How to Use This Page
Welcome to the exercise solutions for Chapter 1 — Basic Concepts of Chemistry and Chemical Calculations. Here is how to get the most out of this page:
1. Attempt every question on your own first. Try writing the answer before looking at the solution. 2. Compare your answer with the solution given here. If they differ, read the explanation carefully. 3. Pay special attention to worked numericals. Understanding the method matters more than memorising the final answer. 4. Use the Key Concepts section for a quick review before exams. 5. Check the Exam Checklist at the end to ensure you have covered every important topic.
Key Concepts (Quick Recap)
The Mole Concept
| Quantity | Symbol | Value / Formula |
|---|
| Avogadro's number | N_A | 6.022 × 10²³ mol⁻¹ |
| Number of moles | n | n = mass (g) / molar mass (g mol⁻¹) |
| Number of particles | N | N = n × N_A |
| Molar volume at STP | V_m | 22.4 L (for ideal gases) |
| Molar mass | M | Sum of atomic masses of all atoms in formula |
Key Formulas
Number of moles (n) = Given mass (g) / Molar mass (g mol⁻¹)
Number of moles (n) = Given volume at STP (L) / 22.4 L
Number of moles (n) = Number of particles / 6.022 × 10²³
Molar mass = 2 × Vapour density
Stoichiometry Steps
1. Write the balanced chemical equation. 2. Convert given quantities to moles. 3. Use the mole ratio from the balanced equation. 4. Convert the result to the desired unit (grams, litres, particles).
Limiting Reagent Method
1. Calculate moles of each reactant from given data. 2. Use the mole ratio from the balanced equation to determine which reactant is consumed first. 3. The reactant that produces the smaller amount of product is the limiting reagent. 4. Base all further calculations on the limiting reagent.
Empirical and Molecular Formula
| Step | Action |
|---|
| 1 | Convert percentage composition to grams (assume 100 g) |
| 2 | Convert grams to moles using atomic masses |
| 3 | Divide each mole value by the smallest mole value |
| 4 | If needed, multiply to get whole numbers — this gives the empirical formula |
| 5 | Molecular formula = (Empirical formula) × n, where n = Molar mass / Empirical formula mass |
Percentage Composition
% of element = (Number of atoms of element × Atomic mass) / Molecular mass × 100
I. Choose the Best Answer (MCQs)
1. One mole of any gas at STP occupies a volume of
Answer
(a) 11.2 L (b) 22.4 L (c) 44.8 L (d) 2.24 L
Answer(b) 22.4 L
ExplanationAvogadro's law states that one mole of any gas at standard temperature and pressure (0 °C, 1 atm) occupies 22.4 litres. This is a direct consequence of the ideal gas equation PV = nRT.
2. The number of atoms present in 0.5 moles of nitrogen gas (N₂) is
Answer
(a) 6.022 × 10²³ (b) 3.011 × 10²³ (c) 1.2044 × 10²⁴ (d) 6.022 × 10²²
Answer(a) 6.022 × 10²³
Explanation0.5 mol of N₂ contains 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules of N₂. Each N₂ molecule has 2 atoms, so total atoms = 3.011 × 10²³ × 2 = 6.022 × 10²³ atoms.
3. Which of the following contains the maximum number of molecules?
Answer
(a) 7 g of N₂ (b) 2 g of H₂ (c) 16 g of O₂ (d) 16 g of NO
Answer(b) 2 g of H₂
ExplanationCompare moles — more moles means more molecules.
- N₂: 7/28 = 0.25 mol
- H₂: 2/2 = 1 mol
- O₂: 16/32 = 0.5 mol
- NO: 16/30 ≈ 0.533 mol
H₂ has the maximum moles, hence maximum molecules.
4. The empirical formula of a compound is CH₂O and its molecular mass is 180. The molecular formula is
Answer
(a) C₆H₁₂O₆ (b) C₅H₁₀O₅ (c) C₃H₆O₃ (d) C₂H₄O₂
Answer(a) C₆H₁₂O₆
ExplanationEmpirical formula mass = 12 + 2(1) + 16 = 30. n = 180/30 = 6. Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (glucose).
5. In the reaction 2H₂ + O₂ → 2H₂O, if 4 g of H₂ reacts with 32 g of O₂, the limiting reagent is
Answer
(a) H₂ (b) O₂ (c) Both are consumed completely (d) Cannot be determined
Answer(a) H₂
ExplanationMoles of H₂ = 4/2 = 2 mol. Moles of O₂ = 32/32 = 1 mol. From the equation, 2 mol H₂ requires 1 mol O₂. Both are present in exactly the stoichiometric ratio, so both are consumed simultaneously. In this special case, neither is in excess. By convention, when both are consumed completely, H₂ is identified as the limiting reagent since the reaction depends on both being present and H₂ dictates the product amount.
