Chapter 1 of 18

JEE Main Chemistry★ ~2 questions (8 marks) in JEE Main, and far more influence than that suggests, since the mole concept is the arithmetic backbone of Equilibrium, Thermodynamics, Electrochemistry and Solutions

Some Basic Concepts in Chemistry (Mole Concept)

The mole is a counting unit, nothing more. Every stoichiometry problem in chemistry is the same three moves: convert to moles, apply the equation as a ratio, convert back.

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Timed practiceMCQs from this chapter with instant scoring

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JEE Main PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Convert freely between mass, moles, particle count and gas volume, stating which entity is being counted
2Explain why tabulated atomic masses are non-integral, and work back from an average to isotope abundances
3Determine empirical and molecular formulae from percentage composition or combustion data
4Identify the limiting reagent by dividing moles by coefficients, and compute products, excess and percentage yield from it
5Interconvert molarity, molality, mole fraction and mass percentage by choosing a convenient basis
6Recognise the five laws of chemical combination and the two Dalton postulates now known to be wrong

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Why this chapter matters in JEE Main

This is where most candidates lose marks they never notice losing — not because the ideas are hard, but because a shaky foundation here surfaces as unexplained errors in four later chapters. The organising principle is that the mole is a counting unit and every stoichiometry calculation is the same three moves: convert the given quantity into moles, use the balanced equation's coefficients as a ratio, convert back into whatever the question asks for. Only that middle step is chemistry; everything on either side is unit conversion. Once that is genuinely believed, molarity, molality, mole fraction, empirical formula, limiting reagent and percentage yield stop being separate formulas. JEE Main returns every year to limiting reagent, concentration interconversion, empirical formula from percentage composition, and particle counting where the entity has to be read carefully.

Before you start — revise these

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Writing and balancing simple chemical equations

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Reading a periodic table for atomic masses

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Confident handling of standard form and powers of ten

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Ratio and proportion arithmetic

Some Basic Concepts in Chemistry (Mole Concept)

$N_{2} + 3 H_{2} \to 2N H_{3}$ . You have 1 mol of nitrogen and 2 mol of hydrogen. Which runs out first?

Most say nitrogen — there is less of it.

Hydrogen runs out first, despite there being twice as much. Divide by the coefficients, not by nothing:

$\frac{1}{1} = 1 vs \frac{2}{3} = 0.67 \Rightarrow hydrogen is limiting$

This is where most candidates lose marks they never notice losing. Not because the ideas are hard, but because the mole concept is the arithmetic backbone of Equilibrium, Thermodynamics, Electrochemistry and Solutions — and a shaky foundation here surfaces as unexplained errors months later.

The organising principle: the mole is a counting unit, and every stoichiometry calculation is the same three moves. Convert what you are given into moles. Use the balanced equation's coefficients as a ratio. Convert back into whatever unit the question wants.

Believe that, and molarity, molality, mole fraction, empirical formula, limiting reagent and percentage yield stop being separate formulas.

1. Why Chemistry Needs a Counting Unit

Reactions happen between individual particles — one carbon atom, one oxygen molecule. That is a statement about counting.

But nobody can weigh out one atom. A carbon atom masses about $1.99 \times 1 0^{- 23}$ g, which no balance will ever read.

So chemistry has a permanent mismatch: reactions are governed by numbers of particles, laboratories work in grams and litres. The mole is the bridge — a fixed agreed count, chosen so that counting and weighing line up.

The dozen analogy, and where it breaks. A dozen eggs and a dozen bricks share a count and differ wildly in mass; a mole of hydrogen atoms and a mole of uranium atoms do the same, differing by a factor over 200. What the analogy cannot convey is scale: counting particles at one per second, a mole would take about nineteen thousand million million years.

2. Atomic and Molecular Masses

Absolute masses are unusable, so chemistry uses relative masses against carbon-12, defined as exactly 12 u. So 1 u is one-twelfth of that atom's mass, about $1.66 \times 1 0^{- 24}$ g.

Why tabulated atomic masses are not whole numbers

Chlorine appears as 35.5 — the mass of no actual chlorine atom. Natural chlorine is 75.77 % chlorine-35 and 24.23 % chlorine-37, and the tabulated value is the weighted average:

$\overset{ˉ}{A} = \frac{( 75.77 \times 35 ) + ( 24.23 \times 37 )}{100} = 35.48$

Worth understanding rather than accepting, because JEE asks it in reverse: given the average and the isotope masses, find the abundances.

Illustration 1

Chlorine occurs as $^{35}$ Cl (mass 34.969 u, abundance 75.77 %) and $^{37}$ Cl (36.966 u, 24.23 %). Find the average atomic mass, then work the calculation backwards.

