10th Class Chemistry Chapter 2 Notes: Stoichiometry (Punjab Board)

Chapter at a glance

In the new Punjab textbook this is Chapter 2 of the 10th class book. Some websites number the 10th class chapters after the 9th class chapters, so you may also see it listed as Chapter 15. The topics are the same.

Key concepts and definitions

• Stoichiometry is the study of the quantitative relationship between reactants and products in a chemical reaction.
• Mole: the amount of substance that contains 6.022 × 10²³ particles (atoms, molecules or ions). This number is called Avogadro's number, written Nₐ = 6.022 × 10²³ mol⁻¹.
• Molar mass is the mass of one mole of a substance, in g mol⁻¹. It has the same number as the relative atomic or formula mass.
• Molar volume: the volume occupied by one mole of any gas. At room temperature and pressure (RTP, 25 °C and 1 atm) it is 24 dm³. At standard temperature and pressure (STP, 0 °C and 1 atm) it is 22.4 dm³.
• Mass concentration is the mass of solute in one dm³ of solution (g dm⁻³). Molar concentration (molarity) is the number of moles of solute in one dm³ of solution (mol dm⁻³).
• Titration: a method in which a solution of known concentration is added to a measured volume of another solution until the reaction between them is exactly complete.
• Empirical formula: the simplest whole-number ratio of the atoms of each element in a compound. Molecular formula: the actual number of atoms of each element in one molecule.
• Limiting reactant: the reactant that is completely used up first and so limits the amount of product. The other reactant is in excess.
• Theoretical yield is the maximum product calculated from the equation. Actual yield is what is really obtained.

Formulas with symbols and units

Methods and derivations step by step

Conversion between mass concentration and molarity

1. Start with mass concentration ρ = m ÷ V, in g dm⁻³.
2. The number of moles in the mass m is n = m ÷ M.
3. Molarity is c = n ÷ V = (m ÷ M) ÷ V = (m ÷ V) ÷ M.
4. So c = ρ ÷ M, and in the other direction ρ = c × M.

Titration formula

1. For any solution, n = c × V, so the moles of solute are concentration times volume in dm³.
2. For a reaction a A + b B, the moles used are in the ratio n(A) : n(B) = a : b, so n(A) ÷ a = n(B) ÷ b.
3. Putting n = cV gives (c(A) × V(A)) ÷ a = (c(B) × V(B)) ÷ b. For a 1 : 1 reaction like HCl + NaOH this becomes c(A) × V(A) = c(B) × V(B).

Empirical formula

1. Write the percentage of each element as grams in 100 g of compound.
2. Divide each mass by the atomic mass to get moles.
3. Divide all the moles by the smallest number.
4. If a ratio is near 1.5, multiply all ratios by 2; if near 1.33 or 1.67, multiply by 3. This gives whole numbers.

For the molecular formula, the mass of the empirical formula unit is found first. Then n = molecular mass ÷ that mass, and every subscript is multiplied by n.

Mass from a balanced equation

1. Write and balance the equation.
2. Convert the given mass to moles.
3. Use the mole ratio from the coefficients.
4. Convert the moles of the required substance into mass or volume.

Example: CaCO₃ → CaO + CO₂. For 50 g of CaCO₃ (M = 100 g mol⁻¹) the moles are 0.5. The ratio is 1 : 1 : 1, so CaO = 0.5 mol × 56 = 28 g and CO₂ = 0.5 × 44 = 22 g.

Percentage composition example

In water, H₂O (M = 18), hydrogen is 2 ÷ 18 × 100 = 11.1% and oxygen is 16 ÷ 18 × 100 = 88.9%. The two add up to 100%.

Solved numericals

Atomic masses used: H = 1, C = 12, O = 16, Na = 23, Mg = 24, Cl = 35.5, Ca = 40. Molar volume at RTP = 24 dm³ mol⁻¹.

Numerical 1: moles, molecules and volume

For 11 g of CO₂: M = 12 + 2 × 16 = 44 g mol⁻¹. Moles = 11 ÷ 44 = 0.25 mol. Molecules = 0.25 × 6.022 × 10²³ = 1.51 × 10²³. Volume at RTP = 0.25 × 24 = 6 dm³.

Numerical 2: molarity

4.0 g of NaOH is dissolved to make 500 cm³ of solution. M = 23 + 16 + 1 = 40 g mol⁻¹. n = 4.0 ÷ 40 = 0.10 mol. V = 500 ÷ 1000 = 0.50 dm³. Molarity = 0.10 ÷ 0.50 = 0.20 mol dm⁻³. Mass concentration = 4.0 ÷ 0.50 = 8 g dm⁻³, which matches 0.20 × 40 = 8.

Numerical 3: titration

25.0 cm³ of NaOH solution is neutralised exactly by 20.0 cm³ of 0.10 mol dm⁻³ HCl. HCl + NaOH → NaCl + H₂O (1 : 1). Moles of HCl = 0.10 × 0.0200 = 0.0020 mol, so moles of NaOH = 0.0020 mol. Concentration of NaOH = 0.0020 ÷ 0.0250 = 0.080 mol dm⁻³, or 0.080 × 40 = 3.2 g dm⁻³.

Numerical 4: empirical and molecular formula

A compound contains 40.0% C, 6.7% H and 53.3% O. Its molar mass is 180 g mol⁻¹. Moles in 100 g: C = 40.0 ÷ 12 = 3.33; H = 6.7 ÷ 1 = 6.7; O = 53.3 ÷ 16 = 3.33. Dividing by 3.33 gives C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O. Its mass is 12 + 2 + 16 = 30. n = 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆.

