NEET Chemistry - Chapter 1

Some Basic Concepts of Chemistry

Original NEET chemistry notes on the mole concept, stoichiometry, concentration terms, empirical and molecular formulae, gas-volume relations, and redox-linked equivalent ideas.

NEET Chemistry Some Basic Concepts of Chemistry Notes Ad
Some Basic Concepts of Chemistry Notes Sponsor

Premium placement inside the NEET chemistry chapter notes for Some Basic Concepts of Chemistry.

Concept Block

1. Mole Concept, Avogadro Number, and Molar Mass

The mole is the chemist's counting unit. One mole of any substance contains exactly NA=6.022×1023N_A = 6.022\times10^{23} entities (atoms, molecules, ions, electrons — whatever the formula unit specifies). This number is called Avogadro's constant and it bridges the atomic world to the laboratory scale.

The Three-Way Mole Bridge

n=mM=NNA=V22.4 L  (at STP, ideal gas only)n=\frac{m}{M}=\frac{N}{N_A}=\frac{V}{22.4\text{ L}}\;(\text{at STP, ideal gas only})

nn = moles, mm = mass (g), MM = molar mass (g mol1^{-1}), NN = number of particles, VV = volume at STP.

Molar mass numerically equals the relative atomic/molecular mass but carries the unit g mol1^{-1}. Always add up the molar mass from atomic masses: M(H2SO4)=2(1)+32+4(16)=98M(\text{H}_2\text{SO}_4)=2(1)+32+4(16)=98 g mol1^{-1}.

QuantityFormulaUnit
Moles from massn=m/Mn = m/Mmol
Particles from molesN=n×NAN = n \times N_Adimensionless
Volume at STP (gas)V=n×22.4V = n \times 22.4L
NEET trap: 1 mol of O2_2 = 6.022×10236.022\times10^{23} molecules but 2×6.022×10232\times 6.022\times10^{23} atoms. 1 mol of P4_4 contains 4 mol of phosphorus atoms. Always clarify what entity is being counted.
Concept Block

2. Stoichiometry, Limiting Reagent, Yield, and Purity

A balanced chemical equation is a mole-ratio map. The coefficients directly give the molar ratios of reactants consumed and products formed.

The limiting reagent is the reactant that is completely consumed first, fixing the maximum theoretical yield. Every stoichiometry problem has exactly one workflow:

Standard 4-Step Stoichiometry Workflow

  1. Convert given masses/volumes to moles using n=m/Mn = m/M.
  2. Divide each reactant's moles by its coefficient. The smallest ratio identifies the limiting reagent.
  3. Use mole ratio from balanced equation to find product moles.
  4. Convert product moles back to mass or volume as required.
Percent yield=actual yieldtheoretical yield×100%\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%

Pure mass=% purity100×sample mass\text{Pure mass}=\frac{\%\text{ purity}}{100}\times\text{sample mass}
Worked example: N2+3H22NH3N_2+3H_2\to 2NH_3. If 28 g N2N_2 and 3 g H2H_2 react: moles N2=1N_2=1, moles H2=1.5H_2=1.5. Ratio check: N2N_2 needs 1/1=1.0, H2H_2 needs 1.5/3=0.5. H2H_2 gives the smaller ratio — it is the limiting reagent. Product = 1.5/3×2=11.5/3\times2=1 mol NH3=17NH_3=17 g.
NEET trap: Theoretical yield is always calculated from the limiting reagent, NOT from the excess reagent.
Concept Block

3. Concentration Terms: Molarity, Molality, Mole Fraction, Normality

Solutions are described by six major concentration terms in NEET chemistry. Understanding which denominator each uses is the fastest way to distinguish them.

TermFormulaDenominatorTemp. dependent?
Molarity (M)mol solute / L solutionSolution volumeYes
Molality (m)mol solute / kg solventSolvent massNo
Mole fraction (x)ni/(n1+n2+)n_i/(n_1+n_2+\ldots)Total molesNo
Normality (N)equiv / L solutionSolution volumeYes
Mass percent (w/w)mass solute/mass soln×100Solution massNo
ppmmass solute/mass soln×106^6Solution massNo
Dilution: M1V1=M2V2M_1V_1 = M_2V_2 (moles of solute are conserved). Also: N1V1=N2V2N_1V_1 = N_2V_2.

