How much of JEE Main comes from Some Basic Concepts in Chemistry?

JEEGenie estimates about 3% of JEE Main marks come from Some Basic Concepts in Chemistry, based on recent papers. The share varies from year to year.

Which formulas matter most in Some Basic Concepts in Chemistry?

The Some Basic Concepts in Chemistry sheet on this page has 22 formula entries, starting with Mole Concept, Atomic Mass and Molecular Mass, Empirical and Molecular Formula, Limiting Reagent, Percentage Yield and Purity. Each one comes with when to use it and the trap to avoid.

What topics does Some Basic Concepts in Chemistry cover for JEE?

Some Basic Concepts in Chemistry covers Mole Concept, Stoichiometry, Concentration Terms, Empirical and Molecular Formula, Limiting Reagent, Laws of Chemical Combination.

How should I revise Some Basic Concepts in Chemistry before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Some Basic Concepts in Chemistry questions on JEEGenie.

How many previous-year questions are there on Some Basic Concepts in Chemistry?

JEEGenie has 415 previous-year questions on Some Basic Concepts in Chemistry, and this page shows a few with worked explanations.

Is practice on Some Basic Concepts in Chemistry free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Some Basic Concepts in Chemistry for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 3% of JEE Main marks come from Some Basic Concepts in Chemistry. This page covers 22 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing molar mass (g/mol) with molecular mass (amu); they are numerically equal but units differ.

Key points to know

Mole Concept. One mole contains exactly $6.022 \times 10^{23}$ elementary entities (Avogadro's number). It is the SI unit for amount of substance.
Atomic Mass and Molecular Mass. Atomic mass is the average mass of an atom relative to $\frac{1}{12}$ of a carbon-12 atom. Molecular mass is the sum of atomic masses in a molecule.
Empirical and Molecular Formula. Empirical formula is the simplest whole-number ratio of atoms; molecular formula is a multiple of empirical formula.
Concentration Terms: Molarity, Molality, Normality. Molarity (M) = moles of solute per liter of solution. Molality (m) = moles of solute per kg of solvent. Normality (N) = equivalents per liter.
Equivalent Mass and n-Factor. Equivalent mass = molar mass / n-factor. n-factor depends on reaction: for acids, it's number of H+; for bases, OH-; for redox, electrons transferred.
Stoichiometry and Balanced Equations. Stoichiometry uses mole ratios from balanced chemical equations to relate amounts of reactants and products.
Dalton's Law of Partial Pressures. Total pressure of a gas mixture equals sum of partial pressures of individual gases. Partial pressure = mole fraction × total pressure.
Significant Figures and Scientific Notation. Significant figures indicate precision. Rules: non-zero digits are significant; zeros between non-zero digits are significant; leading zeros are not; trailing zeros after decimal are significant.
Law of Conservation of Mass and Definite Proportions. Mass is conserved in chemical reactions. A compound always contains the same elements in fixed proportions by mass.
Average Atomic Mass. Average atomic mass is weighted average of isotopic masses based on natural abundance.
Hardness of Water. Hardness is due to Ca2+ and Mg2+ ions. Expressed as ppm of CaCO3 equivalents.
Raoult's Law and Colligative Properties. Relative lowering of vapor pressure = mole fraction of solute. Colligative properties depend on number of solute particles.

