Is Stoichiometry part of the EAPCET Chemistry syllabus?

Yes. Stoichiometry is taught in Intermediate first year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Chemistry questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.

What mistakes do students make in Stoichiometry?

Ignoring the medium when choosing the n-factor of KMnO₄ — it is 5 in acid, 3 in neutral or weakly basic and 1 in strongly basic solution. Equating moles instead of equivalents in a redox titration — write N₁V₁ = N₂V₂ or use the balanced mole ratio, never M₁V₁ = M₂V₂ unless the n-factors are equal. Using molarity where molality is asked — molality divides by kilograms of solvent, not litres of solution. Picking the reactant present in smaller mass as the limiting reagent — compare moles divided by stoichiometric coefficients instead.

EAPCET Chemistry · Intermediate first year

Stoichiometry

Stoichiometry covers the mole concept, chemical calculations from equations, concentration terms, equivalent weights and balancing redox reactions.

Subject: Chemistry
Taught in: Intermediate first year
Chemistry in the paper: 40 of 160 questions

Key formulas and results

1 mole $= 6.022 \times 10^{23}$ particles
Molarity = moles of solute per litre of solution; molality = moles per kg of solvent
Equivalent weight = molar mass / n-factor; normality = molarity × n-factor
The limiting reagent decides the amount of product
In a redox reaction, total increase in oxidation number = total decrease
Molarity from mass percent and density: $M = \frac{w\% \times d \times 10}{M_{\text{solute}}}$ , with $d$ in g/mL
Molality from molarity: $m = \frac{1000M}{1000d - M \times M_{\text{solute}}}$
Molality from mole fraction: $m = \frac{x_{\text{solute}} \times 1000}{(1 - x_{\text{solute}}) \times M_{\text{solvent}}}$
Molar mass of a gas $= 2 \times$ vapour density
Mixing solutions: $M = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$ ; dilution: $M_1V_1 = M_2V_2$
Parts per million: $\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$ ; hardness of water in ppm $= \frac{\text{mass of } \ce{CaCO3} \text{ equivalent}}{\text{mass of water}} \times 10^6$
Combustion of a hydrocarbon: $\ce{C_{x}H_{y}} + \left(x + \frac{y}{4}\right)\ce{O2} \to x\,\ce{CO2} + \frac{y}{2}\,\ce{H2O}$
Molecular formula $= n \times$ empirical formula, where $n = \frac{\text{molar mass}}{\text{empirical formula mass}}$

What this chapter covers

Mole concept
Empirical and molecular formulae
Limiting reagent
Concentration terms
Equivalent weight
Oxidation number and redox balancing

How it is usually asked

Mole and mass calculations
Limiting reagent
Concentration conversions
Balancing redox equations

Understanding Stoichiometry

Stoichiometry is the bookkeeping of chemistry. The mole links the mass you weigh to the number of particles: moles = mass/molar mass, and one mole of an ideal gas occupies 22.4 L at STP. A balanced equation gives mole ratios, not mass ratios, so every calculation should pass through moles. When reactants are not in the exact ratio, the one that runs out first — the limiting reagent — fixes the amount of product.

Concentration can be stated in several ways. Molarity (mol per litre of solution) changes with temperature because volume does; molality (mol per kg of solvent) and mole fraction do not. Normality = molarity × n-factor, where the n-factor is the number of H⁺ given (acids), OH⁻ given (bases) or electrons exchanged per formula unit (redox). The law of equivalence says reactants always react in equal numbers of equivalents, so N₁V₁ = N₂V₂ in titrations.

Redox reactions are balanced by tracking oxidation numbers: the total increase must equal the total decrease. The n-factor of an oxidant depends on the medium — MnO₄⁻ gains 5 electrons in acid (to Mn²⁺), 3 in neutral or weakly basic solution (to MnO₂) and 1 in strongly basic solution (to MnO₄²⁻). Cr₂O₇²⁻ going to Cr³⁺ has an n-factor of 6. EAPCET frequently tests titration volumes built on these values.

Worked example

EAPCET-style question

What volume of 0.1 M $\ce{KMnO4}$ is needed to completely oxidise 50 mL of 0.2 M $\ce{FeSO4}$ in acidic medium?

(1)20 mL
(2)100 mL
(3)500 mL
(4)40 mL

Solution

In acid, $\ce{MnO4- -> Mn^2+}$ (gain of 5 $\ce{e-}$ ) and $\ce{Fe^2+ -> Fe^3+}$ (loss of 1 $\ce{e-}$ ), so the n-factors are 5 and 1.
Milliequivalents of $\ce{FeSO4} = 50 \times 0.2 \times 1 = 10$ .
Normality of $\ce{KMnO4} = 0.1 \times 5 = 0.5\ \text{N}$ .
Equal equivalents react: $0.5 \times V = 10$ , so $V = 20\ \text{mL}$ .
100 mL results from ignoring the n-factors altogether.

Answer: 20 mL

Shortcuts and results to remember

Millimoles $= M \times V(\text{mL})$ , which avoids converting volumes to litres in titration problems.
For dilute aqueous solutions molarity and molality are nearly equal, because 1 L of solution contains about 1 kg of water.
Normality = molarity × n-factor, where the n-factor is the basicity of an acid, the acidity of a base, or the electrons exchanged per formula unit in a redox reaction.

Common mistakes to avoid

Ignoring the medium when choosing the n-factor of KMnO₄ — it is 5 in acid, 3 in neutral or weakly basic and 1 in strongly basic solution.
Equating moles instead of equivalents in a redox titration — write N₁V₁ = N₂V₂ or use the balanced mole ratio, never M₁V₁ = M₂V₂ unless the n-factors are equal.
Using molarity where molality is asked — molality divides by kilograms of solvent, not litres of solution.
Picking the reactant present in smaller mass as the limiting reagent — compare moles divided by stoichiometric coefficients instead.

How to prepare it

Always start from the balanced equation and write the mole ratio — everything else follows.

Once the formulas feel familiar, test them the way the exam does: practise Stoichiometry questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Chemistry chapters

See every chapter in the EAPCET Chemistry syllabus.

Practise Stoichiometry chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Stoichiometry part of the EAPCET Chemistry syllabus?: Yes. Stoichiometry is taught in Intermediate first year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Stoichiometry for EAPCET?: Always start from the balanced equation and write the mole ratio — everything else follows.
How many Chemistry questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.
What mistakes do students make in Stoichiometry?: Ignoring the medium when choosing the n-factor of KMnO₄ — it is 5 in acid, 3 in neutral or weakly basic and 1 in strongly basic solution. Equating moles instead of equivalents in a redox titration — write N₁V₁ = N₂V₂ or use the balanced mole ratio, never M₁V₁ = M₂V₂ unless the n-factors are equal. Using molarity where molality is asked — molality divides by kilograms of solvent, not litres of solution. Picking the reactant present in smaller mass as the limiting reagent — compare moles divided by stoichiometric coefficients instead.