Section 9.1
The Mole: A Measurement of
Matter
Unit 7:
(Chapter 9)
Chemical
Quantities
Describe how Avogadro’s number is
related to a mole of any substance.
� Calculate the mass of a mole of any
substance.
�
What is a Mole?
An Animal or What?
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GHS Honors Chem
Moles (abbreviated: mol)
What is a Mole?
•
•
•
•
•
You can measure mass,
or volume,
or you can count pieces.
We measure mass in grams.
We measure volume in liters.
• We count pieces in MOLES.
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What are Our Representative, or
Basic Particles?
• Defined as the number of carbon
atoms in exactly 12 grams of carbon12.
• 1 mole is 6.02 x 1023 particles.
• Treat it like a very large dozen
• 6.02 x 1023 is called Avogadro’s
number.
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Let’s Check your Understanding?
•
• They are the smallest pieces of a
substance.
• For a molecular compound: it is the
molecule.
• For an element: it is the atom.
• Remember the 7 diatomic elements (made of
molecules)
• Br I N Cl H O F
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How many oxygen atoms are in each
of the following molecules?
•
•
•
•
CaCO3
Al2(SO4)3
O2
How many oxygen atoms in 1 mole of
the following?
•
•
•
CaCO3
Al2(SO4)3
O2
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1
Mole Questions
• How many molecules of CO2 are
there in 4.56 moles of CO2 ?
• How many moles of water is 5.87 x
1022 molecules?
• How many atoms of carbon are there
in 1.23 moles of C6H12O6 ?
What’s the Significance of the Mole?
Let’s look at Molar Masses …
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Gram Atomic Mass (gam)
• Equals the mass of 1 mole of an element in
grams
• 12.01 grams of C has the same number of atoms
as 1.008 grams of H and 55.85 grams of iron.
• Each of these has 6.02 x 1023 atoms
• We can write this as
12.01 grams C = 1 mole C
• We now have a Factor of 12.01 g/mole of Carbon
• Now we can count things by weighing them.
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Is There a Similar Measure for
Compounds?
• in 1 mole of H2O molecules there are two
moles of H atoms and 1 mole of O atoms
• To find the mass of one mole of a
compound
• determine the moles of the elements they
have
• Find out how much they would weigh
• add them up
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Examples Using GAM
•
•
•
•
•
•
How much would 2.34 moles of carbon
weigh?
How much would 1.50 moles of bromine
weigh?
How many moles of magnesium is 24.31 g
of Mg?
How many moles are in 56 grams of
Nitrogen?
How many atoms of lithium is 1.00 g of Li?
How much would 3.45 x 1022 atoms of U
weigh?
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Gram Molecular Mass (gmm)
• What is the mass of one mole of CH4?
• 1 mole of C = 12.01 g
• 4 mole of H x 1.01 g = 4.04g
• 1 mole CH4 = 12.01 + 4.04 = 16.05g
• The Gram Molecular Mass (gmm) of CH4
is 16.05g/mol
• this is the mass of one mole of a molecular
compound.
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2
Mole-Mass and Mole-Volume
Relationships
Use the molar mass to convert between
mass and moles of a substance.
� Use the mole to convert among
measurements of mass, volume, and
number of particles.
�
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• Molar mass is the generic term for the
mass of one mole of any substance
(in grams)
• The same as: 1) gram molecular
mass, 2) gram formula mass, and 3)
gram atomic mass- just a much
broader term.
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Calculate the molar mass of the
following:
•
•
•
•
•
•
Another New Term … Molar Mass
Na2S
N2O4
C
Ca(NO3)2
C6H12O6
(NH4)3PO4
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Molar Mass
The number of grams of 1 mole of
atoms, ions, or molecules.
� Let’s take a closer look at making
conversion factors to change grams
of a compound to moles of a
compound.
�
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For example
�
How many moles is 5.69 g of NaOH?
For example
�
How many moles is 5.69 g of NaOH?
