EMPIRICAL AND MOLECULAR FORMULAE
The empirical formula is the simplest formula that gives the ratio of the number of different atoms present in a compound. It does not give the exact number of each atom, whereas the molecular formula gives the exact number of atoms present in a molecule of a compound. The molecular formula of a compound is a whole-number multiple of its empirical formula.
Examples:
EMPIRICAL FORMULA: is a formula that simply gives the relative number of atoms of each element present in a molecule. It is the simplest formula.
MOLECULAR FORMULA: is a formula that states the actual number of each kind of atom found in a molecule.
A substance whose empirical formula is CH2 has a molecular formula C2H4, C4H8, and so on. The empirical formula of a compound can be found if the percentage composition and relative atomic masses of each element in the compound are known. Also, the empirical formula can be related to the molar mass of a compound in order to obtain the molecular formula for the compound.
EXAMPLE1: An organic compound has the composition 55% of carbon, 9% hydrogen and 36% oxygen. Calculate the empirical formular for the organic compound. (C= 12, H= 1, O= 16)
NOTE: Always check that the addition of the percentage composition of all elements present in a compound equal to 100.
SOLUTION:
CARBON HYDROGEN OXYGEN
Relative composition: 55 9 36
Divide by atomic mass: 55 9 36
12 1 16
=4.58 = 9.00 = 2.25
Divide by the smallest: 4.58 9 2.25
2.25 2.25 2.25
= 2 = 4 = 1
Hence the empirical formular = C2H4O.
EXAMPLE 2: What is the empirical formula of an organic compound whose percentage composition is carbon, 52.2%, hydrogen, 13.1% and oxygen, 34.7% (C = 12, H = 1, O = 16).
SOLUTION:
CARBON HYDROGEN OXYGEN
Relative composition: 52.2 13.1 34.7
Divide by atomic mass: 52.2 13.1 34.7
12 1 16
=4.35 = 13.1 = 2.17
Divide by the smallest: 4.35 13.1 2.17
2.17 2.17 2.17
= 2 = 6 = 1
Hence, the empirical formula = C2H6O.
EXAMPLE 3: An organic compound contains carbon, 62.1%, hydrogen, 10.3% and oxygen, 27.6% by mass.
(i) Find the empirical formula of the compound
(ii) If the molar mass of the compound is 58.0g, find its molecular formula. (C = 12, H = 1, O = 16).
SOLUTION:
CARBON HYDROGEN OXYGEN
Relative composition: 62.1 10.3 27.6
Divide by atomic mass: 62.1 10.3 27.6
12 1 16
=5.1 = 10.3 = 1.8
Divide by the smallest: 4.35 13.1 1.8
1.8 1.8 1.8
= 3 = 6 = 1
Hence the empirical formula = C3H6O.
(ii) To calculate the molecular formula, relate the empirical formula to the molar mass.
(Empirical formula)n = Molar mass
(C3H6O)n = 58
(3×12 + 1×6 + 16×1)n = 58
(58)n = 58
n = 1
The molecular formula = (C3H6O)n = C3H6O.
EXAMPLE 4: A hydrocarbon contains 20.80% of hydrogen and has a relative molar mass of 30. What is the
(i) Empirical formula
(ii) molecular formula (C = 12, H = 1).
SOLUTION:
Hydrocarbon is known to contain carbon and hydrogen only. Since the percentage composition of all elements in a compound must be equal to 100. Therefore, the percentage composition of carbon which is the second element contained by a hydrocarbon, equals 79.20%. i.e. 100 – percentage composition of hydrogen (20.80%).
CARBON HYDROGEN
Relative composition: 79.20 20.80
Divide by atomic mass: 79.20 20.80
12 1
=6.60 = 20.80
Divide by the smaller: 4.60 20.80
6.60 6.60
= 1 = 3
Hence, the empirical formula = CH3.
(ii) To calculate the molecular formula, relate the empirical formula to the molar mass.
(Empirical formula) n = Molar mass
(CH3) n = 30
(12 + 1×3)n = 30
(15)n = 30
n = 2
The molecular formula = (CH3)n = C2H6.
EXAMPLE 5: A carbohydrate contains 40% and hydrogen 6.72%, Calculate its empirical formula and the molecular formula, if the molar mass is 180 (C = 12, H = 1, O = 16).
SOLUTION:
Carbohydrate contains the elements carbon, hydrogen and oxygen, but from the question above oxygen is missing, hence the percentage composition of oxygen equals 100 – (percentage composition of carbon and hydrogen)
= 100 – (40 + 6.72)
= 100 – 46.72 = 53.30%
CARBON HYDROGEN OXYGEN
Relative composition: 40 6.72 53.3
Divide by atomic mass: 40 6.72 53.3
12 1 16
=3.33 = 6.72 = 3.33
Divide by the smallest: 3.33 6.72 3.33
3.33 3.33 3.33
= 1 = 2 = 1
Hence, the empirical formula = CH2O.
(ii) To calculate the molecular formula, relate the empirical formula to the molar mass.
(Empirical formula) n = Molar mass
(CH2O) n = 180
(12 + 1×2 + 16) n = 180
(30) n = 180
n = 180/30
n = 6
The molecular formula = (CH2O) n = (CH2O6)
= C6H12O6.
