Why do we include water of crystallisation in molar mass?

In stoichiometric calculations, water of crystallisation is part of the reactant like CuSO4 dot 5 H2O. Include its mass in molar mass. Use anhydrous molar mass only for the anhydrous salt.

Stoichiometry

Q: Why is molality preferred over molarity in colligative property calculations?

Molality depends only on masses which do not change with temperature. Molarity depends on solution volume which changes with temperature. For freezing point and boiling point calculations, molality stays valid throughout.

Q: Why does concentrated sulphuric acid have such a high molarity of about 18 M?

Because it is 98 percent pure by mass with density near 1.84 g per mL, so 1 L contains about 1800 g of which 1764 g is H2SO4, giving about 18 moles.

Q: Can percentage yield ever exceed 100 percent?

No. If a calculation gives more than 100 percent, the actual yield contains impurities or the theoretical yield was miscalculated. Purify the product or recheck the calculation.

Q: How do I choose between molarity and molality?

Use molarity when the problem involves volumes of solution: titrations, dilutions. Use molality when the problem involves colligative properties: boiling point elevation, freezing point depression.

Q: Are coefficients in a balanced equation moles or molecules?

Both. The coefficients give the ratio of molecules at the microscopic scale and moles at the macroscopic scale. For gases at same T and P they also give volume ratios.

Q: Why is ppm used only for very dilute solutions?

For concentrated solutions ppm gives huge numbers. A 1 percent solution is 10000 ppm. ppm becomes convenient when concentrations are much less than 1 percent, like trace pollutants.

Q: Does the density of a solution matter when converting between molarity and molality?

Yes. Molarity is per litre of solution, molality is per kilogram of solvent. To convert between them you need to know the mass of solvent in a given volume of solution, which requires density.

ChemistrySome Basic Concepts of ChemistryFor JEE aspirants

Stoichiometry is the arithmetic of chemical reactions. From a balanced equation you can predict exactly how much of each reactant is consumed, how much product forms, which reactant runs out first (the limiting reagent), and the concentration of any solution involved. In this concept you will learn to balance equations, calculate mass-mass and mass-volume relationships, work out percentage composition, identify limiting reagents, and use every standard concentration unit for JEE and NEET: mass percent, volume percent, mole fraction, molarity, molality, and parts per million.

Key Formulas - Quick Reference

Moles: $n = \frac{mass}{molar mass} = \frac{number of particles}{6.022 \times 1 0 ^{23}} = \frac{V ( at STP )}{22.4 L}$
Mass percent of element $= \frac{mass of element in 1 mol compound}{molar mass of compound} \times 100$
Mass percent of solute: $w / w % = \frac{mass of solute}{mass of solution} \times 100$
Volume percent: $v / v % = \frac{volume of solute}{volume of solution} \times 100$
Mole fraction: $x_{A} = \frac{n _{A}}{n _{A} + n _{B}}$ , with $x_{A} + x_{B} = 1$
Molarity: $M = \frac{moles of solute}{volume of solution in L} = \frac{w \times 1000}{M _{solute} \times V _{ml}}$
Molality: $m = \frac{moles of solute}{mass of solvent in kg} = \frac{w \times 1000}{M _{solute} \times w _{solvent, g}}$
ppm: $ppm_{A} = \frac{mass of A}{mass of solution} \times 1 0^{6}$
M and m relation: $M = \frac{m \cdot d}{1 + \frac{m \cdot M _{2}}{1000}}$ where $d$ is solution density and $M_{2}$ is solute molar mass

1. What Stoichiometry Means

Stoichiometry is the quantitative study of the amounts of reactants and products in a chemical reaction. It relies on balanced chemical equations and the mole concept.

