Ch. 10 “The Mole”
23
6.02 X 10
SAVE PAPER AND INK!!! When you
print out the notes on PowerPoint,
print "Handouts" instead of
"Slides" in the print setup. Also,
turn off the backgrounds
(Tools>Options>Print>UNcheck
"Background Printing")!
1 mole = 22.4 L
@ STP
1 mol = molar mass
1 mol =
6.02 x 1023 particles
Welcome
to Mole
Island
Chemical equations can be interpreted in many ways!
We will use
the mole
interpretation!
Which
interpretations
support the
Law of
Conservation
of Mass?
Atoms & Mass!
Only mass &
Atoms are
conserved in
every reaction!
Chemical Equations are simple.
2H2 + O2  2H2O
Chemical equations are quite easy. Here we count out 4
hydrogens and 2 oxygen atoms and we can make two
water molecules.
But how many hydrogens and oxygens would we count
out to make enough water to fill a one liter bottle of water?
Imagine the number of grains of
sand in a handful of sand:
thousands or maybe a million? It
would be difficult to count that
many. Now imagine the number
of grains of sand that would be
present if we covered all of the
Earth (land and sea) 10 miles
deep with sand. That number of
grains of sands is the same
number of water molecules
needed to fill a 1 liter bottle!
How Scientists Keep Track of Atoms
• One way to measure the amount of
a substance available is to count
the # of particles in that sample
– However, atoms & molecules are
extremely small
– and, the # of individual particles in
even a small sample is very large
– Therefore, counting the # of
particles is not a practical measure
of amount
• To solve this problem, scientists
developed the concept of the mole
– It’s the “chemical counting unit”
• Just as a dozen eggs equals 12 eggs,
• a Mole = 6.02 x 1023
602,000,000,000,000,000,000,000
– It is equal to that number no matter
what kind of particles you’re counting!
– It could represent donuts, pens,
bikes, atoms, molecules, etc.
The word “mole” was introduced about
1896 by Wilhelm Oswald, who derived
the term from the Latin word moles
meaning a “heap” or “pile.”
• The mole is the SI base unit for
measuring the
amount of a substance.
Avogadro’s Number
• 6.02 X 1023
• This number is named in honor of
Amadeo Avogadro (1776 – 1856),
who studied quantities of gases
and discovered that no matter
what the gas was, there were the
same number of molecules present
under certain conditions.
The Mole
 1 mole of anything = 6.022  1023 units
of that thing
6.022  1023 is equal to the number of
carbon atoms in exactly 12 grams of pure
carbon-12.
Oxygen
32.00 g
One mole of
common
substances.
CaCO3
100.09 g
Water
18.02 g
Copper
63.55 g
The Mole
This photograph shows one
mole of solid (NaCl), liquid
(H2O), and N2 gas.
Mass and Moles of a Substance
1-octanol (C8H17OH)
Mercury(II) Iodide (HgI2)
One mole of
different
substances.
Sulfur (S8)
Methanol (CH3OH)
1 Mole of Particles
Exactly How Big is a Mole?
Also known as: The ‘Get-a-Life’ Syndrome!
Q: How long would it take to spend a mole of $1
coins if they were being spent at a rate of 1 billion
per second?
A Mollionaire
A: If a mole of $1 coins were being spent
at a rate of 1 billion per second:
$ 6.02 x 1023 / $1 000 000 000
= 6.02 x 1014 payments = 6.02 x 1014
seconds
6.02 x 1014 seconds / 60 = 1.003 x 1013
minutes
1.003 x 1013 minutes /60 = 1.672 x 1011
hours
1.672 x 1011 hours / 24 = 6.968 x 109 days
6.968 x 109 days / 365.25 = 1.908 x 107
years
A: It would take 19 million years
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.
More Examples, please!
One mole of marbles would cover the entire Earth
(oceans included) for a depth of three miles.
One mole of $100 bills stacked one on top of another
would reach from the Sun to Pluto and back 7.5 million
times.
It would take light 9500 years to travel from the
bottom to the top of a stack of 1 mole of $1 bills.
Reminder: The Mole is just a
counting number!
