NEET UG Chemistry · Chapter 1

Some Basic Concepts of Chemistry

Mole concept, molar mass, percentage composition, empirical formula, stoichiometry and concentration terms.

NCERT Class 11NEET: 1–2 Qs per paper 13 min read

1. Chapter at a Glance

This chapter is the foundation of all numerical chemistry. It explains how chemists count particles using the mole, how to find formulas of compounds from percentage data, how to use balanced equations to calculate amounts (stoichiometry), and how to express the concentration of solutions. Almost every physical chemistry question in NEET uses these ideas, and one or two questions come directly from this chapter every year.

What NEET asks most
  • Number of moles, molecules and atoms in a given mass.
  • Percentage composition, empirical and molecular formulas.
  • Limiting reagent and mass of product.
  • Molarity, molality, mole fraction and mass percent.
  • Laws of chemical combination and significant figures.

2. Matter, Measurement and Laws of Combination

  • Matter can be classified as pure substances (elements and compounds) or mixtures (homogeneous or heterogeneous).
  • SI base units include the metre, kilogram, second, kelvin, mole, ampere and candela. Temperature in kelvin: T(K) = t(°C) + 273.15.
  • Significant figures: all non-zero digits are significant; zeros between them are significant; leading zeros are not; trailing zeros after a decimal point are significant. In multiplication, the answer keeps as many significant figures as the least precise value.
LawStatementNEET Note
Conservation of mass (Lavoisier)Mass is neither created nor destroyed in a reactionBasis of balancing equations
Definite proportions (Proust)A compound always has the same elements in the same mass ratioCO₂ from any source has C : O = 3 : 8
Multiple proportions (Dalton)Masses of one element combining with a fixed mass of another are in a simple ratioO in CO and CO₂ is 1 : 2
Gay-Lussac's law of gaseous volumesGases react in simple whole-number volume ratiosH₂ : O₂ : H₂O(g) = 2 : 1 : 2
Avogadro's lawEqual volumes of gases at the same T and p contain equal numbers of moleculesMolar volume

Dalton's atomic theory explained these laws: matter is made of atoms; atoms of an element are identical; compounds form when atoms combine in fixed whole-number ratios; atoms are neither created nor destroyed in reactions.

3. Atomic Mass and the Mole Concept

  • 1 atomic mass unit (u) = 1/12 of the mass of one carbon-12 atom = 1.66 × 10⁻²⁴ g.
  • Average atomic mass is the weighted average of isotopic masses. For example, chlorine (75% ³⁵Cl, 25% ³⁷Cl) has an average atomic mass of 35.5 u.
  • One mole is the amount of substance containing 6.022 × 10²³ particles (Avogadro constant NA).
  • Molar mass is the mass of one mole in grams; it is numerically equal to the atomic or molecular mass in u.
  • At STP (273.15 K and 1 bar), one mole of an ideal gas occupies 22.7 L. At 273 K and 1 atm it occupies 22.4 L (older convention, still seen in questions).
n = mass/molar mass = number of particles/NA = volume of gas at STP/22.7 L
MOLES (n) Mass (g) Particles Gas volume Solution (M×V) ÷ M÷ NA ÷ 22.7 LM × V(L)
The mole is the "hub": convert any given quantity into moles first, then into whatever the question asks for.

4. Percentage Composition and Formulas

  • Mass percent of an element = (mass of the element in 1 mol of compound / molar mass) × 100.
  • Empirical formula: the simplest whole-number ratio of atoms. Steps: convert % to grams (assume 100 g), divide by atomic mass to get moles, divide by the smallest, and round to whole numbers.
  • Molecular formula = n × empirical formula, where n = molar mass/empirical formula mass.
  • Example: glucose C₆H₁₂O₆ and acetic acid C₂H₄O₂ both have the empirical formula CH₂O.

5. Stoichiometry and Limiting Reagent

  • A balanced equation gives mole ratios. For 2H₂ + O₂ → 2H₂O, 2 mol H₂ react with 1 mol O₂ to give 2 mol H₂O.
  • The limiting reagent is used up first and decides how much product forms. The other reactant is in excess.
  • To find it: divide the moles of each reactant by its coefficient; the smallest value marks the limiting reagent.
  • Percentage yield = (actual yield / theoretical yield) × 100.