Formula sheet

Name	Formula	When to use
Mole Concept	$n = \frac{N}{N_{A}} = \frac{m}{M} = \frac{V}{22.4 L (STP)}$	Use to convert between mass, moles, number of particles, and gas volume at STP.
Atomic Mass and Molecular Mass	$Atomic mass = \sum (isotopic mass \times abundance)$	Use to calculate molar mass, percentage composition, and empirical/molecular formulas.
Empirical and Molecular Formula	$Molecular formula = (Empirical formula)_{n}$ , where $n = \frac{Molecular mass}{Empirical formula mass}$	Use when given percentage composition or combustion analysis data to find the formula of a compound.
Limiting Reagent	$Moles of reactant \div stoichiometric coefficient \to smallest value is limiting$	Use in any reaction where amounts of two or more reactants are given.
Percentage Yield and Purity	$% yield = \frac{actual yield}{theoretical yield} \times 100$ ; $% purity = \frac{pure mass}{impure mass} \times 100$	Use when a reaction gives less product than expected or when reactant is impure.
Concentration Terms: Molarity, Molality, Normality	$M = \frac{n}{V (L)}$ ; $m = \frac{n}{mass of solvent (kg)}$ ; $N = \frac{equivalents}{V (L)}$	Use to express concentration in solutions; molarity for reactions in solution, molality for colligative properties.
Mole Fraction and ppm	$χ_{A} = \frac{n_{A}}{n_{A} + n_{B}}$ ; $ppm = \frac{m_{solute}}{m_{solution}} \times 10^{6}$	Use for dilute solutions or when dealing with gas mixtures.
Equivalent Mass and n-Factor	$Equivalent mass = \frac{M}{n}$ ; $Number of equivalents = \frac{m}{Equivalent mass} = n \times moles$	Use in titration problems and for calculating normality.
Titration and Law of Equivalence	$N_{1} V_{1} = N_{2} V_{2}$ (for same reaction); $n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$ (using n-factor)	Use in acid-base, redox, or precipitation titration problems.
Stoichiometry and Balanced Equations	$a A + b B \to c C + d D$ implies $\frac{n_{A}}{a} = \frac{n_{B}}{b} = \frac{n_{C}}{c} = \frac{n_{D}}{d}$	Use for any quantitative chemical reaction problem.
Gas Laws and Ideal Gas Equation	$P V = n R T$ ; $R = 0.0821 {L atm mol}^{- 1} K^{- 1}$ or $8.314 {J mol}^{- 1} K^{- 1}$	Use for gases when pressure, volume, temperature, or moles are involved.
Dalton's Law of Partial Pressures	$P_{total} = P_{1} + P_{2} + . . .$ ; $P_{i} = χ_{i} P_{total}$	Use for gas mixtures, especially when collecting gas over water.
Empirical Formula from Combustion Data	$n_{C} = \frac{m_{C O_{2}}}{44}$ ; $n_{H} = 2 \times \frac{m_{H_{2} O}}{18}$ ; $n_{O} = \frac{m_{sample} - m_{C} - m_{H}}{16}$	Use when given combustion analysis data (masses of CO2, H2O, etc.) for an organic compound.
Significant Figures and Scientific Notation	$Number = a \times 10^{b}$ , where $1 \leq a < 10$	Use in all calculations to report answers with correct precision.
Percentage Composition	$% element = \frac{atomic mass \times number of atoms}{molar mass} \times 100$	Use to find mass of element in a given mass of compound or to verify formula.
Law of Conservation of Mass and Definite Proportions	$\sum m_{reactants} = \sum m_{products}$	Use to check consistency of reaction data or to find unknown masses.
Redox Titration Calculations	$N_{1} V_{1} = N_{2} V_{2}$ ; $n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$	Use for problems involving KMnO4, K2Cr2O7, oxalic acid, etc.
Average Atomic Mass	$\overset{ˉ}{M} = \frac{\sum (m_{i} \times {abundance}_{i})}{100}$	Use when given isotopic masses and abundances to find atomic mass of an element.
Volume Strength of H2O2	$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ ; $Volume strength = 11.2 \times M$ (where M is molarity)	Use for problems involving hydrogen peroxide solutions.
Hardness of Water	$Hardness (ppm) = \frac{{mass of CaCO}_{3} equivalent}{mass of water} \times 10^{6}$	Use in water analysis problems.
Eudiometry (Gas Analysis)	$V_{reacted} = V_{initial} - V_{final}$ ; $Volume of gas \propto moles$	Use for problems involving combustion of hydrocarbons or gas mixtures with volume changes.
Raoult's Law and Colligative Properties	$\frac{P^{0} - P}{P^{0}} = χ_{solute}$ ; $Δ T_{f} = i K_{f} m$ ; $Δ T_{b} = i K_{b} m$ ; $π = i C R T$	Use for solutions, especially to find molar mass or degree of dissociation.

