Section 3.5—Counting
Molecules
So the number of molecules affects pressure of an
airbag…how do we “count” molecules?
What is a Mole? Ted Ed video
http://ed.ted.com/lessons/daniel-dulekhow-big-is-a-mole-not-the-animal-theother-one
What is a mole?
Mole – metric unit for counting
We use it just like we use the terms dozen and ream!
The only acceptable abbreviation for “mole” is
“mol”…not “m”!!
What is a counting unit?
You’re already familiar with one counting unit…a
“dozen”
A dozen = 12
“Dozen”
12
A dozen doughnuts
12 doughnuts
A dozen books
12 books
A dozen cars
12 cars
What can’t we count atoms in “dozens”?
Atoms and molecules are extremely small
We use the MOLE to count particles
A mole = 6.02  1023 particles
(called Avogadro’s number)
6.02  1023 = 602,000,000,000,000,000,000,000
“mole”
6.02  1023
1 mole of doughnuts
6.02  1023 doughnuts
1 mole of atoms
6.02  1023 atoms
1 mole of molecules
6.02  1023 molecules
This number was named after Amadeo Avogadro. He did
not calculate it!
FUNNY!
Representative Particles
Remember, matter is broken down into either SUBSTANCES
or mixtures
Substances are broken down into either ELEMENTS or
COMPOUNDS
Type of Matter
Example
Representative Particle
Element
Fe
Atom
Ionic Compound
NaCl
Formula Unit
Covalent Compound
CO2
Molecule
Example: Particles & Moles
Use the conversion factor (1 mol = 6.02 x 1023)
particles to convert
Example
1:
How many
molecules
of water
are in 1.25
moles?
Example: Molecules & Moles
1 mol = 6.021023 molecules
Example
1:
How many
molecules
of water
are in 1.25
moles?
1.25 mol
H2O
6.02  1023
1
Molecules
H2O
mol H2O
7.531023 molecules H2O
= _______
Let’s Practice #2
Example:
How many
moles are
equal to
2.8 × 1022
formula
units of
KBr?
Let’s Practice #2
1 mol = 6.021023 formula units
Example:
How many
moles are
equal to
2.8 × 1022
formula
units KBr?
2.8 × 1022 formula units
1
mole
6.02  1023 Formula units
0.047
= _______
moles
Let’s Practice #3
Example:
How many
atoms are
equal to
3.56
moles of
Fe?
Let’s Practice #3
1 mol = 6.021023 molecules
Example:
How many
atoms are
equal to
3.56
moles of
Fe?
3.56 moles Fe
6.02 x 10 23 atoms
1
moles
2.14 x 1024 atoms
= _______
Molar Mass
Molar Mass – The mass for one mole
of an atom or molecule.
Other terms commonly used for the same meaning:
Molecular Weight
Molecular Mass
Formula Weight
Formula Mass
Molar Mass for Elements
The average atomic mass = grams for 1 mole
Average atomic mass is found on the periodic table
Element
Mass
1 mole of carbon atoms (C)
12.01 g
1 mole of oxygen atoms (O2)
16.00 g x 2 = 32.00 g O2
1 mole of hydrogen atoms (H2)
1.01 g x 2 = 2.02 g H2
Unit for molar mass: g/mole or g/mol
Molar Mass for Compounds
The molar mass for a molecule = the
sum of the molar masses of all the
atoms
Calculating a Molecule’s Mass
To find the molar mass of a molecule:
1
Count the number of each type of atom
2
Find the molar mass of each atom on the periodic
table
3
Multiply the # of atoms by the molar mass for each
atom
Find the sum of all the masses
4
Example: Molar Mass
Example:
Find the
molar
mass for
CaBr2
Example: Molar Mass
1
Example:
Find the
molar
mass for
CaBr2
Count the number of each type of atom
Ca
1
Br
2
Example: Molar Mass
2
Example:
Find the
molar
mass for
CaBr2
Find the molar mass of each atom on the periodic table
Ca
1
40.08 g/mole
Br
2
79.90 g/mole
Example: Molar Mass
3
Example:
Find the
molar
mass for
CaBr2
Multiple the # of atoms  molar mass for each atom
Ca
1  40.08 g/mole =
40.08 g/mole
Br
2  79.90 g/mole =
159.80 g/mole
Example: Molar Mass
4
Example:
Find the
molar
mass for
CaBr2
Find the sum of all the masses
Ca
1  40.08 g/mole =
40.08 g/mole
Br
2  79.91 g/mole =
+ 159.80 g/mole
199.88 g/mole
1 mole of CaBr2 =199.90 g
Example 2: If you see a Parentheses
in the Formula
Be sure to distribute the subscript outside the
parenthesis to each element inside the parenthesis.
