C142 B (Campbell / Callis)
Chapter 2: The Components of Matter
2.1: Elements, Compounds, and Mixtures: An Atomic
Overview
2.2: The Observations That Led to an Atomic View
of Matter
2.3: The Observations That Led to the Nuclear Atom
Model
2.4: The Atomic Theory Today
2.5: Elements: A First Look at the Periodic Table
2.6: Compounds: Introduction to Bonding
2.7: Compounds: Formulas, Names, and Masses
2.8: Mixtures: Classification and Separation
Fig.2.1
Definitions for Components of Matter
Pure Substances - Their compositions are fixed! Elements and
compounds are examples of Pure Substances.
Element - Is the simplest type of substance with unique physical and
chemical properties. An element consists of only one type
of atom. It cannot be broken down into any simpler
substances by physical or chemical means.
Molecule - Is a structure that is consisting of two or more atoms that
are chemically bound together and thus behaves as an
independent unit.
Compound - Is a substance composed of two or more elements that
are chemically combined.
Mixture - Is a group of two or more elements and/or compounds that
are physically intermingled.
Fig.2.2
Fig2.4
Laws of Mass Conservation &
Definite Composition
Law of Mass conservation: The total mass of substances
does not change during a chemical reaction.
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.
Like Sample Problem 2.1
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 =
Like Sample Problem 2.1
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 = 3.52 g Ca / (9.07 g total) = 0.388
(i.e., 38.8% Ca by mass in ANY sample of compound)
Mass Ca in 1.000 kg =
(1.000 kg total)x(0.388 g Ca/ g total) = 0.388 kg Ca
= 388 g Ca
Fig.2.5
Mass Fraction and Mass %
Mass of Red Balls =
Mass Fraction Red =
Mass % Red =
Mass Fraction Purple =
Similarly, mass fraction yellow =
Check:
Mass Fraction and Mass %
Mass of Red Balls = 3 balls x 3.0 g/ball = 9.0 g
Mass Fraction Red = 9.0 g / 16.0 g total = 0.56
Mass % Red = 0.56 x 100% = 56% red
Mass Fraction Purple =
2 balls x 2.0 g/ball / (16.0 g total) = 0.25 = 25%
Similarly, mass fraction yellow = 3x1.0/16.0 = 0.19
Check: 56% + 25% + 19% = 100%
Mass Percent Composition of Na2SO4
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Elemental masses
2 x Na = 2 x 22.99 = 45.98
1xS=
4xO=
Check
% Na + % S + % O = 100%
Percent of each Element
% Na = Mass Na / Total mass x 100%
% Na =
% S = Mass S / Total mass x 100%
%S=
%O=
%O=
Mass Percent Composition of Na2SO4
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Elemental masses
Percent of each Element
2 x Na = 2 x 22.99 = 45.98
1 x S = 1 x 32.07 = 32.07
4 x O = 4 x 16.00 = 64.00
142.05
Check
% Na = Mass Na / Total mass x 100%
% Na = (45.98 / 142.05) x 100% =32.37%
% S = Mass S / Total mass x 100%
% S = (32.07 / 142.05) x 100% = 22.58%
% O = Mass O / Total mass x 100%
% O = (64.00 / 142.05) x 100% = 45.05%
% Na + % S + % O = 100%
32.37% + 22.58% + 45.05% = 100.00%
Calculating the Mass of an Element in a
Compound: Ammonium Nitrate
How much Nitrogen is in 455 kg of Ammonium Nitrate?
Ammonium Nitrate = NH4NO3
Therefore mass fraction N:
28.02 g Nitrogen
=
80.052 g Cpd
Mass N in sample =
The Formula Mass of Cpd is:
4 x H = 4 x 1.008 = 4.032 g
2 x N = 2 X 14.01 = 28.02 g
3 x O = 3 x 16.00 = 48.00 g
80.052 g
Calculating the Mass of an Element in a
Compound: Ammonium Nitrate
How much Nitrogen is in 455 kg of Ammonium Nitrate?
Ammonium Nitrate = NH4NO3
The Formula Mass of Cpd is:
4 x H = 4 x 1.008 = 4.032 g
2 x N = 2 X 14.01 = 28.02 g
Therefore mass fraction N:
3 x O = 3 x 16.00 = 48.00 g
28.02 g Nitrogen
80.052 g
= 0.35002249 g N / g Cpd
80.052 g Cpd
455 kg x 1000g / kg = 455,000 g NH4NO3
455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 105 g Nitrogen
or: 455 kg NH4NO3
X 28.02 kg Nitrogen = 159 kg Nitrogen
80.052 kg NH4NO4
Fig.2.7
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
Nitrogen Oxide I : 46.68% Nitrogen and 53.32% Oxygen
Nitrogen Oxide II : 30.45% Nitrogen and 69.55% Oxygen
in 100 g of each Cpd: g O = 53.32 g & 69.55 g
g N = 46.68 g & 30.45 g
g O/g N =
Ratio of elemental mass ratios =
&
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
Nitrogen Oxide I : 46.68% Nitrogen and 53.32% Oxygen
Nitrogen Oxide II : 30.45% Nitrogen and 69.55% Oxygen
in 100 g of each Cpd: g O = 53.32 g & 69.55 g
g N = 46.68 g & 30.45 g
g O/g N =
2.284
1.142
=
2
1
1.142 & 2.284
= Ratio of elemental mass ratios
Dalton’s Atomic Theory
1. All matter consists of tiny particles called atoms.
2. Atoms of one element cannot be converted into
atoms of another element.
