8. Using moles to
calculate empirical formula deduce
molecular formula
of a compound/molecule starting with reacting masses or
% composition by mass
The basis of this method, is that a mole of
defined species has the same number of defined 'particles' in it. So calculating
the mole ratio of atoms in the combination, gives you the actual atomic ratio in
the compound. This ratio is then expressed in the simplest whole number
atomic ratio, which is the empirical
formula. (it means the formula 'from experiment', see examples 1 and 2). (see section 3. for some simpler examples not using
moles)
However, there is one problem to resolve for
covalent molecular compounds.
The molecular formula is
the summary of all the atoms in one individual molecule  it shows how many
of each different type of atoms is present in one molecule.
The empirical formula
of a compound is the simplest whole number ratio of atoms present in a compound.
Here, the word 'empirical' means derived from experimental
data.
Do not confuse with molecular formula
which depicts the actual total numbers of each atom in a molecule.
Therefore, the
molecular formula might not be the same as the empirical formula!
In order to
deduce the molecular formula from the empirical formula, you ALSO need to know
the molecular mass of the molecule from another data source. See example 3,
which also illustrates the calculation using % element composition AND the
method is no different for more than two elements!
Examples where molecular formula = empirical
formula
e.g. sodium sulfate Na_{2}SO_{4}
and propane C_{3}H_{8}
You cannot simplify the atomic ratios 2 : 1 :
4 or 3 : 8 to smaller whole number (integer) ratios
Examples of where molecular formula and empirical
formula are different
e.g. butane molecular formula C_{4}H_{10},
empirical formula C_{2}H_{5}
numerically, the empirical formula of
butane is 'half' of its molecular formula
4 : 10 ==> 2 : 5
glucose molecular formula C_{6}H_{12}O_{6},
empirical formula CH_{2}O
numerically, the empirical formula of
glucose is '^{1}/_{6}th' of the molecular formula
6 : 12 : 6 ==> 1 : 2 : 1
Suppose
you start with a molecular structure like dodecane.
If you count the atoms you find the molecular formula
is C_{12}H_{26}.
BUT the simplest ratio formula, that is the empirical
formula, is 'half' of the molecular formula i.e. C_{6}H_{13}.
The molecule on the left is an ester called ethyl
ethanoate.
From the diagram you deduce that the molecular formula
is C_{4}H_{8}O_{2}.
From the molecular formula you deduce that the
empirical formula is C_{2}H_{4}O.
AND don't forget to be able to think, calculate and
deduce the other way round e.g.
(i) Suppose a hydrocarbon molecule has an empirical
formula of C_{2}H_{5} and a molecular mass of 58 (C = 12, H
= 1).
Deduce its molecular formula. The empirical
formula mass = (2 x 12) + 5 = 29.
Dividing 58 by 29 gives 2. So the molecular
formula is 2 x the empirical formula = C_{4}H_{10}
(ii) Suppose a molecule has an empirical formula of
simply CH, but a molecular mass of 78 (C = 12, H = 1).
The empirical formula mass is 12 + 1 = 13. Therefore
78/13 = 6, so the molecular formula is 6 x CH = C_{6}H_{6}
Note the simple numerical factors connecting
the empirical formula and molecular formula
empirical formula x n = molecular formula
where n = 1, 2, 3 etc. If n = 1 then they are
the same formula!
BUT remember, if you have to suggest a structure
for a molecule you must know at least its molecular formula because where
the empirical formula and molecular
formula are different, you need extra information to deduce the molecular
formula from the empirical formula (examples set out further on down the
page).
Simple empirical formula calculations NOT using moles were covered in section 5.
Examples of where the empirical formula is the
same as the molecular formula ...
water H_{2}O, methane CH_{4},
pentane C_{5}H_{12} (these molecular formula cannot be
'simplified')
Examples of where the molecular formula is
different from the empirical formula in () ...
e.g. ethane C_{2}H_{6} (CH_{3}),
phosphorus(V) oxide P_{4}O_{10} (P_{2}O_{5})
Three examples are set out below to illustrate
all the situations and how to calculate an empirical formula and a molecular
formula from reacting masses or % element composition by mass.
For nonmolecular compounds (i.e.
inorganic ionic compounds), the empirical formula is usually, but not always,
quoted as the compound formula.
The relative atomic masses of the elements (A_{r})
are given in the tabular format method of solving the problem.
Empirical formula calculation Example 8.1:
The compound formed between sodium and sulfur
1.15g of sodium reacted with 0.8g of
sulphur. Calculate the empirical formula of sodium sulphide.
You convert the masses to moles i.e.
mass in g divided by the relative atomic mass.
Since one mole of any defined
substance contains the same number of particles (e.g. atoms), it means that the
atomic mole ratio is also the actual ratio of atoms in the compound.
The ratio is then expressed as the
simplest whole ratio from which the empirical formula is derived.
Apart from a 1, other numbers e.g.
2, 3 etc. should be seen as subscripts in the empirical formula.
RATIOS ... 
Sodium Na (A_{r} = 23.0) 
Sulphur S (A_{r }= 32.0) 
Comments and tips 
masses 
1.15g 
0.80g 
not the
real atom ratio 
moles (mass in g / A_{r}) 
1.15 / 23 = 0.05 mol 
0.8 / 32 = 0.025 mol 
can
now divide by smallest ratio number or scale up by x factor to get
simplest whole number ratio 
atom ratio = simplest whole number mole ratio
by trial and error 
0.05 / 0.025 = 2
or 0.05 x 40 = 2

