Why this chapter leaks marks into other chapters
Consider an electrochemistry question asking how long a current must flow to deposit a given mass of metal. The electrochemistry is one line - Faraday's law. Everything else is mole arithmetic: converting mass to moles, accounting for the number of electrons per ion, converting back. A student who is slow or unreliable at that arithmetic will get the chemistry right and the answer wrong.
The same pattern holds for equilibrium, thermodynamics and solutions. That is why this chapter is worth securing completely and early, even though its own question count is modest.
How JEE actually asks Mole Concept
NTA publishes no chapter-wise weightage, so any figure circulating online is a coaching estimate from past papers; confirm your session's syllabus and pattern in the official NTA information bulletin.
Direct questions tend to be on limiting reagents, percentage yield, empirical and molecular formulas, and concentration conversions. Indirect appearances are far more numerous - almost every numerical question in Physical Chemistry opens with a mole calculation, which is why the chapter's real weight is invisible in any breakdown.
Key concepts, compressed
- ✓A mole is Avogadro's number of particles, 6.022 × 10²³, and the molar mass in grams contains exactly that many.
- ✓Moles connect mass, number of particles and gas volume - it is the hub through which all three convert.
- ✓Stoichiometric coefficients give mole ratios, not mass ratios, so equations must be balanced first.
- ✓The limiting reagent is the one that runs out first and therefore fixes the amount of product.
- ✓Molarity depends on temperature because volume does; molality does not, because mass does not.
- ✓The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is a whole-number multiple of it.
Relations you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Moles from mass | n = mass / molar mass | |
| Moles from particles | n = N / Nₐ | Nₐ = 6.022 × 10²³ |
| Moles from gas volume at STP | n = V / 22.4 | V in litres, ideal gas |
| Molarity | M = moles of solute / litres of solution | temperature-dependent |
| Molality | m = moles of solute / kg of solvent | temperature-independent |
| Dilution | M₁V₁ = M₂V₂ | |
| Mole fraction | x(A) = n(A) / total moles | dimensionless, sums to 1 |
| Percentage yield | (actual / theoretical) × 100 | |
| Density of a gas at STP | molar mass / 22.4 | in g/L |
| Empirical to molecular | n = molar mass / empirical formula mass |
The five mistakes that cost the most marks
- ✓Confusing molarity with molality. One divides by solution volume, the other by solvent mass, and only molality is unaffected by temperature.
- ✓Calculating from an unbalanced equation. The coefficients are the mole ratio, so balancing must come first.
- ✓Skipping the limiting-reagent check when two reactant amounts are given. That check is usually the whole question.
- ✓Applying 22.4 L per mole away from STP. Outside those conditions you need the ideal gas equation.
- ✓Reporting an empirical formula when the molecular formula was asked for. The molar mass is given precisely so you can scale it.
Practice set 1: moles and molar mass
1. How many moles are there in 36 g of water?
2 moles. The molar mass of water is 2(1) + 16 = 18 g/mol, so n = 36/18 = 2 mol. This single division - mass over molar mass - is the most frequently used operation in all of Physical Chemistry, and it is worth being able to do without conscious effort.
2. How many water molecules are there in that 36 g?
About 1.2 × 10²⁴. Multiply the moles by Avogadro's number: 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules. Note the order of magnitude - even a small mass of a light molecule contains an enormous number of particles, which is why the mole exists as a unit at all.
3. What is the mass of 0.5 mol of carbon dioxide?
22 g. The molar mass of CO₂ is 12 + 2(16) = 44 g/mol, so the mass is 0.5 × 44 = 22 g. This is the reverse of question 1 - moles times molar mass gives mass - and being fluent in both directions is what makes stoichiometry quick.
4. How many atoms are there in 1 mole of carbon dioxide?
About 1.8 × 10²⁴. Each CO₂ molecule contains three atoms - one carbon and two oxygens - so one mole contains 3 × 6.022 × 10²³ = 1.8066 × 10²⁴ atoms. Questions distinguish carefully between molecules and atoms, and answering with 6.022 × 10²³ means the multiplication by three was missed.
5. What is the molar mass of sulphuric acid, H₂SO₄?
98 g/mol. Add the contributions: 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol. Working through the formula systematically, element by element, is more reliable under time pressure than trying to recall the value - though this one appears often enough to be worth remembering.
6. How many moles of atoms are there in 12 g of carbon?
1 mole. The molar mass of carbon is 12 g/mol, so 12/12 = 1 mol of atoms, which is 6.022 × 10²³ atoms. Carbon-12 is the reference on which the atomic mass scale is defined, which is why this particular number comes out exactly one.