Example:
Find the
molar
mass for
Sr(NO3)2
Example 2: Molar Mass &
Parenthesis
Be sure to distribute the subscript outside the
parenthesis to each element inside the parenthesis.
Example:
Find the
molar
mass for
Sr(NO3)2
Sr
1  87.62 g/mole =
87.62 g/mole
N
2  14.01 g/mole =
28.02 g/mole
O
6  16.00 g/mole =
+ 96.00 g/mole
211.64 g/mole
1 mole of Sr(NO3)2 =211.64 g
Let’s Practice #3
Example:
Find the
molar
mass for
Al(OH)3
Let’s Practice #2
Be sure to distribute the subscript outside the
parenthesis to each element inside the parenthesis.
Example:
Find the
molar
mass for
Al(OH)3
Al
1  26.98 g/mole =
26.98 g/mole
O
3  16.00 g/mole =
48.00 g/mole
H
3  1.01 g/mole
=
+ 3.03 g/mole
78.01 g/mole
1 mole of Al(OH)3 =78.01 g
Using Molar Mass in
Conversions
Example: Moles to Grams
Example:
How many
grams are
in 1.25
moles of
water?
Example: Moles to Grams
When converting between grams and moles, the
molar mass is needed
Example:
How many
grams are
in 1.25
moles of
water?
1.25 mol H2O
H 2  1.01 g/mole =
2.02 g/mole
O 1  16.00 g/mole = + 16.00 g/mole
18.02 g/mole
1 mole H2O molecules = 18.02 g
18.02 g H2O
1
mol H2O
22.5
= _______
g H2O
Example: Grams to Moles
Example:
How many
moles are
in
25.5 g
NaCl?
Example: Grams to Moles
Example:
How many
moles are
in
25.5 g
NaCl
25.5 g NaCl
Na 1  22.99 g/mole = 22.99 g/mole
Cl 1  35.45 g/mole = + 35.45 g/mole
58.44 g/mole
1 moles NaCl molecules = 58.44 g
1
mol NaCl
58.44 g NaCl
.436 moles NaCl
= ____
Example: Grams to Molecules
Example:
How many
formula
units are
in
25.5 g
NaCl?
Example: Grams to Moles
Example:
How many
formula
units are
in
25.5 g
NaCl
25.5 g NaCl
Na 1  22.99 g/mole = 22.99 g/mole
Cl 1  35.45 g/mole = + 35.45 g/mole
58.44 g/mole
1 moles NaCl formula units = 58.44 g
1
mol NaCl
58.44 g NaCl
6.02 x 1023 FU’s
1 mol NaCl
x 1023 FU’s NaCl
=2.63
____
Gases & Moles
Amounts
Molar Volume
The molar volume of a gas is measured at
STP (standard temperature and pressure)
1 mole of any gas = 22.4 L
36
Molar Volume as a Conversion Factor
The molar volume at STP
has about the same volume as 3
basketballs
can be used to form 2 conversion
factors:
22.4 L
1 mole
and
1 mole
22.4 L
37
Let’s Try it Out! Example
When converting between volume and moles, STP
must be a condition to use molar volume
Example:
An experiment
requires .0580
moles of NO.
What volume
would you
need
at STP?
1 mole NO = 22.4 L
.O580 mol NO 22.4
1
L NO
mol NO
1.2992 L NO
= _______
Try Another!
Example:
Suppose you
need 4.22 g
of Cl2. What
volume at
STP would
you use?
4.22 g Cl2
1 moles Cl2 = 70.90 g Cl2
1 mol = 22.4 L at STP
1 mol Cl2
70.90 g Cl2
22.4
L Cl2
1 mol Cl2
1.33
= _________
L Cl2
Percent Composition
Defined as the percent by mass of each element in a
compound
Steps to Finding Percent Composition
1. Add up the mass of each element within the compound to
get the mass of the compound.
