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Atomic Theory
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Atomic Theory
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Dalton’s
Atomic Theory
The Laws
Early Models of
the Atom
All About
Isotopes
The Modern
Atomic Theory
Concept map of the atom
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3
Dalton’s Atomic Theory
In the earliest days of chemistry
the chief ‘chemical’ occupations
were held by the alchemists.
During the middle ages,
alchemists tried to transform
various metals into gold. They
also produced metals from their
ores, made glasses and enamels,
and dyed fabrics. However, there
was no understanding of what
happens in these processes.
Dalton’s Atomic Theory
It wasn’t until 1803, however,
when an English school teacher,
John Dalton, was able to propose
the very first atomic theory. This
theory was the result of much
experimentation into the nature
of matter. It paved the way
towards a deeper understanding
of what chemicals are and what
happens when they react.
Dalton’s Atomic Theory
1. All elements are composed of atoms, which are indivisible and
indestructible particles with no internal structure.
2. All atoms of the same element are exactly alike; in particular,
they all have the same mass.
3. Atoms of different elements are different; in particular, they have
different masses.
4. Compounds are formed by the joining of atoms of two or more
elements. They are joined in a definite whole-number ratio, such as
1 to 1, 2 to 1, 3 to 2, etc.
5. Chemical reactions involve the rearrangement of atoms to make
new compounds.
Dalton’s Atomic Theory
How was Dalton able to develop this atomic theory?
Proceed to the next section on The Laws…..
The Laws
John Dalton based his atomic theory upon the
following laws.
Law of Conservation of Mass (1782)
Law of Definite Proportions (1797)
Law of Multiple Proportions (1803)
Law of Definite Proportions
Joseph Proust (1754 - 1826) found that the proportion by mass of the
elements in pure samples of a given compound is always the same,
regardless of the sample’s origin or size.
Example 1
There are 50 grams of a chemical
in this test tube. Analysis shows it
to contain 26.36 g of chlorine and
23.64 g of copper.
What is the ratio of the mass of
chlorine to the mass of copper in
this compound?
Do this now before you click the answer!
Answer
26.36 g Cl = 1.12 g Cl
23.64 g Cu
1.00 g Cu
For every gram of copper in the
compound, there are 1.12 g of chlorine.
Continued
Law of Definite Proportions
Joseph Proust (1754 - 1826) found that the proportion by mass of the
elements in pure samples of a given compound is always the same,
regardless of the sample’s origin or size.
Example 2
There are 20 grams of a chemical
in this test tube. Analysis shows it
to contain 10.55 g of chlorine and
9.45 g of copper.
What is the ratio of the mass of
chlorine to the mass of copper in
this compound?
Do this now before you click the answer!
Answer
10.55 g Cl = 1.12 g Cl
9.45 g Cu
1.00 g Cu
For every gram of copper in the
compound, there are 1.12 g of chlorine.
Summary
Law of Definite Proportions
Summary
Both of these samples contain the same
substance. Even though there are different
quantities in each tube, the ratios or
proportions of the elements to one another
by mass are the same. Each contains:
1.12 g chlorine
1.00 g copper
This is known as
The Law of Definite Proportions.
End of section.
Law of Multiple Proportions
John Dalton (1766 - 1844) proposed that the same elements that
make up one compound could also make up another compound
based upon the work of French chemist Berthollet in 1790. This
law was proved by the Swedish chemist, Berzelius, after
Dalton published his atomic theory!
Example
These two vials contain
different compounds. They
are, however, composed of
the same elements.
Yellow = K2CrO4
Orange = K2Cr2O7
Law of Multiple Proportions
How was John Dalton able to make such a statement? Let’s
investigate! Click on the red arrow below to proceed.
Suppose the balloons shown below contain two different
oxides of carbon, both of which are known to exist. Let’s
find the mass ratio for each compound.
More
Law of Multiple Proportions
Analysis of this compound shows there are 16.0 g
of oxygen and 12.0 g of carbon. Calculate the ratio
of the mass of oxygen to the mass of carbon.
Answer
16 g O
= 1.33
12 g C
Continue
Law of Multiple Proportions
Analysis of this compound shows there are 32.0 g
of oxygen and 12.0 g of carbon. Calculate the ratio
of the mass of oxygen to the mass of carbon.
