Empirical and Molecular Formulas
In this presentation you will:
 explore the difference between empirical and
molecular formulas
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Introduction
There are two different
types of formulas for a
compound.
These are the
empirical and
molecular formulas.
The empirical formula gives the simplest ratio
of the number of atoms present in a molecule of
a compound.
The molecular formula gives the actual number of
atoms of each element present in a compound. Next >
Empirical Formula
The empirical formula is often used when dealing
with large organic molecules.
These may have many
hundreds of atoms, or,
in ionic or giant
covalent lattice
structures, millions of
atoms, bound together
in a fixed ratio.
Empirical
formula = HO
Empirical
formula = NaCl
For many simple inorganic
substances, the empirical formula is
the same as the molecular formula.
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Example of Empirical and Molecular
Formulas
Phosphorus(V) oxide exists as P4O10 molecules.
This is therefore the molecular
formula.
This can be simplified
to P2O5.
This is the empirical formula.
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Percentage Composition
The idea of percentage composition can be used to
help us calculate empirical formulas.
This is the percentage by mass of a particular
element in a chemical compound.
The chemical formula of the
compound gives the number
of molecules of each element
in the compound, but not
their mass.
So, in methane, CH4, there is
one molecule of carbon and
four molecules of hydrogen.
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Percentage Composition
To calculate the percentage composition:
% of element in compound =
mass of element in sample of compound × 100
mass of sample of compound
However, there is an alternative method:
% of element in compound =
mass of element in 1 mol of compound × 100
molar mass of compound
Using this formula, the percentage composition
can be calculated.
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Percentage Composition
Calculate the percentage composition of
methane (CH4):
% of element in compound =
mass of element in 1 mol of compound × 100
molar mass of compound
For hydrogen:
% of element in compound =
mass of hydrogen in 1 mol of CH4 × 100
molar mass of CH4
= 4 × 1.01 × 100 = 25.2 to 3 s.f.
16.05
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Percentage Composition
For carbon:
% of element in compound =
mass of carbon in 1 mol of CH4 × 100
molar mass of CH4
= 12.01 × 100 = 74.83
16.05
Check: 25.2% + 74.83% ≈ 100%
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Question 1
Find the percentage compositions of PbCl2 given
that the relative atomic masses of Pb and Cl are
207.2 and 35.45 respectively.
A) Pb: 69.54% Cl: 30.46%
B) Pb: 73.99% Cl: 26.01%
C) Pb: 74.51% Cl: 25.49%
D) Pb: 76.05% Cl: 23.95%
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Question 1
Find the percentage compositions of PbCl2 given
that the relative atomic masses of Pb and Cl are
207.2 and 35.45 respectively.
A) Pb: 69.54% Cl: 30.46%
B) Pb: 73.99% Cl: 26.01%
C) Pb: 74.51% Cl: 25.49%
D) Pb: 76.05% Cl: 23.95%
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Percentage Composition
The empirical formula
of a compound can be
calculated from data
which gives the
percentage
composition, by mass,
of each element in the
compound.
50 g of unknown substance X
20 g C : 3.3 g H : 26.7 g O
Mass 20
3.3
26.7

C:
H:
O
Ar
12
1
16
= 1.67 C : 3.3 H : 1.67 O
The molecular formula
can then be deduced if
the relative molecular
mass is known.
= CH2O
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Calculation of Empirical Formula
The empirical formula of a compound can be
calculated, with the help of the periodic table, if its
mass and the percentages of the components are
known.
We are given a 100 g sample of a compound
containing 40% carbon, 6.7% hydrogen and 53.3%
oxygen by mass.
The masses of the elements are therefore:
C = 40 g, H = 6.7 g and O = 53.3 g.
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Calculation of Empirical Formula
The number of moles, according to the
formula mol = mass / Ar is:
Carbon: 40 / 12 = 3.3
Hydrogen: 6.7 / 1 = 6.7
Oxygen: 53.3 / 16 = 3.3
To obtain the empirical formula, divide the number
of moles of each element by the smallest number
and round the results to the nearest integers:
3.3 : 6.7 : 3.3 → 1 : 2 : 1, so the empirical formula
is CH2O.