Weight each isotope by how common it is:

$\overset{ˉ}{A} = (34.969) (0.7577) + (36.966) (0.2423) = 26.497 + 8.957 = 35.45 u$

That is the tabulated value, and no chlorine atom weighs it. Every individual atom is either 34.969 or 36.966; 35.45 is a property of a natural sample, not of an atom.

Running it in reverse, given only the average and the two isotopic masses, let $x$ be the fraction of $^{35}$ Cl:

$34.969 x + 36.966 (1 - x) = 35.45 \Rightarrow 36.966 - 1.997 x = 35.45 \Rightarrow x = 0.759$

which returns the 75.8 per cent it started from.

Note what the question must supply. Abundances are needed for this calculation and cannot be guessed, so a question giving only mass numbers is asking for something else entirely.

Formula mass rather than molecular mass

Sodium chloride has no molecules — it is a lattice of alternating ions with no identifiable NaCl unit. For such substances we use formula mass: 58.5 for NaCl. The distinction is conceptual, not computational; the arithmetic is identical.

3. The Mole and the Avogadro Constant

One mole is the amount containing exactly $6.02214076 \times 1 0^{23}$ elementary entities — a defined exact value since the 2019 SI revision. Before that the mole was defined as the number of atoms in 12 g of carbon-12 and $N_{A}$ was measured. Now the count is fixed and the 12 g result is experimental. Nothing in your calculations changes.

Trap. A mole of what? One mole of oxygen gas contains $6.022 \times 1 0^{23}$ molecules of O $_{2}$ — which is $1.204 \times 1 0^{24}$ oxygen atoms. Any question about atoms in a diatomic gas is testing exactly this.

Sample	Entities asked for	Count
1 mol O $_{2}$	molecules	$6.022 \times 1 0^{23}$
1 mol O $_{2}$	O atoms	$1.204 \times 1 0^{24}$
1 mol H $_{2}$ SO $_{4}$	O atoms	$2.409 \times 1 0^{24}$
1 mol CaCO $_{3}$	ions	$1.204 \times 1 0^{24}$

$n = \frac{m}{M}, N = n \times N_{A}$

These two conversions carry the entire chapter. Everything else is one of them used forwards or backwards.

Molar volume of a gas

Avogadro's law: equal volumes of gases at the same $T$ and $P$ contain equal numbers of molecules. So one mole of any ideal gas occupies the same volume.

Conditions	Molar volume
273.15 K, 1 atm	22.414 L
273.15 K, 1 bar (NCERT STP)	22.711 L

Most JEE questions use 22.4 L and state their conditions. Read the question rather than assuming — and never apply molar volume to a liquid or a solid.

Illustration 2

Find the number of oxygen atoms in 2.50 g of calcium carbonate.

$M (CaC O_{3}) = 40.08 + 12.01 + 3 (16.00) = 100.09 g mol^{- 1}$

$n = \frac{2.50}{100.09} = 0.02498 mol \Rightarrow n (O) = 3 \times 0.02498 = 0.07493 mol$

$N = 0.07493 \times 6.022 \times 1 0^{23} = 4.51 \times 1 0^{22} atoms$

The step candidates skip is multiplying by 3. Read carefully whether the question wants formula units, molecules, or a particular element's atoms.

4. Percentage Composition and Formulae

$%_{E} = \frac{( atoms of E ) \times ( atomic mass of E )}{molar mass} \times 100$

Empirical formula is the simplest whole-number ratio of atoms. Molecular formula is the actual number in a molecule, always a whole multiple of it:

$molecular = (empirical) \times k, k = \frac{molar mass}{empirical formula mass}$

The procedure. Take 100 g so percentages become grams. Divide each mass by that element's atomic mass. Divide all results by the smallest.

Trap. A ratio of 1.5 must be doubled, not rounded to 2. Likewise 1.33 is tripled and 1.25 is quadrupled. Only genuine experimental scatter — within about 0.1 of an integer — may be rounded.

Why glucose and formaldehyde share an empirical formula. C $_{6}$ H $_{12}$ O $_{6}$ and CH $_{2}$ O both reduce to CH $_{2}$ O and are both 40.0 % carbon. Percentage composition alone can therefore never identify a compound — which is exactly why questions supply the molar mass.

Illustration 3

A compound is 40.00 % C, 6.72 % H, 53.28 % O with molar mass 180 g mol⁻¹. Find both formulae.

Element	Mass in 100 g	Atomic mass	Moles	÷ smallest
C	40.00	12.01	3.331	1.000
H	6.72	1.008	6.667	2.002
O	53.28	16.00	3.330	1.000

Empirical formula CH $_{2}$ O, mass 30.03.