Open the Some Basic Concepts in Chemistry formula page

What JEE asks

Mole Concept. JEE often asks to find number of atoms/molecules from given mass or volume, or to compare moles of different substances.
Atomic Mass and Molecular Mass. JEE may give isotopic abundances and ask for average atomic mass, or ask to find molecular mass from a formula.
Empirical and Molecular Formula. JEE often gives combustion data (CO2, H2O masses) to find empirical formula, then molecular formula using molar mass.
Limiting Reagent. JEE often gives masses of reactants and asks for mass of product formed or leftover reactant.
Percentage Yield and Purity. JEE may give impure reactant mass and ask for product mass, or give actual yield and ask for percentage yield.
Concentration Terms: Molarity, Molality, Normality. JEE often asks to convert between concentration terms or calculate moles from given volume and molarity.
Mole Fraction and ppm. JEE may ask to find mole fraction from given masses or calculate ppm of a pollutant.
Equivalent Mass and n-Factor. JEE often asks to find n-factor of a compound in a given reaction or calculate equivalent mass.
Titration and Law of Equivalence. JEE often gives titration data and asks for concentration of unknown solution.
Stoichiometry and Balanced Equations. JEE often requires balancing equations first, then using stoichiometry to find masses or volumes.
Gas Laws and Ideal Gas Equation. JEE often combines gas equation with stoichiometry (e.g., volume of gas produced from a reaction).
Dalton's Law of Partial Pressures. JEE often asks to find partial pressure of a gas produced in a reaction or collected over water.
Empirical Formula from Combustion Data. JEE often gives combustion data and asks for empirical formula, then molecular formula using molar mass.
Significant Figures and Scientific Notation. JEE may ask to report answer with certain number of significant figures or identify correct scientific notation.
Percentage Composition. JEE often asks to calculate percentage of an element in a compound or find mass of element from given mass of compound.
Law of Conservation of Mass and Definite Proportions. JEE may ask to apply these laws to verify experimental data or identify compounds.
Redox Titration Calculations. JEE often asks to find concentration of unknown solution using redox titration data.
Average Atomic Mass. JEE may give two isotopes and ask for abundance or atomic mass.
Volume Strength of H2O2. JEE often asks to find molarity from volume strength or vice versa.
Hardness of Water. JEE may ask to calculate hardness from given ion concentrations.
Eudiometry (Gas Analysis). JEE often gives volume of gas mixture before and after reaction to find composition.
Raoult's Law and Colligative Properties. JEE often asks to find molar mass from colligative property data or calculate i-factor.

Common traps

Confusing molar mass (g/mol) with molecular mass (amu); they are numerically equal but units differ. Mole Concept
Using atomic number instead of atomic mass in calculations. Atomic Mass and Molecular Mass
Assuming empirical formula is same as molecular formula without checking molecular mass. Empirical and Molecular Formula
Comparing masses directly instead of converting to moles. Limiting Reagent
Using actual yield as theoretical yield in calculations. Percentage Yield and Purity
Confusing molarity with molality; molarity depends on temperature, molality does not. Concentration Terms: Molarity, Molality, Normality
Forgetting that sum of mole fractions of all components equals 1. Mole Fraction and ppm
Using same n-factor for different reactions (e.g., H2SO4 in acid-base vs redox). Equivalent Mass and n-Factor
Using molarity instead of normality in $N_{1} V_{1} = N_{2} V_{2}$ without adjusting for n-factor. Titration and Law of Equivalence
Using unbalanced equations for mole ratios. Stoichiometry and Balanced Equations
Using wrong units for R (e.g., using L atm with J value). Gas Laws and Ideal Gas Equation
Confusing partial pressure with vapor pressure. Dalton's Law of Partial Pressures
Forgetting to multiply by 2 for hydrogen atoms from water. Empirical Formula from Combustion Data
Counting leading zeros as significant (e.g., 0.0012 has 2 sig figs, not 4). Significant Figures and Scientific Notation

Topics in Some Basic Concepts in Chemistry

Mole Concept
Stoichiometry
Concentration Terms
Empirical and Molecular Formula
Limiting Reagent
Laws of Chemical Combination

Previous-year questions on Some Basic Concepts in Chemistry

JEEGenie has 415 previous-year questions on Some Basic Concepts in Chemistry. A few worked examples follow.

Year	Questions
2026	19
2025	31
2024	30
2023	37
2022	36
2021	40
2020	22
2019	34
2018	14
2017	7
2016	6
2015	7
2014	6
2013	6
2012	4
2011	3
2010	3
2009	1
2007	3
2006	2
2005	3
2004	3
2003	5
2002	4
2001	5
2000	4
1999	5
1998	2
1997	5
1996	4
1995	2
1994	1
1993	2
1992	4
1991	5
1990	5
1989	2
1988	5
1986	1
1985	4
1984	2
1983	4
1982	1
1981	1
1980	7
1979	10
1978	8

Question 1 easy

The mass of iron converted into

F e_{3} O_{4}

by the action of 18 g of steam is: (Given: Molar mass of H, O and Fe are 1, 16 and 56 g mol−1 respectively) Assume iron is present in excess:

2.1 g
4.2 g
21 g
42 g

Answer: D

Explanation.

