All Stoichiometry Formulas Class 11, JEE, NEET

Here is the list of all Stoichiometry Formulas Class 11. This formula list is very important to revise before your JEE Main, Advanced or NEET Exam.

Visit this page for all important formulas of JEE Mains

Other important links

• Class 11 Notes
• Download NCERT Books
• Formula Bank for Class 11

All Stoichiometry Formulas Class 11

Relative atomic mass (R.A.M) =$\frac{\text { Mass of one atom of an element }}{\frac{1}{12} \times \text { mass of one carbon atom }}=$ Total Number of nucleons

---

Y-map

---

Density:
Specific gravity $=\frac{\text { density of the substance }}{\text { density of water at } 4^{\circ} C }$

For gases:

Absolute density (mass/volume) $=\frac{\text { Molar mass of the gas }}{\text { Molar volume of the gas }}$

$\Rightarrow \quad \rho=\frac{ PM }{ RT }$

Vapour density

$V . D =\frac{ d _{\text {gas }}}{ d _{ H _{2}}}=\frac{ PM _{\text {gas } / RT }}{ PM _{ H _{2} / RT }}=\frac{ M _{\text {gas }}}{ M _{ H _{2}}}=\frac{ M _{\text {gas }}}{2}$

$M _{ gas }=2 V . D$

---

Mole-mole analysis:

---

Concentration terms :

Molarity (M): $\therefore \quad$ Molarity $( M )=\frac{ w \times 1000}{\text { (Mol. wt of solute) } \times V _{ inm 1}}$

Molality (m):

Molality $=\frac{\text { number of moles of solute }}{\text { mass of solvent in gram }} \times 1000=1000 w _{1} / M _{1} w _{2}$

Mole fraction $( x ):$

$\therefore$ Mole fraction of solution $\left(x_{1}\right)=\frac{n}{n+N}$

$\therefore$ Mole fraction of solvent $\left(x_{2}\right)=\frac{N}{n+N}$

$x_{1}+x_{2}=1$

$\%$ Calculation:

(i) $\% w / w =\frac{\text { mass of solute in } gm }{\text { mass of solution in } gm } \times 100$

(ii) $\% w / v =\frac{\text { mass of solute in } gm }{\text { mass of solution in } ml } \times 100$

(iii) $\quad \% v / v =\frac{\text { Volume of solution in } ml }{\text { Volume of solution }} \times 100$

---

Some Important Formulas:

1. Mole fraction of solute into molarity of solution $M =\frac{ x _{2} \rho \times 1000}{ x _{1} M _{1}+ M _{2} x _{2}}$
2. Molarity into mole fraction $x_{2}=\frac{M M_{1} \times 1000}{\rho \times 1000-M M_{2}}$
3. Mole fraction into molality $m =\frac{ x _{2} \times 1000}{ x _{1} M _{1}}$
4. Molality into mole fraction $x_{2}=\frac{m M_{1}}{1000+m M_{1}}$
5. Molality into molarity $M =\frac{ m \rho \times 1000}{1000+ mM _{2}}$
6. Molarity into Molality $m =\frac{ M \times 1000}{1000 \rho- MM _{2}}$

Here, $M _{1}$ and $M _{2}$ are molar masses of solvent and solute. $\rho$ is density of solution $( gm / mL )$

$M =$ Molarity (mole/lit.), $m =$ Molality (mole/kg), $x _{1}$ $=$ Mole fraction of solvent, $x_{2}=$ Mole fraction of solute

---

Average/Mean atomic mass:

$A_{x}=\frac{a_{1} x_{1}+a_{2} x_{2}+\ldots .+a_{n} x_{n}}{100}$

---

Mean molar mass or molecular mass:

$M _{ av g .}=\frac{ n _{1} M _{1}+ n _{2} M _{2}+\ldots \ldots n _{ n } M _{ n }}{ n _{1}+ n _{2}+\ldots n _{ n }}$ or $M _{ avg .}=\frac{\sum_{j=1}^{j=n} n_{j} M_{j}}{\sum_{j=1}^{j=n} n_{j}}$

---

Calculation of individual oxidation number:

Formula: Oxidation Number $=$ number of electrons in the valence shell- number of electrons left after bonding

---

Concept of Equivalent weight/Mass:

For elements, equivalent weight $(E)=\frac{\text { Atomic weight }}{\text { Valency -factor }}$

For acid/base, $E =\frac{ M }{\text { Basicity } / Acidity }$

Where $M=$ Molar mass

For O.A/R.A, $\quad E =\frac{ M }{\text { no. of moles of } e ^{-} \text {gained/lost }}$

Equivalent weight (E) $=\frac{\text { Atomic or moleculear weight }}{\text { v.f. }} \quad(\text { v.f. }=$ valency factor)

---

Concept of number of equivalents:

No. of equivalents of solute $=\frac{ Wt }{ Eq . \text { wt. }}=\frac{ W }{ E }=\frac{ W }{ M / n }$

No. of equivalents of solute $=$ No. of moles of solute $\times$ v.f.

---

Normality (N):

Normality $( N )=\frac{\text { Number of equivalents of solute }}{\text { Volume of solution (in litres) }}$

Normality $=$ Molarity $\times v.f$

---

Calculation of Valency Factor:

$n$ -factor of acid $=$ basicity $=$ no. of $H ^{+}$ ion (s) furnished per molecule of the acid. n-factor of base $=$ acidity $=$ no. of $OH ^{-}$ ion (s) furnised by the base per molecule.

At equivalence point:

$N _{1} V _{1}= N _{2} V _{2}$

$n _{1} M _{1} V _{1}= n _{2} M _{2} V _{2}$

Volume strength of $H _{2} O _{2}$

$20 VH _{2} O _{2}$ means one litre of this sample of $H _{2} O _{2}$ on decomposition gives 20 lt. of $O _{2}$ gas at S.T.P.

Normality of $H _{2} O _{2}( N )=\frac{\text { Valume, strength of } H _{2} O _{2}}{5.6}$

Molarity of $H _{2} O _{2}( M )=\frac{\text { Volume strength of } H _{2} O _{2}}{11.2}$

Measurement of Hardness:

Hardness in ppm $=\frac{\text { mass of } CaCO _{3}}{\text { Total mass of water }} \times 10^{6}$

Calculation of available chlorine from a sample of bleaching powder:

$\%$ of $Cl _{2}=\frac{3.55 \times x \times V ( mL )}{ W ( g )}$ where $x =$ molarity of hypo solution and $v = mL$. of hypo solution used in titration.

---

At “Study Material Center” you will get all the important exam related study material, such as Notes, Study Material, Formula bank, Previous Year Chapterwise Questions etc.

This was the lis of All Stoichiometry Formulas Class 11, JEE, NEET. Up Next: Gaseous State Class 11 Formulas

Class 11 Chemistry Formulas

• Chemical Bonding and Molecular Structure
• Chemical Equilibrium
• Classification of Elements and Periodicity in Properties
• Gaseous State
• Ionic Equilibrium
• P Block
• S Block
• Stoichiometry
• Structure of Atom
• Thermodynamics