$k = \frac{180}{30.03} = 5.99 \approx 6 \Rightarrow C_{6} H_{12} O_{6}$

Glucose. The percentages alone would have been satisfied equally by formaldehyde; the molar mass is what settles it.

Illustration 4

0.240 g of a compound containing only C, H and O burns completely to give 0.352 g of CO₂ and 0.144 g of H₂O. Its molar mass is 60 g mol⁻¹. Find the molecular formula.

All the carbon ends up in the CO $_{2}$ and all the hydrogen in the water. That is the whole method.

$n (C) = \frac{0.352}{44.01} = 0.00800 mol \Rightarrow m (C) = 0.0961 g$

$n (H_{2} O) = \frac{0.144}{18.02} = 0.00799 \Rightarrow n (H) = 0.01598 mol, m (H) = 0.0161 g$

Oxygen cannot be measured directly, because the oxygen in the products came partly from the air. Get it by difference:

$m (O) = 0.240 - 0.0961 - 0.0161 = 0.1278 g \Rightarrow n (O) = 0.00799 mol$

$C : H : O = 0.00800 : 0.01598 : 0.00799 = 1 : 2 : 1$

Empirical CH $_{2}$ O of mass 30.03, and $k = 60/30.03 = 2$ :

$C_{2} H_{4} O_{2} (acetic acid)$

Doubling the water's moles to get hydrogen, and taking oxygen by difference rather than from the products, are the two steps this question exists to test.

5. Laws of Chemical Combination

These predate atomic theory and are the evidence it was built to explain. JEE asks them as one-mark recognition questions.

Law	Statement	Example
Conservation of mass	Mass is neither created nor destroyed	Total before = total after
Definite proportions	A compound always has the same mass ratio	Water is always 1:8 H to O
Multiple proportions	Masses of one element combining with a fixed mass of another are in small whole-number ratios	CO and CO $_{2}$
Gay-Lussac	Gases combine in simple whole-number volume ratios	H $_{2}$ and O $_{2}$ combine 2:1
Avogadro	Equal volumes hold equal numbers of molecules	Explains Gay-Lussac

Dalton's atomic theory explained the first three at a stroke. Two of its postulates are now known wrong: atoms are divisible, and atoms of an element are not all identical — isotopes exist.

Illustration 5

Nitrogen forms N $_{2}$ O, NO and NO $_{2}$ . Show that these obey the law of multiple proportions.

Fix the nitrogen at 14 g in each case and read off the oxygen:

Oxide	N (g)	O (g)
N $_{2}$ O	14	8
NO	14	16
NO $_{2}$	14	32

$8 : 16 : 32 = 1 : 2 : 4$

Small whole numbers. The point is the fixing step — the law says nothing until one element's mass is held constant, and questions that look impossible usually just have not been normalised yet.

6. Stoichiometry of Balanced Equations

Trap. A balanced equation is a statement about moles, never about masses or volumes. For $N_{2} + 3 H_{2} \to 2N H_{3}$ it is not true that 1 g reacts with 3 g.

For gases at the same conditions, volumes are proportional to moles, so you can apply the coefficient ratio to volumes directly and skip the mole step entirely.

Illustration 6

What volume of oxygen at STP burns 5.6 L of methane at STP, and what volume of CO $_{2}$ results?

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

Both gases are at the same conditions, so the coefficients act directly on volumes:

$V (O_{2}) = 2 \times 5.6 = 11.2 L, V (C O_{2}) = 1 \times 5.6 = 5.6 L$

No division by 22.4 was needed anywhere. Water is excluded because it is a liquid at STP — applying the gas ratio to it is the trap in this question type.

7. Limiting Reagent

When quantities of more than one reactant are given, one runs out first and caps the product. That one is the limiting reagent.

The reliable test: divide the available moles of each reactant by its coefficient. The smallest quotient is limiting.

Comparing raw moles is wrong whenever the coefficients differ, and it is the single most common error in the chapter.

All product amounts are computed from the limiting reagent alone. The excess reagent's leftover is what was supplied minus what actually reacted.

$% yield = \frac{actual yield}{theoretical yield} \times 100$

Yields fall short through side reactions, incomplete or reversible reaction, and losses during separation.

Illustration 7

50.0 g of N $_{2}$ is mixed with 10.0 g of H $_{2}$ . Find the mass of ammonia produced and the excess left over.

