ChemistryICSE Class 10 · Mole Concept and Stoichiometry

Empirical and Molecular Formula from Percentage Composition (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of atoms in one molecule. Divide each percentage by its atomic mass, divide by the smallest result, make whole numbers, then multiply by n = molecular mass ÷ empirical formula mass.

Empirical formula is the simplest ratio; multiply it by n (molecular mass ÷ empirical mass) to get the molecular formula, as glucose CH₂O becomes C₆H₁₂O₆.

Illustration drawn for QuestionBuddy.

Understand Empirical and Molecular Formula

Two kinds of formula. The empirical formula shows only the simplest whole-number ratio of the atoms of each element in a compound. The molecular formula shows the actual number of atoms of each element in one molecule. Glucose has the molecular formula C₆H₁₂O₆ but the empirical formula CH₂O. The two are related by a whole number $n$ :

molecular formula $= n \times$ empirical formula, where $n = \frac{molecular mass}{empirical formula mass}$ .

For many simple compounds $n = 1$ (water, H₂O; ammonia, NH₃), so both formulae are the same.

Percentage composition works in the other direction: from a formula, the percentage of an element $= \frac{mass of that element in one mole}{gram molecular mass} \times 100$ . The same formula gives the percentage of water of crystallisation in a hydrated salt.

Finding the empirical formula. Analysis gives either the percentage of each element or the masses of elements that combine. Treat the percentages as grams in a 100 g sample. Then:

1. Divide each mass (or percentage) by the atomic mass of that element: this gives the relative number of moles of atoms.
2. Divide every result by the smallest of them.
3. If any ratio is close to a simple fraction (1.5, 1.33, 2.5), multiply all ratios by 2 or 3 to make whole numbers.
4. Write the formula with these numbers.

If only some percentages are given (say C and H) and the compound also contains oxygen, find oxygen by difference: % O $= 100 -$ (sum of the others).

Getting the molecular formula. You need the molecular mass. It may be given directly, or as the vapour density (VD): the ratio of the mass of a volume of the gas to the mass of the same volume of hydrogen at the same temperature and pressure. Since hydrogen is H₂ (molecular mass 2), molecular mass = 2 × vapour density (the syllabus does not need the proof). Then $n = M / (empirical formula mass)$ , rounded to the nearest whole number, and the molecular formula is the empirical formula multiplied by $n$ .

Empirical and Molecular Formula: formulas and key facts

Percentage of an element: $% element = \frac{mass of element in 1 mol}{gram molecular mass} \times 100$
Relative moles of atoms: $moles = \frac{% (or mass)}{atomic mass}$
Molecular mass and vapour density: $M = 2 \times V D$; Formal proof not required by the syllabus.
Multiplier n: $n = \frac{molecular mass}{empirical formula mass}$; Molecular formula = (empirical formula) × n.
Turning decimals into whole numbers: Ratio ending .5 multiply all by 2
Ratio ending .33 or .67 multiply all by 3
Ratio ending .25 or .75 multiply all by 4
Ratio within 0.1 of a whole number round it
Same empirical formula, different molecules: Ethyne C₂H₂ and benzene C₆H₆ both CH
Ethene C₂H₄ and propene C₃H₆ both CH₂
Methanal CH₂O and glucose C₆H₁₂O₆ both CH₂O

Ratio ending .5	multiply all by 2
Ratio ending .33 or .67	multiply all by 3
Ratio ending .25 or .75	multiply all by 4
Ratio within 0.1 of a whole number	round it

Ethyne C₂H₂ and benzene C₆H₆	both CH
Ethene C₂H₄ and propene C₃H₆	both CH₂
Methanal CH₂O and glucose C₆H₁₂O₆	both CH₂O

How to solve Empirical and Molecular Formula questions

Set out empirical formula working as a table: element, percentage, atomic mass, moles, simplest ratio. Examiners award marks column by column.

