Every later Chemistry calculation — in Equilibrium, Electrochemistry, Thermodynamics, all the way through Class 12 — assumes Mole Concept is completely solid. This is 50 original problems, 5 for every distinct numerical type, and every single calculation on this page was computed and independently verified before publishing.
How to use this page: Attempt each problem fully before revealing the solution — check your working against each step, not just the final number.
Quick reference:
Every Numerical Type Covered
1. Mole-mass-particle conversions
2. Mole-volume at STP
3. Percentage composition
4. Empirical formula
5. Molecular formula from empirical
6. Molarity
7. Molality
8. Mole fraction
9. Limiting reagent
10. Stoichiometry
Mass → Moles
Find the number of moles in 24 g of magnesium. (Atomic mass Mg = 24)
Reveal Solution
n = mass/molar mass = 24/24
✓ n = 1 mol
Moles → Mass
Find the mass of 0.5 mol of CO₂. (Molar mass CO₂ = 44 g/mol)
Reveal Solution
mass = n × molar mass = 0.5 × 44
✓ mass = 22 g
Mass → Particles
Find the number of molecules in 4.4 g of CO₂.
Reveal Solution
n = 4.4/44 = 0.1 mol
Molecules = n × Nₐ = 0.1 × 6.022×10²³
✓ 6.022 × 10²² molecules
Particles → Moles
How many moles are present in 3.011 × 10²³ atoms of sodium?
Reveal Solution
n = particles/Nₐ = 3.011×10²³ / 6.022×10²³
✓ n = 0.5 mol
Particles → Mass
Find the mass of 1.2044 × 10²⁴ molecules of water. (Molar mass H₂O = 18 g/mol)
Reveal Solution
n = 1.2044×10²⁴ / 6.022×10²³ = 2 mol
mass = n × molar mass = 2 × 18
✓ mass = 36 g
Moles → Volume
Find the volume occupied by 2 mol of O₂ gas at STP.
Reveal Solution
V = n × 22.7 = 2 × 22.7
✓ V = 45.4 L
Volume → Moles
How many moles are in 11.2 L of CO₂ at STP?
Reveal Solution
n = V/22.7 = 11.2/22.7
✓ n ≈ 0.493 mol
Volume → Mass
Find the mass of 5.6 L of N₂ gas at STP. (Molar mass N₂ = 28 g/mol)
Reveal Solution
n = 5.6/22.7 ≈ 0.2467 mol
mass = 0.2467 × 28
✓ mass ≈ 6.91 g
Mass → Volume
Find the volume occupied by 8.5 g of NH₃ gas at STP. (Molar mass NH₃ = 17 g/mol)
Reveal Solution
n = 8.5/17 = 0.5 mol
V = 0.5 × 22.7
✓ V = 11.35 L
Volume → Particles
Find the number of molecules present in 0.56 L of a gas at STP.
Reveal Solution
n = 0.56/22.7 ≈ 0.02467 mol
Molecules = 0.02467 × 6.022×10²³
✓ ≈ 1.486 × 10²² molecules
% Composition
Calculate the percentage composition of hydrogen and oxygen in water (H₂O).
Reveal Solution
Molar mass = 18. %H = (2×1/18)×100, %O = (16/18)×100
✓ %H = 11.11%, %O = 88.89%
% Composition
Calculate the percentage composition of carbon and oxygen in CO₂.
Reveal Solution
Molar mass = 44. %C = (12/44)×100, %O = (32/44)×100
✓ %C = 27.27%, %O = 72.73%
% Composition
Calculate the percentage composition of sodium and chlorine in NaCl.
Reveal Solution
Molar mass = 58.5. %Na = (23/58.5)×100, %Cl = (35.5/58.5)×100
✓ %Na = 39.32%, %Cl = 60.68%
% Composition
Calculate the percentage composition of all elements in calcium carbonate, CaCO₃.
Reveal Solution
Molar mass = 100
✓ %Ca = 40.00%, %C = 12.00%, %O = 48.00%
% Composition
Calculate the percentage composition of all elements in ammonium sulphate, (NH₄)₂SO₄.
Reveal Solution
Molar mass = 132
✓ %N = 21.21%, %H = 6.06%, %S = 24.24%, %O = 48.48%
Empirical Formula
A compound contains C = 40%, H = 6.7%, O = 53.3% by mass. Find its empirical formula.
Reveal Solution
Moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33
Divide by smallest (3.33): C = 1, H = 2.01 ≈ 2, O = 1
✓ Empirical formula = CH₂O
Empirical Formula
A compound of iron and oxygen contains 70% Fe and 30% O by mass. Find its empirical formula.
