§ 1.9NCERT Class 11 · Chemistry · Chapter 1

Percentage Composition: Formula, Numericals & Empirical Formula

One formula, two directions — turn any chemical formula into element percentages, and turn percentages back into empirical and molecular formulas. NCERT-aligned theory, the traps NTA sets, and a live % lab that even parses hydrates.

QCC Notes · Class 11 Chemistry NCERT & NTA aligned UPDATED 30 AUG 2026 12 MIN READ

The basis behind every percentage
% of X = grams of X per 100 g of compound
Urea: 46.67 g of N in every 100 g — always

What is Percentage Composition? — Complete Theory

Chemical formulas tell you how many atoms of each element a compound holds; percentage composition tells you how much mass each element contributes. It is the compound written as a recipe by mass: urea’s nitrogen percentage of 46.67 is a contract — whether you weigh 2 g in a test tube or unload a 50 kg fertilizer bag, every 100 g of urea carries exactly 46.67 g of nitrogen.

The definition flows straight from the mole idea of § 1.8: take one mole of the compound, note the mass each element contributes, and divide by the molar mass. For water, one mole is 18 g and hydrogen’s share is 2 × 1 = 2 g, so %H = 2/18 × 100 = 11.11% and %O = 16/18 × 100 = 88.89%. The two shares must add to 100 — that check is free marks.

Core formula% element = n·a×100 / M VerificationΣ all % = 100 Reverse pullm(element) = w·%/100

Here n is the number of atoms of that element in one formula unit, a its atomic mass, and M the molar mass of the compound. The m(element) = w × %/100 chip runs the logic forward through any sample: since 52.17% of ethanol’s mass is carbon, 50 g of ethanol always contains 50 × 52.17/100 = 26.1 g of carbon. One percentage, unlimited sample sizes.

Table 1 — Exam-favourite compounds, element by element
Compound	M (g/mol)	Element → % by mass
Water — H₂O	18	H 11.11 · O 88.89
Carbon dioxide — CO₂	44	C 27.27 · O 72.73
Ammonia — NH₃	17	N 82.35 · H 17.65
Urea — NH₂CONH₂	60	N 46.67 · C 20.00 · H 6.67 · O 26.67
Glucose — C₆H₁₂O₆	180	C 40.00 · H 6.67 · O 53.33
Calcium carbonate — CaCO₃	100	Ca 40.00 · C 12.00 · O 48.00

Run the whole idea backwards and percentages become a formula detective tool. The four-step algorithm: assume a 100 g sample (percentages become grams), convert each element’s mass to moles, divide all mole values by the smallest, and scale to whole numbers. The simplest whole-number ratio is the empirical formula; comparing the empirical-formula mass with the true molar mass gives the multiplier n = M ÷ empirical mass, and the molecular formula = n × empirical formula. Glucose’s data — 40% C, 6.67% H, 53.33% O — collapses to CH₂O, then ×6 to C₆H₁₂O₆.

Table 2 — Ratio multiplier quick-reference (empirical formula step)
Decimal in the ratio	Multiply all by	Worked pattern
0.5 or 1.5	× 2	1 : 1.5 → 2 : 3 → Fe₂O₃
0.33 or 0.67	× 3	1 : 1.33 → 3 : 4 → ascorbic acid’s C₃H₄O₃
0.25 or 0.75	× 4	1 : 1.25 → 4 : 5
0.2 / 0.4 / 0.6 / 0.8	× 5	1 : 0.4 → 5 : 2

One special family deserves its own arithmetic: hydrates. The dot in CuSO₄·5H₂O is an instruction to add the water mass — never multiply, never ignore. M = 63.5 + 32 + 64 + 5(18) = 249.5 g/mol, of which water contributes 90 g, so the water of crystallisation is 90/249.5 × 100 ≈ 36% of every crystal’s mass.

Why does this section matter beyond Chapter 1? Ore quality, fertilizer labels, drug purity — industry runs on percentage composition. And in your syllabus it feeds directly into § 1.10 Stoichiometry, where percent purity and percent yield are percentage composition wearing exam clothes. Master the two directions here and stoichiometry’s numericals lose half their weight. (Need the atomic masses themselves? That data lives in § 1.7.)

Visualising Percentage Composition

Ek compound, ek ledger — the mass ledger of ethanol shows where all 46 g go, and the lab below lets you audit any formula yourself.

FIG. 1 — Read the ledger any way you like: 46 g of ethanol is 24 g C + 6 g H + 16 g O, so carbon is 52.2% by mass — and 50 g of any ethanol sample would carry 26.1 g of carbon. Percentages scale with the sample; the recipe never changes.

Try it live

% Composition Lab

Preset

Formula — type any compound

Element	Atoms (n)	Atomic mass (u)	Mass in 1 mol (g)	% share

Atomic masses follow exam convention (H = 1, C = 12, N = 14, O = 16 …). Brackets work one level deep — (NH4)2SO4. Hydrates: write CuSO4.5H2O with a dot.

Solved Examples (Step-by-Step)

Given → formula → substitute → verify. Write these same steps on your rough sheet — that is what earns full method marks.

