Ch 2_Part 1_100311 - 10/2/2011 Chapter 2 Fundamental...

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10/2/2011 1 Chapter 2 Fundamental Chemical Laws (2.2) Dalton’s Atomic Theory (2.3) Defining the Atom (2.5) Atomic Structure (2.6) Molecules and Ions (2.7) The Periodic Table (2.8) Nomenclature (2.9) This is the outline for the content we will cover in lecture. Please read the entire chapter. Law of Conservation of Mass The total mass of substances does not change during a chemical reaction – mass is neither created nor destroyed . 2
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10/2/2011 2 Law of Definite Proportions No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass . 3 WATER H 2 O No matter what the source water is ALWAYS 1 part hydrogen to 8 parts oxygen (by mass: one molecule is 2 g H and 16 g O) Chemical analysis of a 9.07 g sample of calcium phosphate shows that it contains 3.52 g of Ca. How much Ca could be obtained from a 1.000 kg sample? Mass fraction of Ca = 3.52 g Ca = 0.388 * 100 = 38.8% 9.07 g sample (i.e., ANY sample of calcium phosphate is 38.8% Ca by mass ) Mass of Ca in 1.000 kg of sample = 1.000 kg sample x 38.8 kg Ca = 0.388 kg Ca 100 kg sample or 388 g Ca Law of Definite Proportions 4
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10/2/2011 3 Law of Multiple Proportions If elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers . In a nutshell, two (or more) compounds can contain different relative amounts of the same elements: (Evidence of the existence tiny individual particles…) 5 Mass of oxygen that combines with 1.00 g of carbon: Compound #1 1.33 g O per g C Compound #2 2.66 g O per g C EXACT 2:1 RATIO Law of Multiple Proportions mass of O in compound #2 = 2.66 g = 2 mass of O in compound #1 1.33 g 1 6
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This note was uploaded on 01/18/2012 for the course CHEM 142B taught by Professor John during the Fall '11 term at University of Washington.

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Ch 2_Part 1_100311 - 10/2/2011 Chapter 2 Fundamental...

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