Atomic Theory Michael Bachrodt William Fremd High School Click to begin
Mar 27, 2015
Atomic Theory
Michael BachrodtWilliam Fremd High SchoolClick to begin
Atomic Theory
New to this presentation?New to this
presentation? The Laws
The Laws
Dalton’s
Atomic TheoryDalton’s
Atomic Theory
Early Models of the Atom
Early Models of the Atom
All About Isotopes
All About Isotopes
The Modern Atomic Theory
The Modern Atomic Theory
Click on a topic to begin your lessonClick on a topic to begin your lesson
Main MenuMain Menu
exitexit
Concept map of the atomConcept map of the atom
The Basic ButtonsThis presentation is intended to be interactive. It is best to click on these buttons only. Please do not just click on the mouse to move you along. You may wind up somewhere you do not want to be! Click a button below to proceed. Have fun learning!
This presentation is intended to be interactive. It is best to click on these buttons only. Please do not just click on the mouse to move you along. You may wind up somewhere you do not want to be! Click a button below to proceed. Have fun learning!
Main MenuMain Menu
Previous SlidePrevious Slide
Next SlideNext Slide
Menu or first slide of the current sectionMenu or first slide of the current section
Returns you to this screenReturns you to this screen
Exits the presentationExits the presentation
Represents specific movesRepresents specific movesLast slide viewedLast slide viewed
33
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.
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
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.
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.
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.
1. All elements are composed of atoms, which are indivisible and indestructible particles.
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…..
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)
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.
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.
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!
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!
AnswerAnswerExample 1Example 1
26.36 g Cl23.64 g Cu26.36 g Cl23.64 g Cu
== 1.12 g Cl1.00 g Cu1.12 g Cl1.00 g Cu
For every gram of copper in the
compound, there are 1.12 g of chlorine.For every gram of copper in the
compound, there are 1.12 g of chlorine.
ContinuedContinued
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.
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.
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!
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!
AnswerAnswerExample 2Example 2
10.55 g Cl 9.45 g Cu10.55 g Cl 9.45 g Cu
== 1.12 g Cl1.00 g Cu1.12 g Cl1.00 g Cu
For every gram of copper in the
compound, there are 1.12 g of chlorine.For every gram of copper in the
compound, there are 1.12 g of chlorine.
SummarySummary
Law of Definite Proportions
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 chlorine1.00 g copper
This is known asThe Law of Definite Proportions.
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 chlorine1.00 g copper
This is known asThe Law of Definite Proportions.
SummarySummary
End of section.End of section.
12
Law of Multiple ProportionsJohn 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!
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!
ExampleExampleThese two vials contain different compounds. They are, however, composed of the same elements.
Yellow = K2CrO4
Orange = K2Cr2O7
These two vials contain different compounds. They are, however, composed of the same elements.
Yellow = K2CrO4
Orange = K2Cr2O7
13
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.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.
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.
MoreMore
14
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.
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.
AnswerAnswer 16 g O12 g C16 g O12 g C
== 1.331.33
ContinueContinue
15
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.
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.
AnswerAnswer 32 g O12 g C32 g O12 g C
== 2.662.66
ContinueContinue
16
Law of Multiple Proportions
What do you notice about the two mass ratios below? Click the mouse button to see!What do you notice about the two mass ratios below? Click the mouse button to see!
Mass ratio is 2.66Mass ratio is 2.66
ContinueContinue
Mass ratio is 1.33Mass ratio is 1.33
2.661.33
The mass ratio of one is double the other! What can this mean?
2.661.33
The mass ratio of one is double the other! What can this mean?
= 2.00= 2.00
17
Law of Multiple ProportionsNotice 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.
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.
ContinueContinue
Mass of carbon Mass of oxygen FormulaMass of carbon Mass of oxygen Formula
1 2 g1 2 g
12 g12 g 32 g32 g
16 g16 g COCO
CO2CO2
Note that the ratio of the masses of oxygen in the two compounds is 2 : 1!Note that the ratio of the masses of oxygen in the two compounds is 2 : 1!
18
Law of Multiple Proportions
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.
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.
ProblemProblem
19
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?
B. What is the mass of oxygen in the sample of sulfur trioxide?
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?
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?
B. What is the mass of oxygen in the sample of sulfur trioxide?
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?
AnswerAnswer
AnswerAnswer
AnswerAnswer
20
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?
B. What is the mass of oxygen in the sample of sulfur trioxide?
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?
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?
B. What is the mass of oxygen in the sample of sulfur trioxide?
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?
AnswerAnswer
10.00 g total - 5.00 g Sulfur = 5.00 g Oxygen10.00 g total - 5.00 g Sulfur = 5.00 g OxygenAnswerAnswer
AnswerAnswer
21
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?
B. What is the mass of oxygen in the sample of sulfur trioxide?
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?
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?
B. What is the mass of oxygen in the sample of sulfur trioxide?
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?