The stoichiometric coefficients in a balanced equation are the ratios of moles (and, for gases, ratios of volumes at the same T and P). For methane combustion:

$CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l)$

This equation reads simultaneously in several ways:

$1$ molecule of $CH_{4}$ reacts with $2$ molecules of $O_{2}$ to give $1$ molecule of $CO_{2}$ and $2$ molecules of $H_{2} O$ .
$1$ mole ( $16 g$ ) of $CH_{4}$ reacts with $2$ moles ( $64 g$ ) of $O_{2}$ to give $1$ mole ( $44 g$ ) of $CO_{2}$ and $2$ moles ( $36 g$ ) of $H_{2} O$ .
At STP, $22.4 L$ of $CH_{4}$ reacts with $44.8 L$ of $O_{2}$ to give $22.4 L$ of $CO_{2}$ (water is a liquid, not counted).

Check total mass:

16 + 64 = 80 g

of reactants;

44 + 36 = 80 g

of products. Conservation of mass holds.

2. Balancing Chemical Equations

Balance so the number of atoms of each element is the same on both sides. Change only coefficients, never subscripts.

Write the correct formulas of all reactants and products.
Balance metals first, then non-metals other than H and O, then H, then O last.
If polyatomic ions like $SO_{4}^{2 -}$ appear unchanged on both sides, balance them as a unit.
Multiply through by an integer to clear any fractions in the coefficients.
Verify: count each type of atom on both sides.

Solved Example 1

Balance the combustion of octane:

C_{8} H_{18} + O_{2} \to CO_{2} + H_{2} O

Solution:

Balance C: $1$ octane gives $8$ $CO_{2}$ .

Balance H: $18$ H atoms give $9$ $H_{2} O$ .

Count O on right: $8 \times 2 + 9 \times 1 = 25$ O atoms, needing $\frac{25}{2}$ $O_{2}$ .

$C_{8} H_{18} + \frac{25}{2} O_{2} \to 8 CO_{2} + 9 H_{2} O$

Multiply through by $2$ :

$2 C_{8} H_{18} + 25 O_{2} \to 16 CO_{2} + 18 H_{2} O$

3. Percentage Composition by Mass

The mass percent of an element in a compound is the mass of that element per

100 g

of the compound.

% by mass of X = \frac{mass of X in 1 mole of compound}{molar mass of compound} \times 100

Solved Example 2

Calculate the percentage composition by mass of ethanol,

C_{2} H_{5} OH

Solution:

Molar mass $= 2 (12) + 6 (1) + 16 = 46 g mol^{- 1}$ .

% C $= \frac{24}{46} \times 100 = 52.17%$

% H $= \frac{6}{46} \times 100 = 13.04%$

% O $= \frac{16}{46} \times 100 = 34.78%$

Check: $52.17 + 13.04 + 34.78 = 99.99 \approx 100%$ .

4. Stoichiometric Calculations from Balanced Equations

Once an equation is balanced, four types of calculation cover almost every JEE and NEET question:

Mass to mass: given mass of one reactant/product, find mass of another.
Mass to volume: given mass of a reactant, find volume of a gaseous product at STP.
Volume to volume: for gas-phase reactions, use volume ratios directly (Gay-Lussac).
Mole to mole: the simplest form; use the coefficient ratio directly.

Universal recipe: convert given quantity to moles

\to

use coefficient ratio to find moles of desired substance

\to

convert to the requested unit (mass, volume, or number of particles).

Solved Example 3

(Mass to mass) Calcium carbonate reacts with dilute HCl:

CaCO_{3} + 2 HCl \to CaCl_{2} + H_{2} O + CO_{2}

. What mass of

CaCl_{2}

is produced from

25 g

CaCO_{3}

Solution:

Molar mass of $CaCO_{3} = 100 g mol^{- 1}$ ; of $CaCl_{2} = 111 g mol^{- 1}$ .

Moles of $CaCO_{3} = \frac{25}{100} = 0.25 mol$ .

From the equation, $1$ mol $CaCO_{3}$ gives $1$ mol $CaCl_{2}$ , so moles of $CaCl_{2} = 0.25$ .

