The Mole
Not the type of mole we are talking about
Measuring Matter
• You often measure the amount of
something by one of three different
methods:
1. Count
2. Mass
3. Volume
Conversions
Try the practice problems on page 289!!!
• 12 apples = 1 dozen apples
Count
12 apples
1dozen apples
1 dozen apples = .20 bushel
Volume
1 dozen apples
.20 bushel
1 dozen apples = 2.0 kg
1 dozen apples
2.0 kg
Mass
The Mole
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?
Mollionaire
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: $ 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
What is a Mole?
Just as 12 eggs is a dozen ( a specific number of
particles), a mole is 6.02 x 10 23 particles.
This number of representative particles is called
Avogadro’s number.
*That’s Avogadro, not
Avocado!!!!
What is meant by “representative particles”?
Representative particles are whatever you are talking
about: atoms, molecules, or formula units (ions).
The representative particle of most elements is the atom.
Seven elements exist as diatomic molecules, and, as such,
its representative particle is the molecule.
The seven: H2, N2, O2, F2, Cl2, Br2, I2.
OK, so now we know that a mole is 6.02 x1023
representative particles. So how many atoms are in
one mole of an compound, or how many moles are in
6.02 x 1023 atoms?
Example:
How many moles are in 6.02 x 1023 atoms of silver?
Solution:
1. Determine the conversion factor.
1 mole = 6.02 x 1023 representative particles
1 mol representative particles
6.02 x 1023 particles
6.02 x
1023 Ag
atoms x
1 mol Ag
6.02 x 1023 Ag atoms
To find the number of atoms in a mole of a compound,
you must determine the number of atoms in a
representative formula of that compound.
Example:
How many oxygen atoms are in a mole of CO2?
1 mole CO2 x
6.02 x 1023 molecules CO2
1 mole CO2
x
2 oxygen atoms
1 molecule CO2
= 12.04 x 1023 atoms O2; or, 1.204 x 1024 atoms O2.
Converting Number of Particles to
Moles
• Look on Page 290
and 291
• Try the practice
problems!!!!
How many representative
Particles are in one mole??
When discussing the mole, using atoms leads to very
large numbers. An easier way to discuss moles is to work
with grams of atoms instead.
The gram atomic mass (gam) is the atomic mass of an
element expressed in grams.
For carbon, the gam is 12.0 g. For atomic hydrogen, the
gam is 1.0 g.
What is the gam for iron and mercury?
(55.85 g & 200.6 g)
How many atoms are contained in the gram atomic mass
of an element?
The gam contains one mole of atoms (6.02 x 1023 atoms)
of that element.
Thus, if 12.0 g of carbon is the gam of carbon, 12.0 g is 1
mol of carbon, and has 6.02 x 1023 atoms.
What is the mass of a mole of a compound?
To answer this you must know the formula of the
compound. The formula tells you the number of atoms of
each element in a representative particle of that
compound.
You calculate the mass of a mole of a compound by
adding together the atomic masses of the atoms making
up the compound.
This is called gram molecular mass (gmm)
Example:
What is the molecular mass of SO3?
Molar mass
•
•
•
•
The mass of one mole is called “molar mass”
E.g. 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)2 308.27 g/mol
Calculate molar masses (to 2 decimal places)
CaCl2
Cu x 3 = 63.55 x 3 = 190.65
(NH4)2CO3 B x 2 = 10.81 x 2 = 21.62
O2
O x 6 = 16.00 x 6 = 96.00
308.27
Pb3(PO4)2
C6H12O6
Molar mass
•
•
•
•
The mass of one mole is called “molar mass”
E.g. 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)2 308.27 g/mol
Calculate molar masses (to 2 decimal places)
CaCl2
110.98 g/mol (Cax1, Clx2)
(NH4)2CO3 96.11 g/mol (Nx2, Hx8, Cx1, Ox3)
O2
32.00 g/mol (Ox2)
Pb3(PO4)2 811.54 g/mol (Pbx3, Px2, Ox8)
C6H12O6 180.18 g/mol (Cx6, Hx12, Ox6)
Example:
How many grams are in 7.20 mol of N2O3?
N2 = 28.0 g
O3 = 48.0 g
1 mole N2O3 = 76.0 g
7.20 mol N2O3 x
76.0 g N2O3
= 547.2 g N2O3
1 mol N2O3
= 5.47 x 102 g N2O3
You Try It!
1. How many grams in .720 mol Be?
6.48 g
2. How many moles in 2.40 g N2?
0.086 mol
Easy peasy!!
