels, and
spread them across the United
States of America, the country would
be covered in popcorn to a depth of
over 9 miles.
• If we were able to count atoms at the
rate of 10 million per second, it
would take about 2 billion years to
count the atoms in one mole.
The Mole
• 1 dozen cookies = 12 cookies
• 1 mole of cookies = 6.02 X 1023 cookies
• 1 dozen cars = 12 cars
• 1 mole of cars = 6.02 X 1023 cars
• 1 dozen Al atoms = 12 Al atoms
• 1 mole of Al atoms = 6.02 X 1023 atoms
Note that the NUMBER is always the same,
but the MASS is very different!
Mole is abbreviated mol (gee, that’s a lot
quicker to write, huh?)
4
Molecular Mass in grams/mole
The unit of molecular mass M is grams per
mole.
Hydrogen, H = 1.0 g/mol
H2
= 2.0 g/mol
Helium,
He = 4.0 g/mol
O2
= 32.0 g/mol
Carbon,
C = 12.0 g/mol
H2O = 18.0 g/mol
Oxygen,
O = 16.0 g/mol
CO2 = 44.0 g/mol
Each mole has 6.022 x 1023
molecules
A Mole of Particles
Suppose we invented a new collection unit
called a crideroo. One crideroo contains 8
objects.
1. How many paper clips in 1 crideroo?
a) 1
b) 4
c) 8
2. How many oranges in 2.0 crideroos?
a) 4
b) 8
c) 16
3. How many crideroos in 40 gummy bears?
a) 5
b) 10
c) 20
5
Moles and # of Molecules
Finding the number of moles
n in a given number of N
molecules:
n=
N
NA
Avogadro’s number: NA = 6.022 x 1023
particles/mol
Example 2: How many moles of any gas will contain
20 x 1023 molecules?
N
20 x 1023 molecules
n=
=
N A 6.023 x 1023 molecules mol
nn =
= 3.32
3.32
mol
mol
Moles and Molecular Mass M
Finding the number of
moles n in a given
mass m of a substance:
n=
m
M
Molecular mass M is expressed in grams per
mole.
Example 3: How many moles are there in
200 g of oxygen gas O2? (M = 32 g/mol)
n=
m
200 g
=
M 32 g/mol
nn =
= 6.25
6.25 mol
mol
6
A Mole of Particles
Contains 6.02 x 1023 particles
1 mole C
= 6.02 x 1023 C atoms
1 mole H2O = 6.02 x 1023 H2O molecules
1 mole NaCl = 6.02 x 1023 NaCl “molecules”
(technically, ionics are compounds not
molecules so they are called formula units)
6.02 x 1023 Na+ ions and
6.02 x 1023 Cl– ions
Avogadro’s Number as
Conversion Factor
6.02 x 1023 particles
1 mole
or
1 mole
6.02 x 1023 particles
Note that a particle could be an atom OR a molecule!
7
1. Number of atoms in 0.500 mole of Al
a) 500 Al atoms
b) 6.02 x 1023 Al atoms
c) 3.01 x 1023 Al atoms
2.Number of moles of S in 1.8 x 1024 S atoms
a) 1.0 mole S atoms
b) 3.0 mole S atoms
c) 1.1 x 1048 mole S atoms
Molar Mass
• The Mass of 1 mole (in grams)
• Equal to the numerical value of the average
atomic mass (get from periodic table)
1 mole of C atoms
=
12.0 g
1 mole of Mg atoms
=
24.3 g
1 mole of Cu atoms
=
63.5 g
8
Other Names Related to Molar Mass
• Molecular Mass/Molecular Weight: If you have a single
molecule, mass is measured in amu’s instead of grams.
But, the molecular mass/weight is the same numerical value
as 1 mole of molecules. Only the units are different. (This is
the beauty of Avogadro’s Number!)
• Formula Mass/Formula Weight: mass is measured in
grams, just like with atoms but for compounds. But again,
the numerical value is the same. Only the units are different.
