The empirical formula of a compound is the simplest possible ratio the atoms of the compound can be expressed. It doesn't tell the exact amount of atoms present in the compound.

The molecular formula of a compound is the actual ratio of moles of atoms in the compound. It is the actual formula of the compound.

The relationship between Empirical formula and Molecular formula is: $$ (E.F)_n = \text{Molar mass} $$ $$ (E.F)_n = M.F $$ $$ n = \text{number of moles} $$ $$ \text{also, R.M.M} = 2 × V.D $$ $$ \text{where, V.D = vapor density} $$ $$\text{R.M.M = Relative molecular mass} $$

The structural formula of a compound is the actual arrangement of atoms that constitute the molecule of the compound. It shows the functional group and bonding of the atoms

Example 1: An organic compound with relative molecular mass 136 contains 70.75% carbon, 5.09% hydrogen and 23.55% oxygen. Determine its

1. Empirical formula
2. Molecular formula
(H = 1, C = 12, O = 16)

The Empirical formula is then expressed using the final ratio $$ E.F = C_8H_7O_2 $$

2. $$ (E.F)_n = \text{Molar mass} $$
$$(C_8H_7O_2)_n = 136 $$ $$ (12 × 8) + (7) + (16 × 2)n = 136 $$ $$ 135n = 136 $$ $$ n = \frac{136}{135} $$ $$ n = 1 $$ $$ M.F = {E.F}_n $$ $$ M.F = (C_8H_7O_2)_1 $$ $$ M.F = C_8H_7O_2 $$ The molecular formula = \( C_8H_7O_2 \)

Example 2: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Determine its molecular formula if it's molar mass is 180 [H = 1, C = 12, O = 16]

The empirical formula = \( CH_2O \)
To derive the molecular formula; $$ (E.f)_n = \text{Molar mass} $$ $$ (CH_2O)n = 180 $$ $$ [(12 + (1 × 2) + 16)n = 180 $$ $$ 36n = 180 $$ $$ n = \frac{180}{36} $$ $$ n = 6 $$ $$ M.F = (E.F)_n $$ The molecular formula = \( C_6H_{12}O_6 \)

Example 3: A hydrocarbon Z with molecular mass 78 on combustion gave 3.385g of CO₂ and 0.692g of H₂O. Determine the molecular formula of Z. [H = 1, C = 12, O = 16]

The combustion of an hydrocarbon produces CO₂ and H₂O $$ Z + O_2 ——> CO_2 + H_2O $$ To express the molecular formula we need to find the masses of the elements that make up Z which is hydrogen and carbon (Hydrocarbon) $$ \text{For carbon, C} $$ $$ C + O_2 ———> CO_2 $$ Relating masses $$ 12g = 44g $$ $$ x = 3.385g $$ $$\text{mass of carbon} = \frac{12 × 3.385}{44} $$ $$ \text{mass of carbon} = 0.923g $$ Doing same for hydrogen $$ \text{mass of hydrogen} = \frac{2 × 0.692}{18} $$ $$ \text{mass of Hydrogen} = 0.077g $$ Now derive the empirical formula

Symbols	C	H
reacting mass	0.923	0.077
divide by atomic mass	$$ \frac{0.923}{12} $$ $$ = 0.077 $$	$$ \frac{0.077}{0.077} $$ $$ = 0.077 $$
Divide by smallest mole	$$ \frac{0.077}{0.077} $$ $$ = 1 $$	$$ \frac{0.077}{0.077} $$ $$ = 1 $$

The Empirical formula = \( CH \)
To get the molecular formula $$ (CH)n = \text{molar mass} $$ $$ (12 + 1)n = 78 $$ $$ 13n = 78 $$ $$ n = \frac{78}{13} $$ $$ M.F = (CH)_6 $$ The molecular formula = \( C_6H_6 \)

Example 4: Determine the molecular formula of an alkane whose vapor density is 15 (H = 1, C = 12)

An alkane is a hydrocarbon which has a general molecular \( C_nH_{2n+2} \)
Let the empirical formula of the hydrocarbon be CH $$ (CH)_n = 2 × VD $$ $$ 13n = 30 $$ $$ n = 2 $$ Substituting the value for n in the general formula
The molecular formula = \( C_2H_6 \)

Example 5: What is the empirical formula of a hydrocarbon containing 0.08 moles of carbon and 0.32 moles of hydrogen?