• 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
Mole is abbreviated mol (gee, that’s a lot
quicker to write, huh?)
Yummm!
Donuts!
I love
How Do We Use
The
Mole?
chemistry!
• We never use the mole to describe
macroscopic or real world objects.
– 1 mole (6.02x1023) of watermelon seeds
would be found inside a watermelon the size
of the moon.
– 1 mole (6.02x1023) of donut holes would
cover the earth and would be 5 miles deep.
• Since the mole is such a huge number of
items, it is only used to describe the
amount of things that are very, very
small.
– 1 mole (6.02x1023) of water molecules
would barely fill a shot glass
A Mole of Particles
Contains 6.02 x 1023 particles
1 mole C
= 6.02 x 1023 C atoms
23 H O molecules
=
6.02
x
10
1 mole H2O
2
1 mole NaCl= 6.02 x 1023 NaCl formula
units
(6.02 x 1023 Na+ ions
and
6.02 x 1023 Cl– ions)
The Mole
1 amu = 1.66054 x 10-24 g
1 g = 6.02214 x 1023 amu
Same #, different unit label!
Learning Check
Suppose we invented a new counting unit
called a “Salb” and one Salb contains 8
objects.
1. How many paper clips in 1 Salb?
a) 1
b) 4
c) 8
2. How many oranges in 2.0 Salbs?
a) 4
b) 8
c) 16
3. How many Salbs contain 40 gummy
bears?
a) 5
b) 10
c) 20
We can use Avogadro’s # in
Conversion Factors
6.02 x
1023
particles
1 mole
Oh,
Goody!
or
1 mole
6.02 x 1023 particles
Note that a “particle” can be an atom, molecule,
formula unit, or ion!
REPRESENTATIVE PARTICLES & MOLES
ATOMIC
NITROGEN
Atom
N
6.02x1023
NITROGEN GAS
Molecule
N2
6.02x1023
WATER
Molecule
H20
6.02x1023
CALCIUM ION
Ion
Ca2+
6.02x1023
CALCIUM
FLUORIDE
Formula Unit
CaF2
6.02x1023
Converting Between Moles and Particles
How we measure moles…
• Do you know how to measure out 3
moles of sugar or salt or water?
– The mole is a counting unit, so we
would have to count out 6.02x1023
particles of each substance. (no
thank you)
• There are 2 ways we can use to
measure out a number of moles of
a substance.
– Measure it in grams (a mass)
– Or measure it in liters (a volume)
Learning Check
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
0.500 mol Al
6.02 x 1023 atoms Al
1 mole Al
3.01 X 1023 atoms Al
=
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
1.8 x 1024 atoms S
1 mole S
6.02 x 1023 atoms S
3.0 moles S
=
Let’s look at Mass of Particles
• Atomic Mass: The mass of an atom (measured in
amu’s.)
• Molecular Mass: the sum of atomic masses of
• all atoms in a molecule (measured in amu’s.)
• example: H2O has a formula mass of 18.0 a.m.u
• Formula Mass: Sum of the mass of atoms in a
formula unit of an ionic compound (measured in
amu’s.)
example: NaCl has a formula mass of 58.5 a.m.u.
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
Na
NaCl Cl
1 x 23.0 amu = 23.0
1 x 35.5 amu = 35.5
NaCl
58.5 amu
SO2
S
1 x32.1 amu
= 32.1
O
2 x 16.0 amu
= 32.0
SO2
64.1 amu
Molecular mass is the sum of
the atomic masses (in amu) in a molecule.
Let’s review what we already know……
We already know that 1 atom of C-12 has a
mass of 2.010 x 10-23 g. (This is also equal
to 12 amu!)
We also want the atomic mass of 12.0 amu
to become 12.0 g.
So, the problem becomes this……..
2.010 x 10-23 g/atom C-12 x ? atoms = 12.0 g C-12
Why not use the average
atomic mass that is
already on the periodic
table? It is already a
relative mass, right? Why
reinvent the wheel?
But the atomic mass
on the chart is in
AMU, not grams! We
do not have an AMU
balance!
Then let’s convert
the number on the
periodic chart from
AMU to grams!
