Chapter 7
Moles and Mass
Relationships in
Chemical Reactions
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Atomic mass
the mass of an atom in atomic mass
units (amu)
Micro World
atoms & molecules
Macro World
grams
Standard:
1 atom 12C = 12 amu
H = ______amu
O = ______amu
Average atomic mass (6.941)
Formula mass
is the sum of the atomic masses of all
the atoms in a compound, in atomic
mass units (amu)
Calculate the formula mass for SO2
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
1 molecule SO2 = 64.07 amu
Calculate the formula mass for:
H2O = ________ amu
CO2 = ________ amu
NO!!
Need to be able to convert to
macroscale (grams)
Discovery of the MOLE

Amedeo Avogadro
•Avogadro’s number (NA)
6.02 × 1023
— is the number of
particles in exactly one
mole of a pure substance.
How do we measure an amount?
The mole is the SI unit for the amount of substance.
mole (mol) is the amount of a substance that
contains as many particles as there
are atoms in exactly 12.00 grams of 12C
1 dozen = 12
1 mol = 6.0221367 x 1023
Avogadro’s number (NA)
Scientists use the mole as a unit of measure
• AMU is not practical to use
• Can’t measure individual atoms very easily
• Measuring in grams is desirable
• Ratio of atoms is transferred between amu and
grams
For any element
atomic mass (amu) = molar mass (grams)
1 mole C atoms = 6.022 x 1023 atoms
1 C atom = 12.00 amu
1 mole C atoms = 12.00 g C
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
Molar mass for H = ______________ g
Molar mass for O = ______________ g
Relating Mass to Numbers of Atoms
Molar mass
is the sum of the atomic masses (in grams) in
a molecule.
•Units:
g/mol
1S
2O
SO2
32.07 grams
+ 2 x 16.00 grams
64.07 grams
SO2
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
For any molecule
formula mass (amu) = molar mass (grams)
Calculate the molar mass for:
H2O = ________ g/mol
CO2 = ________ g/mol
Mole Conversions
Mass to Mole Conversions
MOLAR MASS
11.2 g NaCl x 1 mol NaCl
58.44 g NaCl
Na: 1 X 22.99 = 22.99
Cl: 1 X 35.45 = 34.45
58.44 g/mol
0.192 mol NaCl
Mole to Mass Conversions
MOLAR MASS
3.2 mol Zn(NO3)2
Zn: 1 X 65.39 = 65.39
N: 2 X 14.07 = 28.14
O: 6 X 16.00 = 96.00
189.53g/mol
x
189.53 g Zn(NO3)2
1 mol Zn(NO3)2
606 g Zn(NO3)2
Particle to Mole Conversions
AVOGADRO’S NUMBER
8.74 x 1023 atoms CaCO3 x
1 mol CaCO3
6.02 x 1023 atoms CaCO3
1.45 mol CaCO3
Mole to Particle Conversions
AVOGADRO’S NUMBER
0.36 mol Al
x
6.02 x 1023 atoms Al
1 mol Al
2.2 x 1023 atoms Al
Moles and Gases
The conditions 0 0C and 1 atm are
called standard temperature and
pressure (STP).
Define Molar Volume
• Experiments show that
at STP, 1 mole of an
ideal gas occupies 22.4 L
Volume to Moles
Molar Volume
22.4 L/ mol
Volume
of Gas
1.0 L CO2
X
1 mol CO2
22.4 L CO2
.045 mol CO2
Moles to Volume
Molar Volume
22.4 L/ mol
Volume
of Gas
.38 mol He x 22.4 L He
1 mol He
8.5 L He
MASS
Mole Map
MOLE
6.02 x 1023
PARTICLES
VOLUME
What conversion
factor do I
use???
Multistep Conversions
250 g C12H22O11
X
1 mol C12H22O11
342.34 g C12H22O11
C: 12 X 12.01 = 144.12
H: 22 X 1.01 = 22.22
O: 11 X 16.00 = 176.00
342.34 g/mol
1 mol C12H22O11 X 6.02 x 1023 molec. C12H22O11
342.34 g C12H22O11 1 mol C12H22O11
4.4 x 1023 molecules C12 H22 O11
Formulas Express Composition
• Any sample of compound has many atoms and ions…
the formula gives a ratio of those atoms or ions.