Reveal Solution
Moles: Fe = 70/56 = 1.25, O = 30/16 = 1.875
Divide by smallest (1.25): Fe = 1, O = 1.5. Multiply both by 2 to clear the fraction: Fe = 2, O = 3
✓ Empirical formula = Fe₂O₃
⚠️ Common mistake: stopping at Fe : O = 1 : 1.5 and rounding 1.5 to 2. Whenever a ratio ends in .5, double every ratio instead of rounding.
Empirical Formula
A compound contains Na = 32.4%, S = 22.5%, O = 45.1% by mass. Find its empirical formula.
Reveal Solution
Moles: Na = 32.4/23 = 1.41, S = 22.5/32 = 0.70, O = 45.1/16 = 2.82
Divide by smallest (0.70): Na = 2, S = 1, O = 4
✓ Empirical formula = Na₂SO₄
Empirical Formula
A compound contains K = 26.6%, Cr = 35.4%, O = 38.0% by mass. Find its empirical formula. (Atomic mass Cr = 52)
Reveal Solution
Moles: K = 26.6/39 = 0.68, Cr = 35.4/52 = 0.68, O = 38.0/16 = 2.375
Divide by smallest (0.68): K = 1, Cr = 1, O = 3.49 ≈ 3.5. Double to clear the fraction: K = 2, Cr = 2, O = 7
✓ Empirical formula = K₂Cr₂O₇
Empirical Formula
A hydrocarbon contains C = 85.7% and H = 14.3% by mass. Find its empirical formula.
Reveal Solution
Moles: C = 85.7/12 = 7.14, H = 14.3/1 = 14.3
Divide by smallest (7.14): C = 1, H = 2
✓ Empirical formula = CH₂
Molecular Formula
A compound has empirical formula CH₂O and molecular mass 180. Find its molecular formula.
Reveal Solution
Empirical formula mass = 12+2+16 = 30
n = molecular mass/empirical mass = 180/30 = 6
✓ Molecular formula = C₆H₁₂O₆ (glucose)
Molecular Formula
A compound has empirical formula CH and molecular mass 78. Find its molecular formula.
Reveal Solution
Empirical formula mass = 13
n = 78/13 = 6
✓ Molecular formula = C₆H₆ (benzene)
Molecular Formula
A compound has empirical formula CH₂ and molecular mass 84. Find its molecular formula.
Reveal Solution
Empirical formula mass = 14
n = 84/14 = 6
✓ Molecular formula = C₆H₁₂
Molecular Formula
A compound has empirical formula HO and molecular mass 34. Find its molecular formula.
Reveal Solution
Empirical formula mass = 17
n = 34/17 = 2
✓ Molecular formula = H₂O₂ (hydrogen peroxide)
Molecular Formula
A compound has empirical formula NO₂ and molecular mass 92. Find its molecular formula.
Reveal Solution
Empirical formula mass = 46
n = 92/46 = 2
✓ Molecular formula = N₂O₄
Find Molarity
Find the molarity of a solution containing 5.85 g of NaCl in 500 mL of solution.
Reveal Solution
n = 5.85/58.5 = 0.1 mol
M = n/V(L) = 0.1/0.5
✓ M = 0.2 mol/L
Find Mass
Find the mass of NaOH needed to prepare 250 mL of 0.4 M solution. (Molar mass NaOH = 40)
Reveal Solution
n = M × V = 0.4 × 0.25 = 0.1 mol
mass = 0.1 × 40
✓ mass = 4 g
Dilution
100 mL of 2 M HCl is diluted to 500 mL. Find the new molarity.
Reveal Solution
M₁V₁ = M₂V₂ → M₂ = (2 × 100)/500
✓ M₂ = 0.4 mol/L
Find Volume
What volume of 0.1 M H₂SO₄ contains exactly 0.02 mol of H₂SO₄?
Reveal Solution
V = n/M = 0.02/0.1
✓ V = 0.2 L = 200 mL
Find Molarity
Find the molarity of a solution containing 5.6 g of KOH in 200 mL of solution. (Molar mass KOH = 56)
Reveal Solution
n = 5.6/56 = 0.1 mol
M = 0.1/0.2
✓ M = 0.5 mol/L
Find Molality
4 g of NaOH is dissolved in 500 g of water. Find the molality of the solution.
Reveal Solution
n = 4/40 = 0.1 mol; mass of solvent = 0.5 kg
m = 0.1/0.5
✓ m = 0.2 mol/kg
Find Molality
9.2 g of glycerol (C₃H₈O₃, molar mass 92) is dissolved in 200 g of water. Find the molality.
