JEE Mains Chemistry · Some Basic Concepts of Chemistry

Percentage Composition and Empirical Formula

The mass percentage of an element from a formula, and the reverse: percentages or combustion data to the empirical and molecular formula.

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Why this matters

Twenty PYQs, eleven of them multiple choice and three from 2026. Six ask for a mass percentage, or for a molar mass worked back from one. Ten go from percentages to the empirical and molecular formula, and four burn a hydrocarbon and read its formula off the gas volumes. One routine, masses to moles to a ratio, covers all three.

Concept 1 of 3: Mass percentage of an element

A formula fixes how much of each element is in one mole. Divide that element's share by the molar mass. Run it backwards and a mass percentage plus a number of moles gives a molar mass.

Definition

$\%X=\frac{(\text{atoms of }X)\times A_X}{M}\times100$ .
Combustion: the carbon is $\frac{12}{44}$ of the mass of $\mathrm{CO_2}$ ; the hydrogen is $\frac{2}{18}$ of the mass of $\mathrm{H_2O}$ ; oxygen is found by difference.
Backwards: moles of the element $\times$ its atomic mass gives its mass in the sample. Divide by its mass fraction for the sample's mass, then by the sample's moles for $M$ .

Mass percentage

\%X=\frac{n_X\,A_X}{M}\times100

Worked example

Find the percentage of nitrogen in urea,

\mathrm{CO(NH_2)_2}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q148Moderate

Example 1 · Some Basic Concepts of Chemistry · Percentage Composition and Empirical Formula

When

0.01\text{ }mol

of an organic compound containing

60\%

carbon was burnt completely,

4.4\text{ }g

CO_{2}

was produced. The molar mass of the compound is ____

g\text{ }mol^{- 1}

(Nearest integer).

Carbon is 12/44 of CO₂

Convert the

\mathrm{CO_2}

to moles of carbon before anything else. Dividing by the compound's molar mass, or using

\frac{12}{28}

, is the usual slip.

Drill 5 more on mass percentage of an element

Concept 2 of 3: Empirical and molecular formula

Percentages are the masses in 100 g. Turn them into moles, find the simplest whole-number ratio, then scale up by the molar mass.

Definition

Take 100 g, so grams equal percentages. If they add to less than 100, the rest is usually oxygen.
Divide each by its atomic mass, then divide all by the smallest.
Clear fractions: 1.5 means $\times2$ , 1.33 means $\times3$ , 1.25 means $\times4$ .
$\text{MF}=k\times\text{EF}$ , where $k=\frac{M}{\text{EF mass}}$ .
Molar mass from a vapour at STP: $M=\frac{m}{V/22.4}$ with $V$ in litres.
Degree of unsaturation of $\mathrm{C}_x\mathrm{H}_y$ : $\frac{2x+2-y}{2}$ .

Molecular formula

\text{MF}=\left(\frac{M}{\text{EF mass}}\right)\times\text{EF}

Worked example

A compound is

40.0\%

C and

6.7\%

H, and the rest is O. Its molar mass is

180

g mol

^{-1}

. Find its molecular formula.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q26Moderate

Example 2 · Some Basic Concepts of Chemistry · Percentage Composition and Empirical Formula

An oxide of iron contains

69.9\%

iron, its empirical formula, is : (Given : Molar mass of Fe and O are 56 and

16\text{ }g{\text{ }mol}^{- 1}

respectively.)

Rounding 1.5 away

A ratio of

1:1.5

2:3

, not

1:2

1:1

. Round only when the value is within about 0.05 of a whole number; otherwise multiply through.

Empirical mass offered as the molar mass

Options often list the EF mass beside the MF mass. Check whether the question asks for the empirical or the molecular formula.

Drill 9 more on empirical and molecular formula

Concept 3 of 3: Formula from combustion volumes

For gases at one temperature and pressure, volumes are in the ratio of moles. Burn the hydrocarbon, find how much

\mathrm{CO_2}

formed and how much

\mathrm{O_2}

was used, and

x

and

y

follow.

Definition

$\mathrm{C}_x\mathrm{H}_y+\left(x+\frac{y}{4}\right)\mathrm{O_2}\rightarrow x\,\mathrm{CO_2}+\frac{y}{2}\,\mathrm{H_2O}$ .
Cooling condenses the water. KOH absorbs $\mathrm{CO_2}$ . What remains is unused $\mathrm{O_2}$ .
$x=\frac{V(\mathrm{CO_2})}{V(\text{hydrocarbon})}$ .
If water is still a vapour, $\frac{y}{2}=\frac{V(\mathrm{H_2O})}{V(\text{hydrocarbon})}$ . Otherwise use $x+\frac{y}{4}=\frac{V(\mathrm{O_2}\text{ used})}{V(\text{hydrocarbon})}$ .
'Equivalents' of oxygen or water in a stem mean moles per mole of fuel.

Combustion of a hydrocarbon

\mathrm{C}_x\mathrm{H}_y+\left(x+\tfrac{y}{4}\right)\mathrm{O_2}\rightarrow x\,\mathrm{CO_2}+\tfrac{y}{2}\,\mathrm{H_2O}

Worked example

20 mL of a gaseous hydrocarbon burns in 150 mL of

\mathrm{O_2}

. After cooling, the gas occupies 110 mL. KOH reduces it to 50 mL. Find the formula.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q31Moderate

Example 3 · Some Basic Concepts of Chemistry · Percentage Composition and Empirical Formula

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :

The water is gone after cooling

Once the gas is cooled, the water is liquid and drops out of the volume. Only

\mathrm{CO_2}

and leftover

\mathrm{O_2}

remain, so find

y

from the oxygen used, not from the residue.

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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Mass percentage of an element
Mass percentage
$\%X=\frac{n_X\,A_X}{M}\times100$
Empirical and molecular formula
Molecular formula
$\text{MF}=\left(\frac{M}{\text{EF mass}}\right)\times\text{EF}$
Formula from combustion volumes
Combustion of a hydrocarbon
$\mathrm{C}_x\mathrm{H}_y+\left(x+\tfrac{y}{4}\right)\mathrm{O_2}\rightarrow x\,\mathrm{CO_2}+\tfrac{y}{2}\,\mathrm{H_2O}$

Watch out for (4)

Carbon is 12/44 of CO₂
→ Mass percentage of an element
Rounding 1.5 away
→ Empirical and molecular formula
Empirical mass offered as the molar mass
→ Empirical and molecular formula
The water is gone after cooling
→ Formula from combustion volumes

Test yourself on Some Basic Concepts of Chemistry

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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