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Chemical
formulae and equations
Francine
Taylor-Campbell, Contributor
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Students
of Central High School tours
the Gleaner's plant and office,
North Street, Kingston, on Tuesday,
February 7. - Winston Sill Photo
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POINTS
TO NOTE
The formula of a compound shows how
many atoms of each element are present
in a molecule or formula unit.
The empirical formula is the simplest
formula, which represents the composition
of the compound.
The actual formula is called the molecular
formula. It is generally a multiple
of the empirical formula and is calculated
from the molar mass.
PERCENTAGE
COMPOSITION
1.
Calculate the Mr
for the compound
e.g.,
Ammonium phosphate (NH4)3PO4
Mr
= (14 + (4*1))*3 + 31 + (4*16) = 149
1 mole = 149g
2.
Calculate the mass of each element
in one mole
N
= 3*14 = 42g H = 3*4*1 = 12g P = 31g
O = 4* 16 = 64g
3.
Calculate the percentage of each element
N
= (42/149)* 100 = 28.2% H = (12/149)*100
= 8.1% P = (31/149)*100 = 20.8%
O
= (64/149)*100 = 43.0%
EMPIRICAL
AND MOLECULAR FORMULAE
The
formula of glucose is given as C6H12O6.
This shows that the compound glucose
is made of six atoms of carbon, 12
atoms of hydrogen and six atoms of
oxygen. This is the molecular formula.
The
empirical formula of glucose is CH2O
and is the whole number ratio of the
elements in this compound. To find
the empirical formula from combustion
or percentage composition data, the
number of moles must be found.
NOTE:
Number of moles = Mass (g)
Molar
mass (g/mol)
1.
Calculate the empirical formula of
the compounds with the following percentage
composition:
(i)
34.5% Fe, 65.5% Cl
METHOD
| Elements
present |
Iron
(Fe)
|
Chlorine
(Cl)
|
| Percentage
by mass |
34.5%
|
65.5%
|
| Mass
of element in 100g of
the compound |
34.5g
|
65.5g
|
| Relative
atomic mass |
56
|
35.5
|
| Number
of moles of element |
34.5/56
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65.5/35.5
|
| |
=
0.616
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=
1.85
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| Ratio
of moles |
0.616/0.616
|
1.85/0.616
|
| |
=
1
|
=
3
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| Empirical
formula = FeCl3 |
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2.
Calculate the empirical formula of
the compounds formed in the following
reactions:
(ii)
3.40g calcium form 9.435g of the chloride
METHOD
| Elements
present |
Calcium
(Ca)
|
Chlorine
(Cl)
|
| Mass
of each element |
3.40g
|
9.435-3.40
= 6.035g
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| Relative
atomic mass |
40
|
35.5
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| Number
of mole |
3.40/40
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6.035/35.5
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=
0.085
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=
0.17
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| Ratio
of moles |
0.085/0.085
|
0.17/0.085
|
| |
=
1
|
=
2
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| Empirical
formula = CaCl2 |
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In
determining molecular formula the
empirical formula and/or the molar
mass must be known.
Calculate
the molecular formula of the hydrocarbon
containing 85.7% carbon given that
the molar mass is 56g/mol.
METHOD
1: 85.7% C 14.3% H
Method
2: Mass of carbon = 85.7% * 56
= 48g
Mass
in 100g 85.7g 14.3g Mass of hydrogen
= 14.3%*56 = 8g
No.
of moles 85.7/12 14.3/1 1 atom of
carbon = 12g
=
7.14 = 14.3 1 atom of hydrogen = 1g
Ratio
of elements 1 : 2 # of carbon atoms
= 48/12 = 4
Empirical
formula = CH2, Mr = (12+2 = 14g) #
of hydrogen atoms = 8/1 = 8
Molecular
formula: (CH2)n = 56 Molecular formula
= C4H8
14n
= 56 & n = 4
Formula
= (CH2)4
= C4H8
*
Francine Taylor-Campbell is
an independent contributor.
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