Molar Mass
A substance’s molar mass is it’s formula
mass in grams.
CO2 = 44.01 grams per mole
H2O = 18.02 grams per mole
Ca(OH)2 = 74.10 grams per mole
C
=
12.0 g/mole
Calculating molar masses using chemical
formulas
What is the mass of 1 mole of C6H12O6?
• The overall mass of 1 mole of C6H12O6 will
be the molar mass of 6 Carbons + the
molar mass of 12 Hydrogens + the molar
mass of 6 Oxygens.
6 Carbons = 6 * 12.011 g = 72.066 g
12 Hydrogens = 12 * 1.0079 g = 12.095 g
6 Oxygens = 6 * 15.999 g = 95. 994 g
180.16 g/mole
The numbers
under the element
names are the
relative weights.
However, if you
consider one
mole of the
element (6 x 1023
atoms) then these
numbers are read
as grams.
Molar mass
•
•
•
•
The mass of one mole is called “molar mass”
Ex. 1 mol Li = 6.94 g Li
This is expressed as 6.94 g/mol
What are the following molar masses?
S 32.06 g/mol
SO2 64.06 g/mol
Cu3(BO3)2308.1 g/mol
Calculate molar masses (to 2 decimal places)
CaCl2
Cu x 3 =
63.5 x 3 = 190. 5
(NH4)2CO3
B x2=
10.8 x 2 = 21.6
O2
O x6=
16.0 x 6 = 96.0
308.1g
Pb3(PO4)2
C6H12O6
Learning Check!
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
Molar Mass of Molecules and
Compounds
Mass in grams of 1 mole equal numerically to the
sum of the atomic masses
1 mole of CaCl2 = 111.1 g/mol
1 mole Ca x 40.1 g/mol
+ 2 moles Cl x 35.5 g/mol = 111.1 g/mol CaCl2
1 mole of N2O4
= 92.0 g/mol
Learning Check!
A. Molar Mass of K2O = ? Grams/mole
B. Molar Mass of antacid Al(OH)3 = ?
Grams/mole
Learning Check
Prozac, C17H18F3NO, is a widely used
antidepressant that inhibits the uptake of
serotonin by the brain. Find its molar mass.
Calculations with Molar Mass
molar mass
Grams
Moles
Converting between grams and moles
• If we are given the # of grams of a compound we
can determine the # of moles, & vise-versa
• In order to convert from one to the other you must
first calculate molar mass
g = mol given gfm
mol = g given 1 mol
1 mole
gfm
Formula
HCl
H2SO4
NaCl
Cu
gfm
36.46
98.08
58.44
63.55
mass mol (n)
9.1
0.25
53.15 0.5419
207
3.55
1.27 0.0200
Converting between Grams and
Moles
• USE MOLAR MASS as your conversion factor!
Converting Moles and Grams
Aluminum is often used for the structure of lightweight bicycle frames. How many grams of Al
are in 3.00 moles of Al?
3.00 moles Al
? g Al
1. Molar mass of Al
1 mole Al = 27.0 g Al
2. Conversion factors for Al
27.0g Al
1 mol Al
3. Setup
or
1 mol Al
27.0 g Al
3.00 moles Al x 27.0 g Al
1 mole Al
Answer = 81.0 g Al
Learning Check!
The artificial sweetener aspartame (NutraSweet) formula C14H18N2O5 is used to sweeten
diet foods, coffee and soft drinks. How many
moles of aspartame are present in 225 g of
aspartame?
Standard Molar
Volume
Equal volumes of all gases
at the same temperature
and pressure contain the
same number of
molecules.
- Amadeo Avogadro
At STP (Standard Temperature and Pressure):
1 mole of a gas occupies 22.4 liters of volume
Convert Between Moles and Volume
10.2
The Mole–Volume Relationship
Examples
• What is the volume of 4.59 mole of
CO2 gas at STP?
Atoms/Molecules and Grams
• Since 6.02 X 1023 particles = 1 mole
AND
1 mole = molar mass (grams)
• You can convert atoms/molecules to moles and
then moles to grams! (Two step process)
• You can’t go directly from atoms to grams!!!! You
MUST go thru MOLES.
