CHEM 1811 Lecture # 8
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EMPIRICAL FORMULAS
•
a molecular formula shows the exact numbers of atoms in a molecule
•
mass % composition
-
is easy to determine experimentally
-
allows determination of relative numbers of atoms in a molecule
empirical formula :
simplest whole number ratio of atoms in a
compound ; a molecular formula has some multiple of these
numbers (1x, 2x, 3x, etc.)
e.g.
EXAMPLE
fructose
C6H12O6
CH2O
benzene
C6H6
CH
acetylene
C2H2
CH
Stoichiometric analysis
Salicylic acid, a material used to make aspirin, is found in tree bark.
An experiment showed that it contains 60.87 % carbon, 4.38 % hydrogen and
34.75 % oxygen by mass. Determine the empirical formula.
Suppose we have a 100. g sample of salicylic acid;
amount C (mol) ? = (100. g acid) (60.87 g C / 100. g acid) (1 mol C / 12.01 g C)
= 5.068 mol C
amount H ? = (100. g acid) (4.38 g H / 100. g acid) (1 mol H / 1.008 g H)
= 4.34 mol H
amount O ? = (100. g acid) (34.75 g O / 100. g acid) (1 mol O / 16.0 g O)
= 2.172 mol O
CHEM 1811 Lecture # 8
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IDEA:
mol ratio = atom ratio
C : H : O = 5.07 : 4.34 : 2.17
find whole numbers by
1)
dividing all by smallest
2)
multiplying all to smallest significant integer
5.07/2.17 : 4.34/2.17 : 2.17/2.17
2.33 : 2.00 : 1.00
∴
EXAMPLE
or
empirical formula for salicylic acid is
7 : 6 : 3
C7H6O3
Empirical and Molecular Formulas
Given that salicylic acid has a molar mass of 138.12 g/mol, determine
the molecular formula by
a)
using the empirical formula
b)
directly using the mass % composition given previously
empirical is C7H6O3
so molecular is C7nH6nO3n , with n = 1 or 2 or 3 or etc.
∴
molar mass = n (emp. formula mass)
138.12 g/mol = n (138.12 g/mol)
n = 1
and molecular formula is
C7H6O3
Suppose we have 1.00 mol sample:
amount C (mol) = (1.00 mol acid) (138.12 g/mol)
x (60.87 g C / 100. g acid) (1 mol C / 12.01 g C) = 7.00 mol C
(likewise)
amount H = 6.00 mol
and
so molecular formula is
amount O = 3.00 mol
C7H6O3
CHEM 1811 Lecture # 8
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COMBUSTION ANALYSIS
•
burning organic compounds in excess O2 converts all carbon to CO2
and all hydrogen to water
•
can determine empirical formula using masses of sample , CO2 and H2O
•
oxygen composition of sample must be found by difference
EXAMPLE
Combustion analysis of glucose
Glucose, a simple sugar, contains carbon, hydrogen and oxygen.
Complete combustion of a 0.360 mg sample of glucose gave 0.528 mg CO2 and
0.216 mg water. What is the empirical formula?
(aside:
Why is mass of sample less than CO2 and H2O? )
amount C = (528 µg CO2) (1 µmol CO2 / 44.01 µg CO2)
x (1 µmol C / 1 µmol CO2) = 12.0 µmol C
amount H = (216 µg H2O) (1 µmol H2O / 18.02 µg H2O)
x (2 µmol H / 1 µmol H2O) = 24.0 µmol H
amount O = { 360 µg glucose - (12.0 µmol C) (12.01 µg / µmol)
- (24.0 µmol H) (1.01 µg / µmol) } x (1 µmol O / 16.0 µg O)
= 12.0 µmol O
C : H : O
is
12.0 : 24.0 : 12.0
empirical formula is
CH2O