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Chemical
formulae and equations
Francine
Taylor-Campbell, Contributor
REVIEW
OF FACTS
- 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,
for example, 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 6 atoms of carbon, 12
atoms of hydrogen and 6 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)
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 |
65.5/35.5 |
| |
=
0.616 |
=
1.85 |
| Ratio
of moles |
0.616/0.616 |
1.85/0.616 |
| |
=
1 |
=
3 |
| Empirical
formula = FeCl3 |
|
|
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 |
| Relative
atomic mass |
40 |
35.5 |
| Number
of moles |
3.40/40 |
6.035/35.5 |
| |
=
0.085 |
=
0.17 |
| Ratio
of moles |
0.085/0.085 |
0.17/0.085 |
| |
=
1 |
=
2 |
| 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 |
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Francine
Taylor-Campbell is an independent
contributor. Send questions and comments
to kerry-ann.hepburn@gleanerjm.com
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