0.025 / 0.025 = 1
or 0.025 x 40 = 1

therefore the
simplest integer ratio = 2 : 1, so empirical formula for
sodium sulphide = Na_{2}S 
Empirical formula calculation Example 8.2
The empirical formula of aluminium oxide
1.35g of aluminium was heated in
oxygen until there was no further gain in weight. The white oxide ash formed
weighed 2.55g. Deduce the empirical formula of aluminium oxide. Note: to get the
mass of oxygen reacting, all you have to do is to subtract the mass of metal
from the mass of the oxide formed.
RATIOS ... 
Aluminium Al (A_{r} = 27.0) 
Oxygen O (A_{r }= 16.0) 
Comments and tips 
masses 
1.35g 
2.55  1.35 = 1.2g 
not the
real atom ratio 
moles (mass in g / A_{r}) 
1.35 / 27 = 0.05 mol 
1.2 / 16 = 0.075 mol 
can
now divide by smallest ratio number or scale up by x factor to get
simplest whole number ratio 
atom ratio = simplest whole number mole ratio
by trial and error 
0.05 / 0.05 = 1
(then x 2 = 2)
or 0.05 x 40 = 2

0.075 / 0.05 = 1.5
(then x 2 = 3)
or 0.075 x 40 = 3

therefore the
simplest integer ratio is 2 : 3, so empirical formula for aluminium oxide =
Al_{2}O_{3} 
Empirical formula calculation
Example 8.3 for an oxide of iron.
1.448g of iron was heated in air in
a crucible until no further gain in weight was observed.
The final mass of the iron oxide was
found to be 2.001g
Calculate the empirical formula of
the iron oxide.
Atomic masses: Fe = 56 and
O = 16
The mass of oxygen combined with the
iron is deduced by subtracting the original mass of iron from final total mass
of iron oxide.
RATIOS ... 
iron (A_{r} = 56.0) 
Oxygen O (A_{r }= 16.0) 
Comments and tips 
masses 
1.448g 
2.001  1.448 = 0.553g 
not the
real atom ratio 
moles (mass in g / A_{r}) 
1.448 / 56 = 0.0259 mol 
0.553 / 16 = 0.0346 mol 
can
now divide by smallest ratio number or scale up by x factor to get
simplest whole number ratio, in this case you have to make a reasonable
judgement as to the values of the integers 
atom ratio = simplest whole number mole ratio
by trial and error 
0.0259 / 0.0259 = 1
(then x 3 = 3)
3
you arrive at this 
0.0346 / 0.0259 = 1.336 (then x
3 = 4.008)
~4
stage by trial and error! 
therefore the
simplest integer ratio = 3 : 4, so empirical formula for
the iron oxide = Fe_{3}O_{4} 
Example 8.4
A compound of contains 43.4% sodium, 11.3 %
carbon and 45.3 % of oxygen.
Atomic masses: Na = 23.0; C = 12.0; O = 16
(a) Using the % composition by mass, calculate a mole
ratio for the three constituent elements
mole ratio: Na = 43.4/23.0 = 1.887; C =
11.3/12.0 = 0.942; O 45.3/16 = 2.831
Na
: C : O is 1.887 : 0.942 : 2.831
(b) Deduce the empirical formula of the compound
Take the above ratio and divide by the smallest
number gives
1.887/0.942 : 0.942/0.942 : 2.832/0.942
gives 2.003 : 1.0 : 3.006
within a reasonable experimental error, this is
pretty clearly a whole number ratio for Na : C : O of
2 : 1 : 3
Therefore the
empirical formula of the compound is
Na_{2}CO_{3}
Example 8.5
A copper ore compound contains 51.5% copper, 9.7%
carbon and an undetermined % of oxygen.
Atomic masses: Cu = 63.5; C = 12.0; O = 16
(a) Calculate the % oxygen in the compound.
(a) Using the % composition by mass, calculate a
mole ratio for the three constituent elements
(b) Deduce the empirical formula of the compound
Do this one for yourself and you should get
CuCO_{3}
Examples 8.1 to 8.5 involved ionic compounds, so no
molecular formula or molecular mass were involved.