Percent Composition:
 The
mass of each element in a
compound compared to the
entire mass of the compound
and multiplied by 100 percent.
There are 2 ways to determine
percent composition of a
compound:
• From a chemical formula
• From experimental data
Percent composition
of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n = the number of moles of the element in 1
mole of the compound
Find the percent composition of the elements
in water (H2O).
H: 2 x 1.01 = 2.02
O: 1 x 16.00 = 16.00
18.02 g/mol
H:
2 x1.01 g/mol H
18.02 g/mol H2O X 100 = 11.21 % H
O:
16.00 g/mol O
18.02 g/mol H2O
X 100 = 88.79 % O
Find the percent composition of the elements
in Al2(SO4)3.
Calculate Total Mass:
Al: 2 x 26.98 = 53.96
S: 3 x 32.07 = 96.21
O: 12 x 16.00 = 192.00
342.17 g/mol
Find the percent composition of the elements
in Al2(SO4)3. - continued.
Al :
53.96 g/mol Al
X 100 = 15.77 % Al
342.17 g/mol Al2(SO4)3
S:
96.21 g/mol S
X
100
=
28.12
%
S
342.17 g/mol Al2(SO4)3
O:
192.00 g/mol O
342.17 g/mol Al2(SO4)3 X 100 = 56.11 % O
Find the percent composition of a
compound that contains 0.9480 g of C,
0.1264 g of O, and 0.0158 g H.
C = 0.9480 g
O = 0.1264 g
H = 0.0158 g
1.0902 g
C:
0.9480 g
1.0902 g
X 100 = 86.96 %
O:
0.1264 g
1.0902 g
X 100 = 11.59 %
H:
0.0158 g
1.0902 g
X 100 =
1.45 %
Remember Empirical Formula:
A chemical formula that gives the
simplest whole-number ratio of the
elements in the formula.
- Subscripts are used for these ratios.
Example Problem
Determine the empirical formula
of a compound found to have
13.5 g of Ca, 10.8 g of O, and
0.675 g of H.
1.
2.
3.
4.
If given %, assume 100 g; % to mass
Convert mass to mole
Divide by the smallest to get ratio
Multiply ‘til whole if less than .9 or greater
than .1
Ca :
13.5 g Ca x
10.8 g O x
1 mol
16.00 g
= 0.675 mol O
0.675 g H x
1 mol
1.01 g
= 0.668 mol H
O:
H:
1 mol = 0.337 mol Ca
40.08 g
.337mol
Ca :
.337 mol
= 1 Ca
0.675 mol
O : 0.337 mol
=2O
0.668 mol
H : 0.337 mol
=2H
Writing the complete formula:
a) Round to whole numbers if less than .1 and/or
greater than .9
b) Put parentheses around polyatomic ions
c) Re-write the final formula.
Ca(OH)2
Remember Molecular Formula:
A chemical formula that gives the
actual number of the elements in
the molecular compound.
Example:
C2H4
C6H12O6
NOT CH2
NOT CH2O
Comparing Empirical and Molecular Formulas
• There is a direct relationship between
empirical and molecular formulas.
• There is a direct relationship between the
empirical formula mass and the molecular
formula mass.
FIND THE COMMON MULTIPLE!
The correct ratio can be found by:
dividing the experimental formula mass by
the empirical formula mass
Example Problem
The empirical formula of a compound of
phosphorus and oxygen was found to be P2O5.
Experimentation shows that the molar mass of
this actual compound is 283.89 g/mol. What is
the compound’s molecular formula?
1. Find Molar Mass of empirical formula.
P: 2 x 30.97 = 61.94
O: 5 x 16.00 = 80.00
141.94 g/mol
2. Divide the experimental formula mass by
the empirical formula mass.
283.89 g/mol
141.94 g/mol
= 2.00
3. Multiply the subscripts by
the common multiple.
2 × (P2O5) = P4O10