Percent Yield in Reactions

							Percent Yield in Reactions


   Welcome to The Real World
Ideal Conditions
   All reactants actually react.
   No outside influences exist.
   People don’t screw things up.
   There are no acts of nature.
   Dream on.
For whatever reason
   Nothing happens in a vacuum
   No chemical reaction is perfect
   Usually a fraction of the expected yield is
    produced
   Theoretical yield: What the calculation says we
    should get
   Actual yield: What is really produced.
   Percentage Yield = Actual yield/theoretical yield
    x 100
Example 1
   A student conducts a single displacement
    reaction that produces 2.755 grams of
    copper. Mathematically he determines that
    3.150 grams of copper should have been
    produced. Calculate the student's percentage
    yield.
   Solve:
    actual amount of product: 2.755 g
    expected amount of product: 3.150 g
Example 1 continued
                           actual amount of product
  percentage yield = ------------------------------------------- x   100
                            expected amount of product


                                 2.755g
            percentage yield = --------------- x 100
                                    3.150g

                 percentage yield = 87.4603174 %
                    percentage yield = 87.46 %
Example 2
   Solid sodium nitrate decomposes to form
    solid sodium nitrite and oxygen gas
    2NaNO3  2NaNO2 + O2

If the percentage yield is 80.4%, how much
    sodium nitrite will be produced from 955g
    of NaNO3?
Solution
   Divide by molar mass of NaNO3
   Multiply by the coefficient ratio
   Multiply by molar mass of NaNO2
   Multiply by percentage yield
   \\admin2000\facultysettings$\dsandberg\do
    cuments\Example problem 2 percent
    yield.doc
Assignment
   Percentage Yield Problem Set