Chapter 3 - Chapter 3 Stoichiometry Chapter 3 Stoichiometry Calculate the molar mass of Al2(SO4)3 12 6 C 6.02 x 1023 atoms = 1 mole of element = 1

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Chapter 3 Stoichiometry 6 12 C 6.02 x 10 23 atoms = 1 mole of element = 1 molar mass 6.02 x 10 23 molecules = 1 mole of compound = 1 molar mass Chapter 3 Stoichiometry Calculate the molar mass of Al 2 (SO 4 ) 3 ___ mol Al x __________g/mol = ___mol S x ___________ g/mol = ___ mol O x ___________g/mol = Chapter 3 Stoichiometry You have a 6.07 g sample of CH 4 . How many moles is this? How many molecules? How many hydrogen atoms? Chapter 3 Stoichiometry Mass element Mass compound Mass compound Mass element Mole element Mole compound Mole element Mole compound Atoms element Molecules compound Molecules compound Atoms element Volume Volume formula molar ratio formula MW MW MW MW AN AN AN AN MM
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Chapter 3 Stoichiometry Structural Formula Molecular Formula Empirical Formula Name Chapter 3 Stoichiometry Calculate the mass percent C, H, and O in glucose (C 6 H 12 O 6 ) Chapter 3 Stoichiometry Elemental analysis shows an unknown compound to be 65.45% C, 5.45 % H and 29.09% O. What is empirical formula of the compound?
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This note was uploaded on 03/31/2008 for the course BIOL 1105 taught by Professor Jrcowles during the Fall '06 term at Virginia Tech.

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Chapter 3 - Chapter 3 Stoichiometry Chapter 3 Stoichiometry Calculate the molar mass of Al2(SO4)3 12 6 C 6.02 x 1023 atoms = 1 mole of element = 1

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