Reveal Solution
n = 9.2/92 = 0.1 mol; mass of solvent = 0.2 kg
m = 0.1/0.2
✓ m = 0.5 mol/kg
Find Molality
12 g of urea, CO(NH₂)₂ (molar mass 60), is dissolved in 1000 g of water. Find the molality.
Reveal Solution
n = 12/60 = 0.2 mol; mass of solvent = 1 kg
✓ m = 0.2 mol/kg
Find Molality
18 g of glucose (C₆H₁₂O₆, molar mass 180) is dissolved in 250 g of water. Find the molality.
Reveal Solution
n = 18/180 = 0.1 mol; mass of solvent = 0.25 kg
m = 0.1/0.25
✓ m = 0.4 mol/kg
Find Molality
11.7 g of NaCl is dissolved in 500 g of water. Find the molality.
Reveal Solution
n = 11.7/58.5 = 0.2 mol; mass of solvent = 0.5 kg
m = 0.2/0.5
✓ m = 0.4 mol/kg
Mole Fraction
A solution contains 2 mol of ethanol and 8 mol of water. Find the mole fraction of each.
Reveal Solution
Total moles = 10
✓ x(ethanol) = 2/10 = 0.2, x(water) = 8/10 = 0.8
Mole Fraction
A solution contains 18 g of water and 46 g of ethanol (molar mass 46). Find the mole fraction of each.
Reveal Solution
Moles: water = 18/18 = 1, ethanol = 46/46 = 1. Total = 2
✓ x(water) = 0.5, x(ethanol) = 0.5
Mole Fraction
90 g of water is mixed with 18 g of glucose (molar mass 180). Find the mole fraction of glucose.
Reveal Solution
Moles: water = 90/18 = 5, glucose = 18/180 = 0.1. Total = 5.1
x(glucose) = 0.1/5.1
✓ x(glucose) ≈ 0.0196
Mole Fraction
A mixture contains 3 mol of solute and 7 mol of solvent. Find the mole fraction of the solute.
Reveal Solution
✓ x(solute) = 3/10 = 0.3
Mole Fraction
A gas mixture contains 5 g of H₂ and 5 g of He (atomic mass He = 4). Find the mole fraction of each gas.
Reveal Solution
Moles: H₂ = 5/2 = 2.5, He = 5/4 = 1.25. Total = 3.75
✓ x(H₂) ≈ 0.667, x(He) ≈ 0.333
⚠️ Common mistake: using equal mass to assume equal moles. Mole fraction depends on moles, not mass — always convert first.
Limiting Reagent
28 g of N₂ reacts with 9 g of H₂ to form NH₃ (N₂ + 3H₂ → 2NH₃). Identify the limiting reagent and find the mass of NH₃ formed.
Reveal Solution
Moles: N₂ = 28/28 = 1 mol, H₂ = 9/2 = 4.5 mol
1 mol N₂ needs 3 mol H₂; 4.5 mol is available (more than enough) → N₂ is limiting
NH₃ = 1 × 2 = 2 mol = 2 × 17
✓ N₂ is limiting. NH₃ formed = 34 g
Limiting Reagent
5.4 g of Al reacts with 10.65 g of Cl₂ (2Al + 3Cl₂ → 2AlCl₃). Identify the limiting reagent and find the mass of AlCl₃ formed.
Reveal Solution
Moles: Al = 5.4/27 = 0.2 mol, Cl₂ = 10.65/71 = 0.15 mol
0.2 mol Al needs (3/2) × 0.2 = 0.3 mol Cl₂; only 0.15 mol is available → Cl₂ is limiting
AlCl₃ = (2/3) × 0.15 = 0.1 mol = 0.1 × 133.5
✓ Cl₂ is limiting. AlCl₃ formed = 13.35 g
Limiting Reagent
6 g of H₂ reacts with 16 g of O₂ (2H₂ + O₂ → 2H₂O). Identify the limiting reagent and find the mass of water formed.
Reveal Solution
Moles: H₂ = 6/2 = 3 mol, O₂ = 16/32 = 0.5 mol
3 mol H₂ needs 1.5 mol O₂; only 0.5 mol is available → O₂ is limiting
H₂O = 0.5 × 2 = 1 mol = 18 g
✓ O₂ is limiting. H₂O formed = 18 g
Limiting Reagent
6 g of carbon reacts with 32 g of O₂ (C + O₂ → CO₂). Identify the limiting reagent and find the mass of CO₂ formed.
Reveal Solution
Moles: C = 6/12 = 0.5 mol, O₂ = 32/32 = 1 mol
0.5 mol C needs 0.5 mol O₂; 1 mol is available (in excess) → C is limiting
CO₂ = 0.5 mol = 0.5 × 44
✓ Carbon is limiting. CO₂ formed = 22 g
Limiting Reagent
4.6 g of Na reacts with 3.55 g of Cl₂ (2Na + Cl₂ → 2NaCl). Identify the limiting reagent and find the mass of NaCl formed.