Normality and n-factor: Normality = Molarity × n-factor. The n-factor for an acid equals its basicity (replaceable H+^+), for a base equals its acidity, and for a redox agent equals the change in oxidation number per formula unit.

NEET trap: Molarity decreases when temperature rises (solution expands), but molality and mole fraction stay constant. Use molality for colligative property problems.
Concept Block

4. Empirical Formula, Molecular Formula, and Average Atomic Mass

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms in one molecule and is always a whole-number multiple of the empirical formula.

Empirical-to-Molecular Formula Steps

  1. Convert mass % of each element to moles (divide by atomic mass).
  2. Divide all mole values by the smallest to get the simplest ratio.
  3. Multiply through by any integer needed to make all ratios whole numbers — this gives the empirical formula.
  4. Find empirical formula mass, then n=Mmolecule/Mempiricaln = M_{molecule}/M_{empirical}.
  5. Molecular formula = (empirical formula) × nn.
A=i(xiAi)\overline{A} = \sum_i (x_i \cdot A_i)

Average atomic mass = weighted mean of isotopic masses, where xix_i is fractional abundance of each isotope.

Classic example: Glucose C6_6H12_{12}O6_6 has empirical formula CH2_2O. Empirical mass = 30. Molecular mass = 180. So n=180/30=6n = 180/30 = 6. Molecular formula = (CH2_2O)6_6 = C6_6H12_{12}O6_6.
NEET trap: Empirical formula and molecular formula can be the same — e.g., H2OH_2O, NH3NH_3, CO2CO_2 are their own empirical and molecular formulae.
Concept Block

5. Gas Volume Relations, n-Factor, and NEET Exam Traps

At STP (0°C, 1 atm), 1 mole of any ideal gas occupies 22.4 L. This is the standard gas-volume bridge. Note: IUPAC now defines STP as 0°C and 1 bar, giving 22.7 L, but NEET papers continue to use 22.4 L — stick with that.

Moles of gas at STP=Volume (L)22.4\text{Moles of gas at STP}=\frac{\text{Volume (L)}}{22.4}

The n-factor (valence factor) determines equivalents and makes acid-base and redox titration calculations faster than full mole algebra.

Substance / Reactionn-factor
HCl (acid-base)1 (monoprotic)
H2_2SO4_4 (acid-base)2 (diprotic)
KMnO4_4 in acidic medium5 (Mn: +7 → +2)
KMnO4_4 in neutral/basic3 (Mn: +7 → +4)
K2_2Cr2_2O7_7 (acidic)6 (two Cr: +6 → +3)
Na2_2S2_2O3_3 (vs I2_2)1
Equivalents=n×n-factor,Equivalent mass=Mn-factor\text{Equivalents} = n\times n\text{-factor},\qquad \text{Equivalent mass}=\frac{M}{n\text{-factor}}

N1V1=N2V2(at equivalence point of titration)N_1V_1 = N_2V_2 \quad (\text{at equivalence point of titration})
Top 3 NEET traps in this chapter:
  1. Confusing molecules and atoms: 1 mol O2O_2 ≠ 1 mol O atoms.
  2. Using 22.4 L for liquids or solids — it applies only to ideal gases at STP.
  3. Using the same n-factor for KMnO4KMnO_4 regardless of medium — n-factor is medium-dependent.
Practice Tests

5 Chapter Tests of 25 Questions Each

Each test is original, NEET-aligned, and answer-backed. Use them as sectional revision instead of a single long mock so your weak subtopics become easier to identify quickly.

Test 1: Mole Basics

Moles, molar mass, particles, STP volume, and concentration definitions.

Test 2: Stoichiometry and Yield

Limiting reagent, purity, dilution, equivalent mass, and percent yield.

Test 3: Formula and Isotope Logic

Empirical formulae, isotopes, average mass, oxidation number, and n-factor.

Test 4: Gas and State Relations

Ideal-gas connections, diffusion, density, partial pressure, and real-gas basics.

Test 5: Mixed NEET Drill

Redox-linked equivalents, solution calculations, practical mole conversions, and traps.

Open Practice Tests
Finished this topic?

Keep the practice loop moving

Move straight from chapter-wise questions into a subject test, then loop back into weaker areas instead of ending the session here.