However examples 8.6 to 8.9 involve covalent molecules and
both empirical formula and molecular formula, and the connection between the
two.
You are likely
to encounter these sort of problems when studying advanced level organic
chemistry.
Relative atomic
masses: C = 12.0; H = 1.0; O = 16.0; Cl = 35.5
Empirical formula and molecular
formula calculation Example 8.6
for a hydrocarbon compound
On analysis a hydrocarbon was found
to consist of 81.8% carbon and 18.2% hydrogen.
Molecular ion measurements in a mass
spectrometer show that the hydrocarbon has a molecular mass of 44  from the
molecular ion peak.
Treat the percentages as if they
were masses in grams, and it all works out fine.
RATIOS ... 
carbon (A_{r} = 12.0) 
hydrogen (A_{r }= 1.0) 
Comments and tips 
masses 
81.8 
18.2 
not the
real atom ratio 
moles (mass in g / A_{r}) 
81.8 / 12 = 6.817 
18.2 / 1 = 18.2 
can
now divide by smallest ratio number or scale up by x factor to get
simplest whole number ratio in this case you have to make a reasonable
judgement as to the values of the integers 
atom ratio = simplest whole number mole ratio
by trial and error 
6.817/6.817 = 1.0
1.0 x 2 = 2
1.0 x 3 = 3.0
you arrive at this 
18.2/6.817 = 2.67
2.670 x 2 = 5.34
2.670 x 3 = 8.01 ~8.0
stage by trial and error! 
therefore the
simplest integer ratio = 3 : 8, so empirical formula for
the hydrocarbon = C_{3}H_{8}
The empirical formula mass = (3 x 12) + 8 =
44
This equals the molecular mass, therefore
the molecular formula is also C_{3}H_{8} 
Empirical formula and molecular
formula calculation Example 8.7
for another hydrocarbon compound
On analysis a hydrocarbon was found
to consist of 83.72% carbon and 16.28% hydrogen.
Molecular ion measurements in a mass
spectrometer show that the hydrocarbon has a molecular mass of 86  from the
molecular ion peak.
Again, treat the percentages as if
they were masses in grams, and it all works out fine.
RATIOS ... 
carbon (A_{r} = 12.0) 
hydrogen (A_{r }= 1.0) 
Comments and tips 
masses 
83.72 
16.28 
not the
real atom ratio 
moles (mass in g / A_{r}) 
83.72 / 12 = 6.977 
16.28 / 1 = 16.28 
can
now divide by smallest ratio number or scale up by x factor to get
simplest whole number ratio, a bit fiddly to find the ratio on this one
to deduce the empirical formula 
atom ratio = simplest whole number mole ratio
by trial and error 
6.977/6.977 = 1.0
1.0 x 2 = 2.0
1.0 x 3 = 3.0
you arrive at this 
16.28/6.977 = 2.333
2.333 x 2 = 4.667
2.333 x 3 = 7.00
stage by trial and error! 
therefore the
simplest integer ratio = 3 : 7, so empirical formula for
the hydrocarbon = C_{3}H_{7}
The empirical formula mass = (3 x 12) + 7 =
43
BUT the molecular mass is double this, so
the molecular formula must be double the empirical formula
Therefore the molecular formula is
C_{6}H_{14} of which there are many structural
isomers! You would need more information to
deduce the structural formula of this hydrocarbon. 
Empirical formula and molecular
formula calculation Example 8.8
for a carbohydrate sugar compound
A carbohydrate compound e.g. a
sugar, was found on analysis to contain 40.00% carbon, 6.67% hydrogen and 53.33%
oxygen.
The molecular mass was 150. From the
information calculate the empirical formula and deduce the molecular formula.
RATIOS ... 
Carbon (A_{r} =
12.0) 
Hydrogen (A_{r }=
1.0) 
Oxygen (A_{r} = 16.0) 
Comments and tips 
masses 
40.00 
6.67 
53.33 
just think of it as based
on 100g 