�
5.69 g�
�
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�
�
�
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For example
�
How many moles is 5.69 g of NaOH?
mole �
�
5.69 g�
�
�
g �
� need to change grams to moles
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For example
�
How many moles is 5.69 g of NaOH?
mole �
�
5.69 g�
�
�
g �
� need to change grams to moles
� for NaOH
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For example
�
How many moles is 5.69 g of NaOH?
mole �
�
5.69 g�
�
�
g �
� need to change grams to moles
� for NaOH
� 1mole Na = 22.99g 1 mol O = 16.00 g 1
mole of H = 1.01 g
GHS Honors Chem
For example
�
How many moles is 5.69 g of NaOH?
mole �
�
5.69 g�
�
�
g �
� need to change grams to moles
� for NaOH
� 1mole Na = 22.99g 1 mol O = 16.00 g 1
mole of H = 1.01 g
� 1 mole NaOH = 40.00 g
GHS Honors Chem
For example
�
How many moles is 5.69 g of NaOH?
1 mole �
�
5.69 g�
�
�
40.00 g �
� need to change grams to moles
� for NaOH
� 1mole Na = 22.99g 1 mol O = 16.00 g 1
mole of H = 1.01 g
� 1 mole NaOH = 40.00 g
GHS Honors Chem
For example
�
How many moles is 5.69 g of NaOH?
�
5.69 g�
�
1 mole �
� = 0.142 mol NaOH
40.00 g �
need to change grams to moles
� for NaOH
� 1mole Na = 22.99g 1 mol O = 16.00 g 1
mole of H = 1.01 g
� 1 mole NaOH = 40.00 g
�
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Examples Using Molar Mass
• How many moles is 4.56 g of CO2?
• 0.104 moles
• How many grams is 9.87 moles of H2O?
• 178 grams
• How many molecules is 6.8 g of CH4?
• 2.55 x 1023 molecules
• 49 molecules of C6H12O6 weighs how
much?
Worksheets on Moles, Mass,
Avogadro’s Number, &
Moles/Molecules/Grams
• 1.47 x 10-20 grams
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What is Standard
Temperature and Pressure?
What About Gases?
• Many of the chemicals we deal with are
gases.
• They are difficult to weigh.
• Need to know how many moles of gas we
have.
• Two things effect the volume of a gas
• Temperature and pressure
• We need to compare them at the same
temperature and pressure.
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• 0ºC (or 273 K) and 1 atm pressure is
abbreviated as STP
• At STP 1 mole of gas occupies 22.4
L
• Called the molar volume
• 1 mole = 22.4 L of any gas at STP
@ STP: 1 mole/22.4 Liters
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Density of a gas
Molar Volume Problems
� What is
the volume of 4.59 mole of
CO2 gas at STP?
� 103
Liters
� How
many moles is 5.67 L of O2 at
STP?
� 0.253
moles
� What is
the volume of 8.8 g of CH4
gas at STP?
� 12.3
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Liters
�
D=m/V
� for a
gas the units will be g / L
� We can
determine the density of any gas
at STP if we know its formula.
� To find the density we need the mass
and the volume.
� If you assume you have 1 mole, then the
mass is the molar mass (from PT)
� At STP the volume is 22.4 L.
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“Molar Density” Problems
� Find
the molar density of CO2
at STP.
�
44 grams / 22.4 L = 1.96 g/L
� Find
the molar density of CH4
at STP.
�
16 grams / 22.4 L = 0.714 g/L
Can we Find the Molar Mass,
given the density of 1 Mole of Gas
at STP?
Pretend you have 1 mole at STP, so V =
22.4 L.
� Rearranging, MM = D x MV
� MM is the molar mass of 1 mole, and MV
is the Molar Volume (22.4 L at STP).
� What is the molar mass of a gas with a
density of 1.964 g/L?
�
�
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44.0 grams
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Have We Learned Anything??