Now let’s see if the fog has lifted….
Example
• If I have 2.00 moles of C13H18O2, how many
moles of each atom would I have?
13 mol C
2.00 mol C13H18O 2 
 26.0 mol C
1 mol C13H18O 2
18 mol H
2.00 mol C13H18O 2 
 36.0 mol H
1 mol C13H18O 2
2 mol O
2.00 mol C13H18O 2 
 4.00 mol O
1 mol C13H18O 2
Formula Mass
•
•
•
•
mass of a molecule, ion, or formula unit
sum of mass of all atoms in the chemical formula
in amu
Ex: H2O
1.00794 amu
2 H atoms 
1 H atom
 2.01588 amu
15.9994 amu
1 O atoms 
 15.9994 amu
1 O atom
• 18.01528 amu
• formula mass = molecular mass for molecular
compound
Example
• Find formula mass of potassium chlorate.
• KClO3
• 1(39.0983) + 1(35.4527) + 3(15.9994)
• 122.549 amu
Molar Masses
• mass of one mole of pure substance
• numerically equal to formula mass
• units: g/mol
•
•
•
•
Find molar mass of barium nitrate.
Ba(NO3)2
1(137.327) + 2(14.00674) + 6(15.9994)
261.337 g/mol
Molar Mass in Conversions
• can be used as a conversion factor
• between grams and moles
• What is the mass in grams of 2.50 mol of
oxygen gas?
2(15.9994) g
2.50 mol O 2 
 79.997  80.0 g O 2
1 mol O 2
Example
• Ibuprofen, C13H18O2, is the active
ingredient in many pain relievers.
• Find molar mass:
(13 x 12.011) + (18 x 1.00794) + (2 x 15.9994)
=206.29 g/mol
Example
• If the tablets in a bottle contain a total of 33 g
of ibuprofen, how many moles are in one
bottle?
1 mol C13H18O 2
33 g C13H18O 2 
 0.16 mol C13H18O 2
206.29 g C13H18O 2
• How many molecules of ibuprofen are in the
bottle?
6.022 x 1023 molec. C13H18O 2
0.16 mol C13H18O 2 
 9.6 x 1022 molec. C13H18O 2
1 mol C13H18O 2
Example
• What is the number of moles of carbon in
that bottle?
13 mol C
0.16 mol C13H18O 2 
 2.1 mol C
1 mol C13H18O 2
• What is the total mass in grams of carbon
in the bottle?
12.011 g C
2.1 mol C 
 25 g C
1 mol C
The Amount of A Mole of Gas
We have seen that one-mole amounts of liquids or solids
have different volumes than other solids or liquids. What
about the volume of gas?
Moles of gases have very predictable volumes.
Changing temperature or pressure of a gas can vary the
volume. That is why the volume of a gas is measured at a
standard temperature and pressure (STP).
Standard temperature is 0°C. Standard pressure is
101.2 kPa or 1 atmosphere (atm)
At STP, 1 mol of any gas occupies a volume of
22.4 L.
22.4 L is known as the molar volume of a gas which
means it contains 6.02 x 1023 representative particles of
that gas.
Example:
Determine the volume, in liters, of 0.600 mol of SO2 gas
at STP.
0.600 mol SO2 x
22.4 L SO2
1 mol SO2
= 13.4 L SO2
Assuming STP, how many moles are in 67.2 L SO2?
1 mole SO2
67.2 L SO2 x
= 3 mol SO2
22.4 L SO2
Review:
1. What volume, in liters, will 0.680 mol of a
certain gas occupy?
2. How many moles is 1.33 x 104 mL of O2 at STP?
Mass/Density of a Gas
Would 22.4 L of one gas also have the same mass as 22.4 L
of another gas at STP?
Probably not. A mole of one gas have a mass equal to its
gfm. Different gases usually have different gfm’s.
Measuring the volume of a gas is preferred to measuring
mass. Knowing the volume can also help find the density of
the gas.
Density is found by dividing the mass of a gas by its
volume. Because volume can change with a change in
temperature, density is measured at STP.
molar mass
g/mol
Density (at STP) =
=
molar volume
22.4 L/mol
g
=
L
Example:
What is the density of oxygen gas at STP? (in
grams per liter.)
molar mass = 32 g/mol = 1.43 g/L
D=
molar volume
22.4 L/mol
Percent Composition
To keep your lawn healthy, wealthy, and wise you need to
use fertilizer. You can’t just use any ol’ fertilizer. You need
to use one that has the right mixture of elements or
compounds depending on what you need to do.
