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JEE Main Chemistry Formulas 2026 - The Joint Entrance Exam is one of the most competitive engineering entrance exams. It is divided into three subsections (Physics,Chemistry, and Maths). Chemistry is an important subject for JEE Main preparation. Hence, having a list of JEE Main Chemistry Formulas 2026 is necessary. Every aspirant should have the short notes for JEE Main 2026 that comprise the important formulas. These important JEE formulas for Chemistry help in solving the typical problems easily. The JEE Main Session 1 took place in January, and the JEE Main Session will be conducted from 2 April to 9 April 2026.
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There will be a total of 25 questions from chemistry in JEE Main 2026, all of which have to be attempted. It is further segregated into Physical, Organic, and Inorganic sections. The maximum number of topics has been reduced from this section (chemistry) only. This makes Chemistry easier than before. However, the high-weightage topics for JEE Main remain the same.
Below are important formulas for JEE Main 2026 Chemistry that are helpful for both sessions. Candidates should solve as many chemistry questions as they can and implement these formulas to remember them quickly. To crack the JEE Main exam with good marks, paste these formulas near the study table and memorise them.
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Candidates, while studying chemistry, need to revise and practice the chemical equations and symbols. To some, chemistry is a tough subject, but when candidates practice chemical equations and revise the properties, formulas and symbols, they will have command over the subject. Candidates can check the JEE Main Chemistry formulas below
Some Basic Concepts in chemistry
1. Boyle's Law: $P_1 V_1=P_2 V_2$ (at constant T and n)
2. Charles's Law: $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (at constant P and n)
3. Avogadro's Law: $\frac{v}{n}=$ constant
4. Average Atomic Mass
$$
=\frac{\Sigma(\text { Mass of Isotopes })_i \times(\% \text { abundance })_i}{100}
$$
5. Mole $=\frac{\mathrm{W}}{\mathrm{M}}=\frac{\text { (Wt. of substance in gm.) }{ }^{100}}{\text { (Molar mass of substance(G.m.m)) }}$
6. Mass $1 \%$ of an element
$$
=\frac{\text { Mass of that element in one mole of the compound }}{\text { Molar mass of the compound }} \times 100
$$
7. Equivalent Weight $=\frac{\text { Molecular weight }}{n-\text { factor }(x)}$
Atomic Structure
1. Frequency $v=\frac{1}{T}$
2. Wave number $(\bar{v}) \bar{v}=\frac{1}{\lambda}$
3. $E=h v=\frac{h c}{\lambda}$
4. Line Spectrum of Hydrogen-like atoms
$$
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
$$
5. Bohr radius of nth orbit:
$$
\mathrm{r}_{\mathrm{n}}=0.529 \frac{\mathrm{n}^2}{\mathrm{z}} \mathrm{~A}^0
$$
6. Velocity of electron in nth orbit:
$$
V_n=\left(2.18 \times 10^6\right) \frac{\mathrm{Z}}{\mathrm{n}} \mathrm{~m} / \mathrm{s}
$$
where Z is atomic number
7. Total energy of electron in nth orbit:
$$
E_n=-13.6 \frac{Z^2}{n^2} \mathrm{eV}=-2.18 \times 10^{-18} \frac{Z^2}{n^2} \mathrm{~J}
$$
where Z is atomic number
8. Hisenberg Uncertainity Principle: $\Delta x \cdot \Delta P \geq \frac{h}{4 \pi}$
9. $\mathrm{E}_{\mathrm{n}}=-\frac{1312 \times \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{~kJ} / \mathrm{mol}$
Chemical Thermodynamics
1. Expansion Work $=\mathrm{P} \times \Delta \mathrm{V}=-\mathrm{P}_{\text {ext. }}\left[\mathrm{V}_2-\mathrm{V}_1\right] \mathrm{P}=$ external pressure And $\Delta \mathrm{V}=$ increase or decrease in volume.
2. Work done in a reversible isothermal process
$$
\begin{aligned}
\mathrm{W} & =-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{~V}_2}{\mathrm{~V}_1} \\
\mathrm{~W} & =-2.303 \mathrm{nRTlog} \frac{\mathrm{P}_1}{\mathrm{P}_2}
\end{aligned}
$$
3. Work done in an irreversible isothermal process
$$
\text { Work }=-P_{\text {ext. }}\left(V_2-V_1\right)
$$
That is, Work $=-\mathrm{P} \times \Delta \mathrm{V}$
4. $W=\Delta E=n C_V \Delta T$
5. Enthalpy: $H=U+p V$
6. First Law of Thermodynamics: $\Delta U=q+W$
7. $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta(\mathrm{S})$
8. $\Delta G=-n F E$
Equilibrium
1. For a reaction:
$$
\mathrm{mA}+\mathrm{nB} \rightleftharpoons \mathrm{pC}+\mathrm{qD} \frac{\mathrm{~K}_{\mathrm{f}}}{\mathrm{~K}_{\mathrm{b}}}=\frac{[\mathrm{C}]^{\mathrm{p}}[\mathrm{D}]^{\mathrm{q}}}{[\mathrm{~A}]^{\mathrm{m}}[\mathrm{~B}]^{\mathrm{n}}}=\mathrm{K}_{\mathrm{c}}
$$
2. $\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]$
3. $\mathrm{k}_{\mathrm{W}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$
4. $\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{\text { [Salt] }}{\text { Acid }}$
5. $\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log _{10} \frac{[\text { Salt] }}{[\text { Base }]}$
ELECTROCHEMISTRY
1. $W=\frac{\text { Eit }}{96500}$
$$
\frac{E_1}{E_2}=\frac{M_1}{M_2} \text { or } \frac{W_1}{W_2}=\frac{Z_1}{Z_2}
$$
2.
