Organic Molecules - Get as PowerPoint

Document Sample
Organic Molecules - Get as PowerPoint Powered By Docstoc
					Unit 8: Chemical Formulas
                      Chapter 11
                   Chemistry 1K
       Cypress Creek High School
The Mole
           •   In the same way a dozen is
               worth 12, a mole is worth
               6.02 x 1023
                •   The mole is a large number
                    because particles are so small, it
                    takes many of them make up an
                    amount we can see and
                    understand
           •   We can use it to count
               anything!
   Mole
 1 dozen cookies = 12 cookies
 1 mole of cookies = 6.02 X 1023 cookies


                  1 gross cars = 144 cars
                  1 mole of cars = 6.02 X 1023 cars




1 ream Al particles = 500 Al particles
1 mole of Al particles = 6.02 X 1023 particles
Mole Conversions

                                        23
  22.4 L                         6.02 X 10 atoms
                                 formula units,
                  1 MOLE         molecules




           Molecular mass in grams
The Mass of a Mole
•   How do we “count” a mole?
     •   By measuring mass (just like how they
         count aluminum cans for recycling)
•   You can find atomic masses on the
    periodic table.
     • 1 mole of Mg atoms           =     24.3 g
     • 1 mole of Cu atoms           =     63.5 g
Moles to Mass
•   Calculate the mass of 0.625
    moles of calcium.
     •   According to the periodic
         table, the atomic mass of
         calcium is 40.078 amu, so the
         molar mass (mass of 1 mole) of
         calcium is 40.078 g.
         X g calcium                      40.078 g calcium
                                =
         0.625 mol calcium                   1 mol calcium
                         = 25.0 g calcium
    Mass to Particles
•   Calculate the number of atoms in
    4.77 g lead.
     •   First determine how many moles are in
         4.77 g lead.
            X moles Pb               1 mole Pb
                            =
             4.77 g Pb              2.7.2 g Pb
                      = 0.0230 mol Pb

     •   Now use a second conversion factor to
         convert moles to number of particles.
    X atoms Pb                   6.02 x 1023 atoms Pb
                            =
     0.0230 mol Pb                         1 mol Pb
                 = 1.38 x 1022 atom Pb
Chemical Formula
•   A Chemical Formula is the abbreviated way of
    naming a compound using numbers and
    element symbols.
     •   Subscripts denote number of each element in a
         compound
           • Ex: H2O has 2 hydrogens and 1 oxygen
           • Ex: Al2(SO4)3 has 2 aluminums, 3 sulfurs, and 12 oxygens

     •   Coefficients act as a scalar (also called a multiplier)
           • Ex: 6H2O has 12 hydrogens and 6 oxygens
           • Ex: 3Al2(SO4)3 has 6 aluminums, 9 sulfurs, and 36 oxygens
    Moles of Compounds
•   If the substance is a molecular compound,
    such as ammonia (NH3), a mole is 6.02 x
    1023 molecules of ammonia.
•   A mole of a compound contains as many
    moles of each element as are indicated by
    the subscripts in the formula for the
    compound.
     •   1 mole of NH3
          •1  mole of N
          • 3 moles of H
    Molar Mass of a Compound
•   The molar mass of a compound is the mass
    of a mole of the representative particles of
    the compound.
     •   In the case of NH3.
Molar mass of NH3 = molar mass of N + 3(molar mass of H)
Molar mass of NH3 = 14.007 g + 3(1.008 g) = 17.031 g/mol
•   You can use the molar mass of a
    compound to convert between mass and
    moles.
Molar Mass Practice
•   Find the molar mass for:
     • SO3
     • Na2SO4

     • Prozac, C17H18F3NO, is a widely used
       antidepressant that inhibits the uptake of
       serotonin by the brain.
    Percent Composition
•   Percent Composition is the percentage by
    mass of each element in a compound
     •   The percent composition is found by using the
         following formula:

              Total mass of element
    % Mass                         100
             molar mass of compound
    Calculating Percent Composition
•   What is the percent of each element in sodium sulfite,
    Na2SO3?
     • 2 mole Na x 22.990 g/mol Na =        45.980 g Na
     • 1 mole S x 32.066 g/mol S       =    32.066 g S
     • 3 moles O x 15.999 g/mol O = + 47.997 g O
                     Molar Mass of Na2SO3 = 126.043 g
•   Sodium
     • % = (45.980 g / 126.043 g) x 100 = 36.486 % Na
•   Sulfur
     • % = (32.066 g / 126.043 g) x 100 = 25.441 % S
•   Oxygen
     • % = (47.997 g / 126.043 g) x 100 = 38.080 % O
% Composition Practice

•   What is the percent of
    C & H in C2H6?

•   What is the percent of
    each element in
    Na2SO3?
Law of Definite Proportions
•   Joseph L. Proust was born in
    1754 in France and worked
    as an apothecary.
     •   Proust made artificial copper
         carbonate and compared it to
         natural copper carbonate.
     •   He showed that each had the same
         proportion of weights between the
         three elements involved.
•   This was called the Law of
    Definite Proporitons
Types of Formulas
•   There are 2 types of formulas that can be used
    to describe compounds.
     •   Empirical Formula
     •   Molecular Formula
•   An Empirical Formula is the LOWEST whole
    number ratio of the elements in a compound.
         •   The empirical formula for C6H6 is CH
•   A molecular formula tells the exact number of
    atoms of each element in a molecule or
    formula unit of a compound.
         •   The actual formula
Examples of Empirical & Molecular
Formulas
•   Example 1: Ethene
    • Empirical: CH2
    • Molecular: C2H4



•   Example 2: Caffeine
    • Empirical: C4N2OH5
    • Molecular: C8N4O2H10
                             Back = Carbon
                             Blue = Nitrogen
                             Red = Oxygen
                             Grey = Hydrogen
Empirical Formulas
•   An Empirical Formula is the LOWEST
    whole number ratio of the elements in a
    compound.
     •   The empirical formula for C6H6 is CH
          • C6H6   is a completely different chemical than CH!
•   The subscript cannot be a fraction
     •   Ex: There is no CO2H1.5 - it must be C2O4H3
•   What is the empirical formula for each?
     • C2H6
     • C6H12O6
Empirical Formula Calculations
•   There are 3 types of empirical calculations:
     •   Mole Ratio
     •   Grams
     •   Percent
    Empirical Formula from Mole Ratios
•   To calculate the empirical formula from mole
    ratios, follow these steps:
      1) Divide each mole quantity by the smallest
         mole quantity
      2) Assign ratios as subscripts

      3) If any of the ratios are not even (ie: .5, 1.25,
         1.3), multiply them all by the lowest common
         multiple to achieve whole numbers then place
         these numbers as the subscripts.
    Calculating Empirical Formula from
    Mole Ratio
•   Find the empirical formula for a
    compound containing 2 mole carbon and
    6 mole hydrogen.          LOWEST



•C= 2/2 = 1
•H = 6/2 = 3

               The formula is = CH3
    Empirical Formula from Grams
•   To calculate the empirical formula from grams, follow
    these steps:
      1) Convert the grams to moles (divide grams by molar
         mass).
      2) Divide each mole quantity by the smallest mole
         quantity.
      3) Assign ratios as subscripts.

      4) If any of the ratios are not even (ie: .5, 1.25, 1.3),
         multiply them all by the lowest common multiple to
         achieve whole numbers then place these numbers as
         the subscripts.
 Calculating Empirical Formula from
 Grams
•Calculatethe empirical formula of a compound
that contains 13.5 grams of calcium,4.05 grams of
carbon, and 16.2 grams of oxygen.
calcium               carbon            oxygen
  13.5 g = 0.337        4.05 g  = 0.337 16.2 g   = 1.01
40.08 g/mol mol       12.011 g/mol mol 15.999 g/mol mol
             LOWEST
0.337 = 1             0.337             1.01 = 3
                            =1
0.337                 0.337             0.337
  empirical formula = CaCO3
Empirical Formula from Percent Composition
 •   To calculate the empirical formula from percent composition,
     follow these steps:
      1) Change % sign to grams. You assume 100 g of the
         compound.
      2) Convert the grams to moles (Divide grams by molar
         mass).
      3) Divide each mole quantity by the smallest mole quantity.