Mass of $CaCl_{2} = 0.25 \times 111 = 27.75 g$ .

Solved Example 4

(Mass to volume) What volume of

CO_{2}

at STP is produced when

50 g

CaCO_{3}

decomposes? Reaction:

CaCO_{3} Δ CaO + CO_{2}

Solution:

Moles of $CaCO_{3} = \frac{50}{100} = 0.5 mol$ .

Moles of $CO_{2}$ produced $= 0.5 mol$ (1:1 ratio).

Volume at STP $= 0.5 \times 22.4 = 11.2 L$ .

Solved Example 5

(Volume to volume) In the Haber process,

N_{2} + 3 H_{2} \to 2 NH_{3}

, what volume of ammonia is produced from

10 L

N_{2}

if the reaction goes to completion at the same T and P?

Solution:

Volumes of gases at the same T and P are in the ratio of their coefficients (Gay-Lussac).

$V_{N_{2}} : V_{NH_{3}} = 1 : 2$ , so $V_{NH_{3}} = 2 \times 10 = 20 L$ .

5. Limiting Reagent

The limiting reagent (or limiting reactant) is the reactant that is completely consumed first, thereby stopping the reaction and fixing the maximum amount of product that can form. Reactants left over after the limiting reagent is used up are said to be in excess.

Figure 1: Limiting reagent analogy. Only 3 complete bicycles can be built because wheels run out first; the extra frames are the excess reagent.

How to Identify the Limiting Reagent

Convert each reactant's given mass to moles.
Divide each mole value by that reactant's coefficient in the balanced equation.
The smallest ratio identifies the limiting reagent. Product amount is computed from that reagent alone.

Solved Example 6

50.0 g

N_{2}

is reacted with

10.0 g

H_{2}

N_{2} + 3 H_{2} \to 2 NH_{3}

. Which is the limiting reagent, and how much ammonia is produced?

Solution:

Moles of $N_{2} = \frac{50}{28} = 1.79 mol$ .

Moles of $H_{2} = \frac{10}{2} = 5.00 mol$ .

Divide by coefficients: $\frac{1.79}{1} = 1.79$ for $N_{2}$ ; $\frac{5.00}{3} = 1.67$ for $H_{2}$ .

Smaller ratio belongs to $H_{2}$ , so $H_{2}$ is the limiting reagent.

Moles of $NH_{3}$ formed $= \frac{2}{3} \times 5.00 = 3.33 mol$ .

Mass of $NH_{3} = 3.33 \times 17 = 56.6 g$ .

Unreacted $N_{2}$ : consumed $= \frac{1}{3} \times 5.00 = 1.67 mol$ , so leftover $= 1.79 - 1.67 = 0.12 mol = 3.4 g$ .

For single-reactant decompositions (e.g.

2 KClO_{3} \to 2 KCl + 3 O_{2}

), the limiting-reagent concept does not apply - there is only one reactant.

6. Concentration of Solutions

A solution is a homogeneous mixture of a solute (usually smaller amount) in a solvent (usually larger amount). Concentration expresses how much solute is in a given amount of solution or solvent. NCERT lists six standard methods.

6.1 Mass Percent (w/w)

Mass % = \frac{mass of solute}{mass of solution} \times 100

A $10%$ (w/w) NaCl solution contains $10 g$ of NaCl in every $100 g$ of solution ( $90 g$ water).

6.2 Volume Percent (v/v)

Volume % = \frac{volume of solute}{volume of solution} \times 100

Used mainly for liquid-in-liquid solutions. A $35%$ (v/v) ethanol solution contains $35 mL$ ethanol in every $100 mL$ solution.