Answer
32 g O
= 2.66
12 g C
Continue
Law of Multiple Proportions
What do you notice about the two mass ratios below? Click
the mouse button to see!
Mass ratio is 1.33
Mass ratio is 2.66
2.66 = 2.00
1.33
The mass ratio of one is double the other!
What can this mean?
Continue
Law of Multiple Proportions
Notice that the mass of carbon in each sample is the same.
This implies there are the same number of carbon atoms in
each sample. Since the mass of the oxygen atoms doubles
from one sample to the next, could this mean that the number
of oxygen atoms also doubles? If we assume the elements in
the first compound combine in a 1:1 ratio then the formula of
the second compound can be predicted.
Mass of carbon
Mass of oxygen Formula
12g
16 g
CO
12 g
32 g
CO2
Note that the ratio of the masses of oxygen in the two compounds is 2 : 1!
Continue
Law of Multiple Proportions
Problem
In summary, the Law of Multiple Proportions states that:
• The same elements that make up one compound can also
make up another compound. Example: CO and CO2.
• The mass ratio of the first compound compared to the mass
ratio of the second compound will be different by a factor of a
whole number.
• The masses of the element that combine with a fixed mass of
the other element are themselves a whole number ratio.
A sample of sulfur dioxide with a mass of 10.00 g contains
5.00 g of sulfur. A sample of sulfur trioxide with a mass of 8.33
g contains 3.33 g of sulfur.
A. What is the mass of oxygen in the sample of sulfur dioxide?
Answer
B. What is the mass of oxygen in the sample of sulfur trioxide?
Answer
C. For a fixed mass of oxygen in each sample, what is the small
whole-number ratio of the mass of sulfur in sulfur dioxide to
the mass of sulfur in sulfur trioxide?
Answer
A sample of sulfur dioxide with a mass of 10.00 g contains
5.00 g of sulfur. A sample of sulfur trioxide with a mass of 8.33
g contains 3.33 g of sulfur.
A. What is the mass of oxygen in the sample of sulfur dioxide?
Answer
10.00 g total - 5.00 g Sulfur = 5.00 g Oxygen
B. What is the mass of oxygen in the sample of sulfur trioxide?
Answer
C. For a fixed mass of oxygen in each sample, what is the small
whole-number ratio of the mass of sulfur in sulfur dioxide to
the mass of sulfur in sulfur trioxide?
Answer
A sample of sulfur dioxide with a mass of 10.00 g contains
5.00 g of sulfur. A sample of sulfur trioxide with a mass of 8.33
g contains 3.33 g of sulfur.
A. What is the mass of oxygen in the sample of sulfur dioxide?
Answer
B. What is the mass of oxygen in the sample of sulfur trioxide?
Answer
8.33 g total - 3.33 g Sulfur = 5.00 g Oxygen
C. For a fixed mass of oxygen in each sample, what is the small
whole-number ratio of the mass of sulfur in sulfur dioxide to
the mass of sulfur in sulfur trioxide?
Answer
A sample of sulfur dioxide with a mass of 10.00 g contains
5.00 g of sulfur. A sample of sulfur trioxide with a mass of 8.33
g contains 3.33 g of sulfur.
A. What is the mass of oxygen in the sample of sulfur dioxide?
Answer
B. What is the mass of oxygen in the sample of sulfur trioxide?
Answer
C. For a fixed mass of oxygen in each sample, what is the small
whole-number ratio of the mass of sulfur in sulfur dioxide to
the mass of sulfur in sulfur trioxide?
5.00 g S / 3.33 g S = 1.5 / 1 or 3 / 2
Answer Remember, 3 : 2 is the ratio because it represents a small
whole number ratio. 1.5 : 1 is not a whole number ratio!
End of section.
Chemical reactions produce a variety of interesting ‘products’.
Perhaps you have seen a clip o f a bomb exploding. Maybe you
remember combining vinegar with baking soda. Wow! That
produces lots of gas! Light can also be the product of a chemical
reaction as shown below.
Antoine Lavoisier (1743 - 1794) wanted to know how the masses
of the reactants and products of a chemical reaction compared. He
carefully determined the masses of reactants and products in many
different chemical reactions. Let’s investigate this ourselves!
