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✦Reverse Engineering a Molecule

An unknown white powder burns completely and produces $\ce{CO2}$ and $\ce{H2O}$ . From the masses collected, a chemist calculates the percentage of C, H, and O. From those percentages alone — with no other information — they can reconstruct the molecular formula of the compound. This is percentage composition analysis, and it's how organic chemists identified the structures of natural compounds for centuries.

Percentage Composition

The percentage composition of a compound tells you how much of each element is in it. We can express this in two parallel ways:

Mole percent — the fraction of total atoms (counted in moles) contributed by each element:

$\% \text{ by mole of element} = \frac{\text{Moles of that element in the formula}}{\text{Total moles of all atoms in the formula}} \times 100$

For a generic compound $\ce{A2B3C}$ , 1 mole contains 2 mol A + 3 mol B + 1 mol C = 6 mol of atoms total. So mole % of A = (2/6) × 100 = 33.3 %.

Mass percent — the fraction of total mass contributed by each element:

$% \text{ by mass of element} = \frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100$$

Example — Water ( $\ce{H2O}$ , molar mass = 18 g/mol):

$\% \text{ H} = \frac{2}{18} \times 100 = 11.11\%$

$\% \text{ O} = \frac{16}{18} \times 100 = 88.89\%$

Check: 11.11 + 88.89 = 100% ✓

Example — $\ce{CuSO4.5H2O}$ (molar mass = 63.5 + 32 + 4×16 + 5×18 = 249.5 g/mol):

$\% \text{ Cu} = \frac{63.5}{249.5} \times 100 = 25.45\%$

Empirical Formula

The empirical formula represents the simplest whole number ratio of the atoms of each element present in a compound.

Molecular Formula	Empirical Formula
$\ce{H2O2}$	$\ce{HO}$
$\ce{C6H6}$	$\ce{CH}$
$\ce{CO2}$	$\ce{CO2}$ (already simplest)
$\ce{CH3COOH}$	$\ce{CH2O}$
$\ce{C2H6}$	$\ce{CH3}$

Step-by-step method to find empirical formula from percentages:

Assume 100 g of compound → % becomes grams directly
Find moles of each element: moles = mass / atomic mass
Divide all mole values by the smallest mole value → gives molar ratio
If ratio contains a decimal (like 1.5 or 1.33), multiply all by a suitable integer to make whole numbers
Write the empirical formula using these whole number ratios

Molecular Formula from Empirical Formula

The molecular formula shows the actual number of atoms of each element in one molecule. It is always a whole-number multiple of the empirical formula:

$\text{Molecular formula} = n \times \text{Empirical formula}$

where:

$n = \frac{\text{Molar mass (given)}}{\text{Empirical formula mass}}$

The molar mass must be given separately (from mass spectrometry or other experiments) — you cannot derive it from percentage composition alone.

Example: Empirical formula = $\ce{CH}$ , empirical formula mass = 13 g/mol. If the molar mass is 78 g/mol:
$n = \frac{78}{13} = 6 \quad \Rightarrow \quad \text{Molecular formula} = \ce{C6H6} \text{ (benzene)}$

Step-by-step flowchart for finding empirical and molecular formulas from percentage composition — 📸 From percentage composition to molecular formula — the standard 5-step method

Example 1

NCERT

A compound contains 4.07% H, 24.27% C, and 71.65% Cl. Its molar mass is 98.96 g/mol. What are the empirical and molecular formulas of the compound?

Example 2Empirical formula → molecular formula

SOLVED

The empirical formula of boron hydride is $\ce{BH3}$ . Calculate the molecular formula when the measured molecular mass of the compound is 27.66 g/mol.

(Atomic masses: B = 10.81, H = 1)

Example 3Empirical formula from elemental analysis

AIIMS 1998

A compound, on analysis, was found to contain C = 18.5 %, H = 1.55 %, Cl = 55.04 %, O = 24.81 %. What is its empirical formula?

(Atomic masses: C = 12, H = 1, Cl = 35.5, O = 16)

(a) $\ce{C2H2OCl}$ (b) $\ce{CH2ClO}$ (c) $\ce{CHClO}$ (d) $\ce{ClCH2O}$

✦JEE / NEET Exam InsightJEE / NEET

◆% composition adds up to 100%: If H, C, and O percentages are given and don't add to 100%, the remainder is oxygen (a common trick — oxygen is sometimes not given explicitly).

◆EF vs MF: Both express composition but at different scales. EF = simplest ratio; MF = actual. For

\ce{C6H12O6}

(glucose) and

\ce{CH2O}

(formaldehyde), the EF is the same (

\ce{CH2O}

) but MF is different.

◆Standard exam trap: If the molar ratio step gives 1 : 1.5, multiply everything by 2 → 2 : 3. If it gives 1 : 1.33, multiply by 3 → 3 : 4. Never round 1.5 to 1 or 2 directly.

◆% by mass formula on-the-fly:

\% X = \frac{n_X \times M_X}{M_{\text{compound}}} \times 100

where

n_X

= number of X atoms per formula unit,

M_X

= atomic mass of X.

Quick Check

Q1.Calculate the mass percentage of oxygen in H₂SO₄ (H=1, S=32, O=16):