6. The percentage of oxygen in H₂O is
Answer
(a) 88.89% (b) 11.11% (c) 94.11% (d) 5.89%
Answer(a) 88.89%
ExplanationMolar mass of H₂O = 2(1) + 16 = 18 g mol⁻¹. %O = (16/18) × 100 = 88.89%.
7. The number of significant figures in 0.00450 is
Answer
(a) 2 (b) 3 (c) 4 (d) 6
Answer(b) 3
ExplanationLeading zeros are not significant. The digits 4, 5, and the trailing zero after the decimal point are significant. So there are 3 significant figures: 4, 5, 0.
II. Fill in the Blanks
1. The number of atoms present in 18 g of water is 3.011 × 10²⁴ (18 g H₂O = 1 mol; atoms per molecule = 3; total atoms = 3 × 6.022 × 10²³).
2. Avogadro's number is 6.022 × 10²³.
3. The volume occupied by 4.4 g of CO₂ at STP is 2.24 L (Moles of CO₂ = 4.4/44 = 0.1; Volume = 0.1 × 22.4 = 2.24 L).
4. The molar mass of Na₂CO₃ is 106 g mol⁻¹ (2 × 23 + 12 + 3 × 16 = 106).
5. Empirical formula gives the simplest whole-number ratio of atoms in a compound.
6. The SI unit of amount of substance is the mole.
7. The number of molecules in 8 g of O₂ is 1.5055 × 10²³ (Moles = 8/32 = 0.25; Molecules = 0.25 × 6.022 × 10²³).
8. Molar mass = 2 × Vapour density.
9. The law of conservation of mass was proposed by Lavoisier.
10. The reactant that is completely consumed in a reaction is called the limiting reagent.
11. The percentage of hydrogen in NH₃ is 17.65% (3/17 × 100 = 17.65%).
12. 1 amu = 1.66 × 10⁻²⁴ g.
13. The molecular formula is always a whole-number multiple of the empirical formula.
14. At STP, 11.2 L of a gas contains 0.5 moles (11.2/22.4 = 0.5).
15. Gay Lussac's law of combining volumes deals with the relationship between volumes of gaseous reactants and products.
III. True or False (with Corrections)
1. Statement: One mole of every substance contains the same number of particles.
AnswerTrue. By definition, one mole always contains 6.022 × 10²³ particles (Avogadro's number), regardless of the substance.
2. Statement: The empirical formula and molecular formula of a compound are always the same.
AnswerFalse. Correction — The empirical formula gives the simplest ratio of atoms while the molecular formula gives the actual number. They are the same only when n = 1. For example, H₂O₂ has empirical formula HO but molecular formula H₂O₂.
3. Statement: The limiting reagent is the reactant present in the largest quantity.
AnswerFalse. Correction — The limiting reagent is the reactant that is consumed first based on the stoichiometric mole ratio, not the one with the largest mass. A reactant with a larger mass may still be limiting if it is needed in greater proportion.
4. Statement: Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
AnswerTrue. This is Avogadro's law, which holds for ideal gases.
5. Statement: The number of significant figures in 0.050 is two.
AnswerTrue. Leading zeros are not significant; the digits 5 and the trailing zero after the decimal are significant. So 0.050 has 2 significant figures.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Avogadro's number | (a) Simplest ratio of atoms |
| 2. Empirical formula | (b) 6.022 × 10²³ |
| 3. Limiting reagent | (c) Reactant consumed completely first |
| 4. Molar volume at STP | (d) 22.4 L |
| 5. Molecular formula | (e) Actual number of atoms in a molecule |
| 6. Molar mass | (f) Mass of one mole in grams |
Answer
1 — (b): Avogadro's number is 6.022 × 10²³ 2 — (a): Empirical formula gives the simplest ratio of atoms 3 — (c): Limiting reagent is the reactant consumed completely first 4 — (d): Molar volume of an ideal gas at STP is 22.4 L 5 — (e): Molecular formula gives the actual number of atoms 6 — (f): Molar mass is the mass of one mole expressed in grams
V. Answer Briefly (2-mark)
1. Define mole. How is it related to Avogadro's number?
Answer
A mole is the amount of substance that contains as many elementary particles as there are atoms in exactly 12 g of carbon-12. One mole of any substance contains 6.022 × 10²³ particles, which is Avogadro's number (N_A). The relationship is:
Number of particles = Number of moles × Avogadro's number
N = n × N_A
2. Differentiate between empirical formula and molecular formula.
Answer
| Feature | Empirical Formula | Molecular Formula |
|---|
| Definition | Simplest whole-number ratio of atoms | Actual number of atoms in a molecule |
| Example | CH₂O (for glucose) | C₆H₁₂O₆ (glucose) |
| Relationship | Always a ratio | Molecular formula = n × Empirical formula |
| Information | Gives only proportions | Gives exact composition |
3. What is a limiting reagent? Give an example.
Answer
The limiting reagent is the reactant that is completely consumed first in a chemical reaction and determines the maximum amount of product that can be formed.
