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Section 3.1 Atomic Mass ,[object Object],[object Object],[object Object],[object Object],[object Object]
Figure 3.1 Schematic Diagram of a Mass Spectrometer
Figure 3.2 Neon  Gas
They are not whole numbers  ,[object Object],[object Object],[object Object],[object Object]
Examples of average mass ,[object Object],[object Object]
Examples of average atomic mass   pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
Examples   pp ,[object Object],[object Object],[object Object],[object Object]
Finding An Isotopic Mass   pp ,[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Learning Check ,[object Object],[object Object]
Solution AT8 ,[object Object],[object Object],[object Object],[object Object],[object Object]
3.2 The Mole ,[object Object],[object Object],[object Object],[object Object],[object Object]
The Mole ,[object Object]
More Stoichiometry
3.3 Molar mass ,[object Object],[object Object],[object Object],[object Object]
Find the molar mass of ,[object Object],[object Object],[object Object],[object Object],[object Object],16.043 134.862 226.095 701.926 172.12
 
Calculating atoms, moles and mass ,[object Object],[object Object],[object Object]
Calculating atoms, moles and mass ,[object Object],[object Object],[object Object]
3.4 Percentage Composition ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example   pp ,[object Object],[object Object],[object Object],[object Object]
Getting % Composition from the formula ,[object Object],[object Object]
Example   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
3.4 Percent Composition Review   pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
3.4 Percent Composition Review ,[object Object],[object Object],[object Object],[object Object],[object Object]
The Empirical Formula ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Calculating Empirical ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Calculating Empirical   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example   pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
Example   pp ,[object Object],[object Object],[object Object],[object Object]
Short cut way   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Clearing the Fractions   pp ,[object Object],[object Object]
Clearing the Fractions   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Clearing the Fractions   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Working backwards ,[object Object],[object Object],[object Object],[object Object],[object Object]
CO 2  is absorbed H 2 O is absorbed Sample is burned completely to form CO 2  and  H 2 O Pure O 2  in
Figure 3.5 Schematic Diagram of the Combustion Device Used to Analyze Substances for Carbon  and Hydrogen
Empirical Example ,[object Object],[object Object]
3.5 Determinine the Formula of a Compound: Empirical To Molecular Formulas ,[object Object],[object Object],[object Object],[object Object],[object Object]
Example ,[object Object],[object Object]
3.6 Chemical Equations ,[object Object],[object Object],[object Object],[object Object],[object Object]
3.7 Balancing equations   pp CH 4  +  O 2     CO 2  +  H 2 O Reactants Products C 1 1 O 2 3 H 4 2
Balancing equations   pp CH 4  +  O 2    CO 2  +  2 H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4
Balancing equations   pp CH 4  +  O 2    CO 2  +  2H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4 4
Balancing equations   pp CH 4  +  2 O 2    CO 2  +  2H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4 4 4
Another Example H 2  + H 2 O O 2  Make a table to keep track of where you are at
Example   pp H 2  + H 2 O O 2  Need twice as much O in the product R P H O 2 2 2 1
Example   pp H 2  + H 2 O O 2  Changes the O R P H O 2 2 2 1 2
Example   pp H 2  + H 2 O O 2  Also changes the  H R P H O 2 2 2 1 2 2
Example   pp H 2  + H 2 O O 2  Need twice as much H in the reactant R P H O 2 2 2 1 2 2 4
Example   pp H 2  + H 2 O O 2  Recount R P H O 2 2 2 1 2 2 4 2
Example   pp H 2  + H 2 O O 2  The equation is balanced, has the same  number of each kind of atom on both sides R P H O 2 2 2 1 2 2 4 2 4
Example H 2  + H 2 O O 2  This is the answer R P H O 2 2 2 1 2 2 4 2 4 Not this
Abbreviations ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Practice ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Practice ,[object Object],[object Object],[object Object],[object Object]
Meaning ,[object Object],[object Object],[object Object],[object Object],[object Object]
3.8 Stoichiometry ,[object Object],[object Object],[object Object],[object Object],[object Object]
For example...   pp ,[object Object],[object Object],[object Object],10.1 g Fe 55.85 g Fe 1 mol Fe = 0.181  mol  Fe
2Fe + 3CuSO 4     Fe 2 (SO 4 ) 3  + 3Cu   pp 0.181 mol  Fe 2 mol Fe 3 mol Cu = 0.272 mol  Cu 0.272 mol Cu 1 mol Cu 63.55 g Cu = 17.3  g  Cu
Could have done it via “spider”   pp 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu = 17.3 g Cu
pp
Examples   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Examples   pp ,[object Object],[object Object],[object Object],[object Object],[object Object]
Examples   pp ,[object Object],[object Object]
Examples if time   pp ,[object Object],[object Object],[object Object],[object Object]
3.9 Limiting Reagent   pp ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Limiting reagent   pp ,[object Object],[object Object],[object Object]
[object Object],[object Object],10.6 g  Cu   63.55g Cu   1 mol Cu 2 mol Cu   1 mol Cu 2 S   1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g  Cu 2 S 3.83 g  S   32.07g S   1 mol S 1 mol S   1 mol Cu 2 S   1 mol Cu 2 S 159.16 g Cu 2 S =  19.0 g  Cu 2 S = 13.3 g  Cu 2 S Cu is Limiting Reagent
Your turn ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Your turn ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Your turn ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example - Class does ,[object Object],[object Object],[object Object],[object Object]
Excess Reagent ,[object Object],[object Object],[object Object],[object Object]
Percent Yield ,[object Object],[object Object]
Example   pp  ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Another Example ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Examples ,[object Object],[object Object],[object Object],[object Object],[object Object]
Examples ,[object Object],[object Object],[object Object],[object Object],[object Object]
Example (lab 2) ,[object Object],[object Object],[object Object],[object Object]