Let’s Investigate!
The test tube contains a solution
of potassium iodide.
The erlenmeyer flask contains
a solution of lead (II) nitrate.
When these chemicals are mixed,
a yellow precipitate (solid) forms in
the liquid!
That’s pretty cool! Do you think the total mass of the products
is greater than the total mass of the original reactants? We’re
going to repeat this experiment but this time in a more
quantitative fashion to find out.
Here is picture showing the initial
mass of our ‘reactants’. Click on
the picture to see the reading in a
larger view. Please make a note
of the mass.
When you are ready, click on the
red arrow to watch the chemicals
being mixed. On some computers
the movie may not work. Just
click the white button on the
movie screen to continue.
Hmmm…Write down what you think happened to the mass of the
system. Try and give a reason for your answer. Remember, two
liquids did make a solid! Click a button below to continue.
Increase
Decrease
Stay the same
Well how about that!
The mass stayed the same! Did you
guess correctly? Good for you if you
did. Can you explain why the mass
stayed the same? If you can’t explain
why the mass is constant or if you
guessed wrong, don’t worry. That’s
why you are taking chemistry. This
course is designed to help you
understand what really goes on
during a chemical reaction. In time,
you will be able to come back to this
demonstration and explain it fully!
We discovered that the total mass of the reactants is equal to the
total mass of the products in a chemical reaction. Lavoisier discovered
this too. This simple statement is now known as the: (click)
Law of
Conservation
of Mass
Problem
The Law of Conservation of Mass
(Click until buttons appear)
Try the following problem to see if you understand this law.
Suppose you have 15 grams of substance A and 13 grams of
substance B. How much substance C will be formed if you also
produce 9 grams of substance D as shown below?
A
+
15 g
B
13g
Reactants
C
?g
+
D
9g
Products
Solution
A
+
15 g
B
13g
Reactants
C
?g
+
D
9g
Products
The reactants side of the chemical equation shows a total of 28
grams of chemical being used. The products side of the
equation shows 9 grams of D produced. Since the law states
that the total mass of the reactants must equal the total mass of
the products, the amount of C that must be produced is: (click)
28 g reactants
- 9 g of D
19 g of C
End of section.
Early Models of Atoms
Significant discoveries occurred since Dalton’s time that seriously
changed the way people envisioned the atom. A summary of the early
models of the atom will appear when you click the mouse button.
Dalton’
Billiard Ball
Model
Thomson’s
Plum Pudding
Model
Click on a model to learn more or…..
Rutherford’s
Planetary
Model
Billiard Ball Model 1803
Dalton’s Billiard Ball model was the accepted model for about
100 years. It was a direct result from his atomic theory.
You may review Dalton’s Atomic Theory if you wish by
clicking on the picture above.
Plum Pudding Model 1898
Sir J.J. Thomson (1856-1940) proposed that the structure of
the atom could be compared to a plum or raisin pudding. The
pudding would consist of a positive charged matrix. Embedded
in it would be electrons that would neutralize the positive charge.
Thus, the atom would be neutral overall. This was the first model
that incorporated the experimental evidence that showed atoms
must be composed of electrical charges. Click red dots below
for more info before you leave this section!
Click here to see
a gas discharge
tube.
Hey!
Read more
about me in
your book!
Cathode Ray Tube
J.J. Thomson was experimenting with electricity using a gas
discharge tube like that shown below. No air was present in the
tube. Thomson discovered that an electric current flowed from
the cathode to the anode by means of a ray of some kind. These
rays later became known as electrons. Your instructor may give
you more information about Thomson’s experiments.
anode
cathode
Cathode rays
Glass tube
Discovery of the Electron
In 1897, J.J. Thomson used a cathode ray tube to deduce the
presence of a negatively charged particle.
Cathode ray tubes pass electricity through a gas that is
contained at a very low pressure.
Conclusions from the
Study of the Electron
 Cathode rays have identical properties regardless of
the element used to produce them. All elements must
contain identically charged electrons.