�
These four items are all equal:
a) 1 mole
b) molar mass (in grams)
c) 6.02 x 1023 representative particles
d) 22.4 L at STP
Thus, we can make conversion factors
from them.
Grams/mole
Is There an Easy Way to Remember
these Mole Conversions?
Introducing Moletown!
6.02 x 1023 molecules/mole
1 mole/22.4 L 6.02 x 1023 molecules/22.4 L
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GHS Honors Chem
Mole Town
Molar
Mass
The Mole
X
Gram
Town
X
X
Volume
City (L)
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Avacado’s
Number
22.4 L/mole
Particle City
(molecules)
We Can use
Moletown for all of
our conversions!
The Mole and Volume, the
Comprehensive Mole (1-46), &
Mixed-Mole Worksheets
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6
Percent Composition and
Chemical Formulas
Calculating Percent Composition
of a Compound
�
�
Part
whole
Calculate the percent composition of a
substance from its chemical formula or
experimental data.
� Derive the empirical formula and the
molecular formula of a compound from
experimental data.
�
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Like all percent problems:
Find the mass of each component,
� then divide by the total mass.
�
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What is the Percent
Composition?
�
x 100 %
Calculate the percent composition of
a compound that is 29.0 g of Ag with
4.30 g of S.
= 29.0g/(29.0g + 4.30g) x 100%
= 87.1 % Ag
What is the Percent
Composition?
�
8.20 grams of Magnesium makes up
60.3% composition of a compound of
Mg and Oxygen. What is the weight
of Oxygen in the sample?
x
0.603 = 8.20 g/(8.20 g + )
x = 5.4 grams
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Finding the Percent
Composition from the Formula
Finding the Percent
Composition from the Formula
Use the ratio from the formula, and
assume that you have 1 mole of each
element.
� Then you would use the Molar Mass
of each element to determine the
percent composition of each.
� For example: What is the percent
composition of hydrogen in Methane,
CH4?
�
What is the percent composition of
hydrogen in Methane, CH4?
�
1 mole of Carbon = 12 grams
There’s 1 C, so the mass of C = 12 grams
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�
�
�
1 mole of Hydrogen = 1.01 grams
There are 4 H’s, so the total mass of H is 4.04
grams.
�
% C = [12 g/(16.04 g)] x 100% = 74.8 %
�
7
Percent Composition As a
Conversion Factor
More Percent Composition
Problems
�
�
Calculate the percent composition of
Hydrogen & Carbon in C2H4?
�
�
C: 85.6 % H: 14.4%
Calculate the percent composition of
each element in Aluminum
carbonate?
�
Al: 23.1 %
For Example … Sulfur makes up 26.7 % of the mass
of NaHSO4. What is the mass of the sulfur in 16.8
grams of NaHSO4?
Mass of Sulfur = 16.8 grams NaHCO3 x (26.7 grams S/100 grams NaHCO3)
Mass of Sulfur = 4.49 grams
C: 15.4 % O: 61.5 %
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Percent Composition As a
Conversion Factor
A bit more difficult:
Calculate the mass of carbon in 82.0 grams of C3H8?
The Solution:
% Composition of C in C 3H8 = (3x12g)/[(3x12g)+(8x1g)]
% Composition of C in C 3H8 = 81.8 %
Mass of Carbon = 82.o grams C 3H8 x (81.8 grams C/100 grams C 3H8)
Mass of Carbon = 67.1 grams
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And the Last Tidbit in
Chapter 7:
The Empirical Formula vs.
the Molecular Formula
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We can also use the percent as a conversion
factor:
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Percent Composition Problems for
those times when you have
difficulty getting to sleep
Please try these problems in the textbook:
� Page 191, practice problems 31, 32, 33, & 34
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The Empirical Formula is
the lowest whole number
ratio of elements in a
compound
CH2
vs.
The Molecular Formula is
the actual ratio of elements
in a compound
C2H4
C8H16
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The Molecular Formula and
Empirical Formula can be the
same
H2O CO
2
CO
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How Can we Calculate the
Empirical Formula?
