New Jersey Center for Teaching and Learning AP Chemistry - PDF document

Slide 1 / 121 Slide 2 / 121 New Jersey Center for Teaching and Learning AP Chemistry 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 The Atom 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 www.njctl.org Slide 3 / 121 Slide 4 / 121 Birth of Atomic Theory Deducing the structure of the atom took a lot of brainpower. . . 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 and even more funky hair styles Slide 5 / 121 Slide 6 / 121 Law of Definite Composition 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. 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 . Sample Location Size Analysis Composition 20.0 g Ca 40% calcium Example: calcium carbonate Eastern 24.0 g O 48% oxygen 1 50.0 g Pennsylvania 6.0 g C 12% carbon 100.0 g Ca 40 % calcium 2 Wyoming 250.0 g 120.0 g O 48% oxygen 30.0 g C 12% carbon

Slide 7 / 121 Slide 8 / 121 Law of Definite Composition Law of Definite Composition Example: Pure water (pure substance) vs. Salt water (mixture) Some substances are not pure and do not obey the law of definite Salt water composition. These are called mixtures. Pure water Sample % mass Sample Size Sample % mass Sample Size location composition location composition 85.3% O Example: Pure water (pure substance) vs. Salt water (mixture) 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 Slide 10 / 121 Using the Law of Definite Using the Law of Definite Composition Composition The law of definite composition can be used mathematically to see how much of a given substance is present in a sample. 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 Example 1: Water is known to be oxygen, how many grams of hydrogen and water would be present? 88.9% oxygen and 11.1 % 34.1 g O x 0.111 g H = 4.26 g H hydrogen by mass. How many grams of oxygen would be present 0.889 g O in a 400 gram sample of pure water? 34.1 g O x 1 g water = 38.4 g water 400 grams of water x 0.889 g O = 355 g O 0.889 g O 1 g water Slide 11 / 121 Slide 11 (Answer) / 121 1 Hydrogen peroxide is known to be 94.1 % oxygen by 1 Hydrogen peroxide is known to be 94.1 % oxygen by mass with the rest being hydrogen. How many grams mass with the rest being hydrogen. How many grams of hydrogen would be present in a pure 230. gram of hydrogen would be present in a pure 230. gram sample of hydrogen peroxide? sample of hydrogen peroxide? Answer Answer 13.6 g [This object is a pull tab]

Slide 12 / 121 Slide 12 (Answer) / 121 2 Calcium oxide is 71.4% calcium by mass with the rest 2 Calcium oxide is 71.4% calcium by mass with the rest being oxygen. How many grams of calcium would be being oxygen. How many grams of calcium would be present if a sample of calcium oxide sample was present if a sample of calcium oxide sample was found to contain 12.3 grams of oxygen? found to contain 12.3 grams of oxygen? Answer Answer 30.7 g [This object is a pull tab] Slide 13 / 121 Slide 13 (Answer) / 121 3 Calcium carbonate is known to be 40% Ca, 48% O, 3 Calcium carbonate is known to be 40% Ca, 48% O, and 12% carbon by mass. When a 200 gram sample and 12% carbon by mass. When a 200 gram sample of what is thought to be pure calcium carbonate is of what is thought to be pure calcium carbonate is decomposed, 18 grams of carbon are found in the decomposed, 18 grams of carbon are found in the sample. Is this substance pure? sample. Is this substance pure? Yes Yes Answer Answer No No NO [This object is a pull tab] Slide 14 / 121 Slide 15 / 121 Law of Definite Composition and Law of Definite Composition and the Atomic Theory Classification of Matter This law allows us to classify the different types of matter The easiest way to explain the law of definite proportions is if we picture matter as being made of atoms. MATTER Does the material obey the Law of Definite composition? If a given substance consists of a certain number of each Yes No type of atom, this would explain why the % composition Mixture Pure Substance would always be the same for that substance. Can the material be broken down into different elements with For example, since water always has distinct properties? two oxygens for every hydrogen, it's % composition by mass must always No Yes be the same. Compounds Elements

Slide 16 / 121 Slide 17 / 121 Law of Multiple Proportions and Law of Definite Composition and Atomic Theory Atomic Theory When we examine the RATIO of these % amounts of each element, Things got particularly interesting when they examined pure we find something very interesting. substances that were made out of the same elements. Water Hydrogen peroxide Water Hydrogen peroxide 88.9 % by mass O 94.1 88.9 % by mass O 94.1 11.1 % by mass H 5.9 11.1 % by mass H 5.9 88.9/11.1 = 8/1 ratio of O/H 94.1/5.9 = 16/1 Since the % composition of each substance was different, atomic WHOA!!!! Hydrogen peroxide has exactly twice the ratio of oxygen theory would suggest this was due to each substance consisting by mass compared to water. of different numbers of atoms of each element. 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 Slide 19 / 121 Law of Multiple Proportions and Law of Multiple Proportions and Atomic Theory Atomic Theory The law of multiple proportions states that when a fixed amount of one Example: Below is some data from a laboratory experiment in which two element is reacted with another to form different compounds of the different oxides of copper were produced. Demonstrate they obey the same two elements, the ratio of the masses of the second element that law of multiple proportions. reacts to form each compound, can always be expressed as a whole g Cu reacted g oxygen reacted g of copper oxide made number. oxide A 4.3 g ? 4.84 g Example: Carbon dioxide and Carbon monoxide oxide B 7.5 g ? 9.38 g Notice that the ratio 1. Use the law of conservation of mass to find the g of O Substance carbon (g) oxygen (g) of oxygen required oxide A = 4.84 - 4.3 = 0.54 g O oxide B = 9.38 - 7.5 = 1.88 g O to form the carbon dioxide 1 2.66 compounds was 2. Find the ratio of Cu to O for both carbon monoxide 1 1.33 2.66/1.33 = 2/1. oxide A = 4.3/0.54 = 8:1 oxide B = 7.5/1.88 = 4:1 Exactly twice the amount of oxygen was needed to make carbon The ratio of copper found in both oxides is clearly a whole number dioxide compared to carbon monoxide! 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 Slide 21 / 121 Law of Multiple Proportions and 4 Two samples of a material are taken and the composition of each sample is given below. Is this Atomic Theory material a pure substance? The law of multiple proportions allowed scientists to hypothesize as Sample A Sample B Yes to how many atoms of each kind may be in a compound. Let's look at the oxides of copper we just examined: 45 % Cu, 12% Si, 43 % O 34% Cu, 19% Si, 47 % O No Oxide A had either twice the copper atoms or half the oxygen atoms. Based on this, one can propose a series of possible formulas! Answer 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.