Some Basic Concepts of Chemistry Formula Sheet — JEE Main Chemistry

Every key Some Basic Concepts of Chemistry formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 10 sub-topics

General Introduction: Importance and scope of Chemistry
Nature of matter
Laws of chemical combination
Dalton's atomic theory: Concept of elements, atoms and molecules
Atomic and molecular masses
Mole concept and molar mass
Percentage composition
Empirical and molecular formula
Chemical reactions
Stoichiometry and calculations based on stoichiometry

Matter, Classification & Measurement

Matter: Anything with mass that occupies space. Three states —

solid

(definite shape & volume),

liquid

(definite volume, takes shape),

gas

(neither) — interconverted by T/P.

Comparative: classification by composition

Type	Composition	Separation
Element	one kind of atom	—
Compound	fixed mass ratio	chemical
Homo. mixture	uniform, variable	physical
Hetero. mixture	non-uniform	physical

Tree classifying matter into pure substances (elements, compounds) and mixtures (homogeneous, heterogeneous) — Matter $\to$ pure substances vs mixtures.

Useful relations

ρ = \frac{m}{V}; 1 L = 1000 mL = 1000 cm^{3} = 1 dm^{3}

SI density kg

m^{- 3}

; chemists use g c

m^{- 3}

Temperature scales

K =^{\circ} C + 273.15;^{\circ} F = \frac{9}{5}^{\circ} C + 32

Kelvin is SI; no negative values

Significant figures

▸Non-zero digits & zeros between them count; $leading$ zeros don't ( $0.0052 \to 2$ s.f.); $trailing$ after a decimal do ( $0.200 \to 3$ ).
▸ $\times / \div$ : keep fewest s.f.; $+ / -$ : keep fewest decimal places.

🚫 Examiner Trap · Matter & measurement

(1) A

compound

has a fixed ratio & new properties; a

mixture

has variable composition & keeps component properties. (2) Use

K =^{\circ} C + 273.15

(NOT

+ 273

) for precise work. (3) Changing units never changes the s.f. count. (4) The 7 SI base units: m, kg, s, A, K, mol, cd — everything else is derived.

Laws of Combination & Dalton's Theory

The five laws of chemical combination

Law	Statement
Conservation of mass	reactant mass $=$ product mass (Lavoisier)
Definite proportions	a compound has a fixed mass ratio (Proust)
Multiple proportions	masses of B per fixed A in small whole-number ratio (Dalton)
Gay-Lussac (gases)	gas volumes react in whole-number ratios
Avogadro	equal V (same T,P) $\Rightarrow$ equal molecules

Bars showing 2 g H with 16 g O in water versus 2 g H with 32 g O in hydrogen peroxide, oxygen in ratio 1 to 2 — Multiple proportions: O is $1 : 2$ for fixed H.

Dalton's atomic theory (1808)

▸Matter is made of indivisible $atoms$ .
▸Atoms of an element are identical; differ between elements.
▸Compounds $=$ atoms in fixed whole-number ratios.
▸A reaction only $rearranges$ atoms (none created/destroyed).

🚫 Examiner Trap · Laws of combination

(1)

Definite

proportions

=

one compound's fixed ratio;

multiple

proportions

=

two compounds of the same elements (

H_{2}

O vs

H_{2} O_{2}

). (2) Avogadro's law resolved the atom/molecule confusion (diatomic gases). (3) Gay-Lussac is about

volumes

, Avogadro about

number

. (4) Dalton's 'identical atoms' fails for

isotopes

Atomic, Molecular & Average Masses

Atomic mass unit (u):

1 u = \frac{1}{12}

mass of one

^{12}

C atom

= 1.66 \times 1 0^{- 24}

g. Atomic masses are relative to

^{12} C = 12

= 1.008

, O

= 16.00

Average atomic mass

\overset{ˉ}{M} = \sum f_{i} M_{i}

f_{i} =

fractional abundance,

M_{i} =

isotope mass; periodic-table values are these averages

Bar chart of carbon isotopes: 12C at 98.9 percent and 13C at 1.1 percent, weighted average 12.011 u — C: $0.989 (12) + 0.011 (13.003) = 12.011$ u.

Molecular vs formula mass

M_{m o l} = \sum (atoms \times atomic mass)

molecular for covalent (

H_{2} O = 18.02

);

formula

mass for ionic solids (NaCl

= 58.5

, no discrete molecule)

✎ Example · Molecular mass of glucose

C_{6} H_{12} O_{6}

$6 (12.01) + 12 (1.008) + 6 (16.00)$
$= 72.06 + 12.10 + 96.00$

∴

= 180.16

u (

180.16

g/mol)

🚫 Examiner Trap · Atomic & molecular masses

(1) Average atomic mass is a

weighted

mean by abundance — not a simple average of isotope masses. (2) 'Atomic mass' (u, relative)

\neq =

'molar mass' (g/mol) but they're

numerically equal

. (3) Ionic compounds have

formula

mass, not molecular mass (no molecule). (4)

f_{i}

are fractions (sum

= 1

), not percentages — divide by 100.