EXAMPLE 01Foundation · NEET

Forward: formula → element percentages

Calculate the mass per cent of each element in urea, NH₂CONH₂. (N = 14, C = 12, H = 1, O = 16)

GivenUrea has 2 N, 1 C, 4 H, 1 O per molecule.
Molar massM = 2(14) + 12 + 4(1) + 16 = 60 g mol⁻¹
Substitute%N = 28/60 × 100 = 46.67 · %C = 12/60 = 20.00 · %H = 4/60 = 6.67 · %O = 16/60 = 26.67
Verify46.67 + 20.00 + 6.67 + 26.67 = 100.01 ≈ 100 ✓

N 46.67% · C 20% · H 6.67% · O 26.67%

EXAMPLE 02JEE Main · Reverse

Reverse: percentages → empirical → molecular formula

A compound contains 40.00% carbon, 6.67% hydrogen and 53.33% oxygen by mass. Its molar mass is 180 g mol⁻¹. Find the empirical and molecular formula.

AssumeTake 100 g: C = 40 g, H = 6.67 g, O = 53.33 g.
Molesn(C) = 40/12 = 3.333 · n(H) = 6.67/1 = 6.67 · n(O) = 53.33/16 = 3.333
RatioDivide by 3.333 → C : H : O = 1 : 2 : 1 → empirical formula CH₂O (mass 30)
Scale upn = 180/30 = 6 → molecular = 6 × CH₂O

Empirical CH₂O · Molecular C₆H₁₂O₆ (glucose)

EXAMPLE 03JEE Main · Application

Fertilizer face-off: who delivers more nitrogen?

Which supplies more nitrogen per 100 g — urea, NH₂CONH₂, or ammonium sulfate, (NH₄)₂SO₄? (N = 14, H = 1, S = 32, O = 16)

UreaM = 60, N atoms = 2 → %N = 28/60 × 100 = 46.67% → 46.67 g N per 100 g
Am. sulfateM = 2(18) + 32 + 4(16) = 132 — the bracket’s 2 doubles the nitrogen: N = 2 × 14 = 28
Substitute%N = 28/132 × 100 = 21.21% → 21.21 g N per 100 g
Compare46.67 g vs 21.21 g — urea delivers more than double the nitrogen per gram of fertilizer.

Urea wins — 46.67 g N vs 21.21 g N per 100 g

Key Formulas & Takeaways

Formula card

Eight lines that solve this topic

% element = (n × a × 100) / MCore formula — n atoms of the element, atomic mass a, molar mass M.

m(element) = w × %/100Mass of the element inside w grams of compound.

w(compound) = m(element) × 100/%Reverse — compound needed for a target element mass.

% water = (x × 18 × 100) / M(hydrate)x waters of crystallisation per formula unit.

% → ÷ a → ÷ smallest → ratioThe empirical-formula algorithm, in one line.

n = M / (empirical mass)Molecular formula = n × empirical formula.

% purity = (pure / sample) × 100Bridge to § 1.10 — same arithmetic, new name.

Σ all percentages = 100Free verification — catch arithmetic slips instantly.

Atomic masses (exam convention): H 1 · C 12 · N 14 · O 16 · S 32 · Cl 35.5 · Ca 40 · Cu 63.5 · Fe 56 · 1 u = 1.6605 × 10⁻²⁴ g · Water = 18 g per mole (11.11% H, 88.89% O)

Percentage composition is a recipe by mass — every 100 g of compound carries exactly these element masses, at any scale.
One formula, two directions — forward (formula → %) and reverse (% → empirical → molecular) cover the entire topic.
The hydrate dot means ADD — water mass joins M and earns its own percentage; brackets multiply everything inside them.
Verify with Σ% ≈ 100 — ten seconds of checking that examiners’ keys rely on, and toppers never skip.

FAQs

What is percentage composition in chemistry?

Percentage composition tells you the mass contributed by each element in a compound, expressed as a percentage of its molar mass. For water (M = 18), hydrogen contributes 2 × 1 = 2 g out of 18 g, so %H = 2/18 × 100 = 11.11% and %O = 88.89%.

How do you calculate the percentage of an element in a compound?

Use % element = (number of atoms × atomic mass ÷ molar mass of the compound) × 100. Example: nitrogen in urea, NH₂CONH₂ (M = 60): there are 2 nitrogen atoms, so %N = (2 × 14 ÷ 60) × 100 = 46.67%.

What is the difference between empirical formula and molecular formula?

The empirical formula gives the simplest whole-number ratio of atoms — CH₂O — while the molecular formula gives the actual count in one molecule, e.g. C₆H₆ for benzene. Molecular formula = n × empirical formula, where n = molar mass ÷ empirical-formula mass.

How do you find the empirical formula from percentage composition?

Assume a 100 g sample, convert each percentage into moles by dividing by atomic mass, divide every value by the smallest one to get a ratio, and scale to whole numbers — that ratio is the empirical formula. Multiply by n = molar mass ÷ empirical-formula mass to reach the molecular formula.

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QCC Notes — Class 11 Chemistry

Strictly NCERT-aligned notes for JEE Main & NEET, prepared by QCC Notes (Padho Likho JEE). Content follows the latest NCERT edition and current NTA exam pattern.

Last updated
30 Aug 2026