• That’s like asking 2 dozen cookies weigh how
many ounces if 1 cookie weighs 4 oz? You have to
convert to dozen first!
ons
volume
22.4 L
molar mass
Grams
Avogadro’s number
Moles
particles
Everything must go through Moles!!!
All Conversions So Far TOGETHER
Interconverting Moles, Mass, and Number of Chemical Entities
no. of grams
Mass (g) = no. of moles x
1 mol
1 mol
No. of moles = mass (g) x
no. of grams
6.022x1023 particles
No. of particles = no. of moles x
1 mol
1 mol
No. of moles = no. of particles x
6.022x1023 particles
Atoms/Molecules and Grams
How many atoms of Cu are present in
35.4 g of Cu?
35.4 g Cu
1 mol Cu
63.5 g Cu
6.02 X 1023 atoms Cu
1 mol Cu
= 3.36 X 1023 atoms Cu
Learning Check!
How many atoms of K are present in 78.4 g of
K?
Learning Check!
What is the mass (in grams) of 1.20 X 1024
molecules of glucose (C6H12O6)?
Learning Check!
How many atoms of O are present in 78.1 g of
oxygen?
78.1 g O2 1 mol O2 6.02 X 1023 molecules O2 2 atoms O
32.0 g O2 1 mol O2
1 molecule O2
2.94 x 10
24
HINTS:
• If you see atoms, molecules, formula units,
or ions, you will use Avogadro’s number
• If you see liters, you will use Molar Volume
(22.4L)
• If you see grams, you will use Molar Mass
(gfm)
Density and the Mole
• If we know the density of a gas, we can use it to
calculate its GFM.
• Remember: Density = mass/volume
• Ex. Let’s say we know the density of a gas is
• 1.964 g/L, and we know it consists of C & O.
• We don’t, however, know if it is CO or
CO2.
1st Convert Liters to moles: density 22.4L
1.964 g
22.4 L
1L
1 mole
1L
1 mole
= 44.0 g/mole
2nd Compare to the GFM of CO & CO2
What’s the compound?
CO2
28 g/mole & 44 g/mole
Density & Moles, Con’t.
• Ex. The densities of gases A, B, & C are
1.25, 2.86, & .714 g/L, respectively.
calculate the GFM for each. Then, identify
each substance as NH3, SO2, Cl2, N2, or
CH4.
(multiply each density by 22.4L/1 mole)
Density & Moles, Con’t.
• If you are given the formula of a gas, you
can find it’s density at STP.
• GFM
• 1 mole
1 mole
22.4 L
• What’s the density of CO gas at STP?
28 g
1 mole
= 1.25 g/L
1 mole
22.4 L
Percent composition
• The percent by mass of each element in a
sample of a compound
• To determine % composition from a
formula
Percent Composition
•
•
•
•
•
Like all percents
Part divided by the Whole x 100 %
Find the mass of each component,
divide by the total mass.
Multiply by 100
Mass percent from the chemical
formula
Mass % of element X =
(atoms of X in formula) x (atomic mass of X)
formula mass of compound
x 100
Percent Composition
What is the percent carbon in C5H8NO4
(the glutamic acid used to make MSG
monosodium glutamate), a compound used
to flavor foods and tenderize meats?
a)
b)
c)
8.22 %C
24.3 %C
41.1 %C
1. Find the gfm of the compound:
C
H
N
O
5 x
8x
1x
4x
12.0 = 60.0
1.0
= 8.0
14.0 = 14.0
16.0 = 64.0
146.0
2. Part/Whole x 100
60.0
146.0
X 100
= 41.1 % Carbon by mass
Simplest and molecular formulas
Consider NaCl (ionic) vs. H2O2 (covalent)
Na Cl Na Cl
Cl Na Cl Na
• Chemical formulas are either “simplest” (a.k.a.
“empirical”) or “molecular”. Ionic compounds are
always expressed as simplest formulas.