You need to know the relative amount of each nutrient.
This is the same in the laboratory. When you make a new
compound, you need to determine its formula by finding
the relative amounts of elements in the compound.
The relative amounts are expressed as the percent
composition, the percent by mass of each element in a
compound.
There are as many percent values as there are elements in
the compound.
The percentages must add up to 100%.
grams of element X
% mass of element X =
grams of compound
x 100%
Example:
An 8.20-g piece of magnesium combines with a
5.40-g sample of oxygen completely to form a
compound. What is the percent composition of
this compound?
1. Find mass of compound.
13.60 g = mass of compound (8.20 g + 5.40 g)
2. Find % of each element
mass Mg
8.20 g
% Mg =
=
mass cmpd.
13.6 g
= 60.3%
mass O
5.40 g
%O=
=
mass cmpd.
13.6 g
= 39.7 %
Once you determine the percent composition of a
compound you can determine the number of grams of an
element in a specific amount of compound.
Example:
Calculate the mass of carbon in 82.0 g of propane,
C3H8.
1. Determine % composition of C3H8.
C3H8 = 44.0 g
36 g
C:
= 81.8 %
44 g
H: 8 g = 18.2 % H
44 g
2. Use conversion factor based on percent by mass
of carbon in ethane.
a. 81.8 % C means that for every 100 g
C3H8, 81.8 g will be C.
81.8 g C
82 g C3H8 x
= 67.1 g C
100 g C3H8
Check to see if 67.1/82 = 81.8 %
Problems:
Calculate the amount of hydrogen in:
a. 350 g C2H6.
b. 20.2 g NaHSO4
c. 124 g Ca(C2H3O2) 2
d. 378 g HCN
e. 100 g H2O
Empirical Formulas
Once you make a new compound in the laboratory, you
can determine the percent composition information. Once
you know the percent composition, you can determine the
empirical formula of the compound.
The empirical formula gives the lowest whole number
ratio of the atoms of the elements in a compound.
CO2 is an empirical formula because it is the lowest
whole-number ratio. N2H4 (an explosive) has an empirical
formula of NH2.
What is the empirical formula of a compound that is
25.9% N and 74.1% O?
1. Remember % composition means that in 100 g
of that compound each % equals that many
grams.
2. Change g to moles
1 mol N
25.9 g N x
= 1.85 mol N
14.0 g N
74.1 g O x
1 mol O
= 4.63 mol O
16.0 g O
N1.85O4.63 = mole ratio
Not empirical formula because needs
to be whole-numbers.
3.
Divide both molar values by smallest
value to give you a “1” for element with
smallest value.
1.85 mol N
= 1 mol N
1.85
4.63 mol O
= 2.50 mol O
1.85
Is N1O2.5 correct? No, not a whole number.
Just multiply both values by a number to make a
whole number.
1 x 2 = 2 mol N; 2.50 x 2 = 5 mol O
Now you have whole numbers.
empirical formula: N2O5
If grams are already given to you, just convert to moles.
Example:
Analysis of a compound indicates it contains 2.08 g K,
1.40 g Cr, and 1.74 g O. Find its empirical formula.
1 mol K
= .053 mol K
39.1 g K
1 mol Cr
1.40 g Cr x
= .027 mol Cr
52 g Cr
2.08 g K x
1.74 g O x
1 mol O
= .109 mol O
16.00 g O
Then divide by smallest mole number to get mole ratio.
Multiply if you need to.
Molecular Formula
Determining the empirical formula does not always tell
you the actual molecular formula.
An example is methanol and glucose.
CH2O = methanol (empirical and molecular)
C6H12O6 = glucose (molecular)
Same empirical; different molecular formula.
Well, then, how do you know if you have empirical or
molecular?
You need to know the molar mass of the compound.
The molecular formula is some multiple of the empirical
formula based on their masses.
(simplest formula)x = molecular formula,
where “x” = whole-number multiple
(simplest-formula mass)x = molecular-formula mass
Example:
The simplest formula of a compound containing
phosphorus and oxygen was found to be P2O5. The
molar mass of this compound is 283.889 g/mol. What
is the molecular formula of this compound?
(simplest-formula mass)x = molecular-formula mass
molecular-formula mass
x=
simplest-formula mass
283.889 g/mol
=
= 1.999
141.945 g/mol
(P2O5)2 = molecular formula
= P4O10