7. What volume does 1 mole of an ideal gas occupy at STP?
22.4 litres. This molar volume applies to any ideal gas regardless of its identity, which is Avogadro's law in practice. It holds only at standard temperature and pressure - at other conditions you must use PV = nRT, and applying 22.4 L per mole anyway is a standard error.
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Practice set 2: stoichiometry
Questions 8 to 13 use the reaction 2H₂ + O₂ → 2H₂O, which is already balanced.
8. How many moles of water are produced from 4 moles of hydrogen with excess oxygen?
4 moles. The balanced equation shows 2 moles of hydrogen give 2 moles of water - a one-to-one ratio. With oxygen in excess, hydrogen determines the outcome entirely, so 4 moles of hydrogen give 4 moles of water. The phrase 'excess oxygen' is what removes the need for a limiting-reagent check.
9. What mass of water is produced from 4 g of hydrogen with excess oxygen?
36 g. First convert to moles: 4 g of H₂ divided by 2 g/mol gives 2 moles. The one-to-one ratio gives 2 moles of water, which is 2 × 18 = 36 g. Note the three-step structure - mass to moles, mole ratio, moles back to mass - which is the template for every stoichiometry question.
10. If 2 moles of hydrogen react with 2 moles of oxygen, which is the limiting reagent and how much water forms?
Hydrogen is limiting, and 2 moles of water form. Divide each amount by its coefficient: hydrogen gives 2/2 = 1, oxygen gives 2/1 = 2. The smaller result identifies the limiting reagent, so hydrogen runs out first. It produces 2 moles of water. Assuming oxygen is limiting because there is less of it in some other sense is the trap.
11. Why must an equation be balanced before any mole calculation?
Because the coefficients are the mole ratio, and an unbalanced equation gives the wrong ratio. Mass is conserved in a chemical reaction, and balancing is what enforces that. Note that the coefficients give mole ratios, not mass ratios - 2 moles of H₂ and 1 mole of O₂ are 4 g and 32 g respectively.
12. If the theoretical yield is 36 g and only 18 g is obtained, what is the percentage yield?
50%. Percentage yield is (actual/theoretical) × 100 = (18/36) × 100 = 50%. The theoretical yield is what perfect stoichiometry predicts; the actual is what the experiment gives. Real yields fall short because of side reactions, incomplete conversion and losses during separation.
13. In question 10, how much oxygen remains unreacted?
1 mole. Two moles of hydrogen require one mole of oxygen by the balanced equation, so of the 2 moles supplied, 1 is consumed and 1 remains. Reporting the leftover excess reagent is a common follow-up, and it requires going back to the limiting reagent to see how much was actually used.
14. What law guarantees that the total mass is unchanged in a chemical reaction?
The law of conservation of mass. Atoms are rearranged but neither created nor destroyed, which is exactly what balancing an equation expresses. It also provides a check on any stoichiometry answer: the total mass of products must equal the total mass of reactants consumed.
Practice set 3: concentration terms
15. What is the molarity of a solution containing 0.5 mol of solute in 250 mL of solution?
2 M. Molarity is moles per litre of solution: 0.5 / 0.25 = 2 mol/L. Converting the volume from millilitres to litres before dividing is essential - leaving it as 250 gives an answer wrong by a factor of a thousand.
16. What is the molality of a solution containing 0.5 mol of solute in 500 g of solvent?
1 m. Molality is moles per kilogram of solvent: 0.5 / 0.5 = 1 mol/kg. Note the two differences from molarity - it uses the solvent rather than the whole solution, and mass rather than volume. Both differences matter.
17. Why is molality preferred over molarity for colligative-property calculations?
Because molality is independent of temperature. Volume expands when heated, so a solution's molarity changes with temperature while its molality does not - mass is unaffected. Since colligative properties are measured across a range of temperatures, the temperature-independent quantity is the appropriate one.
18. What is the concentration when 50 mL of 2 M solution is diluted to 500 mL?
0.2 M. Use M₁V₁ = M₂V₂: (2)(50) = M₂(500), so M₂ = 100/500 = 0.2 M. The moles of solute are unchanged by dilution - only the volume increases - which is exactly what the relation expresses. A tenfold dilution gives a tenth of the concentration, which is a useful sanity check.
19. What is mass percentage, and how does it differ from mole fraction?
Mass percentage is the mass of solute as a percentage of the total mass of solution. Mole fraction is the moles of one component divided by the total moles. They answer different questions and are numerically different unless the components happen to have equal molar masses.