The Mole Concept

Mole (mol): Amount of substance with exactly

N_{A} = 6.022 \times 1 0^{23}

entities (atoms/molecules/ions) — the number of atoms in

12

g of

^{12}

Always state the entity

Core interconversions

n = \frac{mass}{M} = \frac{no. of particles}{N _{A}} = \frac{V _{g a s} ( STP )}{22.4 L}

M =

molar mass (g/mol);

1

mol ideal gas

= 22.4

L at STP (0

^{\circ}

C, 1 atm)

Hub-and-spoke diagram with 1 mole at the centre and conversions to mass, number of particles, gas volume and molarity — The mole links mass, number, volume & molarity.

✎ Example · Conversion

How many molecules in

36

g of water?

$n = 36/18 = 2$ mol
$2 \times 6.022 \times 1 0^{23}$

∴

= 1.2 \times 1 0^{24}

molecules

🚫 Examiner Trap · The mole

(1)

Always state the entity

: 1 mol

H_{2}

= 2

mol H atoms

+ 1

mol O atoms. (2)

22.4

L/mol holds only for an

ideal gas at STP

— not for solids/liquids or non-STP. (3) Molar mass (g/mol)

=

molecular mass (u) numerically. (4)

N_{A}

counts entities, so 1 mol of electrons

= N_{A}

electrons.

% Composition & Empirical/Molecular Formula

Mass % of an element

% = \frac{mass of element in 1 mol}{molar mass} \times 100

H in

H_{2}

= \frac{2 ( 1.008 )}{18.02} \times 100 = 11.18%

Empirical vs molecular formula:

Empirical

=

simplest whole-number atom ratio (C

H_{2}

Cl).

Molecular

=

actual atoms (

C_{2} H_{4} C l_{2}

). Molecular

=

(empirical

)_{n}

Finding

n

n = \frac{molar mass}{empirical formula mass} \Rightarrow Molecular = (Empirical)_{n}

✎ Example · Empirical & molecular formula

4.07%

24.27%

71.65%

Cl; molar mass

98.96

moles/100 g: H $= 4.04$ , C $= 2.02$ , Cl $= 2.02$
$\div$ smallest $\Rightarrow$ C $H_{2}$ Cl (mass $49.48$ )
$n = 98.96/49.48 = 2$

∴

C_{2} H_{4} Cl_{2}

🚫 Examiner Trap · Formulae

(1) Empirical formula gives only the

ratio

, never the actual count — you need the molar mass for n. (2) Convert

% \to

moles (divide by atomic mass), then divide by the smallest. (3) If a ratio comes to e.g.

1 : 1.5

multiply all

by 2 to clear fractions. (4) Ionic compounds are always written as empirical (formula unit).

Stoichiometry & Limiting Reagent

Balanced equation = recipe

▸Same number of atoms of each element on both sides (conservation of mass).
▸Coefficients give the $mole ratio$ (and volume ratio for gases).
▸Balance by changing $coefficients$ only — NEVER subscripts.

Mole bridge

\frac{n _{A}}{a} = \frac{n _{B}}{b} (a A + b B \to \dots)

given

\to

moles

\to

coefficient ratio

\to

moles of target

\to

mass/volume

Limiting reagent: The reactant consumed first — it

limits

the product. Find it by dividing each reactant's moles by its coefficient; the

smallest

value is limiting.

For N2 plus 3 H2 makes 2 NH3, with 1 mol N2 and 5 mol H2: N2 is limiting, 2 H2 left over — $N_{2}$ runs out first $\to$ limiting; $H_{2}$ in excess.

Percentage yield

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

theoretical yield computed from the LIMITING reagent

🚫 Examiner Trap · Stoichiometry

(1) Find the

limiting reagent first

— divide each reactant's moles by its coefficient, smallest wins. (2) Balance by coefficients, NEVER change a formula's subscript. (3) Theoretical yield is from the limiting reagent, not the excess one. (4) For gases at the same T,P, coefficient ratio

=

volume ratio (Avogadro).

Concentration Terms

Comparative: concentration terms

Term	Definition	$T$ -dependent?
Molarity $M$	mol solute / L solution	yes
Molality $m$	mol solute / kg solvent	no
Mole fraction $x$	$n_{A} / (n_{A} + n_{B})$	no
Mass %	mass solute / mass soln $\times 100$	no

Molarity / Molality

M = \frac{mol solute}{L soln}; m = \frac{mol solute}{kg solvent}

M

uses solution volume;

m

uses solvent mass

Mole fraction & dilution

x_{A} = \frac{n _{A}}{n _{A} + n _{B}} (x_{A} + x_{B} = 1); M_{1} V_{1} = M_{2} V_{2}

dilution: moles of solute stay constant when solvent is added

🚫 Examiner Trap · Concentration

(1)

Molarity

is T-dependent (volume expands);

molality

/mole-fraction are not — use the latter for colligative properties. (2) Molality uses kg of

solvent

, not solution. (3)

M_{1} V_{1} = M_{2} V_{2}

only for dilution (same solute moles), not for reactions. (4) ppm

= \frac{mass solute}{mass soln} \times 1 0^{6}

for very dilute solutions.

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This Some Basic Concepts of Chemistry formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across General Introduction: Importance and scope of Chemistry, Nature of matter, Laws of chemical combination, Dalton's atomic theory: Concept of elements, atoms and molecules, Atomic and molecular masses — each shown with the key result and, where useful, a worked example.

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