Chapter 2 PPt - Chem I

Chapter 2 PPt - Chem I - Definitions for Components of...

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Definitions for Components of Matter Pure Substances - Their compositions are fixed! Elements and compounds are examples of Pure Substances. Element - Is the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means. Molecule - Is a structure that is consisting of two or more atoms that are chemically bound together and thus behaves as an independent unit. Compound - Is a substance composed of two or more elements that are chemically combined. Mixture - Is a group of two or more elements and/or compounds that are physically intermingled.
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(p. 43)
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Fig. 2.3 Separating a Mixture
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Fig. 2.4
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Laws of Mass Conservation & Definite Composition Law of Mass conservation: The total mass of substances does not change during a chemical reaction. Law of Definite ( or constant ) composition: No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass.
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Mass Percent Composition of Na 2 SO 4 Na 2 SO 4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen Elemental masses 2 x Na = 2 x 22.99 = 45.98 1 x S = 1 x 32.07 = 32.07 4 x O = 4 x 16.00 = 64.00 142.05 Percent of each Element % Na = Mass Na / Total mass x 100% % Na = (45.98 / 142.05) x 100% =32.37% % S = Mass S / Total mass x 100% % S = (32.07 / 142.05) x 100% = 22.58% % O = Mass O / Total mass x 100% % O = (64.00 / 142.05) x 100% = 45.05% Check % Na + % S + % O = 100% 32.37% + 22.58% + 45.05% = 100.00%
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Calculating the Mass of an Element in a Compound Ammonium Nitrate Ammonium Nitrate = NH 4 NO 3 How much Nitrogen is in 455 kg of Ammonium Nitrate? The Formula Mass of Cpd is: 4 x H = 4 x 1.008 = 4.032 g 2 x N = 2 X 14.01 = 28.02 g 3 x O = 3 x 16.00 = 48.00 g 80.052 g Therefore gm Nitrogen/ gm Cpd 28.02 g Nitrogen 80.052 g Cpd = 0.35002249 g N / g Cpd 455 kg x 1000g / kg = 455,000 g NH 4 NO 3 455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 10 5 g Nitrogen 28.02 kg Nitrogen 80.052 kg NH 4 NO 4 = 159 kg Nitrogen 455 kg NH 4 NO 3 X or:
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Fig. 2.7
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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: Nitrogen Oxide I : 46.68% Nitrogen and 53.32% Oxygen Nitrogen Oxide II : 30.45% Nitrogen and 69.55% Oxygen in 100 g of each Compound: g O = 53.32 g & 69.55 g g N = 46.68 g & 30.45 g g O /g N = 1.142 & 2.284 2.284 2 1.142 1 =
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Dalton’s Atomic Theory 1. All matter consists of atoms. 2. Atoms of one element cannot be converted into atoms of another element. 3. Atoms of an element are identical in mass and other
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Chapter 2 PPt - Chem I - Definitions for Components of...

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