• THE POINT: You may hear all of these terms which
mean the SAME NUMBER… just different units
Find the molar mass
(usually we round to the tenths place)
A. 1 mole of Br atoms =
B. 1 mole of Sn atoms =
79.9 g/mole
118.7 g/mole
9
Molar Mass of Molecules and
Compounds
Mass in grams of 1 mole equal numerically to
the sum of the atomic masses
How much does 1 mole of CaCl2 weigh
40.1g
+ 2 moles Cl x 35.5 g/mol = 71.0 g
1 mole Ca x 40.1 g/mol
=
111.1 g/mol CaCl2
How much does 1 mole of N2O4 =
92.0 g/mol
A. Molar Mass of K2O =
94.20 g/mole K2O
B. Molar Mass of antacid Al(OH)3 =
78.01 g/mole Al(OH)3
C. Molar Mass of water H2O =
18.02 g/mole H2O
10
Prozac, C17H18F3NO, is a widely used
antidepressant that inhibits the uptake of
serotonin by the brain. Find its molar
mass.
309.36 g / mol
Calculations with Molar Mass
molar mass
Grams
Moles
11
The artificial sweetener aspartame
(Nutra-Sweet) formula C14H18N2O5 is
used to sweeten diet foods, coffee and
soft drinks. How many moles of
aspartame are present in 225 g of
aspartame?
.764 moles
294.29 g/mol
Converting Moles and Grams
Aluminum is often used for the structure
of light-weight bicycle frames. How
many grams of Al are in 3.00 moles of
Al?
3.00 moles Al
? g Al
12
Molar mass of Al 1 mole Al = 26.98 g Al
Setup
3.00 moles Al
Answer
x
26.98 g Al
1 mole Al
= 80.94 g Al
Calculations
molar mass
Grams
Avogadro’s number
Moles
particles
Everything must go through
Moles!!!
13
Atoms/Molecules and Grams
How many atoms of Cu are present
in 35.4 g of Cu?
35.4 g Cu
1 mol Cu
63.54 g Cu
6.02 X 1023 atoms Cu
1 mol Cu
= 3.35 X 1023 atoms Cu
How many atoms of K are present in 78.4
g of K?
2.01 moles
1.21 X 1024 atoms Cu
14
Atoms/Molecules and Grams
How much does one molecule of
water weigh?
18.0 g H20
1 mol H20
1 mol H20
6.02 X 1023 atoms H20
= 2.99 X 10-23 g for each molecule
15
What is the mass (in grams) of 1.20 X 1024
molecules of glucose (C6H12O6)?
How many atoms of O are present in 78.1 g
of oxygen?
78.1 g O
1 mol O 6.022 X 1023 molecules O
16.00 g O 1 mol O
= 2.94 X 1024 molecules
16
RELATIVE MASS (ISOTOPE)
HOW DO WE COME UP WITH THE MASS
OF EACH ELEMENT ON THE PERIODIC
TABLE?
Average Atomic Masses (as shown
on the periodic table: a Weighted
average of all the Isotopes of an
element.
Grades are an example of weighted averages
If you had a "C" on a 10 point paper
And an "A" on a 100 point paper
Would you want the teacher to do a simple average
of "A" and "C" to make a "B"?
The points for work act as a like weight on the periodic
table.
So the 100 point assignment counts more than the 10
point assignment in the averaging of the grade.
A 99% isotope has more impact on the recognized
atomic weight than the 1% isotope
17
With element masses, the % Abundance is the
weight, in the weighted average.