1List the elements with their percentages (find any missing one by difference from 100).
2Divide each percentage by its atomic mass (not the other way round).
3Divide all the answers by the smallest one.
4Convert near-fractions to whole numbers by multiplying all ratios by the same factor.
5Write the empirical formula and calculate its mass.
6Find the molecular mass (given, or 2 × VD) and divide by the empirical formula mass to get n.
7Multiply each subscript by n to write the molecular formula.

Empirical and Molecular Formula in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Percentage composition of a fertiliser

Suppose a gardener buys a 2 kg bag of potassium nitrate, KNO₃, to supply both potassium and nitrogen to plants. (K = 39, N = 14, O = 16)

Calculate the percentage of potassium and of nitrogen in KNO₃, and the mass of nitrogen in the bag.

Try it first, then show the solutionSolution

Gram molecular mass of KNO₃ $= 39 + 14 + 48 = 101$ g.
% K $= \frac{39}{101} \times 100 = 38.61%$ ; % N $= \frac{14}{101} \times 100 = 13.86%$ .
Mass of N in 2000 g $= 2000 \times \frac{14}{101} = 277.2$ g.

Answer: K 38.61%, N 13.86%; the bag supplies about 277 g of nitrogen.

Memory hook: Percentage is the element's share of one mole's mass.

Tricks to remember Empirical and Molecular Formula

Short memory aids. Say each one aloud once, then use it on a question the same day.

Per cent over atomic, then over smallest
'Percent on top, atom below, smallest divides the row.'
Use it when: The two divisions in every empirical formula question.
VD doubles to M
Vapour density is half the molecular mass because hydrogen is H₂.
Use it when: Converting VD to molecular mass.
The n check
n must come out close to a whole number. If it does not, recheck the empirical formula mass.
Use it when: Catching arithmetic slips.

Empirical and Molecular Formula: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Molecular formula from VD · 1 mark · MCQBoard level

A hydrocarbon has the empirical formula CH and a vapour density of 39. Its molecular formula is: (C = 12, H = 1)
(a) C₂H₂
(b) C₄H₄
(c) C₆H₆
(d) C₈H₈

Try it first, then show the solutionSolution

Molecular mass $= 2 \times 39 = 78$ .
Empirical formula mass (CH) $= 13$ ; $n = 78/13 = 6$ , giving C₆H₆.

Answer: (c) C₆H₆.

Example 2: Water of crystallisation · 2 marks · Short answerBoard level

Calculate the percentage of water of crystallisation in Epsom salt, MgSO₄·7H₂O. (Mg = 24, S = 32, O = 16, H = 1)

Try it first, then show the solutionSolution

Gram molecular mass $= 24 + 32 + 64 + 7 \times 18 = 246$ g.
Mass of water $= 126$ g; percentage $= \frac{126}{246} \times 100 = 51.22%$ .

Answer: 51.22%.

Example 3: Empirical formula from combining masses · 3 marks · Short answerHarder board level

1.08 g of aluminium combines completely with 0.96 g of oxygen. Find the empirical formula of the oxide. (Al = 27, O = 16)

Try it first, then show the solutionSolution

Moles of Al $= 1.08/27 = 0.04$ ; moles of O $= 0.96/16 = 0.06$ .
Divide by the smaller (0.04): Al = 1, O = 1.5.
Multiply by 2: Al = 2, O = 3.

Answer: Al₂O₃.

Example 4: Empirical and molecular formula of a liquid · 3 marks · Short answerHarder board level

A compound contains 52.17% carbon, 13.04% hydrogen and 34.78% oxygen. Its vapour density is 23. Find its empirical and molecular formulae. (C = 12, H = 1, O = 16)

Try it first, then show the solutionSolution

Moles: C $= 52.17/12 = 4.348$ ; H $= 13.04/1 = 13.04$ ; O $= 34.78/16 = 2.174$ .
Divide by 2.174: C = 2, H = 6, O = 1. Empirical formula C₂H₆O; mass $= 24 + 6 + 16 = 46$ .
Molecular mass $= 2 \times 23 = 46$ ; $n = 46/46 = 1$ .

Answer: Empirical and molecular formula both C₂H₆O.

