Transcript Empirical and Molecular Formulas

Percentage
Composition:
is the percent mass of
each element
present in a
compound.
Percent Composition
Can be calculated if given:
the chemical formula
OR
masses of elements in
compound
By Chemical Formula
% mass =
molar mass of an element
X 100%
total molar mass of the compound
Example: What is the % composition
of CaCO3? CHEMICAL FORMULA
 Step 1:
Find the molar mass of
CaCO3 : Ca x 1 = 40.1 g/mol

C x 1 = 12.0 g/mol

O x 3 = 48.0 g/mol

CaCO3 = 100.1 g/mol
Step 2: Find the % composition:
% Ca = 40.1 g/mol x 100 % = 40.1 % Ca
100.1 g/mol
% C = 12.0 g/mol x 100% = 12.0 % C
100.1 g/mol
% O = 48.0 g/mol x 100 % = 48.0 % O
100.1 g/mol
Example:
Calculate the percent composition of the
compounds that is formed from this
reaction: 29.0g of Ag combines completely
with 4.30g of S.
Masses of elements in compound
STEP 1: find the total mass of the
elements. 29.0g + 4.30g = 33.30g
STEP 2: find the % composition
Ag = 29.0g x 100% = 87.1%
33.30g
S = 4.30g x 100% = 12.9%
33.30g
Try These:
1) Find the percent composition of
KMnO4.
2) Calculate the % composition of
the compound that results from
9.03g Mg reacting completely with
3.48g N.
ANSWERS:
 1) K = 24.7%
Mn = 34.8%
O = 40.5%
2) Mg = 72.2%
N = 27.8%
1) Do Problem 17 on page 131
2) Do Problem 18 & 19 on page 133
CHECK YOUR ANSWERS!
Example 2 How much carbon
is present in 15.2 g of carbon
dioxide gas?
% carbon = 12.0 g/mol x 100 % = 27.3 %
44.0 g/mol
Xg carbon = 15.2 g of CO2 X 27.3 g C
100. g CO2
= 4.15 g of Carbon
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….we assume that it is a
100g sample:
25.9% N = 25.9g
74.1% O = 74.1g
2. Convert the mass of each element to
moles, by dividing by the molar mass.
 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
1.85
:
4.63
=
1
:
2.5
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 using
the numbers you obtained.
N2 O5
NOTE:
 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
Check your
answers
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