Transcript Chapter #2-Newest CPub

Lecture #4
Fundamental Chemical Laws
and
Dalton’s Atomic Theory
and
Cannizzaro’s Contribution
Chemistry 142 A
James B. Callis, Instructor
Winter Quarter, 2006
Some Definitions
• Matter- The ‘stuff’ of the universe. That which
occupies space.
• Substance - A material that cannot be separated by
physical means into two or more materials with
different properties.
• Element - A substance that that cannot be
decomposed into simpler substances by chemical
means.
• Compound – A substance containing two or more
chemical elements.
Three Laws that Led to the
Atomic Theory
• Law of Mass Conservation: The total mass of substances
does not change during a chemical reaction (Lavoisier).
• Law of Definite (or Constant) Composition: No matter
what its source, a particular chemical compound is
composed of the same elements in the same parts
(fractions) by mass (Proust).
• The Law of Multiple Proportions: When two elements
form a series of compounds, the masses of one element
that combine with a fixed mass of the other element are in
the ratio of small integers to each other (Dalton).
Conservation of Mass
• Matter is neither created nor destroyed in a
chemical reaction.
• In every chemical operation an equal quantity of
matter exists before and after the operation.
• Mass is conserved in a chemical reaction.
• Moreover, in chemical change, the mass of the
elements is conserved, element by element.
• Development of this law was made possible by the
analytical balance.
Problem 4-1: Definite Composition
Chemical analysis of a 9.07 g sample of calcium
phosphate shows that it contains 3.52 g of Ca.
How much Ca could be obtained from a 1.000 kg
sample?
Mass fraction Ca =
Mass Ca in 1.000 kg =
Problem4-2: Law of Multiple Proportions
If elements A and B react to form two compounds,
the different masses of B that combine with a fixed
mass of A can be expressed as a ratio of small whole
numbers.
Example: Nitrogen Oxides I & II
Cmpd Mass % Mass % Mass Ratio
N
O
O/N
I
46.68
53.32
II
30.45
69.55
Ratio of
Ratios
Dalton’s Atomic Theory
• Matter is composed of indivisible atoms.
• All atoms of a given chemical element are
identical in mass and in all other properties.
• Different chemical elements are composed of
different atoms of different masses.
• Atoms are indestructible. They retain their
identities in a chemical reaction.
• A compound forms from its elements through the
combination of atoms of unlike elements in small
whole number ratios.
Compounds and Molecules
The composition of a compound is shown by
its chemical formula. The symbol "CO2" for
the substance carbon dioxide indicates that it
contains two atoms of oxygen for every atom
of carbon. In most cases, the chemical formula
also defines a molecule. Here, the atoms in the
formula are linked together by attractive forces
strong enough to keep the group together for a
reasonable period of time.
The Theory’s First Challenge:
The Relative Masses of Atoms
• Dalton incorrectly assigned a mass of 8 to
elemental oxygen because he thought that
elemental hydrogen was monatomic (mass =1). He
concluded that the formula of water was HO.
• Gay-Lussac found that 2 volumes of hydrogen and
1 volume of oxygen combined to form 2 volumes
of water. This implied that the formula of water
was H2O.
However, another of Gay-Lussac’s results was
confusing: 2 L of water vapor were produced for
every 2 L of hydrogen gas that reacted. Here are the
expected and actual results:
Expected Results:
If water is HO: 1 L hydrogen gas + 1 L oxygen gas -> 1 L
water vapor
If water is H2O: 2 L hydrogen gas + 1 L oxygen gas -> 1 L
water vapor
Actual Result:
2 L hydrogen gas + 1 L oxygen gas -> 2 L water vapor
Implication: If is assumed that oxygen is monatomic, the
oxygen must split in half!!?
Gay – Lussac’s Law of Combining Volumes
The volumes of two reacting gases (at the
same temperature and pressure) are in the ratio
of simple integers. Moreover, the ratio of the
volume of each product gas to the volume of
either reacting gas is the ratio of simple
integers.
Avogadro’s Hypothesis
Equal volumes of different gases (at the same
temperature and pressure contain equal numbers
of particles.
Cannizzaro’s Contribution
• Cannizzaro analyzed many gaseous compounds
and showed that their chemical formulas could be
established with a consistent scheme that used
Avogadro’s hypothesis but avoided any extra
assumptions about molecular formulas.