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Experimental Determination
Experimental data can also be used to find the
molecular mass of compounds and hence their
empirical formula.
A useful result of this is the ability to determine the
amount of water of crystallization present in a
salt sample; this is water bound within the crystal
lattice structure of the compound in a solid state.
20.00 g of a sample of iron(II) sulfate, FeSO4·xH2O,
is found by analysis to contain 4.03 g of pure iron
(Fe). What is the water of crystallization (x) of the
sample of iron(II) sulfate?
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Experimental Determination
Mass of anhydrous iron sulfate =
mass (Fe) × (Mr (FeSO4) / Ar (Fe))
Mass of anhydrous iron sulfate =
4.03 × (152 / 56) = 10.94 g
Mass of water of crystallization =
20.00 g – 10.94 g = 9.06 g
Moles water = 9.06 / 18 =
0.50 mol; Moles iron = 4.03 / 56 = 0.072 mol
Ratio = 0.072 Fe : 0.5 H2O =
1:7 by quantity, so formula is FeSO4·7H2O
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Question 2
The molar ratio of two elements in a sample of an
ionic substance was found to be 2.4 : 4.8. If the two
elements in question are Mg and Cl, what is the
empirical formula for this magnesium chloride
compound?
A) MgCl
C) MgCl2
B) Mg2Cl
D) Mg2Cl4
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Question 2
The molar ratio of two elements in a sample of an
ionic substance was found to be 2.4 : 4.8. If the two
elements in question are Mg and Cl, what is the
empirical formula for this magnesium chloride
compound?
A) MgCl
C) MgCl2
B) Mg2Cl
D) Mg2Cl4
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Question 3
What information does the empirical formula of a
compound give?
A) The absolute number of atoms in a compound
B) The simplest ratio of atoms in a compound
C) The number of atoms in one mole of a compound
D) The number of moles of a compound in a sample
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Question 3
What information does the empirical formula of a
compound give?
A) The absolute number of atoms in a compound
B) The simplest ratio of atoms in a compound
C) The number of atoms in one mole of a compound
D) The number of moles of a compound in a sample
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Determination of Molecular Formula
Once we know the empirical formula, we can go on
to calculate the molecular formula.
Suppose the empirical formula comes to AB2.
We know that the molecular formula must be a
whole number multiple of this formula: AB2 or A2B4
or A3B6 and so on.
Molecular formula = x(empirical formula) where x is
a whole number.
Alternatively:
Molecular formula mass = x(empirical formula mass)
where x is a whole number.
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Determination of Molecular Formula
Molecular formula mass = x(empirical formula mass)
where x is a whole number.
The empirical formula
mass is found by adding
together the masses for x = molecular formula mass
empirical formula mass
each of the atoms in the
empirical formula.
The molecular formula x =
molar mass
mass is the molar mass
empirical formula mass
of the compound.
Dividing the molar mass by the empirical formula
mass should give x.
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Example
Determine the molecular formula of the compound
with an empirical formula of CH and a formula mass
of 78.110 atomic mass units.
The empirical formula mass is
12.01 + 1.01 = 13.02.
x = 78.11
13.02
x=6
The molecular formula is C6H6.
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Question 4
A sample of a compound, with a formula mass of
34.01 atomic mass units, is found to consist of
0.44 g H and 6.92 g O.
Find its molecular formula, given the relative atomic
mass of hydrogen is 1.01 and of oxygen is 16.00.
A) HO
B) H2O2
C) H2O
D) HO2
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Question 4
A sample of a compound, with a formula mass of
34.01 atomic mass units, is found to consist of
0.44 g H and 6.92 g O.
Find its molecular formula, given the relative atomic
mass of hydrogen is 1.01 and of oxygen is 16.00.
A) HO
B) H2O2
H + O = 17.01
C) H2O
34.01/17.01= 2 ignoring rounding up errors
D) HO2
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Summary
In this presentation you have seen:
 the difference between empirical and
molecular formulas
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