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Stoichiometry

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					Solutions are homogenous mixtures.
      When a substance is mixed with a liquid and it
      disintegrates into sub microscopic particles such as
      molecules atoms or ions, the mixture is a homogenous
      mixture and therefore a solution.

                                  Solute
   Solutions                          The substance dissolved
                                      in the other.


                                  Solvent
                                       The substance present in
                                       excess that dissolves the
                                       solute.
                                Why is water called the
                                universal solvent?




        because it can dissolve most ionic and
        polar compounds.

Example of the dissolving process of an ionic
compound in water
http://www.northland.cc.mn.us/biology/Biology1111/animations/dissolve.html
Vocabulary
Solubility: amount of substance that can be dissolved in a
            certain solvent.
Diluted: Contains small amount of solute for the given
       amount of solution.
Concentrated: Contains large amounts of solute for the
                 given amount of solution.


  Concentrated
  solution                                 Diluted solution
Saturated solution: It cannot contain more solute at the
                    given temperature.

Supersaturated solution: Contains more solute that the
amount it can contain at the given temperature.

This is an unstable solution and the excess of solute will
eventually separate, with any disturbance, or with the use of
a seed crystal, from the solution forming a precipitate.

This process is called crystallization
 http://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/11solutionspag
 e/crystallization_from_super.htm
Concentration:


  Is the amount of substance
  contained within a given
  volume of solution.




            c = Amount of substance
               Solution volume in dm3
MOLARITY

   Molarity is the concentration of
   a solution in moles of solute per
   liter of solution.



           M= n       or
                           M= n
              VL             Vdm3
•Other forms of concentration used
are: % composition (parts per
hundred) and ppm (parts per million)

•ppm is used when the concentration
is very small (like the addition of
fluorine to tap water)

            % by mass =    g solute    X   100
                          g solution

                     ppm = mg dm-3
      •Sample Problems

   Calculate the molarity of a solution prepared by dissolving
11.5g of solid NaOH in enough water to make 1.50 L of solution.


         What do we need to find first?

         The moles of NaOH…

                       1 mol NaOH
    11.5g NaOH X                      = 0.288 mol NaOH
                    40.00 g NaOH
    Molar mass
    Na: 22.99
    O: 16.00        M=n=       0.288 mol NaOH = 0.192 mol L-1
    H: 1.01           V(L)      1.50L solution
       40.00g/mol
 •Sample Problems

 What is the concentration in ppm of the solution in
               the previous problem?.


   What do we need to find first?

   The mg of NaOH…

               1,000 mg NaOH
11.5g NaOH X                = 11,500 mg NaOH
                   1 g NaOH

          ppm = mg =      11,500 NaOH         = 7,670 ppm
                 V(L)      1.50L solution
                                    Solve the problems in your notes
                 Dilution
              When a solution is prepared
              dissolving a concentrated solution
              the dilution formula can be used.



Formula:   VcMc = VdMd

                 Remember that you can       only use
                 this formula for dilution
                    Stoichiometry

               Stoichiometry refers to the use of a
               balanced equation to calculate the
               mass of products and/or reactants.


Example: How many moles and what mass
of carbon dioxide and water can be obtained
when 5.00g of ethane (C2H6) react with enough
oxygen (Combustion reaction).
Steps   1. Write and balance the equation for the
           reaction
        2. Convert the known (given) mass of the
           reactant or product to moles of the
           substance
        3. Use the balanced equation to set the
           appropriate mole ratio
        4. Use the ratio to calculate the number of
           moles of the desired reactant or product
        5. Convert from moles back to grams if
           required by the problem.

                               Solve the problems in your notes
             Limiting and excess
             reactants
Limiting reactant is the reactant that is totally
consumed during the reaction and therefore
controls the amount of products that can be
formed.

Excess reactant is the reactant that doesn’t
react completely during the reaction and
therefore there is an excess of it remaining after
the reaction.

Mg(s) + FeSO4(aq) ----- MgSO4 (aq) + Fe(s)
2.00g   2.00g     Which one will react completely?
  Example
Mg(s) + FeSO4(aq) ----- MgSO4 (aq) + Fe(s)
    2.00g Mg X 1 mol Mg = 0.0823 mole Mg
               24.31g Mg
 0.0823 mole Mg X 1 mole FeSO4 = 0.0823 mol FeSO4
                    1 mol Mg    Moles of FeSO4 needed to
                                react completely with
                                0.0823 moles Mg
 2.00g FeSO4 X    1mole FeSO4 = 0.0132 moles FeSO4
                  151.92 g FeSO4 Moles of FeSO4 that we have

     We need 0.0832 moles FeSO4 to react with 2.00 g of Mg
     and we have only 0.0132 moles of FeSO4

 Which reactant is going to be totally consumed in the reaction?
       FeSO4 is the limiting reactant
       It will be totally consumed during this reaction
       It determines the amount of products that can
       be obtained
How much iron (Fe) will be obtained?    (explain)
                                                     Answer: 0.737g Fe

      Mg is the reactant in excess. This means that at the end
      of the reaction there will be some magnesium left.

How much Mg will be left after the reaction?        (explain)

                                                                Answer: 0.320g Mg




                          Do prob.15 p 177. Students do 16
Molarity (review)

Molarity is the concentration of a
solution in moles of solute per liter
of solvent.
                M= n
                   Vin dm3
      M= Molarity
      N= number of moles of solute
      VL= Volume in liters of solution.
                              Do prob.17and 18 p 179.
                             Percent YIELD
           The real amounts of products formed during a
           chemical reaction are usually different from
           the expected amounts.
          The real amount of one product obtained
          (experimental) is called the ACTUAL YIELD
The expected amount, the amount of products calculated
through the stoichiometry of the reaction, from the
limiting reactant, is the THEORETICAL YIELD
                        Actual yield    X 100 %
     Percent Yield =
                       Theoretical Yield

                                        Do prob.19and 20 p 182.
       combustion reaction to
       find the Empirical formula
A 0.496 g sample of an unknown hydrocarbon was
completely burned in oxygen. The sample produced
   1.56 g of carbon dioxide and 0.638 g of water.
   a) How many moles of carbon dioxide were
   formed?
   b) How many moles of water were formed?
   c) What is the empirical formula of the
   hydrocarbon?
   d) A 26.0 g sample of another unknown
   compound containing only carbon and hydrogen
   was burned in excess oxygen and 88g of CO2
   were produced. What is the possible molecular
   formula of this compound?

				
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