C142A Ch 3 2011 Lecture Notes

C142A Ch 3 2011 Lecture Notes - Chapter#3 Stoichiometry...

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Chapter #3 : Stoichiometry - Mole - Mass Relationships in Chemical Systems 1. Atomic Masses 2. The Mole 3. Determining the Formula of an Unknown Compound 4. Writing and Balancing Chemical Equations 5. Calculating the amounts of Reactant and Product 6. Limiting Reagent Calculations:
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MOLE – just a number, like a dozen • The Mole is based upon the definition: The amount of substance that contains the same number of elementary particles (atoms, molecules, ions, or other?) as there are atoms in exactly 12 grams of carbon-12 (i.e., 12 C). That number (i.e. 1 Mole ) = 6.022045 x 10 23 particles (atoms, molecules, ions, electrons, apples, hairs, etc…) = N A particles (called Avogadro’s Number ) ~million x million x million x million
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Avogadro’s Number ( N A ) N A = 6.022045 x 10 23 = # of particles (atoms, molecules, ions, electrons, or…) in one mole of that thing. Think of it as a units conversion factor: N A ! 6.022045 x 10 23 = 1 Example: 1.00 moles of CO 2 molecules x (6.022x10 23 CO 2 molecules / mol CO 2 molecules ) = 6.02x10 24 CO 2 molecules whatevers mol whatevers
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2 H 2 (g) + O 2 (g) ! 2 H 2 O(g) 2 dozen H 2 molecules react with exactly 1 dozen O 2 molecules to give exactly 2 dozen H 2 O molecules. 2 moles of H 2 molecules react with exactly 1 mole of O 2 molecules to give exactly 2 moles of H 2 O molecules. Why do we do this? Because these last sizes are in the gram range and were easy to weigh in 1800s. Conventions: 1 mole of 12 C atoms weighs 12 g exactly. 1 atom of 12 C weighs 12 amu exactly. (amu = atomic mass unit = ~mass of a proton or neutron)
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2 H 2 (g) + O 2 (g) ! 2 H 2 O(g) 2 H 2 molecules react with exactly 1 O 2 molecule to give exactly 2 H 2 O molecules. 2 moles of H 2 molecules 4 x 1.008 g = 4.032 g react with exactly 1 mole of O 2 molecules 2 x 16.00 g = 32.00 g to produce exactly 2 moles of H 2 O molecules 2 x (2x1.008 g + 16.00 g) = 36.03 g or 4 mol H x = 4.032 g H 1.008 g H mol H
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Number – mass relationships Mole – mass relationships 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms S
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Mole - Mass Relationships of Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 10 23 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 10 23 atoms 1 atom of S = amu 1 mole of S = g = atoms 1 atom of O = amu 1 mole of O = g = atoms Molecular mass: 1 molecule of O 2 = amu 1 mole of O 2 = g = molecules 1 molecule of S 8 = amu 1 mole of S 8 = g = molecules
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Mole - Mass Relationships of Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 10 23 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 10 23 atoms 1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 10 23 atoms 1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 10 23 atoms Molecular mass: 1 molecule of O 2 = 16.00 x 2 = 32.00 amu 1 mole of O 2 = 32.00 g = 6.022 x 10 23 molecule 1 molecule of S 8 = 32.07 x 8 = 256.56 amu 1 mole of S 8 = 256.56 g = 6.022 x 10 23 molecules
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Isotopes of Hydrogen Natural Abundance 1 1 H 1 Proton 0 Neutrons 99.985 % 1.00782503 amu 2 1 H ( D ) 1 Proton 1 Neutron 0.015 % 2.01410178 amu 3 1 H (T) 1 Proton 2 Neutrons -------- ---------- The average mass of Hydrogen is 1.008 amu WATER H-ISOTOPE DEMO: 2 H 2 O (or D 2 O, >$200 / glass) vs. normal H 2 O Which is more dense?
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Element #8 : Oxygen, Isotopes 16 8 O 8 Protons
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This note was uploaded on 01/18/2012 for the course CHEM 142A taught by Professor Campbell during the Fall '11 term at University of Washington.

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C142A Ch 3 2011 Lecture Notes - Chapter#3 Stoichiometry...

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