The Law of Combining Volumes of Gases: When two gases react, the volumes that combine are in a ratio of small whole numbers. The ratio of the volume of each product, if a gas, is also in the ratio of small whole numbers.

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The Law of Combining Volumes of Gases When two
gases react, the volumes that combine are in a
ratio of small whole numbers. The ratio of the
volume of each product, if a gas, is also in the
ratio of small whole numbers. Example 1
Liter of hydrogen 1 Liter of chlorine 2
Liters of Hydrogen Chloride 2 Liters of hydrogen
1 Liter of oxygen 2 Liters of Water 3 Liters
of hydrogen 1 Liter of nitrogen 2 Liters of
Ammonia Mass is always conserved but the volume
of a gas is not. Avogadros Hypothesis Equal
volumes of different gases contain the same
number of particles. The particles of a gas may
be atoms or molecules. One liter of hydrogen
one liter of chlorine one liter of hydrogen
chloride in terms of particles (read atoms or
molecules)
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Avogadros Law and the reaction of hydrogen and
oxygen to produce water 2 liters of hydrogen
1 Liter of oxygen -gt 2 Liters of water Dalton
H O -gt HO Simplicity of atoms Avogadro
2 H2 O2 -gt 2 H2O Equal volumes equal
particles Both equations are balanced. Which is
correct (if either?) Daltons hypothesis
provided no link between the relationships
between the volumes of the reactants and
products. Avogadros hypothesis provided a
direct link between the relationships between
the volumes of the reactants and products.
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(1) Avogadros Number (N0) is defined a the
number of atoms in exactly 12 g (0.012 kg) of
12C (2) Experimentally, N0 is found to be equal
to 6.02 x 1023 (3) The mass (expressed in
grams) of Avogadros number of atoms of an
element is equal to the relative atomic mass of
that element (4) A mole of a substance equals
the amount of that substance that contains
Avogadros number of atoms (element) or molecules
(compound)
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Lorenzo Romano Amedeo Carlo Avogadro, conte di
Quaregna e di Cerreto (1776 - 1856) Avogadros
number is 6.02 x 1023 (unlisted)

• An Avogadro's number of standard soft drink cans
  would cover the surface of the earth to a depth
  of over 200 miles.
• If you had Avogadro's number of unpopped popcorn
  kernels, and spread them across the United States
  of America, the country would be covered in
  popcorn to a depth of over 9 miles.
• If we were able to count atoms at the rate of
  10 million per second, it would take about 2
  billion years to count the atoms in one mole.