2. Divide each element’s mass by the mass of the compound.
3. Multiply by 100
=
massofeleme
% composition=
mass of element
mass of compound
x 100
Unit 3.6 Formula Calculations
Percent Composition by Mass of Air
Example: Calculate the % composition of each
element in calcium carbonate.
CaCO3
Molar mass = 100.09 g
% C = 12.01/100.09 x 100 = 12.00 %
%Ca = 40.08/100.09 x 100 = 40.04%
%O = 48.00/100.09 x 100 = 47.96%
Example: What is the % of each element in a
compound that contains 29.00g Ag and 4.30g S
only?
Total mass of compound = 33.30 g
% Ag = 29.00/33.30 x 100 = 87.09 %
%S = 4.30/33.30 x 100 = 12.9%
Hydrates
 A HYDRATE is an ionic compound with water trapped
in its crystal.
Examples are:
CuSO4 5H2O
MgSO4 7 H2O
CoCl2 H2O
 Heating a hydrate removes the water and leaves
behind just the salt which is called the anhydrate.
Example: What is the % water in the hydrate,
CuCl2  2H2O
Molar mass of hydrate = 170.48 g
% water = 36.04/170.48 x 100 = 21.14%
Heating of A Hydrate Animation
Calculating the experimental %
composition of water in a hydrate.
http://group.chem.iastate.edu/Greenb
owe/sections/projectfolder/flashfiles/st
oichiometry/empirical.html
Empirical Formula
A chemical formula showing the simplest
whole number ratio of moles of elements
(subscipts)
Hydrogen Peroxide has an actual formula
(molecular formula) of H2O2
but an
empirical formula of HO
How to Calculate Empirical Formula
RHYME: Percent to Mass
Mass to Mole
Divide by Small
Multiply til Whole
1.Assume 100 grams of the sample of compound. Switch the percent
sign to grams
2.Convert each element’s mass into moles.
3.Divide each element’s mole amount by the smallest mole amount
in the entire problem. The answer is the subscript of the element
within the compound.
4.OPTIONAL: If mole ratio is not within .1 of a whole number,
multiply each amount by the smallest whole number that will produce
either a whole number itself or a number within .1 of a whole number.
Example: What is the empirical formula for
40.05% S and 59.95% O?
1. Switch the percent sign to grams & convert each element’s
mass into moles
40.05 g S / 32.01g = 1.250 mol S
59.95 g O / 16.00 g = 3.747 mol O
2. Divide each element’s mole amount by the smallest mole amount in the
entire problem.
1.250 mol S = 1
1.250 mole
3.747 mol O = 2.99 = 3
1.250 mol
S1O3  SO3
Example: What is the empirical formula for
43.64% P and 56.36% O?
1. Switch the percent sign to grams & convert each element’s mass
into moles
43.64 g P / 30.97g = 1.409 mol S
56.36 g O / 16.00 g = 3.522 mol O
2. Divide each element’s mole amount by the smallest mole
amount in the entire problem.
1.409 mol S = 1
1.409 mole
3.522 mol O = 2.49 ≠ 3
1.409 mol
3. If mole ratio is not within .1 of a whole number, multiply each amount by
the smallest whole number that will produce either a whole number itself
or a number within .1 of a whole number.
1x2=2
2.49 x 2 = 4.998 = 5
P2O5
Molecular Formula
Is the ACTUAL, true formula of the
compound.
They are usually multiples of their empirical
formula
N2O4 is the molecular formula; the empirical
formula is NO2
Notice that the molecular formula is 2 times
larger than the empirical formula
Molecular Formula
How to Calculate Molecular Formula
1. You need to find the empirical formula and
calculate its molar mass. Call this empirical formula
mass EFM.
2. Find the mass of the actual formula which will most
likely be given to you in grams. Call this molecular
formula mass MFM.
3. Divide the MFM by the EFM to get a factor.
4. Multiply the factor by the empirical formula to get
the MOLECULAR FORMULA
Example:
What is the molecular formula of a compound
whose empirical formula is CH4N and the
molecular mass is 60.12 g/mol?
1. Empirical Formula Mass (EFM) = 12.01 + 4.04 + 14.01 = 30.06 g
2. Molecular Formula Mass (MFM) = 60.12 g
3. 60.12 / 30.06 = 2
4. 2(CH4N) = C2H8N2