Example
Carbon:
12C, 13C, 14C
Carbon Isotopes:
Carbon-12 =
% Abundance in Nature
12C
Carbon-13 =
13C
Carbon-14 =
14C
+
= 0.01%
= 12 x 0.99
=
X
= 13 x 0.0099
=
Y
=
Z
12C
13C
= 0.99%
12C, 13C, 14C
Carbon:
+
= 99%
14C
= 14 x 0.0001
Carbon Average Mass =
X+Y+Z
18
RELATIVE MASS (ISOTOPE)
The General Formula for
Calculating Atomic Masses Is…
(mass of isotope 1)(% abundance/100)
+(mass of isotope 2)(% abundance/100)
+(mass of isotope 3)(% abundance/100)
+’keep going if there are more isotopes’
RELATIVE MASS (ISOTOPE)
Calculate the Atomic Mass
of Carbon
Isotope
Mass
% abundance
#1
13.003355
1.10
#2
12.000000
98.90
19
RELATIVE MASS (ISOTOPE)
12.000000 (98.90 / 100)
+ 13.003355 (1.10 / 100)
11.868000 + .143036 =
12.01
RELATIVE MASS (ISOTOPE)
Isotope
#1
#2
Mass
10.0
11.0
% abundance
20.0
80.0
20
RELATIVE MASS (ISOTOPE)
10.0 (20.0/100) + 11.0 (80.0 / 100)
2.00
+
8.80
= 10.80 AMU
BORON
RELATIVE MASS (ISOTOPE)
Calculate the Atomic Mass
Isotope
#1
#2
#3
#4
Mass
% abundance
53.938882
52.940651
51.940510
49.946046
2.36
9.50
83.79
4.35
21
RELATIVE MASS (ISOTOPE)
For Chromium…
49.94… (4.35/100)
51.94… (83.79/100)
52.94… (9.50/100)
+ 53.94…(2.36/100)
2.17
43.5
5.02
1.27
51.96 AMU
CHROMIUM
RELATIVE MASS (ISOTOPE)
Calculate the Atomic Mass
Isotope
#1
#2
Mass
68.9256
70.9247
% abundance
60.11%
39.89%
22
RELATIVE MASS (ISOTOPE)
For Gallium…
68.9256 (60.11 /100)
70.9247 (39.89 /100)
41.43
28.29
69.72 AMU
GALLIUM
PERCENT COMPOSITION
Always Means By Mass
• A sample contains 26.0 grams of
carbon, 6.5g of Hydrogen, and
17.5g of oxygen.
• What is the mass percentage (AKA
PERCENT COMPOSITION) of each
23
PERCENT COMPOSITION
• % C: 26.0 g = .52 X 100 = 52% C
50.0g
• % H: 6.5 g = .13 X 100 = 13% H
50.0g
• % O: 17.5 g = .35 X 100 = 35% O
50.0g
100%
PERCENT COMPOSITION
• A 11.6 gram sample of Iron reacts with 4.86g
of oxygen to make FeO. What is the Percent
composition of the compound?
24
PERCENT COMPOSITION
11.6 + 4.86 = 16.5g
• % Fe: 11.6 g = .703 X 100 = 70.3% Fe
16.5g
• % O: 4.86 g = .295 X 100 = 29.5% O
16.5g
99.8%
If you know two of the three you can find the other one. But if you Know all three, do all three to limit computational error.
PERCENT COMPOSITION
• An 8.44g sample of Silver (Ag) reacts
with Sulfur that weights 9.69g. What is
the Percent composition of the
compound?
25
PERCENT COMPOSITION
• % Ag: 8.44g = .466 X 100 = 46.6% Ag
18.13g
• % S: 9.69g S = .534 X 100 = 53.4 % S
18.13g
100%
PERCENT COMPOSITION
• A compound has a formula C3H6O2. What
is the Percent Composition of the
compound?
• use one mole to find molar mass
26
Percent Composition
3C = 3(12.01) = 36.03g
6H = 6(1.01) = 6.06g
2O = 2(16) = 32.00g
Total
74.09g/mol
MOLAR MASS
• % C: 36.0 g = .486 X 100 = 48.6% C
74.0g
• % H: 6.5 g = .081 X 100 = 8.1% H
74.0g
• % O: 32.0 g = .432 X 100 = 43.2% O
74.0g
99.9%
Percent Composition
Always Means By Mass
• A compound has a formula K2CrO4. What
is the Percent composition of the
potassium chromate?
• Don’t forget: use one mole to find molar
mass
27
Percent Composition
2K = 2(39.1) = 78.2g
Cr = 1(52.0) = 52.0g
4O = 4(16) = 64.0g
Total
194.2g/mol
MOLAR MASS
• % K: 78.2 g = .402 X 100 = 40.2% K
194.2g
• % Cr: 52.0 g = .267 X 100 = 26.7% Cr
194.2g
• % O: 64.0 g = .329 X 100 = 32.9% O
194.2g
99.8%
EMPIRICAL VS. MOLECULAR
Oreo Cookies can be combined in many
ways but as long as we add in whole
numbers (CAN’T SPLIT AN ATOM) the
ratio of cream to cookie wont change.