6. Concentration Terms

TermDefinitionUnitNEET Note
Mass percent (w/w)(mass of solute/mass of solution) × 100%Independent of temperature
Mole fraction (x)moles of component/total molesnoneSum of all x = 1
Molarity (M)moles of solute/volume of solution in Lmol L⁻¹Changes with temperature (volume changes)
Molality (m)moles of solute/mass of solvent in kgmol kg⁻¹Independent of temperature
ppm(mass of solute/mass of solution) × 10⁶ppmFor very dilute solutions
  • Dilution: M₁V₁ = M₂V₂. Mixing solutions of the same solute: M = (M₁V₁ + M₂V₂)/(V₁ + V₂).
  • From mass % and density d (g mL⁻¹): M = (10 × d × %)/molar mass.

7. Formula Master Table

QuantityFormula
Moles from massn = w/M
Number of particlesN = n × 6.022 × 10²³
Moles of gas at STPn = V/22.7 L
Mass percent of element(mass in 1 mol/M) × 100
Molecular formulan = M/(empirical formula mass)
MolarityM = n/V(L) = (w × 1000)/(M × V mL)
Molalitym = (w × 1000)/(M × W solvent g)
Mole fractionxA = nA/(nA + nB)
DilutionM₁V₁ = M₂V₂
Molarity from % and densityM = 10 × d × %/Msolute
Average atomic massΣ(% abundance × isotopic mass)/100

8. Master Summary Table

ConceptKey ideaNEET Note
Mole6.022 × 10²³ particlesAlways convert to moles first
Molar volume22.7 L at STP22.4 L at 273 K, 1 atm
Empirical formulaSimplest ratioDivide by smallest, then round
Limiting reagentUsed up firstDecides product amount
Molarity vs molalityPer litre of solution vs per kg of solventMolality is temperature-independent
Significant figuresLeast precise value decides0.0050 has 2 significant figures
Law of multiple proportionsSimple ratioCO / CO₂; N₂O / NO

9. Mind Map

Basic Concepts of Chemistry
Laws
  • Conservation of mass
  • Definite and multiple proportions
  • Gay-Lussac, Avogadro
Mole concept
  • NA
  • Molar mass
  • Molar volume
Formulas
  • Mass percent
  • Empirical
  • Molecular
Stoichiometry and solutions
  • Limiting reagent
  • M, m, x
  • Dilution

10. Tips and Tricks

Shortcuts
  • Number of atoms = moles × NA × atoms per molecule (e.g. 1 mol H₂O has 3NA atoms).
  • For a 1 : 1 reaction, the reactant with fewer moles is the limiting reagent.
  • If the empirical-formula ratio gives x.5, multiply all subscripts by 2; if x.33, multiply by 3.
  • Molality from molarity: m = 1000M/(1000d − M × Msolute).
Common NEET traps
  • Counting molecules instead of atoms (or the reverse).
  • Using the mass of solution instead of the mass of solvent for molality.
  • Assuming the reactant with the smaller mass is limiting (compare moles, not grams).
  • Forgetting that molarity changes with temperature.

11. Important One-liners

  • 1 mole of any substance contains 6.022 × 10²³ particles.
  • The atomic mass unit is defined using the carbon-12 isotope.
  • Molality and mole fraction do not depend on temperature.
  • Glucose, formaldehyde and acetic acid share the empirical formula CH₂O.
  • The limiting reagent is completely consumed in a reaction.
  • The law of multiple proportions was given by Dalton.
  • 18 g of water contains 6.022 × 10²³ molecules and 3 × 6.022 × 10²³ atoms.

12. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, Ca = 40. Click to reveal each answer and working.

Average level

N1Average

How many moles, molecules and oxygen atoms are present in 22 g of carbon dioxide?

Show answer & solution

Concept: n = w/M; then multiply by NA.

  1. M(CO₂) = 12 + 32 = 44 g mol⁻¹; n = 22/44 = 0.5 mol.
  2. Molecules = 0.5 × 6.022 × 10²³ = 3.011 × 10²³.
  3. Each molecule has 2 O atoms: 6.022 × 10²³ oxygen atoms.

Answer: 0.5 mol; 3.011 × 10²³ molecules; 6.022 × 10²³ O atoms

N2Average

Find the percentage composition of glucose, C₆H₁₂O₆.

Show answer & solution

Concept: mass % = (mass of element in 1 mol/molar mass) × 100.