$E_1=$ equivalent weight
$E_2=$ equivalent weight
W or M = mass deposited
3.
$$
\begin{gathered}
E_{\text {cell }} \text { or } E M F=\left[E_{\text {red }}(\text { cathode })-E_{\text {red }}(\text { anode })\right. \\
E_{\text {eell }}^{\circ} \text { or } E M F^{\circ} \\
=\left[E_{\text {red }}^{\circ}(\text { cathode })-E_{\text {red }}^{\circ}(\text { anode })\right]
\end{gathered}
$$
$$
\begin{aligned}
& \text { 4. } \mathrm{E}=\mathrm{E}^{\circ}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln Q \\
& \text { 5. } \mathrm{xA}+\mathrm{yB} \rightarrow \mathrm{mC}+\mathrm{nD}
\end{aligned}
$$
The emf can be calculated as
$$
\text { Ecell }=E^{\circ} \text { cell }-\frac{0.059}{n} \log \frac{[\mathrm{C}]^{\mathrm{m}}[\mathrm{D}]^{\mathrm{n}}}{[\mathrm{~A}]^{\mathrm{x}}[\mathrm{~B}]^{\mathrm{y}}}
$$
6.
$$
\Lambda_{\mathrm{m}}=\kappa \times \frac{1000}{\mathrm{c}}
$$
7. $\Lambda_{\mathrm{eq}}=\frac{1000 \times \kappa}{\mathrm{N}}$
Solutions
1.
Mass \% of a component $=\frac{\text { Mass of the component in the solution }}{\text { Total mass of the solution }} \times 100$
2. Volume $\%$ of a component $=\frac{\text { Volume of the component }}{\text { Total volume of solution }} \times 100$
3. Mass by Volume $\%$ of a component $=\frac{\text { Total valume of solution }}{\text { Total volume of solution }} \times 100$
4. Parts per million
$$
=\frac{\text { Number of parts of the component }}{\text { Total number of parts of all components of the solution }} \times 10^6
$$
5. Mole fraction of a component
$$
=\frac{\text { Number of moles of the component }}{\text { Total number of moles of all the components }}
$$
6. Molarity: $(M)=\frac{\text { No. of Moles of Solutes }}{\text { Volume of Solution in Liters }}$
7. Molality: $(m)=\frac{\text { No. of Moles of Solutes }}{\text { Mass of solvent inkg }}$
$$
P_A=P_A^o X_A
$$
8. $\left(P_T\right)=P_A^o X_A+P_B^o X_B\left(P_B=P_B^o X_B\right)$
9. $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \frac{\mathrm{W}}{\mathrm{M}} \times \frac{1000}{\mathrm{~W}}$
10. $\Delta T_f=K_f \times \frac{w}{M} \times \frac{1000}{W}$
11. $\Pi=C R T$
$$
\mathrm{i}=\frac{\text { Observed number of solute particles }}{\text { Number of particles initially taken }}
$$
12.
$$
\mathrm{i}=\frac{\text { Observed value of colligative property }}{\text { Theoretical value of colligative property }}
$$
Chemical kinetics
1. Unit of average velocity $=\frac{\text { Unit of concentration }}{\text { Unit of time }}=\frac{\text { mole }}{\text { litre second }}=$ mole litre -1 second -1
2. $a A+b B \rightarrow c C+d D$
Rate w.r.t. $[\mathrm{A}]=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}} \times \frac{1}{\mathrm{a}}$
Rate w.r.t. $[B]=-\frac{d[B]}{d t} \times \frac{1}{b}$
Rate w.r.t. $[\mathrm{C}]=-\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}} \times \frac{1}{\mathrm{c}}$
Rate w.r.t. $[\mathrm{D}]=-\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}} \times \frac{\mathrm{c}}{\mathrm{d}}$
3. $\mathrm{R} \propto[\mathrm{A}]^{\mathrm{p}}[\mathrm{B}]^{\mathrm{q}}$
4. Unit of Rate Constant-
The differential rate expression forn th order reaction is as follows:
$$
\begin{gathered}
-\frac{d x}{d t}=k(a-x)^n \\
\text { ork }=\frac{d x}{(a-x)^n d t}=\frac{(\text { concentration })}{(\text { concentration })^n \text { time }}=(\text { conc. })^{1-n} \text { time }^{-1}
\end{gathered}
$$
5. For the first-order reaction,
$$
\begin{aligned}
& k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_0}{[\mathrm{R}]} \\
& \text { 6. } t_1 / 2=\frac{0.693}{k}
\end{aligned}
$$
7. For any general nth order reaction it is evident that,
$$
\mathrm{t}_{\frac{1}{2}} \propto[\mathrm{~A}]_0^{1-\mathrm{n}}
$$
It is to be noted that the above formula is applicable for any general nth-order reaction except $\mathrm{n}=1$.
