Transcript Empirical and Molecular Formulas

Percent Composition
 Can be calculated if given:
masses of elements in compound
OR
the chemical formula
Percent Composition
 Can be used to:
 calculate the mass of elements in a
compound
 determine the empirical formula of a
compound
 determine the molecular formula of a
compound
Empirical Formula
 shows the simplest mole ratio of the
elements.
 CO is a 1:1 ratio of carbon to oxygen
 H2O is a 2:1 ratio
 CO2 is a 1:2 ratio
 Empirical formulas can’t be reduced.
Molecular Formula
 shows the actual number of atoms in a
molecule.
 The molecular formula for hydrogen
peroxide is H2O2. Its empirical formula
would be HO.
 Often the molecular formula is the same
as the empirical formula: H2O, CO2
Empirical?
 CH4O
– yes, cannot be reduced further
 C 2H 6
– no, empirical would be CH3
 C3H10O
– yes
 C6H6O2
– no. What would empirical be?
– C3H3O
Calculating Empirical Formulas
 A chemist with an unknown compound
can easily figure out its percent
composition, but it is much more
meaningful to know its formula.
 EXAMPLE: What is the empirical
formula for a compound that is 25.9%
nitrogen and 74.1% oxygen?
Method
1. Write the mass
(g) of each
element in the
compound.
So….
25.9% N = 25.9g
74.1% O = 74.1g
2. Convert the mass of each element to
moles.
 N = 25.9g
= 1.85 mol
14.0g/mol
 O = 74.1g
16.0g/mol
= 4.63 mol
3. Calculate the simplest whole number
ratio by dividing the number of moles
by the smallest number of moles.
1.85 : 4.63
= 1 : 2.5
1.85
1.85
(If the result is not within 0.1 of a whole
number, multiply all numbers by a whole
number)
2(1
: 2.5)
=
2
:
5
4. Write the empirical formula.
N2 O5
 For inorganic compounds, write
the most positive element first.
 For organic compounds, write C
first, H second and all others
alphabetically.
A special present just for
you……..
Page
135,
Problems
#20 & 21
Molecular Formula
Given the empirical formula and the gram
formula mass (gfm)
OR
Given the percent composition and the
gram formula mass (gfm)
Example #1
Calculate the molecular formula for NaO
having a gfm of 78g.
 Determine the efm (empirical formula
mass).
NaO = 23.0g + 16.0g = 39.0
Divide the efm into the gfm.
78.0 = 2
39.0
This is the conversion factor used to
determine the molecular formula.
Na2O2
Example #2
Find the molecular formula for a
compound having a composition of
58.8% C, 9.8% H and 31.4% O and a
gmm of 102g/mol.
Determine the mass of each
component.
C = 102g/mol x 58.8% = 60.0g/mol
H = 102g/mol x 9.8% = 10.0g/mol
O = 102g/mol x 31.4% = 32.0g/mol
  convert to moles
C = 60.0g/mol = 5
12.0g
H = 10.0g/mol = 10
1.0g
O = 32.0g/mol = 2
16.0g
 Use moles as subscripts for components of
compound
C5H10O2
 Check the gmm of this compound…does it
equal 102.0g/mol?
 5(12.0) + 10(1.0) + 2(16.0) = 102.0g/mol
 YES!
And Now…..
Oh Yeah! And
there’s more…
Page 136,
Problems #22 &
23
Now Try
page 139,
#41 44