Chemistry You Need to Know

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					         Section 6.2—Concentration



How do we indicate how much of the electrolytes are in the drink?
Concentrated versus Dilute

                   solute       solvent




      Lower                           Higher
   concentration                   concentration
                                 More solute (what’s being
  Not as many solute (what’s
                                   dissolved) particles
   being dissolved) particles
Concentration

Concentration gives the ratio of amount
 dissolved to total amount
There are several ways to show
 concentration
Percent Weight/Volume
 This is a method of showing concentration
  that is not used as often in chemistry
 However, it’s used often in the food and
  drink industry
  For example, your diet drink can might say you
   have less than 0.035 g of salt in 240 mL.
  That would give you a concentration of
   0.035 g / 240 mL, which is 0.015% solution

                    grams solute
        %(W / V )               100
                     mL solvent
%(W/V) Example

                                     grams solute
                         %(W / V )               100
                                      mL solvent
        Example:
  If you dissolve 12 g
   of sugar in 150 mL
      of water, what
         percent
    weight/volume is
      the solution?
%(W/V) Example

                                     grams solute
                         %(W / V )               100
                                      mL solvent
        Example:
  If you dissolve 12 g
   of sugar in 150 mL
                                      12g
      of water, what
         percent         %(W / V )        100
    weight/volume is                 150mL
      the solution?


                                     8.0% (W/V)
%(W/V) Example #2

                                   grams solute
                       %(W / V )               100
       Example:                     mL solvent
  You want to make
   200 mL of a 15%
   (W/V) solution of
  sugar. What mass
    of sugar do you
  need to add to the
         water?
%(W/V) Example #2

                                   grams solute
                       %(W / V )               100
       Example:                     mL solvent
  You want to make
   200 mL of a 15%
   (W/V) solution of                   ?g
  sugar. What mass     15%(W / V )        100
    of sugar do you                  200mL
  need to add to the
         water?

                                   30 g of sugar
Concentration using # of molecules
 When working with chemistry and molecules,
  it’s more convenient to have a concentration
  that represents the number of molecules of
  solute rather than the mass (since they all
  have different masses)
 Remember, we use moles as a way of
  counting molecules in large numbers
Quick Mole Review (Remember the
mole road map ?!?)
1 mole = 6.02 × 1023 molecules
The molecular mass of a molecule is
 found by adding up all the atomic masses
 in the atom
Molecular mass in grams = 1 mole of that
 molecule
Quick Mole Example



Example:
How many
moles are
 in 25.5 g
   NaCl?
Quick Mole Example



Example:                 Na 1  22.99 g/mole = 22.99 g/mole
How many                 Cl 1  35.45 g/mole = + 35.45 g/mole
moles are                                        58.44 g/mole
 in 25.5 g
   NaCl?                   1 mole NaCl molecules = 58.44 g


   25.5 g NaCl   1   mole NaCl       0.44
                                 = _______ mole NaCl
                 58.44 g NaCl
Molarity
 Molarity (M) is a concentration unit that uses
  moles of the solute instead of the mass of the
  solute


                   moles solute
                M
                    L solvent


 Molarity Simulation
Molarity Example
       Example:
If you dissolve 12 g
 of NaCl in 150 mL
   of water, what is
     the molarity?
Molarity Example
         Example:                Na 1  22.99 g/mole = 22.99 g/mole
  If you dissolve 12 g           Cl 1  35.45 g/mole = + 35.45 g/mole
   of NaCl in 150 mL                                     58.44 g/mole
     of water, what is
       the molarity?                1 mole NaCl molecules = 58.44 g

     12 g NaCl           1   mole NaCl       0.21
                                         = _______ mole NaCl
                         58.44 g NaCl


Remember to change mL to L! 150 mL of water = 0.150 L

   moles solute                    0.21moles
M                            M                        1.4 M NaCl
    L solvent                       0.150L
Converting between the two

If you know the %(W/V), you know the
 mass of the solute
You can convert that mass into moles
 using molecular mass
You can then use the moles solute to find
 molarity
Converting from % to M Example
     Example:
 What molarity is a
 250 mL sample of
 7.0 %(W/V) NaCl?
Converting from % to M Example
                                                   ?g
      Example:                  7.0%(W / V )          100   ? = 17.5 g NaCl
  What molarity is a                             250mL
  250 mL sample of
  7.0 %(W/V) NaCl?               Na 1  22.99 g/mole = 22.99 g/mole
                                 Cl 1  35.45 g/mole = + 35.45 g/mole
                                                         58.44 g/mole

                                       1 mole NaCl molecules = 58.44 g

     17.5 g NaCl       1    mole NaCl          0.30
                                           = _______ mole NaCl
                       58.44 g NaCl

Remember to change mL to L! 250 mL of water = 0.250 L

                           0.30moles
                   M                      1.2 M NaCl
                            0.250L
Let’s Practice #2
    Example:
What is the %(W/V)
   of a 500. mL
sample of a 0.25 M
  CaCl2 solution?
Let’s Practice #2
     Example:                         ? moles   ? = 0.125 moles CaCl2
 What is the %(W/V)         0.25M 
                                      0.500L
    of a 500. mL
 sample of a 0.25 M
   CaCl2 solution?      Ca 1  40.08 g/mole = 40.08 g/mole
                        Cl 2  35.45 g/mole = + 70.90 g/mole
                                                110.98 g/mole

                            1 mole CaCl2 molecules = 110.98 g

0.125 moles CaCl2 110.98 g CaCl2     13.9
                                 = _______ g CaCl2
                  1    mole CaCl2


              13.9 g
  %(W / V )         100
              500mL               2.8 %(W/V) CaCl2
Concentration of Electrolytes

An electrolyte breaks up into ions when
 dissolved in water
You have to take into account how the
 compound breaks up to determine the
 concentration of the ions

       CaCl2  Ca+2 + 2 Cl-1

For every 1 CaCl2 unit that dissolves, you will produce 1 Ca+2 ion and
2 Cl-1 ions
If the concentration of CaCl2 is 0.25 M, the concentration of Ca+2 is 0.25
M and Cl-1 is 0.50 M
Let’s Practice #3
       Example:
     What are the
 molarities of the ions
  made in a 0.75 M
 solution of Ca(NO3)2
Let’s Practice #3
       Example:
     What are the
 molarities of the ions
  made in a 0.75 M
 solution of Ca(NO3)2



                   Ca(NO3)2  Ca+2 + 2 NO3-1
For every 1 Ca(NO3)2, there will be 1 Ca+2 and 2 NO3-1 ions


                             Ca+2 = 0.75 M
                             NO3-1 = 1.5 M

				
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