CALCULATION of EMPIRICAL
FORMULA
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5.
Simple empirical formula and formula mass from reacting masses
or % composition
(easy start, no moles involved !)
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empirical formula from reacting masses or % composition by mass of a
compound. This page describes and
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a compound. The empirical formula of a compound is defined and
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5.
Empirical formula and formula mass from reacting masses (easy start, no moles!)
The EMPIRICAL FORMULA of a compound can be worked out by knowing the exact masses of the elements that combine to form a given mass of a compound.
The empirical formula
of a compound is the simplest whole number ratio of atoms present in a compound. (see section 3. for some simpler
examples). Here the word 'empirical' means from experimental
data.
Do not confuse with molecular formula
which depicts the actual total numbers of each atom in a molecule.
The molecular
formula and empirical formula can be different or the same.
They are the same if
the molecular formula cannot be simplified on a whole number basis.
Examples where molecular formula = empirical
formula
e.g. for 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
Similarly for
pentane, the diagram shows it's molecular formula is C_{5}H_{12},
which cannot be simplified, so the molecular formula = empirical formula.
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 full molecular formula
6 : 12 : 6 ==> 1 : 2 : 1
Suppose
you start with a molecular structure like octane (left diagram)
If you count the atoms you find the molecular formula
is C_{8}H_{18}.
BUT the simplest ratio formula, that is the empirical
formula is 'half' of the molecular formula i.e. C_{4}H_{9},
but remember, if you have to suggest a structure for a molecule you must know at
least its molecular formula.
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}
Where the empirical formula and molecular
formula are different, you need extra information to deduce the molecular
formula from the empirical formula (see link below).
This page is only concerned with
calculating empirical formula.
For more advanced students see
Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
The following examples illustrate the ideas using numbers more easily appreciated than in real experiments.
In real laboratory experiments only a fraction of a gram or a few grams of elements would be used, and a more 'tricky' mole calculation method is required than shown here (dealt with later for higher students
in section 8).
However the examples below show in principal how formulae are worked out from experiments.
Any calculation method must take into account the different relative atomic masses of the elements in order to get to the actual ratio of the atoms in the formula.
For example, just because 10g of X combines with 20g of Y,
it does not mean that the formula of the compound is
XY_{2 }!
If you divide the mass of each element
by its atomic mass, you actually get the atomic ratio.

Empirical formula calculation Example 5.1
The compound formed between lead and sulfur
 It is found that 207g of lead combined with 32g of sulphur to form 239g of lead sulphide.
 From the data work out the formula of lead sulphide. (Relative atomic masses: Pb = 207 and S = 32)
 In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of
1 atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass)
 so the formula is simply
PbS
 You can set out the calculation
in a simple table format, in this case the numbers are very easy
to deal with!

RATIOS ... 
lead (A_{r} = 207) 
sulphur S (A_{r }= 32) 
Comments and tips 
Reacting mass 
207g 
32g 
not the
real atom ratio 
atom ratio from mass / atomic
mass values 
207/207 = 1 
32/32 = 1 
work out the
simplest whole number ratio 
simplest whole number atom ratio
by trial & error 
1 
1 
therefore the
integer simplest ratio of 1 : 1 gives the empirical formula
for lead sulphide as PbS 
 

Empirical formula calculation Example 5.2
The empirical formula of a lead oxide

Empirical formula calculation Example 5.3
The empirical formula of aluminium sulfide
 It is found that 54g of aluminium forms 150g of aluminium sulphide.
 Work out the formula of aluminium sulphide. (Relative atomic masses: Al = 27 and S = 32).
 Amount of sulphur combined with the aluminium = 150  54 = 96g
 By atomic ratio, the 54 of aluminium is equivalent to
2 atoms of aluminium and the
96 of sulphur is equivalent to 3 atoms of sulphur.
 Therefore the atomic ratio is 2 to 3,
 so the formula of aluminium sulphide is
Al_{2}S_{3}

RATIOS ... 
aluminium (A_{r} =
27) 
sulfur S (A_{r }=
32) 
Comments and tips 
Reacting mass 
54g 
15054 = 96g 
not the
real atom ratio 
atom ratio from mass / atomic
mass values 
54/27 
96/32 
work out the
simplest whole number ratio 
simplest whole number atom ratio
by trial & error 
2 
3 
therefore the
simplest integer ratio of 2 : 3 gives the empirical formula
for aluminium sulphide as Al_{2}S_{3} 
 

Empirical formula calculation Example 5.4 From now on, questions
just using
the table method to work out empirical formula from more awkward
numbers! In this case a compound formed between copper and
chlorine.