• Many gaseous elements (hydrogen, oxygen,
nitrogen, fluorine, chlorine, etc.) did indeed
consist of diatomic molecules under ordinary
conditions.
Stoichiometry
• The study of mass relationships in
chemical compounds and chemical
reactions.
• The mass balance makes possible
the elucidation of stoichiometry.
The Atomic Theory Explains Stoichiometry
According to the atomic theory, the mass of one molecule of
a compound, mcpd, is related to the number and masses of its
constituent atoms as follows:
It is assumed that one molecule of the compound is comprised of N
types of atoms, identified by the subscript i. The number of atoms
of type i in the molecule is given the symbol niatom and the mass of
each type of atom is given the symbol miatom. According to the fifth
statement of the atomic theory, the niatoms are integers.
Note also that the above equation is equivalent to a statement of the
conservation of mass.
The Catch
The masses of individual atoms and the
molecules formed from them are vanishingly
small, on the order of 10-23 - 10-21 grams. The
laboratory balance can measure masses only as
small as milligrams. This forces the chemist to
work with gram amounts of compounds. How
can such measurements relate to and be used
to confirm the atomic theory?
The Solution
According to atomic theory, the mass of a sample of a
compound, mcpdsam is simply the mass of one molecule
of the compound, mcpd, multiplied by the number of
molecules in the sample, ncpdsam:
In Context
Newton:
F = ma
Einstein:
2
mc
Dalton:
E=
m
cpd
sam
n
N
cpd
sam
n
i 1
atom
i
m
atom
i
Working in Mass Per Cent Eliminates the Need to Know ncpdsam
The percent by mass of the ith atom in a compound
is given by the ratio:
The Law of Multiple Proportions
• When two elements form a series of
compounds, the masses of one element that
combine with a fixed mass of the other
element are in the ratio of small integers to
each other.
• Now we can make a table of calculations
using the symbolic notation developed
above.
How Dalton Predicts the Law of Multiple Proportions
Ratio of
Mass of B
that
combines
with a fixed
Mass of A
Ratio of
Ratios:
Using
value for
AiBn as
denomina
tor
Compd.Mass % A Mass % B
AiBn
i*M A/M AiBn n*M B/M AiBn (n*M B)/(i*M A)
1
AjBo
j*M A/M AjBo o*M B/M AjBo (o*M B)/(j*M A) (o*i)/(j*n)
AkBp
k*M A/M AkBp p*M B/M AkBp (p*M B)/(k*M A) (p*i)/(k*n)
These are ratios
of integers.
Problem 4-3: Multiple Proportions
A chemist obtains the following composition data for
four compounds of nitrogen and oxygen. Compound a
contains 63.65 % nitrogen and 36.35 % oxygen.
Compound b contains 46.68 % nitrogen and 53.32 %
oxygen. Compound c contains 30.45 % nitrogen and
69.55 % oxygen. Compound d contains 25.94 %
nitrogen and 74.06 % oxygen. (1) calculate the mass
of oxygen per gram of nitrogen in each compound. (2)
Calculate the ratio of ratios using the smallist ratio.
How do the ratio of ratios support Dalton’s atomic
theory?
Multiple Proportions Problem
Cmp
Ratio of
(symbolic Ratios
rep)
Amt N, Amt O, mO/mN
g
g
(from Dalton)
Ratio of
Ratios
(from masses)
N2O
(NiOn)
63.65
36.35
NO
(NjOo)
46.68
53.32
NO2
(NkOp)
30.45
69.55
N2O5
(NlOq)
25.94
74.06
Caution! There are limitations to the
information in Dalton’s interpretation of
the multiple proportions table.
• You can use Dalton’s equation to go from known
composition to predicted ratios.
• But you cannot go from the experimental ratios to
prediction of molecular composition (inverse
direction).
• Examination of the multiple proportions table
shows that for N (N>1) compounds, it provides N1 equations in 2N unknowns. For a table of two
compounds there is one equation in four
unknowns. For three compounds there are two
equations in six unknowns, etc.
Answers to Problems in Lecture #4
1. 388 g Ca
2. 2/1
3. 1:2:4:5