3 Fe + 4 H_{2} O \to {Fe}_{3} O_{4} + 4 H_{2}

Moles of steam:

n_{H_{2} O} = \frac{18}{18} = 1 mol

From the equation, 4 mol steam react with 3 mol Fe, so:

n_{Fe} = \frac{3}{4} \times 1 = 0.75 mol

Mass of iron:

m_{Fe} = 0.75 \times 56 = 42 g

D) 42 g

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 medium

On complete combustion 1.0 g of an organic compound

(X)

gave 1.46 g of

{C O}_{2}

and 0.567 g of

H_{2} O

. The empirical formula mass of compound

(X)

is __________ g. (Given molar mass in

g {m o l}^{- 1} C : 12, H : 1, O : 16

)

Answer: C

Explanation. Moles of

{C O}_{2} = \frac{1.46}{44} = 0.03318 \to

moles C = 0.03318, mass C =

0.03318 \times 12 = 0.3982

g. Moles of

H_{2} O = \frac{0.567}{18} = 0.0315 \to

moles H =

2 \times 0.0315 = 0.0630

, mass H =

0.0630 \times 1 = 0.0630

g. Mass O =

1.0 - (0.3982 + 0.0630) = 0.5388

g, moles O =

\frac{0.5388}{16} = 0.03368

. Divide by smallest (0.03318): C:

\frac{0.03318}{0.03318} = 1

, H:

\frac{0.0630}{0.03318} \approx 1.9 \approx 2

, O:

\frac{0.03368}{0.03318} \approx 1.015 \approx 1

. Empirical formula:

C H_{2} O

, mass =

12 + 2 + 16 = 30

g. Answer is option C.

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 1

Question 3 hard

A closed vessel contains

10

g of an ideal gas X at

300

K, which exerts

2

atm pressure. At the same temperature,

80

g of another ideal gas Y is added to it and the pressure becomes

6

atm. The ratio of root mean square velocities of X and Y at

300

K is

$2 \sqrt{2} : \sqrt{3}$
$2 \sqrt{2} : 1$
$1 : 2$
$2 : 1$

Answer: D

Explanation. Given the ideal gas law, at constant

V

and

T

P \propto n

Initially,

n_{X} = \frac{10}{M_{X}}

and

P_{X} = 2

atm. After adding Y, total pressure

= 6

atm, so pressure due to Y alone

= 4

atm. Thus,

\frac{n_{Y}}{n_{X}} = \frac{4}{2} = 2

\frac{80 / M_{Y}}{10 / M_{X}} = 2 ⟹ \frac{8 M_{X}}{M_{Y}} = 2 ⟹ M_{Y} = 4 M_{X}

Root mean square velocity:

v_{rms} = \sqrt{\frac{3 R T}{M}}

At same

T

\frac{v_{X}}{v_{Y}} = \sqrt{\frac{M_{Y}}{M_{X}}} = \sqrt{4} = 2

So,

v_{X} : v_{Y} = 2 : 1

D

From NTA/IIT papers: JEE Advanced 2024

Question 4 easy

The volume of

0.02 M

aqueous

H B r

required to neutralize

10.0 m L

0.01 M

aqueous

B a (O H)_{2}

is (Assume complete neutralization)

7.5 mL
5.0 mL
10.0 mL
2.5 mL

Answer: C

Explanation. The neutralization reaction is:

2 H B r + B a (O H)_{2} \to B a B r_{2} + 2 H_{2} O

Moles of

B a (O H)_{2}

0.01 M \times 0.010 L = 1.0 \times 10^{- 4} m o l

From stoichiometry, moles of

H B r

needed:

2 \times 1.0 \times 10^{- 4} = 2.0 \times 10^{- 4} m o l

Volume of

0.02 M H B r

V = \frac{2.0 \times 10^{- 4}}{0.02} = 0.010 L = 10.0 m L

Thus, the answer is option C.

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 2

Frequently asked questions

How much of JEE Main comes from Some Basic Concepts in Chemistry?: JEEGenie estimates about 3% of JEE Main marks come from Some Basic Concepts in Chemistry, based on recent papers. The share varies from year to year.
Which formulas matter most in Some Basic Concepts in Chemistry?: The Some Basic Concepts in Chemistry sheet on this page has 22 formula entries, starting with Mole Concept, Atomic Mass and Molecular Mass, Empirical and Molecular Formula, Limiting Reagent, Percentage Yield and Purity. Each one comes with when to use it and the trap to avoid.
What topics does Some Basic Concepts in Chemistry cover for JEE?: Some Basic Concepts in Chemistry covers Mole Concept, Stoichiometry, Concentration Terms, Empirical and Molecular Formula, Limiting Reagent, Laws of Chemical Combination.
How should I revise Some Basic Concepts in Chemistry before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Some Basic Concepts in Chemistry questions on JEEGenie.
How many previous-year questions are there on Some Basic Concepts in Chemistry?: JEEGenie has 415 previous-year questions on Some Basic Concepts in Chemistry, and this page shows a few with worked explanations.
Is practice on Some Basic Concepts in Chemistry free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.