3. Atoms of an element are identical in mass and other
properties and are different from atoms of any other
element.
4. Compounds result from the chemical combination of
a specific ratio of atoms of different elements.
Scanning Tunneling Microscope Image
Reveals Individual Atoms on Silicon Surface
MoS2(0001)
Fig.2.8
Next:
Structure of Atoms
electrons (-) in cloud around
nucleus = protons(+) + neutrons
Expts. which led to this picture.
Neon Signs
Low pressure neon gas in evacuated
tube, between electrodes.
High voltage separates +, - charges
in Ne atoms:
+ goes to - electrode.
- goes to + electrode.
Measure current between electrodes:
Proves atoms made of +, - charges.
Fig.2.9
Deflection -> mass/charge
Fig.2.11
Millikan’s Expt.
1. Measured rate of droplet’s fall without
voltage: gave its mass.
2. Voltage across plates influenced speed, due
to charge of droplet.
3. Quantitative effect of voltage w/ laws of
physics -> amt. of charge on droplet.
4. RESULT: Different droplets had different
charges, but always a multiple of same
number ->
elementary charge on electron:
e = 1.602x10-19 coulombs (negative).
5. Mass/charge ratio x e = mass of e-
Rutherford Experiment
• Alpha (I.e., subatomic) particles bombarding the
atom.
• Rationale - to study the internal structure of the
atom, and to know more about the mass
distribution in the atom!
• Bombarded a thin Gold foil with Alpha particles
from Radium.
Fig.2.12
Ernest Rutherford (1871-1937)
• Won the Nobel Prize in Chemistry
in 1908
• “It was quite the most incredible
event..... It was almost as if a gunner
were to fire a shell at a piece of tissue
and the shell bounced right back!!!!! ”
Imagine throwing balls over a house.
Most go right over to a friend on other side,
just where you tried to throw them.
A few come back to your side of house.
WHY?
Fig.2.13
Moving electron cloud
surrounding nucleus.
Almost all the mass in __________________
the nucleus!
Notes: mass of e- tiny relative to p+, n.
p+, n have same mass (almost).
e-, p+ have same charge, opposite sign.
Atomic Definitions I: Symbols, Isotopes,Numbers
A
X
Z
The Nuclear Symbol of the Atom, or Isotope
X = Atomic symbol of the element, or element symbol
A = The Mass number; A = Z + N
Z = The Atomic Number, the Number of Protons in the Nucleus
(All atoms of the same element have the same no. of protons.)
N = The Number of Neutrons in the Nucleus
Isotopes = atoms of an element with the same number of protons,
but different numbers of Neutrons in the Nucleus
Fig.2.14
Neutral ATOMS
If neutral, then # e-s = # p+s.
Numbers of each particle:
•
•
•
•
•
51 Cr
= p+ ( ), e- ( ),
n( )
239 Pu = p+( ), e-( ),
n( )
15 N = p+( ), e-( ), n( )
56 Fe = p+( ), e-( ),
n( )
235 U =p+( ), e-( ),
n( )
Neutral ATOMS
If neutral, then # e-s = # p+s.
Numbers of each particle:
•
•
•
•
•
51 Cr
= p+ (24), e- (24 ),
n ( 27)
239 Pu = p+(94), e-(94),
n (145)
15 N = p+(7), e-(7), n(8)
56 Fe = p+(26), e-(26),
n (50)
235 U =p+(92), e-(92),
n (143)
Atomic Definitions II: AMU, Dalton, 12C Std.
Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom
on this scale Hydrogen has a mass of 1.008 AMU.
Dalton (D) = The new name for the Atomic Mass Unit,
one dalton = one Atomic Mass Unit
on this scale, 12C has a mass of 12.00 daltons.
Isotopic Mass = The relative mass of an Isotope relative to the
Isotope 12C the chosen standard.
Atomic Mass = “Atomic Weight” of an element is the average of the
masses of its naturally occurring isotopes weighted
according to their abundances.
Isotopes of Hydrogen
•
•
•
1 H
1
2 H
1
3 H
1
1 Proton
0 Neutrons 99.985 % 1.00782503 amu
(D) 1 Proton
1 Neutron
0.015 % 2.01410178 amu
(T) 1 Proton
2 Neutrons
----------------The average mass of Hydrogen is 1.008 amu
• 3H is Radioactive with a half life of 12 years.
• H2O Normal water “light water “
•
mass = 18.0 g/mole , BP = 100.000000C
• D2O Heavy water
•
mass = 20.0 g/mole , BP = 101.42 0C