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The Mole
Mass Relationships and Avogadro’s Number
Relative Mass
It is possible to determine the mass of an atom without
knowing the mass of a single atom.
 This procedure involves comparing the masses of equal
numbers of atoms.
Ex: Oranges: 2160g = 3 = 0.600
Grapefruit: 3600g 5
 Since there are an equal number of oranges and
grapefruit, the mass of one orange is 3/5 or 0.600 that
of the mass of one grapefruit.
 0.600 is the Relative mass of the orange.
 Relative mass of any object is expressed by comparing it
mathematically to the mass of another object.
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The mass of an atom is called its atomic mass.
At first the masses of atoms were compared to
hydrogen, as it was the lightest atom.
The nitrogen was 14 times as heavy, and so on.
Later, Oxygen with a mass of 16 was used.
Eventually Carbon with a mass of 12.0000 was chosen
as the standard.
The masses of individual atoms are assigned a unit of
relative measurement known as the atomic mass unit
(amu).
The atomic mass unit is defined as 1/12 of the mass of a
carbon-12 atom.
Combining Volumes of Gases and
Avogadro’s Hypothesis
Joseph Louis Gay-Lussac (1778-1850) was a French
chemist who performed experiments to investigate how
gases combined to form compounds that were also
gases.
 He did much of his work with nitrogen and oxygen.
 For all compounds formed, the ratios of the volumes of
gases used were simple, whole-number ratios. (Table 41, p. 96)
 The volumes of gases that react to form each
compound can be expressed as a simple ratio: 2 to 1, 1
to 1, or 1 to 2.
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The importance of Gay-Lussac’s results was recognized by
Amadeo Avogadro, an Italian scientist.
 In 1811 Avogadro wrote what became known as
Avogadro’s Hypothesis: Equal volumes of gases (at the same
temperature and pressure) contain equal numbers of particles.
 The table contains data obtained from equal volumes of
several gases
 The relative mass of one atom of nitrogen (14) compared to
one atom of hydrogen is the same as the relative mass found
when comparing all of the molecules in a liter of each gas.
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Gas
Mass of 1L of
Gas
Mass of Gas Relative to
Hydrogen
hydrogen
0.08g
1
nitrogen
1.12g
14
oxygen
1.31g
16
fluorine
1.52g
19
chlorine
2.80g
35
How Many is a Mole?
A mole (mol) is simply the amount of a substance that
contains 6.02 x 1023 particles.
 6.02 x 1023 is known as Avogadro’s number.
 The particle can be anything: atoms, molecules, or
baseballs.
 Relative masses of atoms do not change when you
consider individual atoms or moles of atoms.
 One mole of carbon-12 atoms has a mass in grams that
equals the atomic mass, in amu’s, of a single atom of
carbon-12: 12.00 grams.
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Formula Calculations
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Actual formulas of compounds are determined by
laboratory analysis.
Experiments are done to measure the amount of each
element in the compound.
The interpretation makes use of molar masses
The analysis provides the simplest ratio of atoms in the
compound
Finding an Empirical Formula
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Empirical means based on experiment.
An empirical formula is one that is obtained from
experimental data and represents the smallest whole number
ratio of atoms in a compound.
CO2 represents one carbon atom for every two oxygen
atoms
A mole of CO2 has 6.02 X 1023 molecules. There are 6.02 X
1023 carbon atoms in a mole of CO2, and 2(6.02 X 1023)
oxygen atoms.
44g of CO2 contains 12g of carbon and 32g of oxygen
One molecule CO2
mass= 44amu
1 atom C
2 atoms O
mass=12amu
mass=32amu
One mole of CO2
Mass=44g
1mole C
2 moles O
Mass=12g
mass=32g
 To determine the formula for a compound, it is not
necessary to count the atoms in a single molecule.
 The information is obtained by finding the number of
moles of each element in a mole of the compound
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Example 1:
 A charcoal briquette of carbon has a mass of 43.2g. It is
burned and combines with oxygen and the resulting
compound has a mass of 159.0g. What is the empirical
formula for the compound?
 159.0g – 43.2g carbon = 115.8g oxygen
 Now find the number of moles of C and O in the
compound.
 43.2g C x 1mol C = 3.60 mol C
12.0g C
115.8g O x 1mol O = 7.24 mol O
16.0 g O
 There are 2.01 moles of oxygen for every 1.0 mole of
carbon (7.24mol O/3.60mol C=2.01mol O/1mol C)
 We can assume that the formula is CO2
Summary of Steps
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The mass of each element in a sample of the
compound is determined.
The mass of each element is divided by its molar mass
to determine the number of moles of each element in
the sample of the compound.
The number of moles of each element is divided by
the smallest number of moles to give the ratio of
atoms in the compound.
Example 2
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Charcoal is mixed with 15.53g of rust and heated in a covered
crucible to keep air out until all of the oxygen atoms in the rust
combine with carbon. When this process is complete, a pellet of
pure iron with a mass of 10.87g remains. Empirical formula for rust?
Mass of rust:
15.53g
Mass of pure iron:
10.87g
Mass of oxygen in rust:15.53g-10.87g=4.66g
10.87g Fe x 1mol Fe = 0.195 mol Fe
55.8g Fe
4.66g O x 1mol O = 0.291 mol O
16.0g O
 0.195mol Fe = 1.00 0.291mol O = 1.49 mol O/mol Fe
0.195mol Fe
0.195mol Fe
 Multiply both numbers to get a whole number ratio:
 Fe2O3
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Molecular Formula/Percent
Composition
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Molecular Formula: Always some multiple of the
empirical formula
Divide the molar mass of the compound by the molar
mass of the empirical formula. (see Ex. 4-14)
Percent Composition: comparison of the elements in a
compound by percentage, rather than by masses (Ex 417)
Molarity
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Many compounds are stored, measured, and used as
solutions.
Concentration describes how much solute is in a given
amount of solution.
% by mass or volume is a convenient way of expressing
concentration.
Chemists commonly describe concentration by
indicating the number of moles of solute dissolved in
each liter of solution.
Molarity is the concentration of a solution in moles per
litre.
 The symbol for molarity is M.
 4.90M solution of NaCl means that there are 4.90
moles of NaCl in one litre of the solution.
Example:
 How many moles of HCl are contained in 1.45L of a
2.25M solution?
2.25M = 2.25 mol/ 1L of solution
1.45L X 2.25mol = 3.26 mol HCl
1L soln
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