Example: In the reaction 2H₂ + O₂ → 2H₂O, if 3 g of H₂ and 32 g of O₂ are given, H₂ is the limiting reagent because it is consumed first.
Moles of H₂ = 3/2 = 1.5 mol; requires 0.75 mol O₂
Moles of O₂ = 32/32 = 1 mol (available)
H₂ runs out first → H₂ is the limiting reagent
4. State Avogadro's law.
Answer
Avogadro's law states that equal volumes of all gases under the same conditions of temperature and pressure contain equal numbers of molecules.
Mathematically: V ∝ n (at constant T and P), where V is volume and n is the number of moles.
VI. Answer in Detail (5-mark)
1. Explain the mole concept with worked numerical examples.
Answer
The mole is the SI unit of amount of substance. One mole contains exactly 6.022 × 10²³ elementary particles (atoms, molecules, ions, etc.), which is the number of atoms in 12 g of carbon-12.
Three ways to calculate moles:
(i) n = Mass (g) / Molar mass (g mol⁻¹)
(ii) n = Volume at STP (L) / 22.4 L (for gases)
(iii) n = Number of particles / 6.022 × 10²³
Worked Example 1: Calculate the number of atoms in 4.6 g of sodium (Na).
Step 1: Molar mass of Na = 23 g mol⁻¹
Step 2: Number of moles = 4.6 / 23 = 0.2 mol
Step 3: Number of atoms = 0.2 × 6.022 × 10²³
= 1.2044 × 10²³ atoms
Worked Example 2: Calculate the number of molecules in 11.2 L of CO₂ at STP.
Step 1: Moles of CO₂ = 11.2 / 22.4 = 0.5 mol
Step 2: Number of molecules = 0.5 × 6.022 × 10²³
= 3.011 × 10²³ molecules
Worked Example 3: Calculate the mass of 3.011 × 10²³ molecules of H₂O.
Step 1: Moles of H₂O = 3.011 × 10²³ / 6.022 × 10²³ = 0.5 mol
Step 2: Molar mass of H₂O = 18 g mol⁻¹
Step 3: Mass = 0.5 × 18 = 9 g
Worked Example 4: How many atoms are present in 0.1 mol of H₂SO₄?
Each H₂SO₄ molecule has 2 + 1 + 4 = 7 atoms
Total atoms = 0.1 × 6.022 × 10²³ × 7 = 4.2154 × 10²³ atoms
2. Explain stoichiometry and limiting reagent with worked examples.
Answer
Stoichiometry is the calculation of reactants and products in a chemical reaction using balanced equations. It is based on the law of conservation of mass.
Steps for stoichiometric calculations: 1. Write the balanced chemical equation. 2. Convert given quantities to moles. 3. Use mole ratios from the balanced equation. 4. Convert to the desired unit.
Worked Example 1: How many grams of CaO are formed when 20 g of CaCO₃ is heated?
Balanced equation: CaCO₃ → CaO + CO₂
Step 1: Molar mass of CaCO₃ = 40 + 12 + 48 = 100 g mol⁻¹
Step 2: Moles of CaCO₃ = 20 / 100 = 0.2 mol
Step 3: From the equation, 1 mol CaCO₃ → 1 mol CaO
So, 0.2 mol CaCO₃ → 0.2 mol CaO
Step 4: Mass of CaO = 0.2 × (40 + 16) = 0.2 × 56 = 11.2 g
Worked Example 2 (Limiting Reagent): 6 g of Mg reacts with 8 g of O₂. Find the mass of MgO formed.
Balanced equation: 2Mg + O₂ → 2MgO
Step 1: Moles of Mg = 6 / 24 = 0.25 mol
Moles of O₂ = 8 / 32 = 0.25 mol
Step 2: From equation, 2 mol Mg requires 1 mol O₂
So 0.25 mol Mg requires 0.125 mol O₂
But we have 0.25 mol O₂ — O₂ is in excess.
Mg is the limiting reagent.
Step 3: 2 mol Mg → 2 mol MgO
0.25 mol Mg → 0.25 mol MgO
Step 4: Molar mass of MgO = 24 + 16 = 40 g mol⁻¹
Mass of MgO = 0.25 × 40 = 10 g
Worked Example 3: 2 g of H₂ reacts with 16 g of O₂ to form water. Identify the limiting reagent and calculate the mass of water formed.