8. Arrhenius Equation: $\mathrm{k}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}}$
9. $\log \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{Ea}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]$
Coordination Compounds
1. $$
\mathrm{EAN}=Z-O+2 L
$$
Where:
$Z=$ Atomic number of the central metal atom/ion
0= Oxidation state of the metal atom/ion
$\mathrm{L}=$ Number of ligands (or donor atoms) × number of electrons donated per ligand
2.
Crystal Field Stabilization Energy (CFSE):
Tetrahedral:
$$
\text { CFSE }=(-0.6 x+0.4 y) \Delta_t
$$
where $x=t_2 g$ electrons, $y=$ e g electrons
d- & f-Block Elements
1. Magnetic Moment:
$$
\mu=\sqrt{n(n+2)} \mathrm{BM}
$$
Chemical Bonding and Molecular Structure
1. Formal Charge:
$$
\text { F.C. }=V-N-\frac{B}{2}
$$
( $\mathrm{V}=$ valence electrons, $\mathrm{N}=$ non-bonding, $\mathrm{B}=$ bonding electrons)
2. Bond Order (Molecular Orbital Theory):
$$
\text { Bond Order }=\frac{\left(N_b-N_a\right)}{2}
$$
3. Dipole Moment:
$$
\begin{gathered}
\mu=q \times d \\
(q=\text { charge }, d=\text { distance between charges })
\end{gathered}
$$
Some Basic Principles of Organic Chemistry
1. Degree of Unsaturation (DU or IHD):
$$
\mathrm{DU}=\frac{2 C+2-H+N-X}{2}
$$
( $\mathrm{C}=$ carbon, $\mathrm{H}=$ hydrogen, $\mathrm{N}=$ nitrogen, $\mathrm{X}=$ halogen )
Hydrocarbons
1. Alkanes: $\mathrm{C}_n \mathrm{H}_{2 n+2}$
2. Alkenes: $C_n H_{2 n}$
3. Alkynes: $\mathrm{C}_n \mathrm{H}_{2 n-2}$
Carboxylic Acids and Derivatives
Method of Preparation of Carboxylic Acid

Given below are some tips to help you prepare for JEE Main and score good marks in the exam:
1. First, students need to understand the JEE Main syllabus and Exam Pattern so that they get familiar with the syllabus and the exam pattern.
2. Try to identify the important and high-scoring topics of JEE Main Chemistry and prepare according to that.
3. Create an effective JEE Main study plan according to your preparation level. Divide your preparation into monthly, weekly, and daily targets and allocate more time to difficult subjects or topics.
4. Students must focus on conceptual clarity; they must understand the logic and derivations behind every formula.
5. Try to solve questions regularly. Solve JEE Main previous year question papers and attempt mock tests and sample papers regularly.
Students find it difficult to learn formulas for JEE Main, but with the right approach, they can remember them. Given below are some points to remember:
1. Students must try to understand why a formula works and how chemical reactions occur and their mechanism.
2. Then break down formulas into chapters or topics.
3. To learn these formulas easily, try to make a formula notebook.
4. Sometimes students must try to make mnemonics and short tricks, as it helps in quick revision.
5. Try to solve as many questions and revise
6. Try to use diagrams and flowcharts.
Frequently Asked Questions (FAQs)
Chemistry has three subparts (Physical, inorganic and organic). Physical chemistry has many numericals based on different formulas.
Yes, Class 12 syllabus carries more than 50% of weightage, so it is important for JEE Main Exam.
JEE Main is an exam conducted for those aspirants who want to take admission in NIITs, IITs and other engineering colleges, and it is also an eligibility test for JEE advance.
On Question asked by student community
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Please download the Hindi medium JEE Mains question paper with solution from the link given below:
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Hi,
High - weightage chapters in JEE Mains are based on past paper analysis, not official NTA rules. Some topics come up more often, so students use them for smart revision. You can refer to this link for details:
JEE Main Chapter-wise Weightage 2026: Important Topics & Marks Distribution
Hi Sasika,
Careers360 provides JEE study material, including books, notes, previous year papers, and revision PDFs for Physics, Chemistry, and Maths.
You can check subject-wise study material here:
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