      4) Assign ratios as subscripts.

      5) If any of the ratios are not even (ie: .5, 1.25, 1.3), multiply
         them all by the lowest common multiple to achieve
         whole numbers then place these numbers as the
         subscripts.
Calculating Empirical Formula from % Composition
  •   Calculate the empirical formula of a
      compound containing 18.8% nickel and 81.2%
      iodine.

  nickel                     iodine
             = 0.320                   = 0.640
  18.8 g                       81.2 g
               mol                       mol
  58.69 g/mol     LOWEST   126.905 g/mol
  0.320 = 1                  0.640 = 2
  0.320                      0.320
      empirical formula = NiI2
Practice Empirical Formula
•   Find the empirical formula given 65.2%
    Sc and 34.8 % O
Formula for a Hydrate
•   Many compounds have a certain
    number of water molecules that are
    associated with them. These are called
    hydrates.
     •   In the formula for a hydrate the number of
         water molecules associated with 1 unit of
         the compound is written following a dot.
          •   CaCl2  3H2O
Determining the Formula of a Hydrate
1)   Start with a known mass of a hydrate.
2)   Heat the hydrate to remove the water.
3)   Subtract the mass of the compound
     without water from the original mass.
     1)   This is the mass of water.
4)   Covert the masses to moles by dividing
     by the molar mass of each compound.
5)   Calculate the mole ratio by dividing by
     the smallest number of moles.
    Calculating Molecular Formula
•   A hydrate of BaCl2  XH2O has a mass of 5.00g. After
    heating until all the water is lost, the compound has a
    mass of 4.26g. What is the formula of the hydrate?
     • Mass before heating     5.00g
     • Mass after heating      4.26g
     • Mass of water        = 0.74g
BaCl2                              H2O
          4.26 g                         0.74 g
          208.23 g/mol                   18.02 g/mol
           0.0205 mol = 1                0.041 mol     = 2
           0.0205                        0.0205
     •   Formula of Hydrate = BaCl2  2H2O
Practice Molecular Formula
•    A mass of 2.50g of blue, hydrated
    CuSO4  XH2O is placed in a crucible and
    heated. After heating, 1.59g of white
    anhydrous CuSO4 remains. What is the
    formula for the hydrate?
•   CuSO4  5H2O
Molecular Formulas
•   For many compounds, the empirical
    formula is not the true formula.
     •   A molecular formula tells the exact number
         of atoms of each element in a molecule or
         formula unit of a compound.
          • The   actual formula
     •   The molecular formula for a compound is
         always a whole-number multiple of the
         empirical formula.
Determining Molecular Formulas
1)   Determine the empirical formula
2)   Find the empirical formula mass (you
     are given the molecular formula mass)
3)   Divide the molecular formula mass by
     the empirical formula mass. This is the
     scalar (multiplier).
4)   Multiply the subscripts in the empirical
     formula by the multiplier.
    Calculating Molecular Formula
•   A compound has an empirical formula of Sc2O3.
    The molecular mass is 414 g/mol. What is the
    molecular formula?
     • Empirical Formula = Sc2O3
     • Empirical Formula Mass =
          •2  mol Sc x 44.956 g/mol Sc =        89.912 g Sc
          • 3 mol O x 15.999 g/mol O   =       + 47.997 g O
                            molar mass =       137.909 g Sc2O3
     •   Find the multiplier…
          • 414   g / 137.909 g = 3 (the multiplier is 3)
     •   Molecular Formula = Sc6O9
Practice Molecular Formula
•   A molecule has 85.6% carbon and
    14.5% hydrogen. If the molecule has a
    mass of 42.1 grams, what is the
    molecular formula?
        End of Unit 8
Be Prepared for Unit 8 Test.

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:77
posted:8/8/2012
language:
pages:35