Percent Yield & Percent Purity

A: Percent Yield
-A percent yield in chemical equations is used because chemical reactions generally                        
  do not produce the predicted amount of a substance predicted by chemists. 
  The percent yield is generally determined by the masses that are used in a chemical reaction as  
   well as by determining the mole ratio

>Ratio of amount of product obtained to amount of product expected by calculation, expressed as %
                                            ________________________________
                           %Yield  = |grams of actual product recovered            |   X   100%
                                            |grams of product expected from stoichi    |
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Example: If you burn 12 grams of carbon to make CO₂, then amount of carbon dioxide expected is one mol of CO₂ or 44 grams of CO₂. Sadly the amount you will get will probably be less than 44 grams and more like 34 grams of CO₂. The problem is a competing reaction that happens. Some carbon reacts to make CO.

2 C(s) + O₂(g) --- > 2 CO(g)

The carbon participating in this "side" reaction will not be able to make CO₂. The reaction will not yield 100% of the expected CO₂.
The amount of carbon dioxide produced, 34 grams of CO₂ is only 77% and not 100 % of the expected 44 grams.   


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B: Percent Purity
-Calculate how much reactant that actually is available to react        

                           %Purity  = Mass of Pure Substance    =x 100%

                                            Mass of Impure sample

  

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Excess and Limiting

A: Excess Quantity
   -A balanced equation describes what should happen in a chemical reaction
    Sometimes necessary to add more of one reactant than the equation
   
    -One reactant is the Excess quantity and some of it will be left over,the second   of reactant is used up to -----limiting Quantity
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Example:
How many grams of OCl2 will be formed when 44.0g of O2 react with 97.0g of Fl2?
Step1: balance the equation
             2Cl2+O2--->2OCl2
Step2: convert both reactants to the desired product 
             44g  mol O2 x 2molOCl2      x    87g =       239g OCl2
             32g                 1mol O2        1mol OCl2


            97.0gmol Cl2 x 2mol OCl2 x       87g         =119gOCl2
            71g                    2Cl2           1mol OCl2


According to the answer, Cl2 is limiting reactant and O2 is excess reactant 
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Stoichiometry-Calculation involving Particle-moles-Mass

E.g
Zn+2HCl---->H2+ZnCl2
1how many grams of Zn are required to produce 6mole of hydrogen 
                   6molH2 x 1mol Zn    x  65.4g          =392g Zn
                                      1mol H2      1mole Zn




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C3H8+5O2---->3CO2+4H2O
a) What mass of CO2 is produced by reacting 3.00mole of O2?


3.00mol O2 x 3mol CO2   x  44.0g CO2    =79.2gCO2
                         5mol O2          1mol CO2
b)What mass of C3H8 is required to produce 100.0g of H2O?


100.0gH2O x 1mol H2O x 1mol C3H8 x 44.0gC3H8  =61.1g
                         18.0gH2O    4mol H2O     1mol C3H8


c) If a sample o propane is burned what mass of h2o is produced if the reaction also produces 50.0L of CO2 at STP? 

50.0L CO2 x 1mol CO2 x 4mol H2O x 18.0g H2O = 53.6g
                        22.4LCO2   3molCO2     1mol H2O
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CH6:Stoichiometry-Calculation involving Reaction

-Stoichio: element
 Metry: measurement


-It's the study of calculating the amount of reactant used in a chemical reaction and how many product produced by the reaction
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4NH3+5O2---->6H2O+4NO
4:5:6:4--the mole ratio
-Balanced chemical equation are required


*Coefficient in balanced equation tell us the number of moles reacted or produced and also be  used a conversion factors


-N2+3H2---->2NH3
N2:H2---1:3
NH3:H2---2:3
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Example:
CH4+2O2---CO2+2H2O


-0.15molCH4 x 1molCO2  =0.15molCO2
                           1molCH4


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