Atomic Theory. Michael Bachrodt William Fremd High School. Click to begin. Atomic Theory. Main Menu. New to this presentation?. Dalton’s Atomic Theory. The Laws . Early Models of the Atom. All About Isotopes. The Modern Atomic Theory.
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Atomic Theory Michael Bachrodt William Fremd High School Click to begin
Atomic Theory Main Menu New to this presentation? Dalton’s Atomic Theory 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 John Dalton based his atomic theory upon the following laws. Click on each law’s red arrow to learn more about it. Law of Conservation of Mass (1782) Law of Definite Proportions (1797) Law of Multiple Proportions (1803)
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. 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! Continued
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. 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! 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. Go back to laws
These two vials contain different compounds. They are, however, composed of the same elements. Yellow = K2CrO4 Orange = K2Cr2O7 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
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 How was John Dalton able to make such a statement? Let’s investigate! Click on the red arrow below to proceed.
Answer 16 g O 12 g C = 1.33 Continue 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 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
2.66 1.33 The mass ratio of one is double the other! What can this mean? = 2.00 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
Mass of carbon Mass of oxygen Formula 1 2 g 16 g CO 12 g 32 g CO2 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. Note that the ratio of the masses of oxygen in the two compounds is 2 : 1!
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? 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
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? 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? 10.00 g total - 5.00 g Sulfur = 5.00 g Oxygen Answer Answer
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? 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 8.33 g total - 3.33 g Sulfur = 5.00 g Oxygen Answer
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? 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! Go back to laws
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
A + B C + D 15 g 13g ? g 9g Reactants Products 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 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 Go back to laws
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. Also look at your book for more information.
Hey! Read more about me in your book! 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.
anode cathode Glass tube Cathode rays 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.
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 see a diagram of the gold foil experiment. 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 the this model and its flaws.
Gold atoms Detector Undeflected alpha particles Deflected alpha particles Gold Foil Experiment(use your book to further explanation of this experiment)
Electron ( - charge) Nucleus Protons (+ charge) Great distance The Model and It’s Flaws Click until buttons appear. 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!
(Postulated by William Wein in 1898. In 1920, Rutherford announced its existence.) The Proton 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. Click here for a summary of the proton, electron, and neutron
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
Neutron Proton Protium Deuterium Tritium 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.
Hydrogen-1 Hydrogen-2 Hydrogen-3 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
Isotope Notation A simple isotope notation is often used to convey information about an isotope. Sample isotope notations are shown below. 1 H 1 6 Li 3 24 Mg 12 Hydrogen-1 Lithium-6 Magnesium-24 Which number in each notation represents the mass number?
Mass Number Mass Number Click here until buttons appear. 24 Mg 12 (Isotope notation for Mg-24) The # of protons + the # of neutrons What does the number 12 stand for?
Atomic Number Atomic Number (Henry Moseley in 1915) Click here until buttons appear. 24 Mg 12 The # of protons (p+) in the atom or…. The # of electrons (e-) if the atom is electrically neutral ( i.e., not an ion. Click hereto learn more about ions).