• Covalent compounds can either be molecular
formulas (I.e. H2O2) or simplest (e.g. HO)
Q - Write simplest formulas for propene (C3H6),
C2H2, glucose (C6H12O6), octane (C8H14)
Q - Identify these as simplest formula, molecular
formula, or both H2O, C4H10, CH, NaCl
Answers
For more lessons, visit
www.chalkbored.com
Q - Write simplest formulas for propene (C3H6),
C2H2, glucose (C6H12O6), octane (C8H14)
Q - Identify these as simplest formula, molecular
formula, or both H2O, C4H10, CH, NaCl
A - CH2 CH CH2O
C 4H 7
A - H2O is both simplest and molecular
C4H10 is molecular (C2H5 would be simplest)
CH is simplest (not molecular since CH can’t
form a molecule - recall Lewis diagrams)
NaCl is simplest (it’s ionic, thus it doesn’t
form molecules; it has no molecular formula)
Types of Formulas
• Empirical Formula
The formula of a compound that
expresses the smallest whole number ratio of
the atoms present.
• Formula Units for ionic compounds are always
empirical (lowest whole number ratio).
• Examples:
• NaCl
MgCl2
Al2(SO4)3
K2CO3
Molecular Formula shows the actual number of each kind of atom
found in one molecule of the compound.
Examples: C6H12O6
N2O4
10.3
Empirical Formulas
• Ethyne (C2H2) is a gas
used in welder’s torches.
Styrene (C8H8) is used
in making polystyrene.
• These two hydrocarbons
have the same empirical
formula (CH) but
different molecular
formulas.
Formulas
(continued)
Formulas for molecular compounds MIGHT
be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Formulas of Compounds
• Formulas give the relative numbers of
atoms or moles of each element in a
formula unit - always a whole number ratio
(the law of definite proportions).
NO2
2 atoms of O for every 1 atom of N
1 mole of NO2 : 2 moles of O atoms to every 1
mole of N atoms
• If we know or can determine the relative
number of moles of each element in a
compound, we can determine a formula for
the compound.
Finding an Empirical Formula
1. Determine the mass in grams of each
element present.
2. Convert mass to moles for each
element.
3. Divide each # of moles by the smallest
number of moles to obtain the simplest
whole number ratio.
4. If whole numbers are not obtained* in
step 3), multiply each by the smallest
number that will give all whole numbers
*
Be careful! Do not round off numbers prematurely
Empirical Formula Calculations
Q- a compound is found to contain the
following % by mass: 69.58% Ba, 6.090% C,
24.32% O. What is the empirical formula?
Step 1: Convert each mass to moles. (calculate
the # of moles (mol = g x mol/gfm) of each
element.)
Step 2: Divide each answer from step 1 by the
lowest number of moles (to get simplest
whole # ratio)
Step 3: Use the #’s from step 2 as
subscripts.
(Make sure the elements are in the correct
order!)
Simplest formula: sample problem
Q- 69.58% Ba, 6.090% C, 24.32% O.
What is the empirical formula?
1: 69.58 g Ba, 6.090 g C, 24.32 g O
2: Ba: 69.58 g x 1 mol/ 137.3g = 0.507 mol Ba
C: 6.090 g x 1 mol/ 12.0g = 0.508 mol C
O: 24.32 g x 1 mol/ 16.0g =
1.52 mol O
3:
Ba
C
O
mol
0.507
0.508
1.52
mol
0.507/
0.508/
1.52/
(reduced) 0.507 = 0.507 =
0.507 =
1.000
1.002
3.000
4: the simplest formula is BaCO3 (1:1:3 ratio)
Mole ratios and simplest formula
Given the following mole ratios for the
hypothetical compound AxBy, what
would x and y be if the mol ratio of A
and B were:
A = 1 mol, B = 2.98 mol AB3
A = 1.337 mol, B = 1 mol A4B3
A = 2.34 mol, B = 1 mol A7B3
A = 1 mol, B = 1.48 mol A2B3
Ex. A compound consists of
29.1 % Na, 40.5 % S, and 30.4 % O.
Determine the empirical formula.