6.3 Mole Fraction

For a binary solution with solute A (

n_{A}

moles) and solvent B (

n_{B}

moles):

x_{A} = \frac{n _{A}}{n _{A} + n _{B}}, x_{B} = \frac{n _{B}}{n _{A} + n _{B}}, x_{A} + x_{B} = 1

Mole fraction is dimensionless and independent of temperature. It is used in ideal-solution laws (Raoult's law) and in gas mixtures (Dalton's law of partial pressures).

6.4 Molarity (M)

Molarity is the number of moles of solute per litre of solution.

M = \frac{n _{solute}}{V _{solution (L)}} = \frac{w \times 1000}{M _{solute} \times V _{ml}}

Unit:

mol L^{- 1}

Molarity depends on temperature because volume changes with temperature.

6.5 Molality (m)

Molality is the number of moles of solute per kilogram of solvent.

m = \frac{n _{solute}}{w _{solvent (kg)}} = \frac{w \times 1000}{M _{solute} \times w _{solvent, g}}

Unit:

mol kg^{- 1}

Molality is temperature-independent (mass does not change with T), so it is preferred for colligative-property calculations.

6.6 Parts Per Million (ppm)

For very dilute solutions, use ppm:

ppm_{A} = \frac{mass of A}{mass of solution} \times 1 0^{6}

For atmospheric pollution (e.g. $SO_{2}$ in air), ppm is often measured by volume: $cm^{3}$ of pollutant per $1 0^{6} cm^{3}$ of air.

Comparison of Units

Unit	Definition	Temperature dependence	Best used for
Mass %	mass ratio $\times 100$	No	Commercial, everyday
Volume %	volume ratio $\times 100$	Yes (V of solution)	Liquid mixtures
Mole fraction $x$	$n_{A} / n_{total}$	No	Vapour pressure, gas mixtures
Molarity $M$	mol solute / L solution	Yes	Titrations, reaction volumes
Molality $m$	mol solute / kg solvent	No	Freezing/boiling point calc.
ppm	ratio $\times 1 0^{6}$	No (mass) / Yes (volume)	Very dilute solutions

Relations Between the Units

For a solution of density $d (g mL^{- 1})$ , solute molar mass $M_{2}$ , and solvent molar mass $M_{1}$ :

Molality from molarity:

$m = \frac{1000 M}{1000 d - M M _{2}}$

Molarity from molality:

$M = \frac{m d}{1 + \frac{m M _{2}}{1000}}$

Mole fraction from molality:

$x_{solute} = \frac{m M _{1}}{1000 + m M _{1}}$

Molarity from mole fraction:

$M = \frac{1000 d x _{solute}}{x _{solvent} M _{1} + x _{solute} M _{2}}$

Solved Example 7

Calculate the molarity of a solution containing

0.5 g

of NaOH dissolved in

500 cm^{3}

of solution.

Solution:

Moles of NaOH $= \frac{0.5}{40} = 0.0125 mol$ (molar mass NaOH $= 40$ ).

Volume $= 500 cm^{3} = 0.500 L$ .

$M = \frac{0.0125}{0.500} = 0.025 mol L^{- 1} = 0.025 M$

Solved Example 8

A solution is prepared by dissolving

18 g

of glucose (

C_{6} H_{12} O_{6}

M = 180

) in

180 g

of water. Calculate (a) mass %, (b) mole fraction of glucose, (c) molality.

Solution:

(a) Total mass $= 18 + 180 = 198 g$ . Mass % of glucose $= \frac{18}{198} \times 100 = 9.09%$ .

(b) Moles of glucose $= \frac{18}{180} = 0.100$ . Moles of water $= \frac{180}{18} = 10.00$ .

$x_{glucose} = \frac{0.100}{0.100 + 10.00} = \frac{0.100}{10.10} = 9.90 \times 1 0^{- 3}$ .

Solved Example 9

Concentrated sulphuric acid is

98%

(w/w)

H_{2} SO_{4}

with density

1.84 g mL^{- 1}

. Calculate its molarity.

Solution:

Take $100 g$ of solution, which contains $98 g$ of $H_{2} SO_{4}$ .