AnswerAnswer
8.33 g total - 3.33 g Sulfur = 5.00 g Oxygen8.33 g total - 3.33 g Sulfur = 5.00 g Oxygen
AnswerAnswer
AnswerAnswer
22
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?
B. What is the mass of oxygen in the sample of sulfur trioxide?
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?
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?
B. What is the mass of oxygen in the sample of sulfur trioxide?
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?
AnswerAnswer5.00 g S / 3.33 g S = 1.5 / 1 or 3 / 2Remember, 3 : 2 is the ratio because it represents a small whole number ratio. 1.5 : 1 is not a whole number ratio!
5.00 g S / 3.33 g S = 1.5 / 1 or 3 / 2Remember, 3 : 2 is the ratio because it represents a small whole number ratio. 1.5 : 1 is not a whole number ratio!
AnswerAnswer
AnswerAnswer
End of section.End of section.
23
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.
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.
24
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!
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!
25
The test tube contains a solution of potassium iodide. The test tube contains a solution of potassium iodide.
The erlenmeyer flask contains a solution of lead (II) nitrate.The erlenmeyer flask contains a solution of lead (II) nitrate.
When these chemicals are mixed, When these chemicals are mixed, a yellow precipitate (solid) forms in the liquid!a yellow precipitate (solid) forms in the liquid!
Let’s Investigate!
26
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.
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.
27
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.
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.
28
29
30
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.
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.
IncreaseIncrease
DecreaseDecrease
Stay the sameStay the same
31
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!
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!
32
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)
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 ofConservation
of Mass
Law ofConservation
of Mass
ProblemProblem
33
Try the following problem to see if you understand this law. 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?
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?
ReactantsReactants ProductsProducts
A + B C + D15 g 13g ? g 9g A + B C + D15 g 13g ? g 9g
The Law of Conservation of Mass(Click until buttons appear)(Click until buttons appear)
34
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)
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)
ReactantsReactants ProductsProducts
A + B C + D15 g 13g ? g 9g A + B C + D15 g 13g ? g 9g
28 g reactants- 9 g of D 19 g of C
28 g reactants- 9 g of D 19 g of C
Solution
End of section.End of section.
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.
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
Dalton’ Billiard Ball
Model
Thomson’s Plum Pudding
Model
Thomson’s Plum Pudding
ModelPlanetary
ModelPlanetary
Model
Click on a model to learn more or…..Click on a model to learn more or…..
Early Models of Atoms
Dalton’s Billiard Ball model was the accepted model for about 100 years. It was a direct result from his atomic theory. 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.You may review Dalton’s Atomic Theory if you wish by clicking on the picture above.
Billiard Ball Model 1803
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!
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!
Hey! Read more about me in your book!
Hey! Read more about me in your book!
Plum Pudding Model 1898
Click here to see a gas discharge
tube.
Click here to see a gas discharge
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.
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.
Cathode Ray Tube
cathodecathode anodeanode
Glass tubeGlass tubeCathode raysCathode rays
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.
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.
FYI (For Your Information)
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.
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.
The Electron
e/m= 1.759 x 108 coulombs/grame/m= 1.759 x 108 coulombs/gram
m = e 1.759 x 108 coulombs/gramm = e 1.759 x 108 coulombs/gram
m = 1.602 x 10-19 coulomb 1.759 x 108 coulombs/gramm = 1.602 x 10-19 coulomb 1.759 x 108 coulombs/gram
m = 9.07 x 19-28 gm = 9.07 x 19-28 g
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.
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 see a diagram of the gold foil experiment.
Click on me to see a diagram of the gold foil experiment.
Click on me to learn about this model and its
flaws.
Click on me to learn about this model and its
flaws.
Planetary Model 1909
DetectorDetector
Undeflected alpha particlesUndeflected alpha particles
Deflected alpha particlesDeflected alpha particles
Gold atomsGold atoms
Gold Foil Experiment
NucleusNucleusElectron ( - charge)Electron ( - charge)
Protons (+ charge)
Protons (+ charge)
Great distanceGreat distance
2. Electrons should attract to the nucleus and collapse the atom. Opposite charges attract!
2. Electrons should attract to the nucleus and collapse the atom. Opposite charges attract!
1. Protons only accounted for half the mass of the atom.
1. Protons only accounted for half the mass of the atom.
The Model and It’s Flaws
FlawsFlaws
Click until buttons appear.Click until buttons appear.
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.
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.
The Proton
Summary of the proton, electron, and
neutron
Summary of the proton, electron, and
neutron
(Postulated by William Wein in 1898. In 1920, Rutherford announced its existence.)(Postulated by William Wein in 1898. In 1920, Rutherford announced its existence.)