The empirical can be expressed by dividing by the smallest mole

Symbols	C	H
Reacting moles	0.08	0.32
Divide by smallest mole	$$ \frac{0.08}{0.08} $$	$$ \frac{0.32}{0.08} $$
	= 1	= 4

Empirical formula = \( {CH}_4 \)

Example 6: A compound contains 52.2% C, 13.1% H and oxygen only. The vapour density of the compound is 23. Determine:

1. Its empirical formula;
2. its molecular formula
(H = 1.0, C = 12.0, O = 16.0)

3. The compound reacts with sodium metal to produce hydrogen gas and when warmed with acidified KMnO₄ gives a solution that turns from purple to colourless. It also forms a sweet-smelling liquid when heated with ethanoic acid in the presence of concentrated H₂SO₄. Name the functional group present in the compound.
4. Draw the structural formula of the compound. (WAEC)

(i) Find empirical formula
Assume 100 g of compound → percentages become masses:
For oxygen = 100 - (52.2 + 13.1)
= 34.7g

Therefore the empirical formula is C₂H₆O.

(ii) Determine molecular formula
$$ (E.f)_n = \text{molar mass} $$ $$ \text{ But, Molar mass }= 2 × VD $$ $$ = 2 × 23 = 46 \text{ g/mol} $$ $$ (C_2H_6O)n = 46 $$ $$ (12 × 2 + 1 × 6 + 16)n = 46 $$ $$ 46n = 46 $$ $$ n = 1 $$ $$ M.F = (E.F)_n $$ $$ M.F = (C_2H_6O)_1 $$ $$ \text{ Molecular formula} = C_2H_6O $$

(iii) Identify functional group and structure from chemical behaviour
Observations given:
• Reacts with sodium metal to produce H₂ → indicates presence of an –OH (alcohol) group (R–OH reacts with Na → R–O⁻Na⁺ + H₂).
• Warmed with acidified KMnO₄, solution turns from purple to colourless → compound is oxidizable (primary or secondary alcohol or an alkene).
• When heated with ethanoic acid in presence of conc. H₂SO₄ forms a sweet-smelling liquid → this is esterification (alcohol + carboxylic acid → ester, sweet smell).
These three facts point to the compound being an alcohol.

Since formula is C₂H₆O and the alcohol functional group is present, the compound is ethanol, with structural formula:
CH₃–CH₂–OH (or in condensed form: C₂H₅OH).

(iii) Functional group: Hydroxyl group (–OH) — Alcohol.
(iv) Structural formula :

    H   H
    |   |
  H–C – C–O–H
    |   |
    H   H
  CH3–CH2–OH

Example 7: An organic compound of relative molecular mass 46, on analysis was found to contain 52.0% carbon, 13.3% hydrogen and 34.7% oxygen. Determine its

1. Empirical formula
2. Molecular formula
3. Draw two possible structures of the compound and name one of them
   ( C = 12.0, H = 1.0, O = 16.0)
   (WAEC )

Therefore the empirical formula is C₂H₆O.

(ii) Determine molecular formula
$$ (E.f)_n = \text{molar mass} $$ $$ \text{ But, Molar mass }= 2 × VD $$ $$ = 2 × 23 = 46 \text{ g/mol} $$ $$ (C_2H_6O)n = 46 $$ $$ (12 × 2 + 1 × 6 + 16)n = 46 $$ $$ 46n = 46 $$ $$ n = 1 $$ $$ M.F = (E.F)_n $$ $$ M.F = (C_2H_6O)_1 $$ $$ \text{ Molecular formula} = C_2H_6O $$

(III) Two possible structures (isomers) and name one
From C₂H₆O there are two common constitutional isomers:

a. Ethanol (an alcohol)

    H   H
    |   |
H - C - C - O - H
    |   |
    H   H

Ethanol (ethyl alcohol)
Functional group: –OH (hydroxyl)

b. Methoxymethane (Dimethyl ether)

    H       H
    |       |
H - C - O - C - H
    |       |
    H       H

Methoxymethane (dimethyl ether)
Functional group: –O– (ether)

Relationship: These two compounds are functional group isomers (same molecular formula but different functional groups).

Will they possess the same chemical properties?
No — they will have different chemical behaviours because they contain different functional groups. For example: ethanol reacts with sodium metal to give H₂, is oxidisable by acidified KMnO₄, and forms esters on reaction with carboxylic acids; dimethyl ether does not show those alcohol-specific reactions and has quite different physical properties (lower boiling point, different solubility).