  1. M = 72 + 12 + 96 = 180 g mol⁻¹.
  2. C = 72/180 × 100 = 40.0%; H = 12/180 × 100 = 6.67%.
  3. O = 96/180 × 100 = 53.33% (the three add to 100% ✓).

Answer: C 40.0%, H 6.67%, O 53.33%

N3Average

4 g of NaOH is dissolved in water to make 250 mL of solution. Find its molarity.

Show answer & solution

Concept: M = moles/volume in litres.

  1. n = 4/40 = 0.1 mol.
  2. M = 0.1/0.250 = 0.4 mol L⁻¹.

Answer: 0.4 M

N4Average

50 g of calcium carbonate is heated strongly: CaCO₃ → CaO + CO₂. Find the mass of CaO formed and the volume of CO₂ at 273 K and 1 atm.

Show answer & solution

Concept: 1 mol CaCO₃ gives 1 mol CaO and 1 mol CO₂.

  1. n(CaCO₃) = 50/100 = 0.5 mol.
  2. Mass of CaO = 0.5 × 56 = 28 g.
  3. Volume of CO₂ = 0.5 × 22.4 = 11.2 L.

Answer: 28 g CaO; 11.2 L CO₂

N5Average

A compound contains 40% C, 6.67% H and 53.33% O by mass. Its molar mass is 60 g mol⁻¹. Find its empirical and molecular formulas.

Show answer & solution

Concept: take 100 g, convert to moles, find the simplest ratio.

  1. Moles: C = 40/12 = 3.33; H = 6.67/1 = 6.67; O = 53.33/16 = 3.33.
  2. Divide by 3.33: C : H : O = 1 : 2 : 1 ⇒ empirical formula CH₂O (mass 30).
  3. n = 60/30 = 2 ⇒ molecular formula C₂H₄O₂ (e.g. acetic acid).

Answer: CH₂O; C₂H₄O₂

Advanced level

N6Advanced

10 g of hydrogen reacts with 64 g of oxygen to form water. Find the limiting reagent, the mass of water formed, and the mass of reactant left over.

Show answer & solution

Concept: 2H₂ + O₂ → 2H₂O; compare moles/coefficient.

  1. n(H₂) = 10/2 = 5 mol (5/2 = 2.5); n(O₂) = 64/32 = 2 mol (2/1 = 2). O₂ gives the smaller value, so it is limiting.
  2. 2 mol O₂ react with 4 mol H₂ to give 4 mol H₂O = 72 g.
  3. H₂ left = 5 − 4 = 1 mol = 2 g.

Answer: O₂ limiting; 72 g water; 2 g H₂ left

N7Advanced

9.8 g of H₂SO₄ is dissolved in 200 g of water. Find the molality of the solution and the mole fraction of H₂SO₄.

Show answer & solution

Concept: molality uses kg of solvent; mole fraction uses total moles.

  1. n(H₂SO₄) = 9.8/98 = 0.1 mol; m = 0.1/0.200 = 0.5 mol kg⁻¹.
  2. n(H₂O) = 200/18 = 11.11 mol.
  3. x(H₂SO₄) = 0.1/(0.1 + 11.11) ≈ 0.0089.

Answer: 0.5 m; about 0.0089

N8Advanced

Concentrated sulphuric acid is 98% H₂SO₄ by mass with a density of 1.84 g mL⁻¹. Find its molarity, and the volume needed to prepare 1 L of 0.1 M acid.

Show answer & solution

Concept: M = 10 × d × %/Msolute; then M₁V₁ = M₂V₂.

  1. 1 L of acid weighs 1840 g, of which 98% = 1803.2 g is H₂SO₄.
  2. Moles = 1803.2/98 = 18.4 ⇒ M = 18.4 mol L⁻¹.
  3. V₁ = (0.1 × 1000)/18.4 ≈ 5.43 mL, diluted to 1 L (always add acid to water).

Answer: 18.4 M; about 5.4 mL

N9Advanced

200 mL of 0.5 M NaCl is mixed with 300 mL of 0.2 M NaCl. Find the molarity of the final solution.

Show answer & solution

Concept: total moles/total volume.

  1. Moles (in millimoles): 200 × 0.5 = 100 and 300 × 0.2 = 60; total 160 mmol.
  2. Total volume = 500 mL.
  3. M = 160/500 = 0.32 M.