Example 5: When n is 2 · 4 marks · Long answerHOTS / competency

An organic compound contains 38.71% carbon and 9.68% hydrogen, the rest being oxygen. Its vapour density is 31. Find (a) its empirical formula, (b) its molecular formula. (C = 12, H = 1, O = 16)

Try it first, then show the solutionSolution

% O $= 100 - (38.71 + 9.68) = 51.61%$ .
Moles: C $= 38.71/12 = 3.226$ ; H $= 9.68/1 = 9.68$ ; O $= 51.61/16 = 3.226$ .
Divide by 3.226: C = 1, H = 3, O = 1. Empirical formula CH₃O; mass $= 12 + 3 + 16 = 31$ .
Molecular mass $= 2 \times 31 = 62$ ; $n = 62/31 = 2$ . Molecular formula $= (C H_{3} O)_{2} = C_{2} H_{6} O_{2}$ .

Answer: (a) CH₃O (b) C₂H₆O₂.

Example 6: Finding x in a hydrated salt · 3 marks · Short answerHOTS / competency

Washing soda, Na₂CO₃·xH₂O, contains 62.94% water of crystallisation by mass. Find x. (Na = 23, C = 12, O = 16, H = 1)

Try it first, then show the solutionSolution

Mass of Na₂CO₃ $= 106$ ; mass of water $= 18 x$ .
$\frac{18 x}{106 + 18 x} = 0.6294$ , so $18 x = 0.6294 \times 106 + 0.6294 \times 18 x$ .
$18 x \times 0.3706 = 66.72$ , giving $x = 10.0$ .
Check: $\frac{180}{286} \times 100 = 62.94%$ .

Answer: x = 10 (Na₂CO₃·10H₂O).

Where students lose marks in Empirical and Molecular Formula

The slips examiners see most often, and what to write instead.

Mistake: Dividing the atomic mass by the percentage
In 2025 (Q6(iii)) examiners reported candidates dividing atomic weights by percentages, giving wrong moles.
Wrong working
C: 12 ÷ 60.0 = 0.2
Right working
C: 60.0 ÷ 12 = 5.0
Do this: Percentage (or mass) on top, atomic mass below.
Usually costs: Most of a 4-mark question.
Mistake: Not dividing by the smallest number
The same 2025 report notes that many stopped after the first division.
Do this: Always divide every mole value by the smallest one before writing a formula.
Usually costs: 1 to 2 marks.
Mistake: Stopping at the empirical formula
Examiners noted that candidates did not multiply the empirical formula by n to get the molecular formula (2025 Q6(iii)).
Do this: Write n = M ÷ empirical formula mass as its own line, even when n = 1.
Usually costs: 1 mark.
Mistake: Mixing up VD and molecular mass
In 2025 (Q1(iv) and Q4(i)) examiners found confusion between vapour density and molecular weight.
Wrong working
VD = 17, so M = 8.5
Right working
M = 2 × VD = 34
Do this: Molecular mass is twice the vapour density, never half.
Usually costs: 1 to 2 marks.
Mistake: Rounding 1.5 to 2
Students round a ratio such as 1 : 1.5 to 1 : 2 and get the wrong formula.
Do this: Only round values within about 0.1 of a whole number; otherwise multiply all ratios by 2, 3 or 4.
Usually costs: 1 to 2 marks.

Exam tips

Draw the table: element | % | atomic mass | moles | ratio | whole number.
Keep moles to three or four significant figures; heavy rounding can turn 1.5 into 1.
If oxygen is not given, find it by difference before anything else.
Check your molecular formula by recalculating its molecular mass.