1: Assume 100 g: 29.1 g Na, 40.5 g S, 30.4 g O
2: Na: 29.1 g x 1 mol/ 23.0 g = 1.27 mol Na
S: 40.5 g x 1 mol/ 32.1 g = 1.26 mol S
O: 30.4 g x 1 mol/ 16.0 g = 1.90 mol O
Na
S
O
mol
1.27
1.26
1.90
mol
1.27/ 1.26 1.26/ 1.26 1.90/ 1.26
(reduced) = 1.01
= 1.00
= 1.51
X2
2
2
3
4: the simplest formula is Na2S2O3
A sample of a brown gas, a major air
pollutant, is found to contain 2.34 g N and
5.34g O.
Determine a formula for this
substance.
Convert grams to moles
moles N
moles N
= 2.34g N
moles O = 5.34g O
moles O
Formula can’t be:
1 mole N
14.0 g N
1mole O
16.00 g O
= 0.167
= 0.334
N0.167 O0.334
So, divide each # moles by the smallest one!
N 0.167 O 0.334  NO 2
0.167
0.167
A compound is composed of 7.20 g carbon,1.20 g
hydrogen, and 9.60 g oxygen. Find the empirical
formula for this compound.
1: 7.20 g C, 1.20 g H, 9.60 g O
2: C: 7.20 g x 1 mol/12.0g = 0.600 mol C
H: 1.20 g x 1 mol/1.00g = 1.20 mol H
O: 9.6 g x 1 mol/16.0g
= 0.600 mol O
C
H
O
3:
mol
0.600
1.20
0.600
mol
0.60/ 0.60 1.20/ 0.60 0.60/ 0.60
(reduced)
=1
=1
=2
4: the simplest formula is CH2O
Finding Molecular formulas
• There is one additional step to solving for the
true molecular formula of a compound.
• First you must be given the molar mass of the
compound. (ex. 150 g/mole)
• 1. Determine molar mass of the empirical
formula. For CH2O it is 30 g/mol (12+2+16).
• 2. Divide the molar mass of the compound by
this to get a factor: 150 / 30 = 5
• 3. Multiply each subscript in the formula by
this factor: C5H10O5 is the molecular formula.
• Q- For OF, give the molecular formula if the
compound is 70 g/mol
70  35 = 2 O2F2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
1. Find the empirical mass of C3H5O2
3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
2. Divide the molecular mass by the
emipirical mass.
3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
146
2
73
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
3. Multiply the empirical formula by this
number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
2
73
(C3H5O2) x 2
=
C6H10O4
Practice Problems
1. Calculate the percentage composition of each
substance: a) SiH4, b) FeSO4
2. Calculate the simplest formulas for the
compounds whose compositions are listed:
a) carbon, 15.8%; sulfur, 84.2%
b) silver,70.1%; nitrogen,9.1%; oxygen,20.8%
c) K, 26.6%; Cr, 35.4%, O, 38.0%
3. The simplest formula for glucose is CH2O and its
molar mass is 180 g/mol. What is its molecular
formula?
4. Determine the molecular formula for each
compound below from the information listed.
substance
simplest formula molar mass(g/mol)
a) octane
C4H9
114
b) ethanol
C2H6O
46
c) naphthalene
C5H4
128
d) melamine
CH2N2
126
5. The percentage composition and approximate
molar masses of some compounds are listed
below. Calculate the molecular formula of each
percentage composition
molar mass(g/mol)
64.9% C, 13.5% H, 21.6% O
74
39.9% C, 6.7% H, 53.4 % O
60
40.3% B, 52.2% N, 7.5% H
80
1 a) Si= 87.43% (28.09/32.13 x 100), H= 12.57%
b) Fe= 36.77% (55.85/151.91 x 100),
(32.06/151.91 x 100), O= 42.13%
S= 21.10%
2 a) Assume 100 g. Thus: 15.8 g C, 84.2 g S.
