# 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.

## 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.

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

## Class 11 Chemistry Formulas

### 16 thoughts on “All Stoichiometry Formulas Class 11, JEE, NEET”

1. the formulae are not easy to understand such a w i’m not sure if its for gravity or something else ,i was wondering if you couldn’t add a key by any chance?
These are very helpful but not all the formulae was here?

2. HELPED BY THE ABOVE TOPIC

Thanks a lot for providing such kinda stuff. I It is so helpful for the revision purpose and it is also explained in a detailed way

3. These are really good for quick revision at the end of chapter one.
And also thanks for keeping these notes/formulas for free ‘cuz most of the sites in which I’ve been always charged money for any kind of notes
So thanks u very much<33

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