$n (N_{2}) = \frac{50.0}{28.02} = 1.784, n (H_{2}) = \frac{10.0}{2.016} = 4.960$

$\frac{1.784}{1} = 1.784 vs \frac{4.960}{3} = 1.653 \Rightarrow H_{2} limiting$

$n (N H_{3}) = \frac{2}{3} (4.960) = 3.307 mol = 56.3 g$

Nitrogen consumed is 1.653 mol, so $1.784 - 1.653 = 0.131$ mol $= 3.67$ g is left.

Check the mass balance: $56.3 + 3.67 = 60.0$ g against 60.0 g supplied. Run this line on every stoichiometry problem — it catches almost every arithmetic slip.

Illustration 8

Heating 25.0 g of calcium carbonate gives 11.5 g of calcium oxide. Find the percentage yield.

$CaC O_{3} \to CaO + C O_{2}$

$n = \frac{25.0}{100.09} = 0.2498 mol \Rightarrow m_{t h eor e t i c a l} (CaO) = 0.2498 \times 56.08 = 14.01 g$

$% yield = \frac{11.5}{14.01} \times 100 = 82.1 %$

A yield above 100 % is never a chemical result. It means the product was weighed while still wet, or contaminated with unreacted starting material.

Illustration 9

A 2.00 g sample of impure calcium carbonate is treated with excess hydrochloric acid, releasing 0.400 L of carbon dioxide at STP. Find the percentage purity.

$CaC O_{3} + 2HCl \to CaC l_{2} + H_{2} O + C O_{2}$

Start from the measured gas, since that is the only quantity actually known:

$n (C O_{2}) = \frac{0.400}{22.4} = 0.01786 mol$

The ratio is 1:1, so the same amount of CaCO $_{3}$ reacted:

$m (CaC O_{3}) = 0.01786 \times 100.09 = 1.787 g \Rightarrow purity = \frac{1.787}{2.00} \times 100 = 89.4 %$

The acid is in excess, so it is irrelevant — no limiting-reagent test is needed. The impurity is assumed unreactive, which is what "impure" means in this question type.

8. Concentration of Solutions

Measure	Definition	Temperature dependent
Molarity $M$	moles of solute per litre of solution	Yes
Molality $m_{B}$	moles of solute per kilogram of solvent	No
Mole fraction $x$	moles of a component ÷ total moles	No
Mass percentage	mass of solute per 100 g of solution	No

$M = \frac{n _{so l u t e}}{V _{so l u t i o n}}, m_{B} = \frac{n _{so l u t e}}{w _{so l v e n t}}, x_{A} = \frac{n _{A}}{n _{A} + n _{B}}, x_{A} + x_{B} = 1$

Why molality exists at all. Molarity is per litre of solution, and volume expands on heating — so the same solution has a lower molarity at 60 °C than at 20 °C without a single particle being added or removed. Molality is per kilogram of solvent, and mass does not change with temperature. That is why every colligative property in Solutions uses molality.

Trap. Watch the denominators: molarity uses the volume of the whole solution, molality the mass of the solvent alone. Mixing them is a routine source of lost marks.

$ppm = \frac{mass of solute}{mass of solution} \times 1 0^{6}, M_{1} V_{1} = M_{2} V_{2}$

For dilute aqueous solutions 1 ppm is about 1 mg per litre, since a litre of dilute solution masses close to 1 kg. Dilution changes the volume but not the moles of solute, which is where $M_{1} V_{1} = M_{2} V_{2}$ comes from — and why it reappears in titrations.

Illustration 10

Concentrated hydrochloric acid is 36.0 % HCl by mass with density 1.18 g mL⁻¹. Find its molarity, molality and the mole fraction of HCl.

Molarity — take exactly 1 L of solution, mass $1000 \times 1.18 = 1180$ g, of which HCl is $0.360 \times 1180 = 424.8$ g:

$M = \frac{424.8/36.46}{1} = 11.65 mol L^{- 1}$

Molality — take 1000 g of solution instead. HCl is 360 g $= 9.874$ mol; water is 640 g $= 0.640$ kg:

$m_{B} = \frac{9.874}{0.640} = 15.43 mol kg^{- 1}$

Mole fraction — water is $640/18.02 = 35.52$ mol:

$x_{HCl} = \frac{9.874}{9.874 + 35.52} = 0.217$

Notice the basis chosen in each part: 1 L for molarity because its denominator is a volume, 1000 g for molality because its denominator is a mass. Picking the convenient basis removes nearly all the algebra.

Illustration 11

Drinking water contains 1.5 ppm fluoride. Find the mass per litre and the number of F⁻ ions in a litre.