THE SAME HAPPENS IN CHEMISTRY
MASSES: WHITE = 1 g
BLACK = 2 g
We can weigh a substance, get the percentage of weights for each substance and
Then determine the mole relationships and then the most basic formula.
28
MOLECULAR FORMULA B16W16MM 48
EMPIRICAL FORMULA B1W1 MM 3
MOLECULAR FORMULA B11W11 MM 33
EMPIRICAL FORMULA B1W1 MM 3
29
MOLECULAR FORMULA B8W8 MM 24
EMPIRICAL FORMULA B1W1 MM 3
MOLECULAR FORMULA B9W9 MM 27
EMPIRICAL FORMULA B1W1 MM 3
30
MOLECULAR FORMULA B18W9 MM 45
EMPIRICAL FORMULA B2W1 MM 5
MOLECULAR FORMULA B12W6 MM 30
EMPIRICAL FORMULA B2W1 MM 5
31
MOLECULAR FORMULA B10W5 MM 25
EMPIRICAL FORMULA B2W1 MM 5
MOLECULAR FORMULA B3W3 MM 9
EMPIRICAL FORMULA B1W1 MM 3
32
MOLECULAR FORMULA B9W3 MM 21
EMPIRICAL FORMULA B3W1 MM 7
MOLECULAR FORMULA B12W3 MM 27
EMPIRICAL FORMULA B4W1 MM 9
33
MOLECULAR FORMULA B8W2 MM 18
EMPIRICAL FORMULA B4W1 MM 9
MOLECULAR FORMULA B6W2 MM 14
EMPIRICAL FORMULA B3W1 MM 7
34
MOLECULAR FORMULA B4W2 MM 10
EMPIRICAL FORMULA B2W1 MM 5
MOLECULAR FORMULA B2W2 MM 6
EMPIRICAL FORMULA B1W1 MM 3
35
MOLECULAR FORMULA B1W1 MM 3
EMPIRICAL FORMULA B1W1 MM 3
MOLECULAR FORMULA B2W1 MM 5
EMPIRICAL FORMULA B2W1 MM 5
36
MOLECULAR FORMULA B5W1 MM 11
EMPIRICAL FORMULA B5W1 MM 11
MOLECULAR FORMULA B12W3 MM 27
EMPIRICAL FORMULA B4W1 MM 9
37
MOLECULAR FORMULA B10W10MM 30
EMPIRICAL FORMULA B1W1 MM 3
MOLECULAR FORMULA B16W8 MM 40
EMPIRICAL FORMULA B2W1 MM 5
38
MOLECULAR FORMULA B24W8 MM 56
EMPIRICAL FORMULA B3W1 MM 7
SO WHAT IF MOLECULES HAVE THE
Molecular Formula
BIG OREO
Empirical Formula
IS AN OREO
Molecular Formula
LITTLE OREO
SAME REL WT, WE JUST HAVE A BIGGER
STRUCTURE THAT APPEARS THE SAME
39
Common Decimals and how to turn
them into whole numbers.
0.20
0.25
0.33
0.50
0.67
FROM DIANE
40
In chemistry, the empirical formula of a chemical compound is a simple expression
of the relative numbers of each type of atom in it. An empirical formula makes no
reference to isomerism, structure, or absolute number of atoms. The empirical
formula is used as standard for most ionic compounds, such as CaCl2, and for
macromolecules, such as SiO2. The term empirical refers to the process of
elemental analysis, a technique of analytical chemistry used to determine the
relative amounts of each element in a chemical compound.
Empirical Formula
• Smallest ratio of atoms of the compound
Actual (Molecular)
H2O2 (hydrogen peroxide)
C6H12O6 (glucose)
C4H8O2 (butanoic acid)
C8H10N4O2
H2O
C12H22O11 (sucrose=sugar)
C12H22O11(NCM)
Empirical
HO
CH2O
C2H4O
C4H5N2O
H2O (NCM)
41
PERCENT COMPOSITION
• Law of Definite Proportions: You can
have five thousand pounds or five pounds
– the percent composition will not change.
