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The Mole and Molar Mass
• Molar mass is the mass of one mole of a
substance.
• Molar mass is numerically equal to atomic mass,
molecular mass, or formula mass. However …
• … the units of molar mass are g/mol.
• Examples:
1 atom Na = 22.99 u
1 mol Na = 22.99 g
1 molecule CO2 = 44.01 u
1 mol CO2 = 44.01 g
1 formula unit KCl = 74.56 u
1 mol KCl = 74.56 g
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We can read formulas
in terms of moles of atoms or ions.
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Molecular Masses and
Formula Masses
• Molecular mass: sum of the masses of the
atoms represented in a molecular formula.
– Simply put: the mass of a molecule.
– Molecular mass is specifically for molecules.
– Ionic compounds don’t exist as molecules; for
them we use …
• Formula mass: sum of the masses of the
atoms or ions present in a formula unit.
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Determining the Formula/Molar
Mass of Ammonium Sulfate
The molar mass of ammonium sulfate is _______ g mol-1
The molar mass of acetic acid (CH3COOH) is ______ g mol-1
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Conversion Factors involving Mass, Moles, and
Number of Atoms/Molecules
1 mol CH3COOH = 6.022 × 1023 CH3COOH molecules ≡ 60.05 g CH3COOH
We can use these equalities to construct conversion factors:
6.022 × 1023
1
6.022 × 1023
molec.
mol
(CH3COOH)
mol
molec.
(CH3COOH)
60.05
g
mol
1
mol
60.05
g
CH3COOH
CH3COOH
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Mass Percent Composition
from Chemical Formulas
The mass percent composition of a compound
refers to the mass proportion of the
constituent elements:
g element
% element = ––––––––––– × 100
g compound
OR …
#grams of the element
per 100 grams of the compound.
X g element
X % element = ––––––––––––––
100 g compound
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Percentage Composition of Butane
g/mol
g/mol
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Elemental Analysis ‫אנליזת יסודות‬
• … is one method of determining empirical
formulas in the laboratory.
• This method is used primarily for simple organic
compounds (that contain carbon, hydrogen,
oxygen).
– The organic compound is burned in oxygen.
– The products of combustion (usually CO2 and H2O)
are weighed.
– The amount of each element is determined from the
mass of products.
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Elemental Analysis Setup
… H2O, which is absorbed
by MgClO4, and …
The sample is
burned in a
stream of oxygen
gas, producing …
… CO2, which is
absorbed by NaOH.
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Elemental Analysis (cont’d)
If our sample
were CH3OH,
every two
molecules of
CH3OH …
… would give two
molecules of CO2 …
… and four
molecules of H2O.
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Elemental analysis of butane (cooking gas) gave the following mass
composition:
C – 82.66%, H – 17.34%.
a) Determine the empirical formula of butane.
b) The molar mass of butane is 58.12 g/mol; what is the molecular
formula?
Determine the empirical formula of phenol (a general disinfectant)
from its elemental analysis:
C – 76.57%
H – 6.43% H.
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Writing Chemical Equations
• A chemical equation is a shorthand description of a
chemical reaction, using symbols and formulas to represent
the elements and compounds involved.
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Writing Chemical Equations
• Sometimes
additional
information
about the
reaction is
conveyed in
the equation.
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Balancing chemical equations ‫איזון משוואות‬
• A balanced equation must conform to the law of
conservation of mass!
‫חוק שימור המסה‬
• To balance an equation we need to adjust the
stoichiometric coefficients ‫מקדמים סטויכיומטריים‬
Balance the following equations:
H2 + O2 → H2O
C2H6 + O2 → CO2 + H2O
Fe + O2 → Fe2O3
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Guidelines for Balancing
Chemical Equations
• If an element is present in just one compound on each side of
the equation, try balancing that element first.
• Balance any reactants or products that exist as the free element
last.
• In some reactions, certain groupings of atoms (such as
polyatomic ions) remain unchanged. In such cases, treat these
groupings as a unit.
• At times, an equation can be balanced by first using a
fractional coefficient(s). The fraction is then cleared by
multiplying each coefficient by a common factor.
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Stoichiometric Equivalence
and Reaction Stoichiometry
• A stoichiometric factor or mole ratio is a conversion
factor obtained from the stoichiometric coefficients in
a chemical equation.
‫יחס סטויכיומטרי‬
• In the equation: CO(g) + 2 H2(g) → CH3OH(l)
1 mol CO
1 mol CO
1 mol CO
1 mol CO was
–––––––––
–––––––––––––
reacted per
consumed per 1 mol
2 mol H2
1 mol CH3OH
2 mol H2
CH3OH that evolved
2 mol H2
–––––––––
1 mol CO
2 mol H2
reacted per
1 mol CO
1 mol CH3OH
–––––––––––––
2 mol H2
1 mol CH3OH was
formed for each 2 mol
of H2 that reacted
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Outline of Simple Reaction Stoichiometry
Stoichiometric factor
(or: mole ratio)
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When 0.105 mol propane is burned in an
excess of oxygen, how many moles of oxygen
are consumed? The reaction is
C3H8 + 5 O2 → 3 CO2 + 4 H2O
Outline of Stoichiometry Involving Mass
To our simple
stoichiometry
scheme …
… and a
conversion to
mass at the end.
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… we’ve added a
conversion from
mass at the
beginning …
Substances A and B
may be two reactants,
two products, or
reactant and product.
Think: If we are
given moles of
substance A initially,
do we need to convert
A to grams?
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The final step in the production of nitric acid involves the reaction of nitrogen
dioxide with water; nitrogen monoxide is also produced.
How many grams of nitric acid are produced for every 5 kg of nitrogen dioxide
that reacts?
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Summary of Concepts
• Mass percent composition can be used to calculate the
empirical formula.
– Additional information on the molecular weight of the material
allows the calculation of the molecular formula
• A balanced equation must conform to the law of
conservation of mass.
• Calculations involving reactions use stoichiometric factors
based on stoichiometric coefficients in the balanced
equation.
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