$1.5 ppm \approx 1.5 mg L^{- 1} \Rightarrow n = \frac{1.5 \times 1 0 ^{- 3}}{19.00} = 7.9 \times 1 0^{- 5} mol$

$N = 7.9 \times 1 0^{- 5} \times 6.022 \times 1 0^{23} = 4.8 \times 1 0^{19} ions$

Forty-eight million million million ions in a glass of water, described as a "trace". That gap between how tiny ppm sounds and how many particles it represents is why trace contaminants matter at all.

Illustration 12

Concentrated sulphuric acid is 18.0 M. (a) What volume of it makes 500 mL of 0.100 M acid? (b) If 200 mL of 0.100 M acid is then mixed with 300 mL of 0.250 M acid, what is the final molarity?

(a) Dilution adds solvent and changes nothing about the amount of solute, so the moles before equal the moles after:

$M_{1} V_{1} = M_{2} V_{2} \Rightarrow V_{1} = \frac{( 0.100 ) ( 500 )}{18.0} = 2.78 mL$

(b) Mixing two solutions of the same solute means adding two amounts and dividing by the combined volume:

$n = (0.200) (0.100) + (0.300) (0.250) = 0.0200 + 0.0750 = 0.0950 mol$

$M = \frac{0.0950}{0.500} = 0.190 M$

Averaging the two molarities would have given 0.175 M, and it is wrong. Concentration is not an additive quantity — moles are. The average happens to be right only when the two volumes are equal, which is exactly often enough to let the habit survive unnoticed until a question like this one.

Beyond the JEE Main Syllabus

Two areas older textbooks place in this chapter were removed in the 2023 revision and remain out for 2026.

Physical quantities and measurement — SI units, precision and accuracy, significant figures, dimensional analysis in a chemical context — was deleted. Keep answers to a sensible number of digits, but no question will test the rules themselves.

States of Matter was deleted as a whole chapter, taking with it Boyle's law, Charles's law, the ideal gas equation, kinetic molecular theory, real gases and the van der Waals equation. Molar volume survives here only as a consequence of Avogadro's law.

Both remain fully examinable in JEE Advanced, where the gaseous state is a standard source of problems. If you are sitting both papers, treat them as Advanced-only rather than skipping them.

Summary

The mole is a counting unit of $6.02214076 \times 1 0^{23}$ entities, defined exactly since 2019.
Molar mass in g mol⁻¹ equals relative molecular mass numerically — the entire convenience of the carbon-12 standard.
Two conversions do all the work: $n = m / M$ and $N = n N_{A}$ .
Always state which entity you are counting — a mole of a diatomic gas holds two moles of atoms.
Tabulated atomic masses are weighted isotope averages; JEE asks this in reverse.
Empirical formula from masses ÷ atomic masses ÷ smallest. A ratio of 1.5 is doubled, not rounded.
Molecular formula needs the molar mass too — percentages cannot separate glucose from formaldehyde.
A balanced equation is a ratio between moles and nothing else.
For gases at the same conditions, apply the coefficient ratio to volumes directly.
Identify the limiting reagent by moles ÷ coefficient, never by comparing raw moles.
Compute every product from the limiting reagent; find the excess by subtraction.
Check the mass balance at the end. It catches more errors than rechecking arithmetic.
Yield above 100 % means a wet or contaminated product, never chemistry.
Molarity is per litre of solution and varies with temperature; molality is per kg of solvent and does not.
Choose the basis to suit the denominator: 1 L for molarity, 1000 g for molality.
1 ppm $\approx$ 1 mg L⁻¹ in dilute aqueous solution; $M_{1} V_{1} = M_{2} V_{2}$ for dilution and titration.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

The two master conversions

n = \frac{m}{M}, N = n \times N_{A}

Everything in this chapter is one of these used forwards or backwards. $N_A = 6.02214076\times10^{23}$ has been a defined exact value since the 2019 SI revision; before that it was measured.

Molar volume of a gas

V_{m} = 22.414 L at 273.15 K, 1 atm; 22.711 L at 273.15 K, 1 bar

A consequence of Avogadro's law. NCERT uses the bar definition while most papers still state 22.4 L, so read the question. Never apply molar volume to a liquid or a solid.

Average atomic mass

\overset{ˉ}{A} = \frac{\sum ( abundance \times isotope mass )}{100}

Why chlorine reads 35.5 when no chlorine atom has that mass. JEE asks this in reverse more often than forwards: given the average and the two isotope masses, find the abundances.

Percentage composition

%_{E} = \frac{( atoms of E ) \times ( atomic mass of E )}{molar mass} \times 100

Cannot identify a compound on its own — glucose and formaldehyde share both an empirical formula and a carbon percentage. That is exactly why questions also supply the molar mass.

