Atomic Theory Michael Bachrodt William Fremd High School Click to begin

Atomic Theory Main Menu New to this presentation? Dalton’s The Laws Early Models of the Atom All About Isotopes The Modern Atomic Theory Concept map of the atom Click on a topic to begin your lesson exit

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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. 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 Law of Conservation of Mass (1782) 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 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! 26.36 g Cl 23.64 g Cu = 1.12 g Cl 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 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! 10.55 g Cl 9.45 g Cu = 1.12 g Cl 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. 16 g O 12 g C = 1.33 Answer 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 12 g C = 2.66 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 1.33 The mass ratio of one is double the other! What can this mean? = 2.00 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 1 2 g 12 g 32 g 16 g CO 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 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? Answer Answer Answer

A sample of sulfur dioxide with a mass of 10. 00 g contains 5 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? Answer 10.00 g total - 5.00 g Sulfur = 5.00 g Oxygen Answer Answer

A sample of sulfur dioxide with a mass of 10. 00 g contains 5 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? Answer Answer 8.33 g total - 3.33 g Sulfur = 5.00 g Oxygen Answer

A sample of sulfur dioxide with a mass of 10. 00 g contains 5 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? Answer Answer 5.00 g S / 3.33 g S = 1.5 / 1 or 3 / 2 Remember, 3 : 2 is the ratio because it represents a small whole number ratio. 1.5 : 1 is not a whole number ratio! Answer End of section.

Chemical reactions produce a variety of interesting ‘products’ 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’ 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 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!

Law of Conservation of Mass 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 + B C + D 15 g 13g ? g 9g Reactants Products

Solution Reactants Products A + B C + D 15 g 13g ? g 9g 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 Planetary Model Click on a model to learn more or…..

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! Hey! Read more about me in your book! Click here to see a gas discharge tube.

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. Cathode rays cathode anode Glass tube

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 = 1.602 x 10-19 coulomb 1.759 x 108 coulombs/gram m = 9.07 x 19-28 g

Click on me to learn about this model and its flaws. 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 see a diagram of the gold foil experiment. Click on me to learn about this model and its flaws.

Gold Foil Experiment Gold atoms Detector Undeflected alpha particles

The Model and It’s Flaws Click until buttons appear. Nucleus Electron ( - charge) Protons (+ charge) Flaws 1. Protons only accounted for half the mass of the atom. Great distance 2. Electrons should attract to the nucleus and collapse the atom. Opposite charges attract!

Summary of the proton, electron, and neutron (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 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

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 Deuterium Tritium Protium

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 Tritium Hydrogen-2 Hydrogen-3 Hydrogen-1

Isotope Notation A simple isotope notation is often used to convey information about an isotope. Sample isotope notations are shown below. 1 H 6 Li 3 24 Mg 12 Hydrogen-1 Lithium-6 Magnesium-24 Which number in each notation represents the mass number?

Click here until buttons appear. 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 (Henry Moseley in 1915) Click here until buttons appear. 24 Mg 12 Atomic Number The # of protons (p+) in the atom or…. The # of electrons (e-) if the atom is electrically neutral ( i.e., not an ion. Click here to learn more about ions).

How many neutrons are there in this isotope? Problem! 24 Mg 12 How many neutrons are there in this isotope? 24 (neutrons + protons) - 12 protons = 12 neutrons

52 Fill in the blanks! 184 W 74 74 110 74 210 Pb +2 82 210 82 80 17 O -2 8 -2 8 17 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 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

Positive Ions (Atoms with a positive electric charge!) Click until button appears Positive Ions (Atoms with a positive electric charge!) 23 Na 11 23 Na+ 11 For this isotope 11 p+ + 11e- 0 net charge For this isotope 11 p+ + 10e- 1+ net charge Notice that only the electrons can change. Changing the protons would change the atom!

Negative Ions (Atoms with a negative electric charge!) Click until button appears Negative Ions (Atoms with a negative electric charge!) 35 Cl 17 37 Cl- 17 For this isotope 17 p+ + 17 e- 0 net charge For this isotope 17 p+ + 18 e- 1- net charge Notice that only the electrons can change. Changing the protons would change the atom!

Modern Atomic Theory Concept 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. Concept map of the atom

Equal to protons if not an ion Atom contains Electrons Nucleus have a has Protons Neutrons Negative charge have a have a Neutral charge Positive charge Equal to protons if not an ion Equal to atomic number Equal to mass number - protons

Thanks for visiting! Presentation Created Photo Credits: by I certainly hope you learned something about the history of Modern Atomic Theory! Presentation Created by Michael Bachrodt Chemistry Teacher William Fremd High School 1000 South Quentin Rd. Palatine, Il. 60067 847.755.2816 Photo Credits: Hands: Mine! Movie: Julie Zinger End Show