20. A mixture contains 2 mol of A and 3 mol of B. What is the mole fraction of A?
0.4. The mole fraction is 2/(2 + 3) = 0.4. Mole fractions are dimensionless and must sum to 1 across all components, so B has mole fraction 0.6 - which is a quick check. This quantity appears in Raoult's law and in partial pressure calculations.
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Practice set 4: empirical and molecular formulas
21. A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
CH₂O. Assume 100 g, so the masses are 40 g, 6.7 g and 53.3 g. Divide each by the atomic mass: 40/12 = 3.33, 6.7/1 = 6.7, 53.3/16 = 3.33. Divide through by the smallest, 3.33, giving 1 : 2 : 1. So the empirical formula is CH₂O. The four-step procedure - assume 100 g, divide by atomic masses, divide by the smallest, round to whole numbers - works every time.
22. That compound has a molar mass of 180 g/mol. What is its molecular formula?
C₆H₁₂O₆. The empirical formula mass of CH₂O is 12 + 2 + 16 = 30. Dividing the molar mass by it gives 180/30 = 6, so the molecular formula is six times the empirical: C₆H₁₂O₆, which is glucose. Answering CH₂O would report the empirical formula when the molecular one was requested.
23. What is the percentage of carbon by mass in carbon dioxide?
About 27.3%. Carbon contributes 12 of the 44 g/mol, so the percentage is (12/44) × 100 = 27.27%. Percentage composition questions always follow this pattern - the mass of the element of interest over the total molar mass. Oxygen therefore makes up the remaining 72.7%.
24. Can two different compounds have the same empirical formula?
Yes. Formaldehyde (CH₂O), acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆) all have the empirical formula CH₂O, because the empirical formula records only the ratio of atoms, not how many there are. This is why the molar mass is required to determine the molecular formula.
25. How does combustion analysis determine an empirical formula?
By burning a known mass of the compound and measuring the carbon dioxide and water produced. All the carbon appears in the CO₂ and all the hydrogen in the H₂O, so the moles of each can be worked back. Any remaining mass is attributed to oxygen, which cannot be measured directly because oxygen is also supplied by the air.
Practice set 5: gases and mixed
26. How many moles of gas occupy 11.2 L at STP?
0.5 mol. Divide by the molar volume: 11.2/22.4 = 0.5 mol. This works for any ideal gas, so 11.2 L of hydrogen and 11.2 L of carbon dioxide at STP both contain half a mole - though their masses differ by a factor of 22.
27. What is the density of oxygen gas at STP?
About 1.43 g/L. Density is mass per volume, and one mole occupies 22.4 L, so the density is the molar mass over 22.4: 32/22.4 = 1.43 g/L. This is why gas densities are proportional to molar mass at fixed conditions, and why hydrogen is the lightest gas.
28. State Avogadro's law.
Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. It is why 22.4 L per mole applies to every ideal gas regardless of identity, and it is what allows gas volumes to be used directly as mole ratios in reactions between gases.
29. What is the mass of a single water molecule?
About 2.99 × 10⁻²³ g. Divide the molar mass by Avogadro's number: 18 / (6.022 × 10²³) = 2.989 × 10⁻²³ g. This is the general method for converting from a per-mole quantity to a per-particle one, and it also works for the energy released per molecule in a reaction.
30. Do 1 mole of hydrogen and 1 mole of carbon dioxide occupy the same volume at STP?
Yes, both occupy 22.4 L. By Avogadro's law the volume depends on the number of molecules, not on their identity or mass. Their masses are very different - 2 g against 44 g - which is precisely why gas density varies with molar mass while molar volume does not.
How to study this chapter efficiently
- ✓Practise the mass-to-moles conversion until it is automatic. It opens almost every numerical question in Physical Chemistry.
- ✓Write the balanced equation as the first line of any stoichiometry answer, even when it is given.
- ✓Whenever two reactant quantities appear, divide each by its coefficient before doing anything else.
- ✓Keep molarity and molality visually separate in your notes - solution volume versus solvent mass.
- ✓Check units at every step. A factor-of-1000 error from millilitres is the single most common arithmetic failure here.
- ✓Secure this chapter before Equilibrium, Thermodynamics and Electrochemistry, since all three assume it.
Turn this into active practice
This chapter rewards volume and speed rather than depth. The individual operations are simple, and the goal is to make them fast enough that they cost no attention when they appear inside a harder question about something else.
The JEE Mole Concept quiz on QUFF generates fresh questions across moles, stoichiometry, limiting reagents, concentration and formulas, marks them instantly and explains each answer. Do timed sets and watch for unit errors specifically - they are the most common failure and the easiest to eliminate.