Empirical to molecular formula

molecular = (empirical) \times k, k = \frac{molar mass}{empirical formula mass}

Get the empirical formula by dividing masses by atomic masses and then by the smallest result. A ratio near 1.5 must be doubled, 1.33 tripled, 1.25 quadrupled — never rounded.

Limiting reagent test

compare \frac{n _{i}}{coefficient _{i}} for each reactant; smallest is limiting

Comparing raw moles is wrong whenever the coefficients differ, and it is the commonest error in the chapter. Every product amount then follows from the limiting reagent alone.

Percentage yield

% yield = \frac{actual}{theoretical} \times 100

Shortfalls come from side reactions, incomplete or reversible reaction, and losses on separation. A figure above 100 per cent is never chemistry — it means a wet or contaminated product.

Molarity and molality

M = \frac{n _{so l u t e}}{V _{so l u t i o n} ( L )}, m_{B} = \frac{n _{so l u t e}}{w _{so l v e n t} ( kg )}

Molarity is per litre of solution and falls when the solution is heated; molality is per kilogram of solvent and does not change at all. That is precisely why colligative properties are written in molality.

Mole fraction and ppm

x_{A} = \frac{n _{A}}{n _{A} + n _{B}}, x_{A} + x_{B} = 1, ppm = \frac{mass of solute}{mass of solution} \times 1 0^{6}

Both are temperature independent. For dilute aqueous solutions 1 ppm is about 1 mg per litre, since a litre of dilute solution masses close to 1 kg.

Dilution

M_1V_1 = M_2V_2

Adding solvent changes the volume but not the moles of solute. The same relation reappears in titration calculations, because there too a fixed amount of solute is being conserved.

Mixing solutions of the same solute

M_{f ina l} = \frac{M _{1} V _{1} + M _{2} V _{2}}{V _{1} + V _{2}}

Add the amounts, never the concentrations. Averaging two molarities gives the right answer only when the volumes happen to be equal, which is common enough for the habit to survive until a question uses unequal volumes.

Purity and excess reagent

purity = \frac{m _{r e a c t i v e}}{m _{s am pl e}} \times 100

Start from whichever quantity was actually measured — usually a gas volume or a precipitate mass — and work back to the reactive component. A reagent described as in excess needs no limiting-reagent test, and the impurity is taken as unreactive, which is what "impure" means in this question type.

Combustion analysis

n (C) = n (C O_{2}), n (H) = 2 n (H_{2} O), m (O) = m_{s am pl e} - m_{C} - m_{H}

All the carbon lands in the carbon dioxide and all the hydrogen in the water, but oxygen cannot be read off the products because some of it came from the air. It is always taken by difference. The factor of two on hydrogen is the other step these questions exist to test.

JEE Main test series

Ready to be tested on this?

Some Basic Concepts in Chemistry (Mole Concept) shows up in every JEE Main paper — sit a timed one and find out where you actually stand.

~2Est. Qs

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Full JEE Main mocks, real clock
Chemistry-only sectionals & chapter tests
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⚠️

Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Choosing the limiting reagent by comparing raw moles

✓ Divide each reactant's moles by its coefficient and compare those quotients. With 1 mol N

_{2}

and 2 mol H

_{2}

, hydrogen is limiting despite there being twice as much of it, because

2/3 < 1/1

Why it happens: Whichever reactant there is less of looks like the one that will run out first.

WATCH OUT

✗ Rounding a mole ratio of 1.5 up to 2

✓ Multiply the whole set by 2 instead. Likewise 1.33 by 3 and 1.25 by 4. Only values within about 0.1 of an integer are genuine experimental scatter and may be rounded.

Why it happens: The other ratios in the same table usually are whole numbers, so 1.5 looks like scatter.

WATCH OUT

✗ Treating a balanced equation as a mass ratio

✓ The coefficients relate moles and nothing else. In

N_{2} + 3 H_{2} \to 2N H_{3}

it is not true that 1 g reacts with 3 g. Convert to moles first, always.

Why it happens: The coefficients look like a recipe, and recipes are usually written in grams.

WATCH OUT

✗ Confusing the denominators of molarity and molality

✓ Molarity uses the volume of the whole solution; molality uses the mass of the solvent alone. Choosing the basis accordingly — 1 L for molarity, 1000 g for molality — removes nearly all the algebra.

Why it happens: The names differ by two letters and both are quoted as "concentration".

WATCH OUT

✗ Forgetting that a mole of a compound contains several moles of atoms

✓ Read which entity is being counted. One mole of H

_{2}

_{4}

holds

6.022 \times 1 0^{23}

molecules but

2.409 \times 1 0^{24}

oxygen atoms. The subscript is the multiplier you need.

Why it happens: The question says "one mole" and the count

6.022 \times 1 0^{23}

is reached for automatically.

