11 notes ChemCalculations 1_ppt [Compatibility Mode]

11 notes ChemCalculations 1_ppt [Compatibility Mode] - 100...

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1 1 1 Chapter 9 Chemical Calculations: The Mole and Chemical Formula 2 Determination of Percent Composition Chapter 9 s Law of Definite Proportions (John Dalton) A given compound always contains the same proportion, by mass, of the elements. Decompose samples of a pure compound to give the constituent elements: NO 2 Sample Mass decompose Mass of nitrogen Mass of oxygen A 5.362 g 1.632 g 3.730 g B 10.613 g 3.230 g 7.383 g 3 Percent Composition What is the percent of nitrogen and percent of oxygen contained in each sample of NO 2 ? Sample A: Sample B: 4 Formula Unit and Formula Mass s Formula Unit s gives proportions of elements in a pure compound NO 2 CaCl 2 molecular ionic s Formula Mass s sum of the atomic masses of the atoms in one formula unit s the atomic masses are obtained from the periodic table in amu (atomic mass units)
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2 2 5 Percent Composition from Formula s NO 2 formula mass ? total mass from each element x
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Unformatted text preview: 100 % formula mass % N? % O? 6 Counting Atoms, Ions and Molecules Consider various reactions forming CO where: 1 atom of C combines with 1 atom of O to give CO, 1 pair of C atoms + 1 pair of O atoms gives 1 pair of CO molecules 1 dozen C atoms + 1 dozen O atoms gives 1 dozen CO molecules 1 gross of C atoms + 1 gross of O atoms gives 1 gross CO molecules 1 mole of C atoms + 1 mole of O atoms gives 1 mole of CO molecules 1 mole = 6.022137 x 10 23 objects Avogadros number 7 The Mole s One mole is the amount of a substance that contains as many elementary units as there are 12 6 C atoms in exactly 12.00000 g of 12 6 C 8 Converting between Moles and number of particles How many atoms are in 1.50 moles of carbon? How many molecules are in 5.35 moles of CO? Moles of substance Particles of a substance...
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11 notes ChemCalculations 1_ppt [Compatibility Mode] - 100...

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