Empirical and Molecular Formula — Complete Notes Definitions · Key Differences

What is the difference between an Empirical Formula and a Molecular Formula? How do you calculate them step by step? This complete guide covers definitions, key differences, the n-factor method, 5 fully solved examples, and practice problems — aligned with Class 11 NCERT, NEET, JEE, and JKSSB exam pattern.

 Chemistry Class 11 · NCERT NEET · JEE JKSSB

Empirical and Molecular Formula — Complete Notes

Definitions · Key Differences · Step-by-Step Method · 5 Solved Examples · Practice Problems

 Key Highlights — Read This First

• ✦Empirical formula gives the simplest whole-number ratio of atoms in a compound
• ✦Molecular formula gives the actual number of atoms in one molecule
• ✦Relationship: Molecular Formula = n × Empirical Formula where n is a whole number
• ✦To find n: n = Molar Mass ÷ Empirical Formula Mass
• ✦This concept appears in NEET, JEE Mains, Class 11 Boards, and JKSSB exam papers regularly

1 What is an Empirical Formula?

The empirical formula of a compound is the simplest chemical formula that shows the ratio of atoms of each element present in the compound using the smallest possible whole numbers.

It does not tell you how many atoms are actually in a molecule — only their ratio. Two completely different compounds can share the same empirical formula.

Empirical Formula

CH₂O

Simplest ratio of C : H : O = 1 : 2 : 1. This is shared by glucose, acetic acid, and lactic acid — all very different molecules.

C : H : O → 1 : 2 : 1

Molecular Formula

C₆H₁₂O₆

Actual atoms in one molecule of glucose. It is 6 times the empirical formula CH₂O, meaning n = 6.

Glucose = 6 × CH₂O

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Remember This!

The empirical formula can be the same as the molecular formula when n = 1. Example: Water (H₂O) — its empirical formula and molecular formula are both H₂O.

2 What is a Molecular Formula?

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

The molecular formula is far more informative than the empirical formula because it tells you the true composition of a molecule, which is essential for understanding chemical reactions, molar mass calculations, and structural chemistry.

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NCERT Connection

This topic is part of Chapter 1 — Some Basic Concepts of Chemistry in Class 11 NCERT Chemistry. It is directly tested in NEET under Physical Chemistry and in JEE Mains under Mole Concept.

3 Key Differences: Empirical vs Molecular Formula

Basis	Empirical Formula	Molecular Formula
Definition	Simplest whole-number ratio of atoms	Actual number of atoms in one molecule
Information	Ratio only — not actual count	Exact count of each atom
Example (Glucose)	CH₂O	C₆H₁₂O₆
Example (Benzene)	CH	C₆H₆
Example (Hydrogen Peroxide)	HO	H₂O₂
Example (Water)	H₂O	H₂O (same, n=1)
Molar Mass	Cannot determine molar mass alone	Directly gives molar mass
Obtained from	Percentage composition data	Empirical formula + molar mass
Relationship	Base formula	MF = n × EF (n ≥ 1)

4 The Master Relationship Formula

The single most important equation in this topic — memorise it cold:

Master Formula

Molecular Formula = n × Empirical Formula

where   n = Molar Mass of Compound ÷ Empirical Formula Mass

n must always be a positive whole number (1, 2, 3, 4...). If your calculation gives a decimal like 1.99 or 3.01, round it to the nearest whole number — small rounding errors come from atomic mass approximations.

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Common Exam Trap

If the molar mass is NOT given in the question, you CANNOT find the molecular formula — only the empirical formula. Many students lose marks by guessing n = 1 without checking.

5 Step-by-Step Method to Find Empirical & Molecular Formula

Follow these 5 steps every time — whether in an exam or practice:

1

Write down the percentage composition

If percentages are not given directly, assume 100 g of compound — percentages become grams directly.

2

Divide each element's mass by its atomic mass

This gives the number of moles of each element. (C = 12, H = 1, O = 16, N = 14, S = 32, Cl = 35.5)

3

Divide all mole values by the smallest mole value

This gives the simplest ratio. The element with the fewest moles gets ratio = 1.

4

Convert to whole numbers if needed

If you get decimals like 1.5, multiply all values by 2. If 1.33, multiply by 3. If 1.25, multiply by 4. Write the empirical formula.

5

Find n and write the molecular formula

Calculate EF mass → Divide given molar mass by EF mass → Multiply subscripts by n → Done.

6 5 Fully Solved Examples

EXAMPLE 1 Find EF from percentage composition — Glucose

Question A compound contains C = 40%, H = 6.67%, O = 53.33%. Find its empirical formula.

