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1.2 Formulas
1.2.5 Define the terms empirical formula and molecular formula
1.2.6 Determine the empirical formula and/or the molecular formula of a
given compound
• empirical formula from the % composition or from experimental data
• % composition from the formula of a compound
• molecular formula when given both the empirical formula and the
molar mass
Terms:




The law of definite proportions describes that
elements in a given compound are always present in
the same proportions by mass.
The Percentage composition of a compound refers to
the relative mass of each element in the compound.
The term molecular mass describes the mass of a
molecule.
The term formula mass describes the mass of the
smallest repeating unit of an ionic compound.
Finding Percentage composition from a chemical
formula:
Find the mass of the compound, then find the mass of the
individual parts (atoms).
2.
Divide the atoms by the compound and multiply by 100%
For example: H2O
H2 = 2 x 1.01 = 2.02
+O =
+16
Total molecular mass = 18.02

Percentage by mass of H in water = 2.02/18.02
=.11 x 100%
= 11%

Percentage by mass of O in water = 16/18.02
=.888 x 100%
= 89%
1.
Percent
Composition
mass
Percentage
Composition
from by
mass
Take the mass of the atom and divide by the total mass
of the compound, then multiply by 100%
Example: Mass of compound is 48.72g, it contains
32.69g of zinc and 16.03g of sulfur
%Zn = (32.69g/ 48.72g) x 100% = 67.10%
%S = (16.03g/ 48.72g) x 100% = 32.90%
Practice:
Calculate the mass percentage of nitrogen
in each compound:
1.
a)
b)
c)
d)
e)
f)
N2O
Sr(NO3)2
NH4NO3
HNO3
NH4
KNO3
Empirical Formulas
 The empirical formula is simply the simplest whole
number ratio of the elements in a compound.
 The molecular formula is the actual number of atoms
that makes up the molecule
 For example,
 the empirical formula of ethene (C2H4) is CH2
 the empirical formula of butene (C4H8) is CH2
Comparing
Name
Molecular Empirica Low
l
ratio
Hydrogen peroxide H2O2
HO
1:1
Glucose
C6H12O6
CH2O
1:2:1
Benzene
C6H6
CH
1:1
Ethene
C2H4
CH2
1:2
Butene
C4H8
CH2
1:2
Aniline
C6H7N
C6H7N
6:7:1
Water
H2O
H2O
2:1
Finding Empirical formula:
 Calculate empirical formula of a compound
that is 85.6% carbon and 14.4% hydrogen.
First, assume that you have 100g of the
substance, so you really have 85.6g of C and
14.4g of H.
Next, convert it to moles using molar mass.
Finally, find the lowest whole ratio between the
two.
85.6g C and 14.4g H
2. mol C = 85.6g x mol = 7.13mol C
12.01g
mol H = 14.4g x mol = 14.3 mol H
1.01g
3. Lowest mole ratio:
C = 7.13 /7.13 = 1
H = 14.3/ 7.13 = 2.01 , so 2
ANSWER: CH2
1.
Practice:
Find the empirical formula for the following:
17.5% hydrogen and 82.5% nitrogen
2. 46.3% lithium and 53.7% oxygen
3. 15.9% boron and 84.1% fluorine
4. 52.52% chlorine and 47.48% sulfur
1.
Finding Molecular Formula
The empirical formula is CH2O, and its molar
mass is 150g/mol. What is molecular
formula?
 Find empirical molar mass of CH2O: 12.01 +
2.02 + 15.99 g/mol =30.02 g/mol
 Divide real molar mass by empirical molar
mass: 150/30.02 = 5
 Multiply subscripts of empirical formula
through by ratio (5): C5H10O5