You can assume any sample size…so
assume you have a mole of the
substance.
Hexane's molecular formula is C6H14, and its empirical formula would be
C3H7 showing a C:H ratio of 3:7.
C6H14, and its empirical formula would be C3H7
if the empirical formula of a compound is C3H8 , its molecular formula may be
C3H8 , C6H16 , etc.
When teaching the method for converting percentage composition to an empirical formula, I have devised the follo
Percent to mass
Mass to mole
Divide by small
Multiply 'til whole
42
If I told you that a compounds percent
composition is C 40.0%, 6.7 %, 53.3 O%
what is the compound’s empirical formula?
The percent composition is only for mass.
Chemical formulas are in moles of atoms.
If I told you that a compounds percent
composition is C 40.0%, 6.7 H %, 53.3 O%
what is the compound’s empirical formula?
The percent composition is only for mass.
Chemical formulas are in moles of atoms.
43
Empirical Formula
40.0% x 100g = 40.0g C
6.7% x 100g = 6.7 g H
53.3% x 100g = 53.3g O
• mol C: 40.0g C = 3.33 mol C
12.0g/mol C
• mol H: 6.7g H = 6.66 mol H
1.007g/mol H
• mol O: 53.3g O = 3.33 mol O
16.0g/mol O
C1H2O1
• Greatest common multiple 3.33
A substance has the following
composition by mass: 60.80 % Na ;
28.60 % B ; 10.60 % H Find the
empirical formula.
44
Empirical Formula
60.80% x 100g = 60.80 g Na
28.60% x 100g = 28.60 g B
10.60 % x 100g = 10.60 g H
• mol Na:
60.80g Na = 2.645 mol Na
22.99 g/mol Na
• mol B: 28.60 g B = 2.646 mol B
10.81 g/mol B
• mol H: 10.60 H =
10.49 mol H
1.007g/mol H
Na1B1H4
• Greatest common multiple 2.645
A compounds mass is 54.1% Ca, 2.70% H,
and 40.8% O. What is the compound’s
empirical formula?
45
Empirical Formula
54.1 g Ca
2.70 g H
40.8 g O
• mol Ca:
54.1 g Ca
= 1.349 mol Ca
1
40.078g/mol Ca
• mol H:
2.7 g H
=
2.68 mol H
1.98 = 2
1.007g/mol H
• mol O:
40.8 g O
= 2.55 mol O
1.89 = 2
16.00g/mol O
• Greatest common multiple 1.349
If we know Empirical Formula (relative amounts of each
by weight and Experimental molar mass (how much
that formula weighs) we can find the molecular formula
The empirical formula of a substance simply gives
the most basic ratio of the combination of the
constituent elements.
Experimental MM = Ratio to Increase EF
Empirical MM
Obviously, an empirical formula does not contain as
much information as does a molecular formula.
46
a. A compound with a molecular mass of 70µ and an empirical formula
of CH2.
b. A compound with a molecular mass of 46.0µ and an empirical
formula
of NO2.
A sample has empirical formula of CH2O
(Remember from previous problem) and a
Molecular Mass (Experimental) = 120.68 g.
What is the molecular formula?
Molecular Weight = 120.68g = 4.019
Empirical Weight
30.03g
C4H8O4
47
We have determined the percentage
composition of benzene: 92.3% C and 7.7% H.
What is the empirical formula of benzene?
Empirical Formula
92.3% x 100 = 92.3 g C
7.7% x 100 = 7.7 g H
• mol C: 92.3g C = 7.69 mol C
12.01 g/mol C
• mol H: 7.7g H = 7.65 mol H
1.007g/mol H
• Least common multiple 7.65
CH
48
Benzene has empirical formula of CH and
a Experimental Weight = 78 g. What is
the molecular formula?
Molecular Weight = 78
= 6.0
Empirical Weight
13.08
C6H6
• A compound was analyzed and found
to contain 13.5 g Ca, 10.8 g O, and
0.675 g H. What is the empirical
formula of the compound?
49
50
NutraSweet is 57.14% C, 6.16% H, 9.52%
N, and 27.18% O. What is the compound’s
empirical formula?