Step 1: Assume 100 g → C = 40 g, H = 6.67 g, O = 53.33 g

Step 2: Divide by atomic masses:

Element	Mass (g)	Atomic Mass	Moles	Ratio
C	40	12	40/12 = 3.33	3.33/3.33 = 1
H	6.67	1	6.67/1 = 6.67	6.67/3.33 = 2
O	53.33	16	53.33/16 = 3.33	3.33/3.33 = 1

Step 3: Smallest mole value = 3.33. Divide all by 3.33.
Step 4: Ratio = C:H:O = 1:2:1 — all whole numbers already.

✓ Empirical Formula = CH₂O

EXAMPLE 2 Find Molecular Formula using Molar Mass

Question The empirical formula of a compound is CH₂O and its molar mass is 180 g/mol. Find the molecular formula.

Step 1: Find Empirical Formula Mass (EFM):
EFM = 12 + 2(1) + 16 = 30 g/mol

Step 2: Find n:
n = Molar Mass ÷ EFM = 180 ÷ 30 = 6

Step 3: Molecular Formula = n × CH₂O = 6 × CH₂O

✓ Molecular Formula = C₆H₁₂O₆ (Glucose)

EXAMPLE 3 Decimal ratio — requires multiplication

Question A compound has C = 85.7% and H = 14.3%. Its molar mass is 42 g/mol. Find EF and MF.

Step 1 & 2: Assume 100 g:

Element	Mass (g)	Atomic Mass	Moles	Ratio
C	85.7	12	7.14	7.14/14.3 = 0.5
H	14.3	1	14.3	14.3/7.14 = 2 → 1

Wait — smallest is 7.14. Ratio: C = 7.14/7.14 = 1, H = 14.3/7.14 ≈ 2

Ratio C:H = 1:2 → Empirical Formula = CH₂

EFM = 12 + 2(1) = 14 g/mol
n = 42 ÷ 14 = 3
MF = 3 × CH₂

✓ Empirical Formula = CH₂  |  Molecular Formula = C₃H₆ (Propene)

EXAMPLE 4 Benzene — classic NEET example

Question Benzene contains 92.3% C and 7.7% H. Its molar mass is 78 g/mol. Find EF and MF.

Element	Mass	At. Mass	Moles	Ratio
C	92.3	12	7.69	7.69/7.69 = 1
H	7.7	1	7.7	7.7/7.69 ≈ 1

EF = CH  |  EFM = 12 + 1 = 13 g/mol
n = 78 ÷ 13 = 6

✓ Empirical Formula = CH  |  Molecular Formula = C₆H₆ (Benzene)

EXAMPLE 5 3-element compound — Acetone (NEET 2022 pattern)

Question A compound contains C = 62.07%, H = 10.34%, O = 27.59%. Molar mass = 58 g/mol. Find EF and MF.

Element	Mass	At. Mass	Moles	÷ Smallest (1.72)	Ratio
C	62.07	12	5.17	5.17/1.72	3
H	10.34	1	10.34	10.34/1.72	6
O	27.59	16	1.72	1.72/1.72	1

EF = C₃H₆O  |  EFM = 3(12) + 6(1) + 16 = 58 g/mol
n = 58 ÷ 58 = 1

✓ Empirical Formula = C₃H₆O  |  Molecular Formula = C₃H₆O (Acetone) — MF = EF here!

7 Practice Problems

Solve these on your own. Answers are revealed on tap.

 Practice Questions — Empirical & Molecular Formula

1

A compound has Na = 43.4%, C = 11.3%, O = 45.3%. Find its empirical formula.
(Na = 23, C = 12, O = 16)
Show Answer

2

EF of a compound is CH. Its molar mass is 26 g/mol. What is the molecular formula?
Show Answer

3

A compound contains S = 50%, O = 50%. Find the empirical formula. (S = 32, O = 16)
Show Answer

4

Hydrogen peroxide has H = 5.88%, O = 94.12%, molar mass = 34 g/mol. Find MF.
Show Answer

5

Acetic acid has C = 40%, H = 6.67%, O = 53.33% and molar mass = 60 g/mol. Find EF and MF.
Show Answer

8 Quick Revision Table — Common Compounds

Compound	Empirical Formula	Molecular Formula	n
Water	H₂O	H₂O	1
Glucose	CH₂O	C₆H₁₂O₆	6
Benzene	CH	C₆H₆	6
Acetylene	CH	C₂H₂	2
Acetic Acid	CH₂O	C₂H₄O₂	2
Acetone	C₃H₆O	C₃H₆O	1
Hydrogen Peroxide	HO	H₂O₂	2
Fructose	CH₂O	C₆H₁₂O₆	6
Ethylene	CH₂	C₂H₄	2
Sodium Carbonate	Na₂CO₃	Na₂CO₃	1

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NEET Exam Tip

Glucose, Fructose, and Galactose all have the same molecular formula C₆H₁₂O₆ AND the same empirical formula CH₂O. They are isomers. This fact is frequently tested as a tricky MCQ in NEET.

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