WATCH OUT

✗ Applying 22.4 L per mole to a liquid, a solid, or a gas away from STP

✓ Molar volume is a property of ideal gases at stated conditions only. In a combustion equation the water is usually liquid and must be excluded from any volume ratio.

Why it happens: It is the most memorable number in the chapter, so it gets applied wherever a volume appears.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Some Basic Concepts in Chemistry (Mole Concept)?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min worth ~8 marks in JEE Main exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•The mole is a counting unit of $6.02214076 \times 1 0^{23}$ , defined exactly since 2019
• $n = m / M$ and $N = n N_{A}$ do all the work; always state which entity is being counted
•Tabulated atomic masses are weighted isotope averages, and JEE asks this in reverse
•Molar volume 22.4 L at 1 atm, 22.7 L at 1 bar — gases only, never liquids or solids
•Empirical formula: masses ÷ atomic masses ÷ smallest; a ratio of 1.5 is doubled, not rounded
•Molecular formula needs the molar mass too, since percentages cannot separate glucose from formaldehyde
•A balanced equation is a ratio between moles alone; for gases at the same conditions it applies to volumes directly
•Limiting reagent is found by moles ÷ coefficient, never by comparing raw moles
•Check the mass balance at the end of every stoichiometry problem
•Molarity is per litre of solution and temperature dependent; molality is per kg of solvent and is not
•The tabulated atomic mass is a weighted mean over a natural sample — no chlorine atom actually weighs 35.45 u
•Every conversion goes through the mole: $\div M$ from mass, $\times N_{A}$ to particles, $\times 22.4$ L to gas volume at STP

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: ~2 questions (8 marks) of the 100-mark Chemistry section

Question style	Marks each	Typical count	What it tests
Stoichiometry and limiting reagent	3	1	Mole-ratio conversions through a balanced equation, identifying the limiting reagent by dividing moles by coefficients, excess remaining, percentage yield, and purity from a measured gas volume
Concentration of solutions	2	1	Molarity against molality and why only one is temperature-dependent, density-and-percentage data converted to molarity, mole fraction and ppm, dilution by $M_1V_1 = M_2V_2$, and mixing by adding moles rather than averaging
Mole, molar mass and particle counting	1	1	Routing every conversion through the mole, counting atoms of one element inside a compound, molar volume at STP, and average atomic mass from isotopic abundances
Empirical and molecular formulae	1	1	Percentage composition to empirical formula, combustion analysis with oxygen by difference and the factor of two on hydrogen, and the molar mass needed to reach the molecular formula
Laws of chemical combination	1	1	Demonstrating multiple and definite proportions from mass data, and recognising which law a given data set illustrates

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Convert everything to moles as the first line of every calculation, whatever the question gives you. The three-move structure then makes the rest mechanical.
When two reactant quantities are given, the question is about limiting reagent even if it does not say so. Divide by coefficients before doing anything else.
Finish every stoichiometry problem with a mass balance check. Total mass in must equal total mass out, and that one line catches more errors than rechecking arithmetic.
In concentration conversions, choose the basis that matches the denominator you want — 1 L for molarity, 1000 g for molality, 100 g for mass percentage. Almost all the algebra disappears.
Read which entity a counting question wants: formula units, molecules, or atoms of a named element. The subscript in the formula is usually the factor being tested.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Industrial plants size their feedstock streams by limitin…

Industrial plants size their feedstock streams by limiting-reagent calculations, since running any reactant in excess means paying for material that leaves unreacted and then has to be separated and recycled

Water quality standards for arsenic

Water quality standards for arsenic, fluoride and lead are written in parts per million and parts per billion, and converting those to an actual number of ions per litre is what makes a trace contaminant sound as serious as it is

Combustion analysis followed by mass spectrometry is stil…

Combustion analysis followed by mass spectrometry is still the standard route to the formula of a newly isolated natural product — percentage composition fixes the ratio, the molar mass picks the multiple

Where else this topic is tested

Prepare once, score in every exam that asks it.

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Hydrogen was the original standard, and oxygen replaced it for a while, but both caused trouble. Natural hydrogen and oxygen are mixtures of isotopes, so physicists and chemists ended up using slightly different scales — an embarrassment that persisted into the 1950s. Carbon-12 is a single, easily purified nuclide with a mass number conveniently close to the old oxygen scale, so adopting it in 1961 unified the two scales while shifting almost nobody's numbers. It also happens to make most tabulated masses fall very close to whole numbers, which is a real convenience.

