PowerPoint Presentation - Formulas, Equations, and Moles - PowerPoint by o4Lx289O

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									             Percentage Composition


•   Percent Composition: Identifies the elements
    present in a compound as a mass percent of the
    total compound mass.

•   The mass percent is obtained by dividing the mass
    of each element by the total mass of a compound
    and converting to percentage.

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                 Empirical Formula

•   The empirical formula gives the ratio of the number
    of atoms of each element in a compound.

  Compound       Formula          Empirical Formula
Hydrogen peroxide H2O2                   OH
    Benzene       C6H6                   CH
    Ethylene      C2H4                   CH2
    Propane       C3H8                  C3H8

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              Percentage Composition


•   Glucose has the molecular formula C6H12O6. What
    is its empirical formula, and what is the percentage
    composition of glucose?

Empirical Formula = smallest whole number ratio
                          CH2O



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     Percentage Composition


                 CH2O

Total mass = 12.01 + 2.02 + 16.00 = 30.03

  %C = 12.01/30.03 x 100% = 39.99%
   %H = 2.02/30.03 x 100% = 6.73%
  %O = 16.00/30.03 x 100% = 53.28%


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           Percentage Composition


 Saccharin has the molecular formula C7H5NO3S.
 What is its empirical formula, and what is the
 percentage composition of saccharin?

Empirical Formula is same as molecular formula

MW = 183.19 g/mole

%C = (7 x 12.011)/183.19 x 100% = 45.89% etc.
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                  Empirical Formula

•   A compound’s empirical
    formula can be determined
    from its percent composition.


•   A compound’s molecular
    formula is determined from
    the molar mass and empirical
    formula.
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                    Empirical Formula

•   A compound was analyzed to be 82.67% carbon and 17.33%
    hydrogen by mass. What is the empirical formula for the
    compound?


•   Assume 100 g of sample, then 82.67 g are C and 17.33 g are
    H




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                   Empirical Formula

•   Convert masses to moles:
82.67 g C x mole/12.011 g = 6.88 moles C
17.33 g H x mole/1.008 g = 17.19 mole H


Find relative # of moles (divide by smallest number)




                                                       8
                  Empirical Formula

Convert moles to ratios:
6.88/6.88 = 1 C
17.19/6.88 = 2.50 H
Or 2 carbons for every 5 hydrogens
                            C2H5




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                  Empirical Formula

Empirical Formula is: C2H5


Formula weight is: 29.06 g/mole
If the molecular weight is known to be 58.12 g/mole
Then the molecular formula must be:
                             C4H10




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                   Empirical Formula

•   Combustion analysis is one of the
    most common methods for
    determining empirical formulas.
•   A weighed compound is burned in
    oxygen and its products analyzed
    by a gas chromatogram.
•   It is particularly useful for analysis
    of hydrocarbons.

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              Combustion Analysis

Combustion Analysis: the technique of finding the mass
 composition of an unknown sample (X) by examining the
 products of its combustion.

                  X + O2 → CO2 + H2O


            0.250 g of compound X produces:
              0.686 g CO2 and 0.562 g H2O



                                                         12
                 Combustion Analysis

                      X + O2 → CO2 + H2O


1. Find the mass of C & H that must have been present in X
   (multiply masses of products by percent composition of the
   products).


C: 0.686 g x 12.01 g/44.01 g = 0.187 g C


H: 0.562 g x (2 x 1.008 g)/18.02 g = 0.063 g H


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               Combustion Analysis

                   X + O2 → CO2 + H2O


0.187 g C + 0.063 g H = 0.250 g total
So compound X must contain only C and H!!


2. Find the number of moles of C and H
C: 0.187 g x mole/12.01 g = 0.0156 moles C
H: 0.063 g x mole/1.008 g = 0.063 moles H


                                             14
               Combustion Analysis

                    X + O2 → CO2 + H2O


3. Find the RELATIVE number of moles of C and H in whole
   number units (divide by smallest number of moles)


C: 0.0156/0.0156 = 1
H: 0.063/0.0156 = 4
If these numbers are fractions, multiply each by the same
     whole number.

                                                            15
               Combustion Analysis

                   X + O2 → CO2 + H2O


3. Write the Empirical Formula (use the relative numbers as
   subscripts)


                            CH4




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                 Combustion Analysis

                           Summary
1.   Find the mass of C and H in the sample.
2.   Find the actual number of moles of C and H in the
     sample.
3.   Find the relative number of moles of C and H in whole
     numbers.
4.   Write the empirical formula for the unknown compound.




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                 Combustion Analysis

                               NOTE!
In step # 1 always check to see if the total mass of C and H adds
    up to the total mass of X combusted. If the combined mass of
    C and H is less than the mass of X, then the remainder is an
    unknown element (unless instructed otherwise).


If a third element is known, calculate the mass of that element by
     subtraction (at the end of step 1), and include the element in
     the remaining steps.



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                 Combustion Analysis

Combustion Analysis provides the Empirical Formula. If a
  second technique provides the molecular weight, then the
  molecular formula may be deduced.


1.   Calculate the empirical formula weight.
2.   Find the number of “formula units” by dividing the known
     molecular weight by the formula weight.
3.   Multiply the number of atoms in the empirical formula by
     the number of formula units.

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                 Combustion Analysis

The molecular weight of glucose is 180 g/mole and its
  empirical formula is CH2O. Deduce the molecular
  formula.


1.   Formula weight for CH2O is 30.03 g/mole
2.   # of formula units = 180/30.03 = 6
3.   Molecular formula = C6H12O6




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