Atoms are neutral, so there must be positive
particles in the atom to balance the negative charge of
the electrons
 Electrons have so little mass that atoms must
contain other particles that account for most of the
mass
FYI (For Your Information)
William Crookes was the first to discover these negative rays
in 1879. Thomson was able to conduct experiments that proved
these rays were actually negatively charged particles. Thomson
is generally given credit for the discovery of the electron.
The Electron
Electrons are very small particles with very little mass and a
fixed amount of negative charge. Thomson was able to calculate
the ratio of the charge on an electron to its mass (e/m).
In 1911, Robert Millikan actually measured the charge on the
electron. With this value and Thomson’s e/m ratio, he was able to
determine the mass of the electron.
e/m= 1.759 x 108 coulombs/gram
m=
e
1.759 x 108 coulombs/gram
m=
m = 9.07 x 19-28 g
1.602 x 10-19 coulomb
1.759 x 108 coulombs/gram
Planetary Model 1909
Ernest Rutherford’s (1871 - 1937) famous gold foil experiment
soon made the plum pudding model obsolete. He and his
researchers found that the atom must be composed of a very dense
positively charged area they called a nucleus. The results of their
experiment further indicated the atom is largely empty space.
Rutherford theorized that the electrons must be located outside the
nucleus at a relatively large distance. Perhaps the electrons circled
about the nucleus like planets around the sun.
Click on me to
learn about this
model and its
flaws.
Click on me to see a
diagram of the gold
foil experiment.
Gold Foil Experiment
Gold atoms
Detector
Undeflected
alpha particles
Deflected alpha
particles
Rutherford’s Gold Foil Experiment
 Alpha () particles are helium nuclei
 Particles were fired at a thin sheet of gold foil
 Particle hits on the detecting screen (film) are recorded
Rutherford’s Findings
 Most of the particles passed right through
 A few particles were deflected
 VERY FEW were greatly deflected
Conclusions:
 The nucleus is small
 The nucleus is dense
 The nucleus is positively charged
The Model and It’s Flaws
Click until buttons appear.
Electron ( - charge)
Nucleus
Protons (+ charge)
Flaws
1. Protons only
accounted for half
the mass of the atom.
2. Electrons should attract to
the nucleus and collapse the
atom. Opposite charges attract!
Great distance
The Proton
(Postulated by William Wein in 1898.
In 1920, Rutherford announced its existence.)
The proton was discovered shortly after the electron using
specially designed gas discharge tubes. The proton has a positive
charge equal in magnitude to the electron’s negative charge.
However, it is about 1800 times heavier than the electron. Every
element has a specific number of protons. Change the number of
protons and you have a new element. Imagine the possibilities!
Rutherford had theorized that there must be another particle
that would have the same mass as the proton, no charge, and
would also be located in the nucleus. In 1932, James Chadwick
proved Rutherford’s theory correct by discovering the neutron.
Summary of the
proton, electron, and
neutron
Protons, Electrons, Neutrons
Find more information in your text!
Particle
Electron
Proton
Neutron
Mass
(0)9.11 x 10-28 g
(1)1.67 x 10-24 g
(1)1.67 x 10-24 g
Charge
11+
0
All About Isotopes
Chadwick’s discovery of the neutron in 1932 helped take
care of one of the flaws of the planetary model. Now, the entire
mass of the atom could be accounted for. While the number of
protons in an element are constant, it was discovered that
atoms of the same element can have different numbers of
neutrons. These atoms became known as isotopes of one
another. Isotopes of hydrogen will appear below upon mouse
clicks.
Neutron
Proton
Protium
Deuterium
Tritium
Naming Isotopes
Not all isotopes have names like the isotopes of hydrogen.
Usually an isotope is identified by it’s Mass Number. This
represents the total number of protons and neutrons in the
nucleus of the atom. For example, protium would also be
known as hydrogen-1. What would deuterium and tritium be
called? Click the mouse button to see if you are correct.
Protium
Deuterium
Hydrogen-1
Hydrogen-2
Tritium
Hydrogen-3
Isotope Notation
A simple isotope notation is often used to convey
information about an isotope. Sample isotope notations are
shown below.
1
6
H
1
Hydrogen-1
24
Li
3
Lithium-6
Mg
12
Magnesium-24
Which number in each notation represents the mass number?