Moles of $H_{2} SO_{4} = \frac{98}{98} = 1.00 mol$ .

Volume of $100 g$ solution $= \frac{100}{1.84} = 54.35 mL = 0.05435 L$ .

$M = \frac{1.00}{0.05435} = 18.4 mol L^{- 1}$

Common Mistakes to Avoid

Watch out

Not balancing the equation before doing any stoichiometric calculation. The coefficient ratios are meaningless if the equation is wrong.
Assuming the reactant with the smaller mass is the limiting reagent. Always divide moles by the coefficient - a large mass of low-molar-mass reactant can still be in excess.
Using molarity for freezing-point or boiling-point calculations. Molality is the correct choice because it is temperature-independent.
Confusing mass of solute with mass of solution in mass %. The denominator is the total solution mass, not the solvent mass.
Multiplying volume in mL by molarity directly. Molarity is per litre; convert mL to L first, or use $n = \frac{M \times V _{mL}}{1000}$ .
Adding volumes of solute and solvent to get volume of solution. Volumes are not strictly additive (they can contract or expand on mixing).
Forgetting that mole fractions of all components must sum to $1$ . This is a useful check.

Frequently Asked Questions

What is the difference between a limiting reagent and an excess reagent?

The limiting reagent is completely consumed and determines how much product forms. The excess reagent has some amount left over after the reaction stops. Identify the limiting reagent by dividing each reactant's moles by its coefficient; the smallest ratio is the limiting one.

Why is molality preferred over molarity in colligative property calculations?

Molality depends only on masses, which do not change with temperature. Molarity depends on solution volume, which expands or contracts with temperature. For freezing-point depression and boiling-point elevation, where solutions are often cooled or heated, molality gives a value that stays valid throughout.

Why does concentrated sulphuric acid have such a high molarity (~18 M)?

Because it is $98%$ pure by mass with a density near $1.84 g mL^{- 1}$ , so $1 L$ of the acid contains $\sim 1800 g$ , of which $\sim 1764 g$ is $H_{2} SO_{4}$ - about $18$ moles.

Can percentage yield ever exceed 100 percent?

No. If a calculation gives more than $100%$ , the reported "actual yield" contains impurities (unreacted starting material, solvent, side product), or the "theoretical yield" was miscalculated. Purify the product and re-weigh, or recheck the calculation.

How do I choose between molarity and molality for a given problem?

Use molarity when the problem talks about volumes of solution (titrations, dilutions, gas-solution reactions). Use molality when the problem involves colligative properties (boiling-point elevation, freezing-point depression, osmotic pressure at extreme temperatures).

Are the coefficients in a balanced equation moles or molecules?

Both. The coefficients give the ratio of molecules (at the microscopic scale) and, equivalently, the ratio of moles (at the macroscopic scale). For gases at the same T and P, they also give the ratio of volumes.

Why is ppm used only for very dilute solutions?

For concentrated solutions, ppm gives huge unwieldy numbers. A $1%$ solution is $10, 000 ppm$ . ppm becomes convenient when concentrations are much less than $1%$ - traces of pollutants, ions in drinking water, or dissolved gases.

Does the density of a solution matter when converting between molarity and molality?

Yes. Molarity is per litre of solution, and molality is per kilogram of solvent. To convert between them, you need to know the mass of the solvent inside a given volume of solution, which requires density. Without density, only mole fraction or mass % can be interconverted directly.

Why do we usually neglect water of crystallisation when computing mass of a compound?

In stoichiometric calculations for reactions in solution, water of crystallisation is often part of the reactant (e.g. $CuSO_{4} \cdot 5 H_{2} O$ ). Include its mass in the molar mass. In gas-phase or fusion calculations for the anhydrous salt, use the anhydrous molar mass.

Previous year questions on Stoichiometry

18 questions from past papers, each with a step-by-step solution.

Show all 18 questions

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