The molecular formula = C6H12O6.
EXAMPLE 6: A hydrocarbon contains 92.40% of carbon. If the vapour density of the hydrocarbon is 39. Find
(i) Empirical formula
(ii) Molecular formula (C = 12, H = 1).
SOLUTION:
Hydrocarbon is known to contain carbon and hydrogen only. Since the percentage composition of all elements in a compound must be equal to 100. Therefore, the percentage composition of hydrogen which is the second element contained by a hydrocarbon equals 7.60%. i.e 100 – percentage composition of carbon (92.40%).
CARBON HYDROGEN
Relative composition: 92.40 7.60
Divide by atomic mass: 92.40 7.60
12 1
=7.70 = 7.60
Divide by the smaller: 7.70 7.60
7.60 7.60
= 1 = 1
Hence the empirical formula = CH.
(ii) To calculate the molecular formula, relate the empirical formula to the molar mass.
(Empirical formula) n = Molar mass
But molecular mass = 2 x vapour density
(CH) n = 2 x 39
(12 + 1) n = 78
(13) n = 78
n = 6
The molecular formula = (CH) n = C6H6.
EXAMPLE 7: Calculate the empirical formular of an organic compound containing 81.8% carbon and 18.2% hydrogen (C = 12, H = 1).
SOLUTION:
CARBON HYDROGEN
Relative composition: 81.8 18.2
Divide by atomic mass: 81.8 18.2
12 1
=6.82 = 18.2
Divide by the smaller : 6.82 18.2
6.82 6.82
= 1 = 2.67
The carbon: hydrogen {C:H} ratio of 1:2.67 is too far from whole numbers and so the lowest multiple of this which gives a whole number ratio is the empirical formula, i.e ,1:2.67, 2:5.34, 3:8.01, 4:10.68 etc.
Hence, the empirical formula = C3H8.
RELATIVE MOLECULAR MASS, MOLECULAR MASS, AND PERCENTAGE COMPOSITION
If the formulae of a substance and the relative atomic Masses of each of the elements are known, then it is possible to determine the relative molecular mass of that substance.
The relative molecular mass refers to the number of times a mole is heavier than one- twelfth the mass of one atom of carbon -12. It has no unit.
The relative molecular mass of a compound is the sum of the masses of all the atoms present in one molecule of the compound .e.g.
For NaCl, the relative molecular mass= (23 +35.5) = 58.5
For ethanol = C2H5OH (carbon=12, H=1, O =16)
The relative molecular mass of ethanol
= C2H5OH
(12×2) + (1×5) + (16) + (1)
24 + 5 + 16 +1 = 46
THE MOLAR MASS
This is the relative molecular mass expressed in grams. E.g. the molar mass of ethanol is 46gmol-1
In 12g of carbon-12, there are 6 × 1023 atoms of carbon. This is one mole of carbon -12.
A mole of any substance is the amount of that substance which contains 6× 1023 particles of that substance e.g. One mole of ethanol has a mass of 46g and contains 6× 1023 ethanol molecules.
NOTE: The relative molecular mass has no units but the molar mass of any substance is expressed in grams.
PERCENTAGE COMPOSITION OF A COMPOUND
To calculate the percentage position of ethanol whose molecular formula is C2H5OH, given that the relative atomic masses of carbon, hydrogen and oxygen are 12, 1, and 16 respectively?
First calculate the molar mass of C2H5OH =(12×2) +(1×5) + (16)+ (1) = 46gmol-1
Then determine the masses of C H and O present;
Mass of carbon= 12×2=24g
Mass of hydrogen= 6×1 = 6g
Mass of oxygen = 16× 1 = 16g
Molar mass of C2H5OH= 46g
Therefore, percentage of C= 2446 × 100 = 52.17%
Percentage of hydrogen=6 46 × 100 = 13.04%
Percentage of oxygen=1646 × 100 = 13.04%
CALCULATION OF THE CHEMICAL FORMULA FROM PERCENTAGE COMPOSITION BY MASS.
We can determine the simplest chemical formula of a compound, given its percentage composition e.g. If the formula for anhydrous disodium trioxocarbonate (iv) is not known, if its percentage composition by mass is known then it chemical formula could be calculated.
For example, the percentage composition of the compound was found to be Na=43.40%, C= 11.32% and O = 45.28%. This would mean that in every 100g of the compound, the masses of Na, C and O were 43.40g, 11.32g and 45.28g respectively.
∴The amount in moles of Na, C and O would be.
43.4023, 11.3212 and 45.2816 respectively.
Therefore, the amount in moles of Na =43.4023 = 1.89.
Amount in mole of C= 11.3012=0.94.
Amount in mole of O= 45.2816=2.83.
Molar ratio of H: C: O is 1.89: 0.94: 2.83.
2: 1 : 3
Na: C: O= Na2C O3
The simplest formula is therefore Na2CO3
GENERAL EVALUATION
ESSAY QUESTION
- Differentiate between valency and oxidation number.
- Determine the empirical formula of an oxide of nitrogen containing 70% oxygen, if the relative molecular mass of the oxide is 92, and deduce its molecular formula.
- Balance this chemical equation NaOH + H2SO4 —> Na2SO4 + H2O