� The
Empirical Formula is not just the
ratio of atoms, it is also the ratio of
moles of atoms.
� In 1 mole of CO2 there is 1 mole of
carbon and 2 moles of oxygen.
� In one molecule of CO2 there is 1 atom
of C and 2 atoms of O.
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Calculating the Empirical
Formula
can get a ratio from the percent
composition.
� Assume you have a 100 g.
� The percentages become grams.
� Convert grams to moles.
� Find lowest whole number ratio by
dividing by the smallest.
Calculating the Empirical
Formula
� We
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Confused ?
Let’s try an example …
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Here’s an Example
Here’s an Example
The Problem: Calculate the empirical
formula of a compound composed of 38.67 %
C, 16.22 % H, and 45.11 %N.
The Problem: Calculate the empirical
formula of a compound composed of 38.67 %
C, 16.22 % H, and 45.11 %N.
1. Assume 100 g so
• 38.67 g C
• 16.22 g H
• 45.11 g N
Now, convert to moles …
GHS Honors Chem
1. 38.67 g C x 1mol C = 3.220 mole C
12.01 g C
2. 16.22 g H x 1mol H
= 16.09 mole H
1.01 g H
3. 45.11 g N x 1mol N = 3.219 mole N
14.01 g N
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Here’s an Example
1. 38.67 g C x 1mol C = 3.220 mole C
12.01 g C
2. 16.22 g H x 1mol H
= 16.09 mole H
1.01 g H
3. 45.11 g N x 1mol N = 3.219 mole N
14.01 g N
Find the lowest whole number ratio:
1. Mole C = 3.220 / 3.219 = 1
2. Mole H = 16.09 / 3.219 = 5
3. Mole N = 3.219 / 3.219 = 1
C1H5N1
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More Examples
A compound is 43.64 % P and 56.36
% O. What is the empirical formula?
Moles of P = 43.64 g/ 31 g/mol
= 1.40 mol
Moles of O = 56.36 g / 16 g/mol
= 3.50 mol
Dividing by 1.40, we get P1O2.5
Multiply by 2 to get P2O5
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More Examples
Caffeine is 49.48% C, 5.15% H,
28.87% N and 16.49% O. What is its
empirical formula?
Moles of C = 4.12
Moles of H = 5.10
• Moles of N = 2.06
• Moles of O = 1.03
•
A Slightly Different Twist?
What is the Empirical Formula of a
compound that has 1.04g K, 0.70g Cr,
and 0.86g O?
Hint: Convert to % first, then follow previous example …
•
C4H5N2O
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Can I Go from Empirical to
Molecular Formulas?
Absolutely
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Ans. K2CrO4
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Can I Go from Empirical to
Molecular Formulas?
� Since
the empirical formula is the
lowest ratio, the actual molecule
would weigh more.
� By a whole number multiple.
� Divide the actual molar mass by
the empirical formula mass.
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Can I Go from Empirical to
Molecular Formulas?
� Caffeine
has a molar mass of 194 g.
what is its molecular formula?
� The
Empirical Formula for Caffeine is
C4H5N2O
� The empirical formula mass is 97 grams.
� 194 g / 97 g = 2
� The molecular formula is 2 x the Empirical
formula, or C8H10N4O2.
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Here’s Another …
�
A compound is known to be composed of
71.65 % Cl, 24.27% C and 4.07% H. Its
molar mass is known (from gas density) to
be 98.96 g. What is its molecular formula?
� Moles
of Cl = 71.65 g / 35.45 g/mol = 2
of C = 24.27 g / 12 g/mol = 2
� Moles of H = 4.07 g / 1 g/mol = 4
� Empirical Formula = CH2Cl
� Empirical Molar Mass = 49.45 g / mol
� 98.96 g / 49.45 g = 2
� Molecular Formula = C2H4Cl2
� Moles
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Percent Composition &
Molecular Formula Worksheet
Chapter 7 Practice Problems
Worksheet
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