molar ratio (mass in g / A_{r}) 
40.00 / 12 = 3.333 mol 
6.67 / 1 = 6.67 mol 
53.33 / 16.0 = 3.333 mol 
can
now divide by smallest ratio number 
atom ratio = simplest whole number mole ratio
by trial and error 
3.333/3.333 = 1.0 
6.67/3.333 ~2.0 
3.333/3.333 = 1.0 
therefore the
simplest integer ratio = 1 : 2 : 1, so empirical formula for
the sugar = CH_{2}O
The empirical formula mass = 12 + 2 + 16 = 30
BUT the molecular mass is 5 x this
(150/30), so the molecular formula must be 5 x the empirical formula
Therefore the molecular formula is
C_{5}H_{10}O_{5} , a pentose, of which there are many
structural isomers! You would need more
information to deduce the structural formula of this sugar molecule. 
Empirical formula and molecular
formula calculation Example 8.9
for a chloroalkane compound
A chlorinated hydrocarbon compound when
analysed, consisted of 24.24% carbon, 4.04% hydrogen, 71.72% chlorine.
The
molecular mass was found to be 99 from another experiment.
Deduce the empirical
and molecular formula.
You can 'treat' the %'s as if they were grams,
and it all works out like examples 1 and 2, i.e. based on a total mass of 100g.
RATIOS ... 
Carbon (A_{r} =
12) 
Hydrogen (A_{r }=
1) 
Chlorine (A_{r} = 35.5) 
Comments and tips 
Reacting mass or % mass 
24.24 
4.04 
71.72 
just think of it as based
on 100g 
molar ratio (mass in g / A_{r}) 
24.24 / 12 = 2.02 mol 
4.04 / 1 = 4.04 mol 
71.72 / 35.5 = 2.02 mol 
can
now divide by smallest ratio number 
atom ratio = simplest whole number mole ratio
by trial and error 
2.02 / 2.02 = 1 
4.04 / 2.02 = 2 
2.02 / 2.02 = 1 
therefore the
simplest atomic ratio = 1 : 2 : 1, so empirical formula for
the chlorinated hydrocarbon = CH_{2}Cl
BUT the molecular mass is 99, and the empirical
formula mass is 49.5 (12+2+35.5)
AND 99 / 49.5 = 2, and so the molecular
formula must be 2 x CH_{2}Cl =
C_{2}H_{4}Cl_{2}
two possible structures, which cannot be
distinguished by the data given or calculation above
1,2dichloroethane and 1,1dichloroethane,
two possible atom arrangements of the same molecular formula known as
structural positional isomers, because one of the chlorine atoms can
be in two different positions.

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Selfassessment Quizzes on determining empirical formula and
molecular formula
type in answer
QUIZ or multiple choice
QUIZ
Above is typical periodic table used in GCSE sciencechemistry specifications in
doing reacting gas volume ratio calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
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OTHER CALCULATION PAGES

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Calculating relative
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Law of Conservation of Mass and simple reacting mass calculations

Composition by percentage mass of elements
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Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)

Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination

Introducing moles: The connection between moles, mass and formula mass  the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)

Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
(this page)

Moles and the molar volume of a gas, Avogadro's Law

Reacting gas volume
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Molarity, volumes and solution
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