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Avogadros interpretation of gas
reactions H2 Cl2 -gt 2 HCl 1 L 1 L 2 L 2
H2 O2 -gt 2 H2O 2 L 1 L 2 L 3 H2 N2 -gt 2
NH3 3 L 1 L 2 L Daltons interpretation of
reactions H Cl -gt HCl H O -gt HO H N -gt H
N
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Conceptual basis of Avogadros Hypothesis The
particles of any gas all occupy the same space at
a given temperature and pressure, independent of
the composition of the gas. Thus, a given number
or particles (atoms or molecules) of a gas will
occupy the same space of same volume. One liter
of hydrogen gas, contains the same number of
hydrogen particles as the number of oxygen
particles in one liter of oxygen gas or the
numbe of nitrogen particles in one liter of
nitrogen gas. Consider the experiment 1 L of
hydrogen 1 L of chloring -gt 2 L of hydrogen
chloride Consider the interpretations H
Cl -gt 2 HCl Impossible! Dalton H2 Cl2 -gt 2
HCl Possible Avogadro
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Amazing consequence of Avogadros hypothesis The
molar volume of a gas is the volume of one mole
of the gas (at standard temperature and
pressure). This molar volume near room
temperature and atmospheric pressure is 22.4
L The weight of 22.4 L of any gas is the
molecular weight the gas!
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In 1757, being at sea in a fleet of 6 sail
bound against Louisburg, I observed the wakes of
two of the ships to be remarkably
smooth, while all the others were ruffled by
the wind, which blew fresh. Being puzzled with
the differing appearance, I at last pointed it
out to our captain, and asked him the meaning
of it? "The cooks," says he, "have, I
suppose, been just emptying greasy water
through the scuppers, which has greased the
sides of those ships a little," and this answer
he gave me with an air of some little contempt
as to a person ignorant of what everybody else
knew. It occurred to me the learned are apt to
slight too much the knowledge of the ancients
and the vulgar. This art of smoothing the
waves with oil is an instance of both! -Benjamin
Franklin, Philosophical Transactions of the
Royal Society of London, 1776
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Clapham Pond
"...at Clapham I observed a large pond very rough
with the wind. I fetched a cruet of oil and
dropt a little of it on the water. The oil,
though not more than a teaspoonful, produced an
instant calm over a space of several yards
square, and then spread amazingly till it filled
a quarter of the pond, perhaps half an acre, as
smooth as a looking glass. .... If a drop of
oil is put on a polished marble table, or on a
looking-glass that lies horizontally, the drop
remains in place, spreading very little. But
when put on water it spreads instantly many feet
around, becoming so thin as to produce the
prismatic colors, for a considerable space, and
beyond them so much thinner as to be invisible,
except in its effect of smoothing the
waves." ---Benjamin Franklin, letter to William
Brownrigg, November 7, 1773.
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This incredibly simple experiment provides a
means of understanding fundamental facts about
molecules and the forces between them. It even
leads to a means of determining molecular size
and shape! For example, if the teaspoon were 2
cc of oil and the area of a half an acre is
approximately 2000 m2, the film thickness
(volume/area) would be ca 10-7 cm (1 nanometer,
10 Å), which is right on the molecular dimensions
of an "olive oil" molecule!!! This is clearly
within the experimental uncertainty of measuring
acreage!
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Chapter 2 Stoichiometry Learning
Goals (1) How to translate the coefficients of
the atoms involved in balanced chemical equations
into moles of elements and compounds and how to
translate moles into mass of elements and
compounds. (2) For gases, how to translate the
coefficients of the atoms involved in balanced
chemical equations into moles of elements and
compounds and how to translate moles into volumes
of elements and compounds which are
gases. (3) How to determine the limiting reagent
in a reaction from the balanced chemical equation
and the available masses of the reagents.
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• Empirical Formulas and Molecular Formulas
• Empirical formula a formula which displays the
  simplest ratios of the number of atoms of
  different elements that make up a substance
• Molecular formula a formula which displays the
  exact number and kinds of atoms that are present
  in one molecule. The molecular formula is the
  compositional structure of a molecule
• Lewis structure a structure which displays how
  the atoms of a molecule are connected. The Lewis
  structure is the constitutional structure of a
  molecule
• (4) Molecular structure a three dimensional
  representation of how the atoms of a molecular
  are arranged in space. The molecular structure
  displays the configurational structure of atoms
  about each other in space.

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Percentage atomic composition from empirical or
molecular formulas Empirical or molecular
formula -gt composition of elements Example
H2O Need relative masses of H and O From
periodic table H 1.00, O 16.0 One mole of
H2O consists of 2 moles of H atoms and one mole
of O atoms. 2 moles of H atoms have a mass of 2
g and 1 mole of O atoms have a mass of 16. The
mass of one mole of water is 18 g. The percent
of H in water is 2/18 x 100 11 the percent of
O in water is 16/18 89 O.
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Section 2.2 Using balanced chemical
equations Hydrocarbon Oxygen -gt Carbon
dioxide Water What is the balanced chemical
equation for complete combustion of C4H10? ?
C4H10 ? O2 -gt ? CO2 ? H2O Answer (by
inspection and molecular weights of reactants and
products) One mol of C4H10 58.12 g one mol
of O2 32.00 g one mol of CO2 44.01 g one
mol of H2O 18.01 g. 2 C4H10 13 O2 -gt 8
CO2 10 H2O 116.2 g C4H10 416.0 g O2
-gt 352.1 g CO2 180.1 g H2O 2 mol of C4H10
13 mol of O2 8 mol CO2 10 mol
H2O Example of the law of mass balance. 116.2 g
416.0 g 352.1 g 180.1 g 532.2
g REACTANTS PRODUCTS SAME MASS