Protons, Electrons, Neutrons
Particle Mass Charge
Electron 9.11 x 10-28 g 1-Proton 1.67 x 10-24 g 1+Neutron 1.67 x 10-24 g 0
Particle Mass Charge
Electron 9.11 x 10-28 g 1-Proton 1.67 x 10-24 g 1+Neutron 1.67 x 10-24 g 0
Find more information in your text!Find more information in your text!
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.
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.
All About Isotopes
ProtiumProtium DeuteriumDeuterium TritiumTritium
ProtonProton
NeutronNeutron
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.
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.
Naming Isotopes
Hydrogen-2Hydrogen-2 Hydrogen-3Hydrogen-3Hydrogen-1Hydrogen-1
ProtiumProtium DeuteriumDeuterium TritiumTritium
A simple isotope notation is often used to convey information about an isotope. Sample isotope notations are shown below.
A simple isotope notation is often used to convey information about an isotope. Sample isotope notations are shown below.
Isotope Notation
1
H 1
Hydrogen-1Hydrogen-1
6
Li 3
Lithium-6Lithium-6
24
Mg 12
Magnesium-24Magnesium-24
Which number in each notation represents the mass number?Which number in each notation represents the mass number?
Mass Number
24
Mg 12
The # of protons + the # of neutrons
What does the number 12 stand for?
Mass Number
Click here until buttons appear.Click here until buttons appear.
(Isotope notation for Mg-24)
(Isotope notation for Mg-24)
50
Atomic Number (Henry Moseley in 1915)
24
Mg 12
The # of protons (p+) in the atom or….
Atomic Number
The # of electrons (e-) if the atom is electrically neutral ( i.e., not an ion. Click here to learn more about ions).
Click here until buttons appear.Click here until buttons appear.
Problem! 24
Mg 12
How many neutrons are there in this
isotope?
How many neutrons are there in this
isotope?
24 (neutrons + protons) - 12 protons = 12 neutrons
Fill in the blanks!Isotopenotation
Atomicnumber
Massnumber
Protons Neutrons Electrons Charge
184 74 0
82 128 +2
8 9 10
7474
-2-2
8080
7474
1717
110110
8282210210
88 17 O -2
8
184 W 74
210 Pb +2
82
Click the mouse button repeatedly to view the order in whichthe boxes are solved. The navigation buttons will appear after the table is filled out.
Click the mouse button repeatedly to view the order in whichthe boxes are solved. The navigation buttons will appear after the table is filled out.
5252
Problem!Write the isotope notation for the
element Chlorine with atomic number 17 on the periodic
table.
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!
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 78.70% 23.985 Mg-25 10.13% 24.986 Mg-26 11.17% 25.983
Average Atomic Mass…Problem
Mg-24 0.7870 x 23.985 = 18.876 Mg-25 0.1013 x 24.986 = 2.5310 Mg-26 + 0.1117 x 25.983 = 2.9023
The average atomic mass of magnesium is 24.309 The average atomic mass of magnesium is 24.309
For this isotope
11 p+
+ 11e-
0 net charge
23
Na 11
23
Na+
11
For this isotope
11 p+
+ 10e-
1+ net charge
Positive Ions (Atoms with a positive electric charge!)
Notice that only the electrons can change. Changing the protonswould change the atom!Notice that only the electrons can change. Changing the protonswould change the atom!
Click until button appearsClick until button appears
35
Cl 17
37
Cl-
17
Negative Ions (Atoms with a negative electric charge!)
Notice that only the electrons can change. Changing the protonswould change the atom!Notice that only the electrons can change. Changing the protonswould change the atom!
For this isotope
17 p+
+ 17 e- 0 net charge
For this isotope
17 p+
+ 17 e- 0 net charge
For this isotope
17 p+
+ 18 e- 1- net charge
For this isotope
17 p+
+ 18 e- 1- net charge
Click until button appearsClick until button appears
Modern Atomic Theory
Concept map of the atomConcept map of the atom
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.
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.
AtomAtom
NucleusNucleus ElectronsElectrons
NeutronsNeutronsProtonsProtons
Positive charge
Positive charge
Equal to atomic number
Equal to atomic number
Neutral chargeNeutral charge
Equal to mass number
- protons
Equal to mass number
- protons
Negative charge
Negative charge
Equal to protons if not
an ion
Equal to protons if not
an ion
containscontains
hashas have a have a
have a have a have a have a
64
Thanks for visiting!I certainly hope you learned something about
the history of Modern Atomic Theory!
Thanks for visiting!I certainly hope you learned something about
the history of Modern Atomic Theory!
Presentation Created by
Michael BachrodtChemistry Teacher
William Fremd High School1000 South Quentin Rd.
Palatine, Il. 60067847.755.2816
Presentation Created by
Michael BachrodtChemistry Teacher
William Fremd High School1000 South Quentin Rd.
Palatine, Il. 60067847.755.2816
Photo Credits:Hands: Mine!
Movie: Julie Zinger
Photo Credits:Hands: Mine!
Movie: Julie Zinger
EndShowEnd
Show