Slide 1 / 121 New Jersey Center for Teaching and Learning Progressive Science Initiative This material is made freely available at www.njctl.org and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Click to go to website: www.njctl.org Slide 2 / 121 AP Chemistry The Atom www.njctl.org Slide 3 / 121 Deducing the structure of the atom took a lot of brainpower. . . and even more funky hair styles
Slide 4 / 121 Birth of Atomic Theory A number of key discoveries led to the current understanding that the atom is the basic building block of matter. 1. Law of Definite Composition 2. Law of Multiple Proportions Slide 5 / 121 Law of Definite Composition If a compound is pure, it will always consist of the same composition no matter where the sample was taken or the size of the sample . Example: calcium carbonate Slide 6 / 121 Law of Definite Composition Example: calcium carbonate If it's calcium carbonate, it's guaranteed to be 40% calcium, 48% oxygen, and 12% carbon by mass. Sample Location Size Analysis Composition 20.0 g Ca 40% calcium Eastern 1 50.0 g 24.0 g O 48% oxygen Pennsylvania 6.0 g C 12% carbon 100.0 g Ca 40 % calcium 120.0 g O 48% oxygen 2 Wyoming 250.0 g 30.0 g C 12% carbon
Slide 7 / 121 Law of Definite Composition Some substances are not pure and do not obey the law of definite composition. These are called mixtures. Example: Pure water (pure substance) vs. Salt water (mixture) Slide 8 / 121 Law of Definite Composition Example: Pure water (pure substance) vs. Salt water (mixture) Salt water Pure water Sample % mass Sample Size Sample % mass Sample Size location composition location composition 85.3% O San 88.9% O 10.7 % H Atlantic 1 500.0 g 1 500.0 g Fransisco Ocean 1.6% Na 11.1 % H Lab 2.4 % Cl 79.5 % O 88.9% O 2 330.0 g Toronto Lab 10.0 % H Indian 11.1% H 2 330.0 g Ocean 4.2 % Na 6.3 % Cl Since the % composition of the water doesn't change, we know it is a pure substance. Since salt water varies in its % composition, it violates the law of definite composition and is a mixture. Slide 9 / 121 Using the Law of Definite Composition The law of definite composition can be used mathematically to see how much of a given substance is present in a sample. Example 1: Water is known to be 88.9% oxygen and 11.1 % hydrogen by mass. How many grams of oxygen would be present in a 400 gram sample of pure water? 400 grams of water x 0.889 g O = 355 g O 1 g water
Slide 10 / 121 Using the Law of Definite Composition The law of definite composition can be used mathematically to see how much of a given substance is present in a sample. Example 2: If a sample of water was found to contain 34.1 grams of oxygen, how many grams of hydrogen and water would be present? 34.1 g O x 0.111 g H = 4.26 g H 0.889 g O 34.1 g O x 1 g water = 38.4 g water 0.889 g O Slide 11 / 121 1 Hydrogen peroxide is known to be 94.1 % oxygen by mass with the rest being hydrogen. How many grams of hydrogen would be present in a pure 230. gram sample of hydrogen peroxide? Answer Slide 11 (Answer) / 121 1 Hydrogen peroxide is known to be 94.1 % oxygen by mass with the rest being hydrogen. How many grams of hydrogen would be present in a pure 230. gram sample of hydrogen peroxide? Answer 13.6 g [This object is a pull tab]
Slide 12 / 121 2 Calcium oxide is 71.4% calcium by mass with the rest being oxygen. How many grams of calcium would be present if a sample of calcium oxide sample was found to contain 12.3 grams of oxygen? Answer Slide 12 (Answer) / 121 2 Calcium oxide is 71.4% calcium by mass with the rest being oxygen. How many grams of calcium would be present if a sample of calcium oxide sample was found to contain 12.3 grams of oxygen? Answer 30.7 g [This object is a pull tab] Slide 13 / 121 3 Calcium carbonate is known to be 40% Ca, 48% O, and 12% carbon by mass. When a 200 gram sample of what is thought to be pure calcium carbonate is decomposed, 18 grams of carbon are found in the sample. Is this substance pure? Yes Answer No
Slide 13 (Answer) / 121 3 Calcium carbonate is known to be 40% Ca, 48% O, and 12% carbon by mass. When a 200 gram sample of what is thought to be pure calcium carbonate is decomposed, 18 grams of carbon are found in the sample. Is this substance pure? Yes Answer No NO [This object is a pull tab] Slide 14 / 121 Law of Definite Composition and the Classification of Matter This law allows us to classify the different types of matter MATTER Does the material obey the Law of Definite composition? Yes No Mixture Pure Substance Can the material be broken down into different elements with distinct properties? No Yes Compounds Elements Slide 15 / 121 Law of Definite Composition and Atomic Theory The easiest way to explain the law of definite proportions is if we picture matter as being made of atoms. If a given substance consists of a certain number of each type of atom, this would explain why the % composition would always be the same for that substance. For example, since water always has two oxygens for every hydrogen, it's % composition by mass must always be the same.
