EMPIRICAL AND MOLECULAR FORMULAE
Empirical formula: - This is the simplest formula of a compound; it shows the elements present and the ratio to which they are combined together.
The empirical formula of a compound can be calculated from the percentage compositions and the relative atomic mass of each element of the compound.
CALCULATION OF THE EMPIRICAL FORMULA FROM PERCENTAGE COMPOSITION BY MASS.h
Te simplest empirical formula of a compound can be calculated from the percentage compositions of the various elements that make up the compound. For example, the formula for anhydrous disodium trioxocarbonate (iv) can be calculated if the percentage composition by mass of each element present is known.
That is, 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.
Na C O
43.40g 11.32 45.28
23 12 16
1.88 0.94 2.83
1.88, 0.94 and 2.83 are the number of moles of each element respectively.
Now we divide by the smallest to get the mole ratio
1.89: 0.94: 2.83.
0.94. 0.94. 0.94
2: 1 : 3
= Na2C O3
The empirical formula is therefore, Na2CO3
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).
It is very important to note that the addition of all the percentage compositions must be equally to 100
SOLUTION:
Carbon Hydrogen Oxygen
52.2 13.1 34.7
divide by the 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
The empirical formular therefore = C2H6O.
Example 2. An organic compound has the following composition 55% of carbon, 9% hydrogen and 36% oxygen. Calculate the empirical formular for the compound. (C= 12, H= 1, O= 16)
SOLUTION:
Carbon Hydrogen Oxygen
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
The empirical formular = C2H4O.
Molecular formula of a compound is the actual formula of a compound, it shows the exact number of atoms present in one molecule of the compound.
Most molecular formulas are actually multiples of their empirical formulas, for example, a substance whose empirical formula is CH2 have a molecular formula C2H4, C4H8 and so on. So the molecular formulas of a compound is calculated from the it's empirical formulary and it's molecular mass.
I.e molecula formula = ( empirical formula)n
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
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
Therefore, the empirical formular = C3H6O.
(ii) To calculate the molecular formula, relate the empirical formula to the molar mass
(Empirical formula) n = Molar mass
(C3H6O)n = 58
(3x12 + 1x6 + 16x1)n = 58
58n = 58
58 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:
Hydrocarbons contain only two elements carbon and hydrogen. And 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
79.20 20.80
12 1
=6.60 = 20.80
4.60 20.80
6.60 6.60
= 1 = 3 Therefore the empirical formular = CH3.
(ii) To calculate the molecular formula, we 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 = 30
15
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
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
Therefore, the empirical formular = CH2O.
(ii) To calculate the molecular formula, we relate the empirical formula to the molar mass
.(Empirical formula)n = Molar mass
(CH2O)n = 180
(12 + 1x2 + 16)n = 180
(30)n = 180
n = 180/30
n = 6
The molecular formula = (CH2O)n =(CH2O)6 = C6H12O6.
The molecular formula = C6H12O6.
OBJECTIVE QUESTIONS
1. The empirical formula of a compound is:
A. The actual number of atoms in a molecule
B. The simplest whole-number ratio of atoms present
C. The molecular mass of the compound
D. The structural arrangement of atoms
2. Which of the following represents an empirical formula?
A. C₂H₄
B. C₆H₁₂O₆
C. CH₂
D. C₄H₈
3. The molecular formula of benzene is C₆H₆. Its empirical formula is:
A. CH
B. CH₂
C. C₂H₂
D. C₃H₃
4. A compound contains carbon and hydrogen in the ratio 1:2. Its empirical formula is:
A. CH
B. CH₂
C. C₂H₄
D. C₃H₆
5. The molecular formula of a compound is:
A. The simplest ratio of atoms present
B. The arrangement of atoms in space
C. The actual number of atoms of each element in a molecule
D. The mass of one mole of the compound
6. The empirical formula of hydrogen peroxide (H₂O₂) is:
A. HO
B. H₂O
C. H₂O₂
D. HO₂
7. If the empirical formula of a compound is CH₂ and its molecular mass is 42, the molecular formula is:
A. CH₂
B. C₂H₄
C. C₃H₆
D. C₄H₈
8. What is the empirical formula of glucose (C₆H₁₂O₆)?
A. CH₂O
B. C₂H₄O₂
C. C₃H₆O₃
D. CHO
9. A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Its empirical formula is:
A. CH₂O
B. CHO
C. C₂H₄O
D. C₂H₂O₂
10. The formula mass of the empirical formula CH₂O is:
A. 12
B. 18
C. 30
D. 32
11. If the molecular mass of a compound is twice the empirical formula mass, then:
A. Molecular formula = empirical formula
B. Molecular formula = 2 × empirical formula
C. Empirical formula = 2 × molecular formula
D. Molecular mass = empirical mass
12. Which of the following pairs have the same empirical formula?
A. CO and CO₂
B. C₂H₄ and C₃H₆
C. H₂O and H₂O₂
D. CH₄ and C₂H₆
13. The empirical formula of ethyne (C₂H₂) is:
A. CH
B. CH₂
C. C₂H₂
D. C₂H₄
14. A compound has the molecular formula C₄H₁₀. Its empirical formula is:
A. CH₂
B. C₂H₅
C. C₄H₁₀
D. CH₃
15. The ratio of carbon to oxygen in carbon monoxide is:
A. 1:1
B. 1:2
C. 2:1
D. 2:2
16. Which statement is correct?
A. Every compound has the same empirical and molecular formula.
B. Molecular formula is always the simplest ratio.
C. Empirical formula may be different from molecular formula.
D. Empirical formula gives the actual number of atoms present.
17. If a compound has an empirical formula NH₂ and a molecular mass of 32, its molecular formula is: A. NH₂
B. N₂H₄
C. NH₃
D. N₂H₂
18. The molecular formula of a compound is C₂H₆O. Which of the following is its empirical formula? A. CH₃O
B. CH₂O
C. C₂H₆O
D. CHO
19. Which of the following compounds has identical empirical and molecular formulae? A. H₂O₂
B. C₆H₆
C. CO₂
D. C₂H₄
20. The empirical formula mass of CH is: A. 12
B. 13
C. 14
D. 15
THEORY QUESTIONS
1. 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).
2. Calculate the empirical formular of an organic compound containing 81.8% carbon and 18.2% hydrogen (C = 12, H = 1).
3. What is the empirical formular of an oxide of phosporius that contains 43.6% phosphorous and 56.4% oxygen (P = 31, O = 16)
4. Determine the empirical formula of an oxide of nitrogen containing 70% oxygen, if the relative molecular mass of the oxide is 92, deduce its molecular formular
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