Quick Revision Notes Chemistry - Some Basic Concepts of Chemistry - JEE, NEET & CBSE

This article compiles the quick revision notes from chemistry chapter "Some Basic Concepts of Chemistry". These notes are best for quick revision if you are preparing for any of the NEET, JEE and CBSE exams. 

Some Basic Concepts of Chemistry - Quick Revision Notes

SOME USEFUL CONVERSION FACTORS

• 1 Å = 10^–10m, 1nm =10^–9 m
• 1 pm = 10^–12m
• 1 litre = 10^–3 m³ = 1 dm³
• 1 atm = 760 mm or torr = 101325 Pa or Nm^–2
• 1 bar = 10⁵Nm^–2 = 10⁵ Pa
• 1 calorie = 4.184 J
• 1 electron volt (eV) = 1.6022 × 10^–19 J
• (1 J = 10⁷ ergs)
• (1 cal > 1 J > 1 erg > 1 eV)

ATOMIC MASS OR MOLECULAR MASS

Mass of one atom or molecule in amu

• C → 12 amu
• H₂O → 18 amu

ACTUAL MASS

mass of one atom or molecule in grams

• C → 12 × 1.66 ×10^–24 g
• H₂O → 18 × 1.66 × 10^–24 g

RELATIVE ATOMIC MASS OR RELATIVE MOLECULAR 

Mass of one atom or molecule w.r.t. 1/12th of 12C atom

• C → 12
• H₂O → 18

It is unitless

GRAMS ATOMIC MASS OR GRAM MOLECULAR MASS

Mass of one mole of atom or molecule

• C → 12 g
• H₂O → 18 g

It is also called molar weight

DEFINITION OF ONE MOLE

One mole is a collection of that many entities as there are a number of atoms exactly in 12 g of C-12 isotope.

1u = 1amu = (1/12 th) of mass of 1 atom of C12 = 1𝑔/𝑁𝐴= 1.66 × 10^–24 g

For elements

• 1 g atom = 1 mole of atoms = NA atoms
• g atomic mass (GAM) = mass of NA atoms in g.
• Mole of atoms = 𝑀𝑎𝑠𝑠 (𝑔) ÷ 𝐺𝐴𝑀 𝑜𝑟 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

For molecule

• 1 g molecule = 1 mole of molecule = NA molecule
• g molecular mass (GMM) = mass of NA molecule in g.
• Mole of molecule = 𝑴𝒂𝒔𝒔 (𝒈) ÷ 𝑮𝑴𝑴 𝒐𝒓 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔

For ionic compounds

• 1 g formula unit = 1 mole of formula unit = NA formula unit.
• G formula mass (GFM) = mass of NA formula unit in g.
• Mole of formula unit = 𝑴𝒂𝒔𝒔 (𝒈) ÷ 𝑮𝑭𝑴 𝒐𝒓 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔

1 mole of substance

• Contains 6.022 × 10²³ particles
• Weight as much as molecular weight/atomic ionic/weight in grams
• If it is a gas, one mole occupies a volume of 22.4 L at 1 atm & 273 K

Average or mean molar mass

The average molar mass of the different substances present in the container 

• 𝑀𝑎𝑣𝑔 =(𝑀₁𝑛₁+𝑀₂𝑛₂+⋯) ÷ (𝑛₁+𝑛₂+⋯)

Here M₁, M₂ are molar masses of substances, and n₁, n₂ are mole of substances present in the container.

DENSITIES

Density = 𝑀𝑎𝑠𝑠 ÷ 𝑣𝑜𝑙𝑢𝑚𝑒, unit : g/cc or g ml-1

Relative Density = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑎𝑛𝑦 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 ÷ 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒

VAPOUR DENSITY

Ratio of density of vapour to the density of hydrogen at similar pressure and temperature.

• Vapor density = 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠/2

STOICHIOMETRY BASED CONCEPT

(Concept of limiting reagent)

aA + bB → cC + dD

• a,b,c,d, represents the ratios of moles, volumes [for gaseous]  and molecules in which the reactants react or products  formed.
• a,b,c,d, does not represent the ratio of masses.

HOW TO FIND Limiting Reagent.

Case I: If data of only one reactant is given then treat that reactant as L.R. and other reactants as excess reagent.

Case-II: If data of more than one reactant are given then first convert all the data into moles then divide the moles ofreactants with their respective stoichiometric coefficient. The reactant having minimum ratio will be L.R. then find the moles of product formed or excess reagent left by comparing it with L.R. through stoichiometric concept.

EQUIVALENT WEIGHT

Equivalent weight = 𝑚𝑜𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 ÷ 𝑣𝑎𝑙𝑒𝑛𝑐𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 (𝑉𝐹)
V.F. for elements = Valency

Example 1: Na = 1, Al=3, N₂=6, O₂ = 4, H₂ = 2
V.F. for ionic compounds (salts)
= total charge on cation/anion

Example 2: Na₂ +1CO₃⁻² → V.F. = +1×2 = 2
 K₄⁺¹[Fe(CN)₆]= V.F. = +1×4 = 4
 V.F. for acids = No. of replaceable H+ ions
 HCl =1, H₂SO₄ = 2, H₃PO₄ = 3
 H₃PO₃=2, H₃PO₂=1
 V.F. for bases = No. of replaceable OH^–
 NaOH =1, Ba(OH)₂=2, Ca(OH)₂=2, Al(OH)₃ = 3

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