Answer: 0.32 M

N10Advanced

46 g of ethanol (C₂H₅OH) is mixed with 36 g of water. Find the mole fraction of ethanol and the molality of ethanol in water.

Show answer & solution

Concept: x = nA/(nA + nB); m = n/kg solvent.

  1. n(ethanol) = 46/46 = 1 mol; n(water) = 36/18 = 2 mol.
  2. x(ethanol) = 1/3 ≈ 0.333.
  3. m = 1/0.036 ≈ 27.8 mol kg⁻¹.

Answer: 0.333; about 27.8 m

13. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

The number of atoms in 0.1 mol of a triatomic gas is:

  1. 6.022 × 10²²
  2. 1.806 × 10²³
  3. 3.011 × 10²³
  4. 1.806 × 10²²
Show solution

The number of molecules is 0.1 × 6.022 × 10²³ = 6.022 × 10²². Each molecule has 3 atoms, so there are 1.806 × 10²³ atoms. (A) counts molecules instead of atoms, and (D) forgets a factor of 10.

Answer: (B)

Q2Easy

Which concentration term does NOT depend on temperature?

  1. Molarity
  2. Normality
  3. Molality
  4. Formality
Show solution

Molality uses the mass of the solvent, and mass does not change with temperature. Molarity, normality and formality all use volume, which expands on heating.

Answer: (C)

Q3Medium

The number of significant figures in 0.00250 is:

  1. 2
  2. 3
  3. 5
  4. 6
Show solution

The leading zeros are not significant. The digits 2 and 5 are significant, and the trailing zero after the decimal point is also significant. That gives 3 significant figures.

Answer: (B)

Q4Medium

Carbon and oxygen form two compounds containing 42.9% and 27.3% carbon. This illustrates the law of:

  1. Definite proportions
  2. Multiple proportions
  3. Conservation of mass
  4. Reciprocal proportions
Show solution

These are CO and CO₂. For a fixed mass of carbon, the masses of oxygen are in the ratio 1 : 2, a simple whole-number ratio. This is the law of multiple proportions. The law of definite proportions describes a single compound.

Answer: (B)

Q5Medium

The mass of 11.35 L of oxygen gas at STP (273.15 K, 1 bar) is:

  1. 8 g
  2. 16 g
  3. 32 g
  4. 11.35 g
Show solution

At STP, n = 11.35/22.7 = 0.5 mol, so the mass is 0.5 × 32 = 16 g. (A) wrongly uses the atomic mass of O (16) instead of the molar mass of O₂.

Answer: (B)

Q6Medium

What is the molarity of a solution containing 5.85 g of NaCl in 500 mL of solution?

  1. 0.1 M
  2. 0.2 M
  3. 0.5 M
  4. 1 M
Show solution

n = 5.85/58.5 = 0.1 mol, and M = 0.1/0.5 = 0.2 M. (A) forgets to convert 500 mL to 0.5 L.

Answer: (B)

Q7Hard

2 g of H₂ and 8 g of O₂ react to form water. The mass of water formed is:

  1. 10 g
  2. 9 g
  3. 18 g
  4. 4.5 g
Show solution

n(H₂) = 1 mol and n(O₂) = 0.25 mol. The ratio needed is 2 : 1, so 0.25 mol of O₂ needs only 0.5 mol of H₂, making O₂ the limiting reagent. It gives 0.5 mol of water = 9 g. (A) assumes that all the mass reacts, and (C) assumes that H₂ is limiting.

Answer: (B)

Q8Hard

An organic compound has the empirical formula CH₂O, and its vapour density is 90. Its molecular formula is:

  1. C₃H₆O₃
  2. C₆H₁₂O₆
  3. C₂H₄O₂
  4. C₅H₁₀O₅
Show solution

Molar mass = 2 × vapour density = 180. The empirical formula mass is 30, so n = 6 and the formula is C₆H₁₂O₆. (A) wrongly takes the vapour density itself (90) as the molar mass.

Answer: (B)

14. Quick Revision Summary

Remember before the exam
  • n = w/M = N/NA = V/22.7 L (STP).
  • Empirical formula from % → moles → simplest ratio; molecular = n × empirical.
  • Limiting reagent: smallest moles/coefficient.
  • M = mol/L solution; m = mol/kg solvent; x = mole ratio.
  • M = 10d%/Msolute; M₁V₁ = M₂V₂.
All Chemistry chaptersNEET Mock Tests