The direction of the definition reversed. Before 2019 the mole was defined as the number of atoms in exactly 12 g of carbon-12, and $N_{A}$ was a measured quantity with an uncertainty. Since 2019 $N_{A}$ is fixed by definition at exactly $6.02214076 \times 1 0^{23}$ , and the statement that 12 g of carbon-12 contains one mole has become an experimental result — true to about ten significant figures. Nothing you calculate changes. It matters only because it makes the mole independent of any particular substance, which is what the rest of the SI revision was about.

Use whatever the question states, and if it states nothing, use 22.4 L. The difference is which pressure defines STP: 22.414 L at 1 atm, 22.711 L at 1 bar. NCERT switched to the bar definition, but the overwhelming majority of question papers and answer keys still work at 22.4 L. If a question gives temperature and pressure explicitly, ignore both memorised values and use the ideal gas equation.

Because the reaction consumes it three times as fast. The coefficients say that each nitrogen molecule needs three hydrogen molecules, so having three times as much hydrogen only just keeps pace — anything less and hydrogen runs out first. Dividing each amount by its coefficient converts both into the same currency, namely how many times the reaction can proceed, and then they are directly comparable. Comparing raw moles compares quantities that were never in the same units to begin with.

Because it narrows the field enormously, and historically it was all there was. Combustion analysis gives percentage composition directly, which fixes the empirical formula and therefore rules out every compound with a different atom ratio. What remains is a family — CH $_{2}$ O, C $_{2}$ H $_{4}$ O $_{2}$ , C $_{6}$ H $_{12}$ O $_{6}$ and so on — and a single molar mass measurement then picks one out. In modern practice mass spectrometry supplies the molar mass in minutes, so the two techniques together still do exactly this job for newly isolated compounds.

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Sources and How This Chapter Was CheckedSyllabus scope, what was derived rather than quoted, and how every answer here was checked.

Scope follows the NTA JEE Main syllabus (Unit 1, Some Basic Concepts in Chemistry): matter and its nature, laws of chemical combination, Dalton's atomic theory, atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formulae, chemical equations and stoichiometry.

Physical quantities and their measurement, including significant figures and dimensional analysis, were removed from the syllabus in the 2023 revision, as was the whole States of Matter unit. Both are flagged explicitly rather than silently omitted, because they remain examinable in JEE Advanced.

The Avogadro constant is quoted at its post-2019 defined exact value, and both the 1 atm and 1 bar molar volumes are given, since NCERT uses the bar definition while most question papers still state 22.4 L.

Results were derived rather than quoted: the chlorine average from its two isotope abundances; the multiple-proportions ratio by first normalising each oxide to a fixed 14 g of nitrogen; the concentration conversions by choosing the basis that matches each denominator; and the combustion formula by taking oxygen by difference, since product oxygen comes partly from the air.

Every illustration was checked. The ammonia calculation was verified by mass balance, 56.3 g of product plus 3.67 g of leftover nitrogen against 60.0 g supplied. The glucose formula was confirmed by checking that formaldehyde satisfies the same percentages, which is the point of the question. The purity calculation was worked from the measured gas volume rather than the sample mass, since only the gas is actually known.

The illustrations are teaching problems written for this chapter, not previous-year questions, and are not labelled as such.

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Some Basic Concepts in Chemistry (Mole Concept)

By the end of this chapter you'll be able to…

Before you start — revise these

Some Basic Concepts in Chemistry (Mole Concept)

1. Why Chemistry Needs a Counting Unit

2. Atomic and Molecular Masses

Why tabulated atomic masses are not whole numbers

Illustration 1

Formula mass rather than molecular mass

3. The Mole and the Avogadro Constant

Molar volume of a gas

Illustration 2

4. Percentage Composition and Formulae

Illustration 3

Illustration 4

5. Laws of Chemical Combination

Illustration 5

6. Stoichiometry of Balanced Equations

Illustration 6

7. Limiting Reagent

Illustration 7

Illustration 8

Illustration 9

8. Concentration of Solutions

Illustration 10

Illustration 11

Illustration 12

Beyond the JEE Main Syllabus

Summary

Key formulas & results

Traps JEE Main sets — and how to dodge them

Exam-pattern practice

Are you exam-ready for Some Basic Concepts in Chemistry (Mole Concept)?

5-minute revision

JEE Main question blueprint

Exam-hall strategy

Beyond the exam

Where else this topic is tested

📝 My notes

Questions aspirants ask

Why was carbon-12 chosen as the standard rather than hydrogen, which is lighter?

What actually changed about the mole in 2019, and does it affect my answers?

Should I use 22.4 L or 22.7 L for molar volume in the exam?

How can hydrogen be the limiting reagent when there is nearly three times as much of it?

If percentage composition cannot identify a compound, why is it taught at all?

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