Mass Number
Click here until buttons appear.
Mass Number
24
Mg
12
(Isotope notation
for Mg-24)
The # of protons + the # of neutrons
What does the number 12 stand for?
Atomic Number (Z)(Henry Moseley in 1915)
Click here until buttons appear.
24
Mg
Atomic Number
The # of protons (p+)
in the atom or….
12
The # of electrons (e-) if the atom is
electrically neutral ( i.e., not an ion.
Click here to learn more about ions).
Problem!
24
How many
neutrons are
there in this
isotope?
Mg
12
24 (neutrons + protons)
- 12 protons
= 12 neutrons
56
Fill in the blanks!
Isotope Atomic Mass Protons Neutrons Electrons Charge
notation number number
184
W
74
210
Pb +2
82
17
O -2
8
74
184
74
110
74
0
82
210
82
128
80
+2
8
17
8
9
10
-2
Click the mouse button repeatedly to view the order in which
the boxes are solved. The navigation buttons will appear after
the table is filled out.
Problem!
Write the isotope
notation for the
element Chlorine
with atomic number
17 on the periodic
table.
Is There a Problem?
(Click)
Yes! Chlorine has a number
that is a decimal! How can
this be? The number on the
periodic table is an Atomic
Mass and not a Mass Number!
Mass Number vs. Atomic Mass
Mass number = # protons + # neutrons
Atomic mass = weighted average mass of
all the isotopes
of the element.
Atomic Mass is shown on the periodic table.
Atomic Mass
Also known as………
Weighted Average Atomic Mass
Average Atomic Mass
Average Atomic Mass
Calculated by…….
[(Mass of Isotope 1) x (% abundance of Isotope 1)]
+ [(Mass of Isotope 2) x (% abundance of Isotope 2)]
+ ………
= The average atomic mass as written on the periodic table
Average Atomic Mass…Problem
Calculate the average atomic mass of
magnesium given the following information:
Isotope
Relative Abundance
Atomic Mass
Mg-24
Mg-25
Mg-26
78.70%
10.13%
11.17%
23.985
24.986
25.983
Average Atomic Mass…Problem
Mg-24
Mg-25
Mg-26
0.7870 x 23.985 = 18.876
0.1013 x 24.986 = 2.5310
+ 0.1117 x 25.983 = 2.9023
The average atomic mass of magnesium is
24.309
Click until button appears
Positive Ions
23
(Atoms with a positive electric charge!)
23
Na
11
Na+
11
For this isotope
For this isotope
11 p+
+ 11e0 net charge
11 p+
+ 10e1+ net charge
Notice that only the electrons can change. Changing the protons
would change the atom!
Click until button appears
Negative Ions
35
(Atoms with a negative electric charge!)
37
Cl-
Cl
17
17
For this isotope
For this isotope
17 p+
+ 17 e0 net charge
17 p+
+ 18 e1- net charge
Notice that only the electrons can change. Changing the protons
would change the atom!
Modern Atomic Theory
1. Atoms are not indivisible. They are made up of protons, electrons, and
neutrons.
2. Atoms of the same element can, and do have different masses. These
atoms are called isotopes. Isotopes have the same number of protons but a
different number of neutrons.
3. Atoms of different elements are different. They differ in their physical
and chemical properties.
4. Compounds are formed by the joining of atoms of two or more
elements. They are joined in a definite whole-number ratio, such as 1 to 1,
2 to 1, 3 to 2, etc.
5. Chemical reactions involve the rearrangement of atoms to make new
compounds.
Concept map of the atom
Modern Atomic Theory
 All matter is composed of atoms
 Atoms cannot be subdivided, created, or destroyed
in ordinary chemical reactions. However, these
changes CAN occur in nuclear reactions!
Atoms of an element have a characteristic average
mass which is unique to that element.
Atoms of any one element differ in properties from
atoms of another element
Atom
contains
Electrons
Nucleus
have a
has
Protons
have a
Positive
charge
Equal to
atomic
number
Neutrons
have a
Negative
charge
Neutral
charge
Equal to
mass number
- protons
Equal to
protons if not
an ion
Thanks for visiting!
I certainly hope you learned something about
the history of
Modern Atomic Theory!
End
Show
69