Empirical Formula
57.14 %
x 100g = 57.14 g C
H
6.16 % x 100g = 6.16 g
9.52 % x 100g = 9.52 g N
27.18 %
x 100g = 27.18 g O
•
•
•
•
57.14 g C = 4.758 mol C
12.01g/mol C
mol H:
6.16 g H = 6.11 mol H
1.007g/mol H
mol N:
9.52 g N = 0.68 mol N
14.01g/mol N
mol O: 27.18 g O = 1.700 mol O
16.00g/mol C
Greatest common multiple 0.68
7
• mol C:
8.97 = 9
1
2.5
51
A sample has empirical formula of C14H18N2O5
and a Molecular Weight = 294.30 g. What is
the molecular formula?
Experimental MM =
Empirical MM
294.30 g = 1.0000
294.278
C14H18N2O5
Write the Empirical Formula for Each of the Following:
a. P2O3
a. P4O6 _________________
b. C2H3
b. C6H9 _________________
c. CH3O
c. CH2OHCH2OH _________
d. BrCl2
d. BrCl2 _________________
e. C3H4O3
e. C6H8O6 _______________
f. C5H11
f. C10H22 _______________
g. CuCO2
g. Cu2C2O4 ______________
h. HgF
h. Hg2F2 _________________
52
The formulas for Ionic compounds
are empirical formulas.
A molecular formula is "a whole
number multiple of an empirical
formula."3
Converting to Empirical Formula
from Molecular Formulas is slightly
more difficult than vice versa.
Hydrogen Peroxide
Step 1: Molecular Formula = H2O2
Step 2: All subscripts are divisible by 2.
Step 3: Empirical Formula = HO
Glucose
Step 1: Molecular Formula = C6H12O6
Step 2: All subscripts are divisible by 6.
Step 3: Empirical Formula = CH2O
Acetic Acid
Step 1: Molecular Formula = C2H3COOH
-- Step 1b: Standard Molecular Formula = C2H4O2
Step 2: All subscripts are divisible by 2.
Step 3: Empirical Formula = CH2O
53
COMBINATION
% COMPOSITION, EMPIRCAL FORMULA
AND MOLECULAR FORMULA
A 9.2 gram sample of a compound is 2.8 grams
nitrogen and 6.4 grams oxygen.
1. Find the percent composition of the compound
2. Find the Empirical Formula of the compound
3. Find Molecular Formula if molecular
(experimental) weight is 92g.
54
PERCENT COMPOSITION
% N: 2.8 g = .304 X 100 = 30.4%=30.% N
9.2 g
% O: 6.4 g = .696 X 100 = 69.6%= 70.% O
9.2 g
100%
Empirical Formula
_30.4_ % x 100g = _30.4_ g N
_69.6_% x 100g = __69.6_ g O
• mol N: 30.4 g N
= 2.17 mol N
14.01g/mol N
• mol O: 69.6 g O
= 4.375 mol O
16.00 g/mol O
• Greatest common multiple 2.17
1
2
• Molar Mass _46g _
NO2
__92__/__46__ = 2.0 = N2O4
55
A compound containing 74.0% carbon (C),
8.65% hydrogen (H), and 17.3% nitrogen (N)
by mass.
4. Determine the Empirical Formula (4)
5. Find the Empirical Molar Mass (3)
6. Find Molecular Formula if molecular (2)
(experimental) mass is 486.78g
Empirical Formula
74.0 % x 100g = 74.0 g C
8.65% x 100g = 8.65 g H
17.3% x 100g = 17.3 g N
• mol C: 74.0 g C
= 6.16 mol C
5
12.01g/mol C
• mol H: 8.65 g H
= 8.56 mol H
7
1.01g/mol H
• mol N: 17.3 g O
= 1.24 mol N
1
14.01g/mol O
C5H7N
• Greatest common multiple 1.24
• Molar Mass 81.13
486.78/81.13 = 6.0
C30H42N6
56
A compound has an empirical formula
of CH2 and a molecular mass of 42g.
12.01 +2.02 = 14.02
42 / 14.02 = 3.0
C3H6
57