Reveal Solution
Moles: Na = 4.6/23 = 0.2 mol, Cl₂ = 3.55/71 = 0.05 mol
0.2 mol Na needs 0.1 mol Cl₂; only 0.05 mol is available → Cl₂ is limiting
NaCl = 0.05 × 2 = 0.1 mol = 0.1 × 58.5
✓ Cl₂ is limiting. NaCl formed = 5.85 g
Mass-Mass
Calculate the mass of CaO produced when 50 g of CaCO₃ decomposes completely (CaCO₃ → CaO + CO₂).
Reveal Solution
n(CaCO₃) = 50/100 = 0.5 mol → n(CaO) = 0.5 mol (1:1 ratio)
mass(CaO) = 0.5 × 56
✓ mass(CaO) = 28 g
Mass-Mass
Calculate the mass of MgO produced when 12 g of Mg burns completely (2Mg + O₂ → 2MgO).
Reveal Solution
n(Mg) = 12/24 = 0.5 mol → n(MgO) = 0.5 mol (1:1 ratio)
mass(MgO) = 0.5 × 40
✓ mass(MgO) = 20 g
Mass-Mass
Calculate the mass of H₂ gas produced when 13 g of zinc reacts completely with excess HCl (Zn + 2HCl → ZnCl₂ + H₂).
Reveal Solution
n(Zn) = 13/65 = 0.2 mol → n(H₂) = 0.2 mol (1:1 ratio)
mass(H₂) = 0.2 × 2
✓ mass(H₂) = 0.4 g
Mass-Mass
Calculate the mass of CO₂ produced when 5.6 g of ethene (C₂H₄) burns completely (C₂H₄ + 3O₂ → 2CO₂ + 2H₂O).
Reveal Solution
n(C₂H₄) = 5.6/28 = 0.2 mol → n(CO₂) = 0.2 × 2 = 0.4 mol
mass(CO₂) = 0.4 × 44
✓ mass(CO₂) = 17.6 g
Mass-Mass
Calculate the mass of O₂ produced when 24.5 g of KClO₃ decomposes completely (2KClO₃ → 2KCl + 3O₂). (Molar mass KClO₃ = 122.5)
Reveal Solution
n(KClO₃) = 24.5/122.5 = 0.2 mol → n(O₂) = 0.2 × (3/2) = 0.3 mol
mass(O₂) = 0.3 × 32
✓ mass(O₂) = 9.6 g
Which of These 10 Types Would You Actually Recognise Cold?
Reading 50 solved problems and being able to solve fresh ones without the type named for you are different skills. The real exam test is spotting which of these 10 patterns a new question belongs to.
What a Genelis weak area map looks like after Mole Concept practice
Next session: limiting reagent problems (31%) — not more basic conversion drilling. Genelis tracks accuracy by type, not just by topic, so it knows exactly which pattern needs more reps.
Genelis is an AI-powered personalized learning platform built on Adaptive Personalized Intelligence. The Genelis learning system generates fresh, unlabelled Mole Concept problems across all 10 types, tracks your accuracy on each specifically, and logs every wrong answer to your wrong-question notebook for reattempt.
Learn smarter. Practice deeper. Improve continuously.
Genelis combines Adaptive Personalized Intelligence, AI-generated notes, targeted practice, mock tests, analytics, and personalised revision to help students improve every study session.
Questions Students Commonly Ask
Quick answers to the most common questions related to this guide.
How do I know whether to use molarity or molality in a given problem?
Molarity is defined per litre of solution and is used when the problem gives or asks for solution volume — it is temperature-dependent because volume changes slightly with temperature. Molality is defined per kilogram of solvent and is used when the mass of solvent is given or needed — it is temperature-independent, since mass doesn't change with temperature. If a problem specifies a solution's total volume, use molarity; if it specifies the mass of solvent used to dissolve a solute, use molality.
What is the difference between empirical formula and molecular formula, and how are they related?
The empirical formula shows the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms in one molecule. The molecular formula is always a whole-number multiple of the empirical formula — found by dividing the compound's actual molar mass by the empirical formula's mass to get that multiple, then multiplying every subscript in the empirical formula by it.
How do I identify the limiting reagent in a reaction?
Convert the given mass of each reactant into moles, then compare how much of one reactant would be needed to completely react with the other, based on the balanced equation's mole ratio. Whichever reactant runs out first — meaning there isn't enough of it to consume all of the other reactant — is the limiting reagent, and it alone determines the maximum amount of product that can form.