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Ch3 z53 stoich

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  • 3. Figure 3.1 Schematic Diagram of a Mass Spectrometer
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  • 39. CO 2 is absorbed H 2 O is absorbed Sample is burned completely to form CO 2 and H 2 O Pure O 2 in
  • 40. Figure 3.5 Schematic Diagram of the Combustion Device Used to Analyze Substances for Carbon and Hydrogen
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  • 45. 3.7 Balancing equations pp CH 4 + O 2  CO 2 + H 2 O Reactants Products C 1 1 O 2 3 H 4 2
  • 46. Balancing equations pp CH 4 + O 2  CO 2 + 2 H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4
  • 47. Balancing equations pp CH 4 + O 2  CO 2 + 2H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4 4
  • 48. Balancing equations pp CH 4 + 2 O 2  CO 2 + 2H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4 4 4
  • 49. Another Example H 2 + H 2 O O 2  Make a table to keep track of where you are at
  • 50. Example pp H 2 + H 2 O O 2  Need twice as much O in the product R P H O 2 2 2 1
  • 51. Example pp H 2 + H 2 O O 2  Changes the O R P H O 2 2 2 1 2
  • 52. Example pp H 2 + H 2 O O 2  Also changes the H R P H O 2 2 2 1 2 2
  • 53. Example pp H 2 + H 2 O O 2  Need twice as much H in the reactant R P H O 2 2 2 1 2 2 4
  • 54. Example pp H 2 + H 2 O O 2  Recount R P H O 2 2 2 1 2 2 4 2
  • 55. Example pp H 2 + H 2 O O 2  The equation is balanced, has the same number of each kind of atom on both sides R P H O 2 2 2 1 2 2 4 2 4
  • 56. Example H 2 + H 2 O O 2  This is the answer R P H O 2 2 2 1 2 2 4 2 4 Not this
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  • 63. 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu pp 0.181 mol Fe 2 mol Fe 3 mol Cu = 0.272 mol Cu 0.272 mol Cu 1 mol Cu 63.55 g Cu = 17.3 g Cu
  • 64. Could have done it via “spider” pp 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu = 17.3 g Cu
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