|
Chemical
formulae and equations
Francine
Taylor-Campbell, Contributor
 |
| Glenmuir's
Onaje La Mont (left), and Campion's Brian Ewbank, contemplate the next move during
their match on board 3 in the RBTT National Schools Chess Championship final at
the Ardenne High School library, recently. - Contributed | Main
points 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)3
PO4 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 formula The
formula of glucose is given as C6H12
O6. 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) 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
= 0.616 | 65.5/35.5
= 1.85 | | Ratio
of moles | 0.616/0.616
= 1 | 1.85/0.616
= 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 element | 3.40g
| 9.435-3.40
= 6.035g | | Relative
atomic mass | 40 | 35.5 | | Number
of moles | 3.40/40
= 0.085 | 6.035/35.5
= 0.17 | | Ratio
of moles | 0.085/0.085
= 1 | 0.17/0.085
= 2 | | Empirical
formula = CaCl2 | | |
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
Mass in 100g 85.7g; 14.3g
No. of moles 85.7/12; 14.3/1 = 7.14
= 14.3
Ratio of elements 1 : 2
Empirical formula = CH2,
Mr = (12+2 = 14g)
Molecular
formula: (CH2)n = 56
14n
= 56 & n = 4
Formula
= (CH2)4 = C4
H8 Method
2 Mass
of carbon = 85.7% * 56 = 48g
Mass of hydrogen = 14.3%*56 = 8g
1 atom of
carbon = 12g
1atom of hydrogen = 1g
# of carbon atoms = 48/12 = 4
# of
hydrogen atoms = 8/1 = 8
Molecular formula = C4 H8
Francine
Taylor-Campbell is an independent contributor. | 