C: 15.8 g  12.01 g/mol = 1.315 mol C
S: 84.2 g  32.06 g/mol = 2.626 mol S
Mol
Mol reduced
C
1.315
1.315/1.315
=1
the simplest formula is CS2
S
2.626
2.626/1.315
= 2.00
2 b) Ag: 70.1 g  107.87 g/mol = 0.6499 mol Ag
N: 9.1 g  14.01 g/mol = 0.6495 mol N
O: 20.8 g  16.00 g/mol = 1.30 mol O
AgNO2
Ag
N
O
Mol
0.6499
0.6495
1.30
Mol
.6499/.6495 .6495/.6495 1.30/.6495
reduced
= 1.0
=1
= 2.00
2 c) K: 26.6 g  39.10 g/mol = 0.6803 mol K
Cr: 35.4 g  52.00 g/mol = 0.6808 mol Cr
O: 38.0 g  16.00 g/mol = 2.375 mol O
K2Cr2O7
K
Cr
O
Mol
0.6803
0.6808
2.375
Mol
.6803/.6803 .6808/.6803 2.375/.6495
reduced
=1
= 1.00
= 3.49
3 C6H12O6 (CH2O = 30 g/mol, 180/30 = 6)
4 a) C8H18 (C4H9 = 57 g/mol, 114/57 = 2)
b) C2H6O (C2H6O = 46 g/mol, 46/46 = 1)
c) C10H8 (C5H4 = 64 g/mol, 128/64 = 2)
d) C3H6N6 (CH2N2 = 54 g/mol, 126/42 = 3)
5 a) C: 64.9 g  12.01 g/mol = 5.404 mol C
H: 13.5 g  1.01 g/mol = 13.37 mol H
O: 21.6 g  16.00 g/mol = 1.35 mol O
C4H10O
C
H
O
Mol
5.404
13.37
1.35
Mol
5.404/1.35 13.37/1.35 1.35/1.35
reduced
= 4.00
= 9.90
=1
C4H10O (C4H10O = 74 g/mol, 74/74 = 1)
5 b) C: 39.9 g  12.01 g/mol = 3.322 mol C
H: 6.7 g  1.01 g/mol = 6.63 mol H
O: 53.4 g  16.00 g/mol = 3.338 mol O
CH2O
C
H
O
Mol
3.322
6.63
3.338
Mol
3.322/3.322 6.63/3.322 3.338/3.322
reduced
=1
= 2.0
= 1.00
C2H4O2 (CH2O = 30 g/mol, 60/30 = 2)
5 c)
B
N
H
Mol
3.728
3.726
7.43
Mol
3.728/3.726 3.726/3.726 7.43/3.726
reduced
= 1.00
=1
= 2.0
B3N3H6 (BNH2 = 26.84 g/mol, 80/26.84= 2.98)
For more lessons, visit
www.chalkbored.com
Volume 22.4 L
PT
Mass
Moles
6.02 x
23
10
Representative
Particles
Atoms
Count
Ions
Comparing sugar (C12H22O11) & H2O
Same
# of moles?
1 gram each
No, they have dif.
densities.
Yes, that’s what
grams are.
No, they have dif.
molar masses
1 mol each
No, molecules
have dif. sizes.
No, molecules
have dif. masses
Yes.
# of
molecules?
No, they have dif.
molar masses
# of atoms?
No
Yes (6.02x1023 in
each)
No, sugar has
more (45:3 ratio)
volume?
mass?
The Molar Mass and Number of
Particles in One-Mole Quantities
Substance
Molar Mass
Number of Particles in One Mole
Carbon (C)
12.0 g
6.02 x 1023 C atoms
Sodium (Na)
23.0 g
6.02 x 1023 Na atoms
Iron (Fe)
55.9 g
6.02 x 1023 Fe atoms
NaF (preventative
for dental cavities)
42.0 g
6.02 x 1023 NaF formula units
CaCO3 (antacid)
100.1 g
6.02 x 1023 CaCO3 formula units
C6H12O6 (glucose)
180.0 g
6.02 x 1023 glucose molecules
C8H10N4O2 (caffeine)
194.0 g
6.02 x 1023 caffeine molecules
Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
Atoms/mole of
compound
6(6.022 x 1023)
atoms
12(6.022 x 1023)
atoms
6(6.022 x 1023)
atoms
Mass/molecule
of compound
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
12.10 g
96.00 g
Mass/mole of
compound
72.06 g
6.02 x 1023 somethings = 1