The empirical formula of a compound
Empirical formula (Simplest Formula): shows the lowest whole
number ratio of the elements in a compound
Molecular formula: shows the actual formula.
Name of
Compound
Molecular
Formula
Lowest Ration
of elements
H2O2
Emprical
Formula
(Simplest)
HO
Hydrogen
Peroxide
Glucose
Benzene
Water
C6H12O6
C6H6
H2O
CH2O
CH
H2O
1:2:1
1:1
2:1
1:1
Determining a Compound’s Empirical Formula:
Example 1:
Determine the empirical formula of methane given that 6.0 g of
methane can be decomposed into 4.5 g of carbon and 1.5 g of
hydrogen.
Example 2:
An inorganic sal is composed of 17.6% sodium, 39.7% chromium,
and 42.8% oxygen.
Element
Mass %
Grams per 100g
Sample
(g)
Molar Mass
(g/mol)
Number of
moles
(mol)
Molar amount \
lowest
molar amount
Na
Cr
O
17.6
39.7
42.8
17.6
39.7
42.8
23.0
52.0
16.0
0.77
0.76
2.675
0.77/0.76 =
1
0.76/0.76 = 1
2.675/0.76 =
3.5
Na1Cr1O3.5 = we cannot have a decimal so we multiply by 2
Emperical Formula = Na2Cr2O7
Determining Molecular Formulas from Empirical Formulas
Example 1:
An artificial sweetener is 57.14% C, 6.16% H, 9.52% N, and
27.18% O. Calculate the empirical formula of the sweetener
and find the molecular formula.
Note: The molar mass of the sweetener is 588.6g/mol
First: repeat step from previous example to find empirical:
Element
Mass %
Grams per 100g
Sample
(g)
Molar Mass
(g/mol)
Number of
moles
(mol)
Molar amount \
lowest
molar amount
Empirical Formula = C14H18N2O5
Molecular Formula = divide Actual Molar Mass by Molar
Mass of Empirical to find ‘n’ (not moles, just a factor)
THEN: multiply empirical formula by ‘n’
MM C14H18N2O5 = (14)12.01 g/mol + (18)1.01 g/mol + (2)14.01 g/mol + (5)16.0 g/mol
= 168.14 g/mol + 18.18 g/mol + 28.02 g/mol + 80.0 g/mol
= 294.34 g/mol
Molecular factor = 588.6 g/mol / 294.34 g/mol
= 1.99
~2
= C(14x2) H(18x2) N(2x2) O(5x2)
Therefore: Molecular Formula = C28H36N4O10