Contributed by:
The highlights are:
1. Percentage formula
2. Empirical formula
3. Molecular formula
1.
Brain Teaser
Using the students present in class
today, complete the data table below
total # # % %
girls boys girls boys
students
Describe how one could calculate the percent of boys
present today using the data in the chart.
2.
Brain Teaser
• Determine the mass and %
composition of all atoms in C6H12O6
3.
Percent Compositon
Empirical Formula
Molecular Formula
4.
Percent Composition
• Percentage of each element in a
compound
– By mass
• Can be determined from
the formula of the compound or
the experimental mass analysis of the
compound
• The percentages may not always total to
100% due to rounding
part
Percentage 100%
whole
5.
Example #1
Determine the Percent Composition
from the Formula C2H5OH
Determine the mass of each element
in 1 mole of the compound
2 moles C = 2(12.01 g) = 24.02 g
6 moles H = 6(1.008 g) = 6.048 g
1 mol O = 1(16.00 g) = 16.00 g
Determine the molar mass of the
compound by adding the masses of
the elements
1 mole C2H5OH = 46.07 g
6.
Example #1
Determine the Percent Composition
from the Formula C2H5OH
Divide the mass of each element by
the molar mass of the compound and
multiply by 100%
24.02g
100% 52.14%C
46.07g
6.048g
100% 13.13%H
46.07g
16.00g
100% 34.73%O
46.07g
7.
Empirical Formulas
• The simplest, whole-number ratio of atoms in a
molecule is called the Empirical Formula
– can be determined from percent composition or
combining masses
• The Molecular Formula is a multiple of the Empirical
Formula
100g MMA
%A mass A (g) moles A
moles A
100g MMB moles B
%B mass B (g) moles B
8.
Example #2
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Convert the percentages to grams
by assuming you have 100 g of the
compound
– Step can be skipped if given masses
47gC
100g 47gC
100g
47gO 6.0gH
100g 47gO 100g 6.0gH
100g 100g
9.
Example #2
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Convert the grams to moles
1 mol C
47g C 3.9 mol C
12.01g
1 mol H
6.0 g H 6.0 mol H
1.008g
1 mol O
47 g O 2.9 mol O
16.00g
10.
Example #2
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Divide each by the smallest number
of moles
3.9 mol C 2.9 1.3
6.0 mol H 2.9 2
2.9 mol O 2.9 1
11.
Example #2
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
If any of the ratios is not a whole number,
multiply all the ratios by a factor to make it a
whole number
– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then
multiply by 3; if ?.25 or ?.75 then multiply by 4
3.9 mol C 2.9 1.3 x 3 4
Multiply all the
6.0 mol H 2.9 2 x 3 6
Ratios by 3
Because C is 1.3 2.9 mol O 2.9 1 x 3 3
12.
Example #2
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
° Use the ratios as the subscripts in
the empirical formula
3.9 mol C 2.9 1.3 x 3 4
6.0 mol H 2.9 2 x 3 6
C4H6O3
2.9 mol O 2.9 1 x 3 3
13.
Brain Teaser
The MnM company
wants to give MnM’s
a new formula!
Using the MnM’s that
were given to you on
the weigh boat,
1. Figure out your
percent by mass
(differentiating by
colors or size)
2. Create an empirical
formula with the
MnM’s.
14.
Molecular Formulas
• The molecular formula is a multiple
of the empirical formula
• To determine the molecular formula
you need to know the empirical
formula and the molar mass of the
compound
15.
Example #3
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Determine the empirical formula
• May need to calculate it as previous
C5H3
Determine the molar mass of the
empirical formula
5 C = 60.05 g, 3 H = 3.024 g
C5H3 = 63.07 g
16.
Example #3
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Divide the given molar mass of
the compound by the molar mass
of the empirical formula
–Round to the nearest whole
number
252 g
4
63.07 g
17.
Example #3
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Multiply the empirical formula by
the calculated factor to give the
molecular formula
(C5H3)4 = C20H12
18.
Example #4
Determine the Empirical Formula of
Benzopyrene, C20H12
Find the greatest common factor (GCF)
of the subscripts
factors of 20 = (10 x 2), (5 x 4)
factors of 12 = (6 x 2), (4 x 3)
GCF = 4
Divide each subscript by the GCF to get
the empirical formula
C20H12 = (C5H3)4
Empirical Formula = C5H3
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