Some Basic Concepts of Chemistry Formula Sheet — NEET Chemistry

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

Syllabus — topics coveredNCERT · 5 sub-topics

Matter and its nature; laws of chemical combination; Dalton's atomic theory
Atomic and molecular masses; mole concept and molar mass; Avogadro number
Percentage composition; empirical and molecular formula
Stoichiometry and stoichiometric calculations; limiting reagent
Concentration terms — molarity, molality, mole fraction

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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What are the most important Some Basic Concepts of Chemistry formulas for NEET?

This Some Basic Concepts of Chemistry formula sheet covers all the high-yield Chemistry formulas, definitions and facts you need for NEET, across Matter and its nature; laws of chemical combination; Dalton's atomic theory, Atomic and molecular masses; mole concept and molar mass; Avogadro number, Percentage composition; empirical and molecular formula, Stoichiometry and stoichiometric calculations; limiting reagent, Concentration terms — molarity, molality, mole fraction — each shown with the key result and, where useful, a worked example.

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