Slide 16 / 121 Law of Definite Composition and Atomic Theory Things got particularly interesting when they examined pure substances that were made out of the same elements. Water Hydrogen peroxide 88.9 % by mass O 94.1 11.1 % by mass H 5.9 Since the % composition of each substance was different, atomic theory would suggest this was due to each substance consisting of different numbers of atoms of each element. Slide 17 / 121 Law of Multiple Proportions and Atomic Theory When we examine the RATIO of these % amounts of each element, we find something very interesting. Water Hydrogen peroxide 88.9 % by mass O 94.1 11.1 % by mass H 5.9 88.9/11.1 = 8/1 ratio of O/H 94.1/5.9 = 16/1 WHOA!!!! Hydrogen peroxide has exactly twice the ratio of oxygen by mass compared to water. The easiest way to explain this is if we assume that matter is composed of atoms, and hydrogen peroxide has either exactly twice the number of oxygen atoms or half the number of hydrogen atoms as water does! Slide 18 / 121 Law of Multiple Proportions and Atomic Theory The law of multiple proportions states that when a fixed amount of one element is reacted with another to form different compounds of the same two elements, the ratio of the masses of the second element that reacts to form each compound, can always be expressed as a whole number. Example: Carbon dioxide and Carbon monoxide Notice that the ratio Substance carbon (g) oxygen (g) of oxygen required to form the carbon dioxide 1 2.66 compounds was carbon monoxide 1 1.33 2.66/1.33 = 2/1. Exactly twice the amount of oxygen was needed to make carbon dioxide compared to carbon monoxide!
Slide 19 / 121 Law of Multiple Proportions and Atomic Theory Example: Below is some data from a laboratory experiment in which two different oxides of copper were produced. Demonstrate they obey the law of multiple proportions. g Cu reacted g oxygen reacted g of copper oxide made oxide A 4.3 g ? 4.84 g oxide B 7.5 g ? 9.38 g 1. Use the law of conservation of mass to find the g of O oxide A = 4.84 - 4.3 = 0.54 g O oxide B = 9.38 - 7.5 = 1.88 g O 2. Find the ratio of Cu to O for both oxide A = 4.3/0.54 = 8:1 oxide B = 7.5/1.88 = 4:1 The ratio of copper found in both oxides is clearly a whole number multiple 8:4 or 2:1. There are exactly twice the copper atoms or half of the oxygen atoms in oxide A compared to oxide B. Slide 20 / 121 Law of Multiple Proportions and Atomic Theory The law of multiple proportions allowed scientists to hypothesize as to how many atoms of each kind may be in a compound. Let's look at the oxides of copper we just examined: Oxide A had either twice the copper atoms or half the oxygen atoms. Based on this, one can propose a series of possible formulas! Formula Set One: (twice the copper atoms) Oxide A = Cu 2 O Oxide B = CuO Formula Set Two: (half the oxygen atoms) Oxide A = CuO Oxide B = CuO 2 It took many more experiments to determine which formulas were correct. We will leave those until later. Slide 21 / 121 4 Two samples of a material are taken and the composition of each sample is given below. Is this material a pure substance? Sample A Sample B Yes 45 % Cu, 12% Si, 43 % O 34% Cu, 19% Si, 47 % O No Answer
Slide 22 / 121 5 A certain material is found to vary in composition by mass. What kind of matter is this? A Element Answer B Compound C Mixture D Pure Substance E A, B, and D Slide 22 (Answer) / 121 5 A certain material is found to vary in composition by mass. What kind of matter is this? A Element Answer B Compound C C Mixture D Pure Substance [This object is a pull E A, B, and D tab] Slide 23 / 121 6 Which of the following is TRUE regarding a pure compound? A It will not obey the law of definite composition B It can be broken down into different elements Answer C The amounts of each element by mass in the compound will not vary D It must contain the same equal mass % of each element in the compound E None of these are true
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