Practice questions on Empirical and Molecular Formula

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A gas has a vapour density of 32. It could be: (S = 32, O = 16, C = 12, N = 14)
(a) O₂
(b) SO₂
(c) CO₂
(d) NO₂
Hint
M = 2 × VD.
Answer
(b) SO₂ (M = 64).
Q2 · 1 mark · Very short answerBoard level
Define empirical formula.
Hint
Simplest ratio.
Answer
The formula that shows the simplest whole-number ratio of the atoms of the different elements present in one molecule of a compound.
Q3 · 2 marks · Short answerBoard level
Calculate the percentage of sulphur in sulphuric acid, H₂SO₄. (H = 1, S = 32, O = 16)
Hint
Mass of S in one mole ÷ molar mass.
Answer
M = 98; % S $= \frac{32}{98} \times 100 = 32.65%$ .
Q4 · 3 marks · Short answerHarder board level
A salt contains Na 29.11%, S 40.51% and O 30.38%. Find its empirical formula. (Na = 23, S = 32, O = 16)
Hint
One ratio comes out as 1.5.
Answer
Moles: Na 1.266; S 1.266; O 1.899. Ratio 1 : 1 : 1.5; × 2 gives 2 : 2 : 3. Empirical formula Na₂S₂O₃.
Q5 · 2 marks · Short answerHarder board level
A compound has the empirical formula CH₂O and a vapour density of 30. Find its molecular formula. (C = 12, H = 1, O = 16)
Hint
Empirical formula mass = 30.
Answer
M $= 2 \times 30 = 60$ ; $n = 60/30 = 2$ ; molecular formula C₂H₄O₂.
Q6 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Ethyne (C₂H₂) and benzene (C₆H₆) have the same empirical formula.
Reason (R): Ethyne and benzene have the same percentage composition by mass.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
The empirical formula is fixed by the percentage composition.
Answer
(a). Both reduce to CH, so both contain 92.3% C and 7.7% H; the same percentages always give the same empirical formula.
Q7 · 3 marks · Short answerHarder board level
3.0 g of a gaseous hydrocarbon contains 2.4 g of carbon. Its vapour density is 15. Find its empirical and molecular formulae. (C = 12, H = 1)
Hint
Mass of hydrogen = 3.0 − 2.4.
Answer
H = 0.6 g. Moles: C $= 0.2$ , H $= 0.6$ ; ratio 1 : 3, empirical formula CH₃ (mass 15). M $= 30$ ; $n = 2$ ; molecular formula C₂H₆.
Q8 · 4 marks · Long answerHOTS / competency
An organic compound contains carbon, hydrogen and oxygen only, with 54.55% carbon and 9.09% hydrogen. Its relative molecular mass is 88. Find the empirical and molecular formulae. (C = 12, H = 1, O = 16)
Hint
Find oxygen by difference first.
Answer
% O $= 36.36$ . Moles: C 4.546; H 9.09; O 2.273. Ratio 2 : 4 : 1; empirical formula C₂H₄O (mass 44). $n = 88/44 = 2$ ; molecular formula C₄H₈O₂.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Empirical formula of glucose (C₆H₁₂O₆)?Show answer
CH₂O.
2.Molecular mass of a gas with VD 14?Show answer
28.
3.If the ratio is C 1 : H 2.5, what is the empirical formula?Show answer
C₂H₅ (multiply by 2).
4.Define vapour density.Show answer
The ratio of the mass of a certain volume of a gas to the mass of an equal volume of hydrogen at the same temperature and pressure.
5.For which compounds are empirical and molecular formulae the same?Show answer
Those with n = 1, e.g. H₂O, NH₃, CO₂, CH₄.

Empirical and Molecular Formula: frequently asked questions

What is the difference between empirical formula and molecular formula?

The empirical formula gives only the simplest whole-number ratio of atoms; the molecular formula gives the actual number of each atom in one molecule. The molecular formula is n times the empirical formula.

How do I find oxygen if its percentage is not given?

Subtract the sum of the other percentages from 100, provided the question says the compound contains only those elements and oxygen.

Why is molecular mass twice the vapour density?

Vapour density compares a gas with hydrogen, whose molecule (H₂) has mass 2. So the gas's molecular mass is 2 × its vapour density. ICSE does not ask for the proof.

What if n is not a whole number?

It should be very close to one. If you get something like 2.5, recheck the empirical formula: you may have rounded a 1.5 ratio wrongly.