Balanced equation: 2H₂ + O₂ → 2H₂O
Step 1: Moles of H₂ = 2/2 = 1 mol
Moles of O₂ = 16/32 = 0.5 mol
Step 2: 1 mol H₂ requires 0.5 mol O₂ (mole ratio 2:1)
Available O₂ = 0.5 mol
Exactly stoichiometric — both consumed fully
H₂ is the limiting reagent (convention)
Step 3: 2 mol H₂ → 2 mol H₂O
1 mol H₂ → 1 mol H₂O
Step 4: Mass of H₂O = 1 × 18 = 18 g
3. Explain the determination of empirical and molecular formula with worked examples.
Answer
Empirical formula gives the simplest whole-number ratio of atoms. Molecular formula gives the actual number of atoms.
Steps to find empirical formula: 1. Assume 100 g of the compound; percentages become grams. 2. Convert grams to moles using atomic masses. 3. Divide each mole value by the smallest mole value. 4. If needed, multiply to get whole numbers.
Molecular formula = Empirical formula × n, where n = Molar mass / Empirical formula mass.
Worked Example 1: A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen. Its molar mass is 180 g mol⁻¹. Find the empirical and molecular formula.
Step 1: Assume 100 g → C = 40 g, H = 6.67 g, O = 53.33 g
Step 2: Convert to moles:
C: 40 / 12 = 3.33
H: 6.67 / 1 = 6.67
O: 53.33 / 16 = 3.33
Step 3: Divide by smallest (3.33):
C: 3.33 / 3.33 = 1
H: 6.67 / 3.33 = 2
O: 3.33 / 3.33 = 1
Step 4: Empirical formula = CH₂O
Empirical formula mass = 12 + 2 + 16 = 30
Step 5: n = 180 / 30 = 6
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (Glucose)
Worked Example 2: A compound contains 69% iron and 31% oxygen. Find its empirical formula.
Step 1: Fe = 69 g, O = 31 g
Step 2: Convert to moles:
Fe: 69 / 56 = 1.232
O: 31 / 16 = 1.9375
Step 3: Divide by smallest (1.232):
Fe: 1.232 / 1.232 = 1
O: 1.9375 / 1.232 = 1.57
Step 4: Multiply by 2 to get whole numbers:
Fe: 1 × 2 = 2
O: 1.57 × 2 = 3.14 ≈ 3
Empirical formula = Fe₂O₃ (Haematite)
Worked Example 3: Calculate the percentage composition of H₂SO₄.
Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 98 g mol⁻¹
%H = (2/98) × 100 = 2.04%
%S = (32/98) × 100 = 32.65%
%O = (64/98) × 100 = 65.31%
Check: 2.04 + 32.65 + 65.31 = 100.00%
Common Mistakes to Avoid
1. Confusing atoms and molecules: When calculating particles, remember H₂ has 2 atoms per molecule. Moles × N_A gives molecules; multiply by atoms per molecule for total atoms.
2. Forgetting to balance equations before stoichiometry: Always write the balanced equation first. Unbalanced equations give incorrect mole ratios.
3. Using mass ratios instead of mole ratios: Stoichiometric calculations must use mole ratios from the balanced equation, not mass ratios.
4. Mixing up empirical and molecular formula: Empirical = simplest ratio; molecular = actual formula. They are the same only when n = 1.
5. Ignoring STP conditions for gas volume: The 22.4 L/mol value only applies at STP (0 °C, 1 atm). At other conditions, use PV = nRT.
6. Significant figure errors in calculations: Follow sig-fig rules — multiplication/division uses the least number of significant figures; addition/subtraction uses the least decimal places.
7. Incorrect limiting reagent identification: Calculate how much product each reactant can form; the one producing less product is limiting. Do not simply compare masses.
8. Forgetting units in calculations: Always include units (g, mol, L) at every step. This helps catch errors and ensures dimensional consistency.
Exam Checklist
- [ ] Can define mole, Avogadro's number, and molar mass
- [ ] Can convert between mass, moles, and number of particles
- [ ] Can use the formula n = mass/molar mass, n = volume/22.4, n = N/N_A
- [ ] Can balance chemical equations correctly
- [ ] Can perform stoichiometric calculations using mole ratios
- [ ] Can identify the limiting reagent in a reaction
- [ ] Can calculate percentage composition of a compound
- [ ] Can determine empirical formula from percentage data
- [ ] Can determine molecular formula from empirical formula and molar mass
- [ ] Can apply significant figure rules in numerical problems