A compound of copper
contained 47.4% copper and 52.6% chlorine.

The atomic masses are:
Cu = 64 and Cl = 35.5

Think of the percentages
as masses in grams to solve the empirical formula problem.

RATIOS ... 
Cu 
Cl 
Comments and tips 
Reacting mass 
47.4 
52.6 
not the
real atom ratio 
atom ratio from mass / atomic
mass values 
47.4/64 = 0.74 
52.6/35.5 = 1.48 
work out the
simplest whole number ratio 
simplest whole number atom ratio
by trial & error 
0.74/0.74 = 1.0 
1.48/0.74 = 2.0 
therefore the
simplest whole number ratio of 1 : 2 gives the empirical
formula for copper chloride as CuCl_{2}
Its actually called
copper(II) chloride 
 

Empirical formula calculation
Example 5.5 The empirical formula of a
compound of carbon and chlorine

Empirical formula calculation Example 5.6 The formula of a
hydrocarbon.

It was that 0.75g of carbon was combined with 0.25g of hydrogen.

Atomic masses: C
= 12 and H = 1

Calculate the
empirical formula of the hydrocarbon

RATIOS ... 
C 
H 
Comments and tips 
Reacting mass 
0.75g 
0.25g 
not the
real atom ratio 
atom ratio from mass /
atomic mass values 
0.75/12 = 0.0625 
0.25/1 = 0.25 
work out the
simplest whole number ratio 
simplest whole number
atom ratio
by trial & error 
0.0625/0.0625 = 1.0 
0.25/0.0625 = 4.0 
therefore the
simplest ratio gives the empirical formula for
the hydrocarbon = 1 : 4, so formula is CH_{4}
This is the simplest
hydrocarbon molecule called methane (main
constituent in natural gas) 
 

Empirical formula
calculation Example 5.7 The analysis of sodium
sulfate, calculating its empirical formula from the % composition by
mass.

On analysis, the
salt sodium sulfate was found to contain 32.4% sodium, 22.5%
sulfur and 45.1% oxygen.

Atomic masses: Na
= 23, S = 32 and O = 16

Calculate the
empirical formula of sodium sulfate

RATIOS ... 
Na 
S 
O 
Comments and tips 
Reacting mass 
32.4 
22.5 
45.1 
not the
real atom ratio 
atom ratio from mass /
atomic mass values 
32.4/23 = 1.41 
22.5/32 = 0.70 
45.1/16 = 2.82 
work out the
simplest whole number ratio, in this case you have to make a reasonable
judgement as to the values of the integers 
simplest whole number
atom ratio
by trial & error 
1.41/7 = 2.01 ~2.0 
0.70/0.70 = 1.0 
2.82/0.70 = 4.03 ~4.0 
the
simplest ratio gives the empirical formula for
sodium sulfate = 2 : 1 : 4, formula is Na_{2}SO_{4} 
 

Empirical formula calculation Example 5.8 The formula of a
hydrocarbon.

Analysis of
hydrocarbon showed it consisted of 83.3% carbon and 16.7%
hydrogen.

Atomic masses: C
= 12 and H = 1

Calculate the
empirical formula of the hydrocarbon (just think of it as 83.3g
C combined with 16.7g H)

RATIOS ... 
C 
H 
Comments and tips 
Reacting mass 
83.3 
16.7 
not the
real atom ratio 
atom ratio from mass /
atomic mass values 
83.3/12 = 6.94 
16.7/1 = 16.7 
work out the
simplest whole number ratio In this
case from the 1 2.4 to the 5:12 ratio, you have to
multiply the 2.4 up until you get a whole number, x2, x3
and x4 don't work, but x5 does! 
simplest whole number
atom ratio
by trial & error 
6.94/6.94 = 1.0 
16.7/6.94 = 2.4 
This is a bit awkward! 
1.0 x 5 = 5.0 
2.4 x 5 = 12.0 
therefore the
simplest ratio gives the empirical formula for
the hydrocarbon = 5 : 12, so formula is C_{5}H_{12}
This is the simplest
hydrocarbon molecule called pentane 
 


See section 8. for more
empirical/molecular formula calculations involving moles.
Selfassessment Quizzes: type in answer
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QUIZ
Above is typical periodic table used in how to deduce and work out empirical
formula sciencechemistry specifications in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
OTHER CALCULATION PAGES

What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass

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

Composition by percentage mass of elements
in a compound

Empirical formula & formula mass of a compound from reacting masses
(easy start, not using moles)
(this page)

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)

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

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ratios, Avogadro's Law
and GayLussac's Law (ratio of gaseous
reactantsproducts)

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