Numerical 5: limiting reactant and percentage yield

4.8 g of magnesium reacts with 7.3 g of HCl: Mg + 2HCl → MgCl₂ + H₂. Moles of Mg = 4.8 ÷ 24 = 0.20. Moles of HCl = 7.3 ÷ 36.5 = 0.20. The equation needs 2 mol HCl for 1 mol Mg. Dividing by coefficients gives Mg = 0.20 and HCl = 0.20 ÷ 2 = 0.10, so HCl is the limiting reactant. Hydrogen formed = 0.10 mol = 0.2 g, and its volume at RTP is 0.10 × 24 = 2.4 dm³. If 2.0 dm³ is collected, percentage yield = 2.0 ÷ 2.4 × 100 = 83.3%.

Common mistakes

• Using an equation that is not balanced, so the mole ratio is wrong.
• Forgetting to change cm³ into dm³ (divide by 1000) before using c = n ÷ V.
• Calling the reactant with the smaller mass the limiting reactant. Compare moles divided by coefficients instead.
• Using 22.4 dm³ when the question gives room temperature and pressure, or the opposite.
• Rounding an empirical ratio like 1.5 to 2 instead of multiplying everything by 2.
• Adding atomic masses wrongly for formulas with brackets, such as Ca(OH)₂ (40 + 2 × 17 = 74).
• Getting a percentage yield above 100%, which is a sign of an error in the theoretical yield.
• Leaving out units in the final answer.

Questions to practise

Short questions

1. Define mole and Avogadro's number.
2. What is molar mass? Find the molar mass of Ca(OH)₂.
3. Define molar volume. State its value at RTP.
4. Differentiate between mass concentration and molar concentration.
5. What is a titration? Why is it used?
6. Define empirical formula. Why is the empirical formula of glucose different from its molecular formula?
7. What is a limiting reactant? Give an everyday example.
8. Differentiate between theoretical yield and actual yield.
9. Why is the percentage yield of a reaction usually less than 100%?
10. Define percentage purity.

Long questions

1. Explain how the empirical formula of a compound is found from percentage composition. Give a worked example.
2. Describe how an unknown concentration is found by titration and derive the formula used.
3. Explain limiting reactant and percentage yield with a numerical based on a balanced equation.
4. Show how mass, moles, number of particles and gas volume are related to each other.

Numericals to try

1. Find the number of moles and molecules in 36 g of water.
2. How many grams of NaCl are needed for 250 cm³ of a 0.20 mol dm⁻³ solution?
3. A compound has 85.7% carbon and 14.3% hydrogen and a molar mass of 56 g mol⁻¹. Find its molecular formula.
4. Find the volume of CO₂ at RTP obtained from 10 g of CaCO₃ with excess acid.
5. A sample of 20 g of impure marble contains 16 g of CaCO₃. Find its percentage purity.

Practice MCQs

Try these first. The answer key is at the end of the article.

1. The number of particles in one mole of a substance is
   A) 6.022 × 10²²
   B) 6.022 × 10²³
   C) 6.022 × 10²⁴
   D) 3.011 × 10²³
2. The molar mass of H₂SO₄ (H = 1, S = 32, O = 16) is
   A) 49 g mol⁻¹
   B) 64 g mol⁻¹
   C) 98 g mol⁻¹
   D) 196 g mol⁻¹
3. Taking the molar volume as 24 dm³ at room temperature and pressure, the volume of 2 mol of a gas is
   A) 12 dm³
   B) 24 dm³
   C) 36 dm³
   D) 48 dm³
4. The empirical formula of acetic acid, C₂H₄O₂, is
   A) CH₂O
   B) CHO
   C) C₂H₂O
   D) CH₄O
5. The reactant that is used up first in a reaction is called the
   A) excess reactant
   B) limiting reactant
   C) catalyst
   D) product
6. A reaction should give 10 g of a product but 8 g is obtained. The percentage yield is
   A) 8%
   B) 20%
   C) 80%
   D) 125%
7. A solution of NaOH has a molar concentration of 0.5 mol dm⁻³. Its mass concentration is (Na = 23, O = 16, H = 1)
   A) 10 g dm⁻³
   B) 20 g dm⁻³
   C) 40 g dm⁻³
   D) 80 g dm⁻³
8. The unit of molar concentration is
   A) g dm⁻³
   B) mol
   C) mol dm⁻³
   D) dm³ mol⁻¹

Preparation strategy

1. Keep one page with all the formulas above and the atomic masses used in your textbook.
2. Learn the four-step method for balanced equations and solve two problems each day for a week.
3. Practise empirical formula, titration and limiting reactant separately, because each has its own steps.
4. Write units in every line of working. It prevents most mistakes.
5. Finish with the chapter MCQs and check which formula you used wrongly.

See the 10th class Chemistry textbook guide and our 10th class Chemistry pairing scheme for the paper pattern.

Diagram ideas

• Mole map. Alt text: "Diagram linking mass, moles, number of particles and gas volume with the formulas used for each step."
• Titration set-up. Alt text: "Burette filled with acid above a conical flask containing alkali and an indicator, on a white tile."
• Limiting reactant picture. Alt text: "Drawing of reactant particles where one type runs out first and the other is left over."
• Empirical formula flow chart. Alt text: "Flow chart from percentage to mass, to moles, to ratio, to empirical formula."
• Gas volume chart. Alt text: "Table comparing the volume of one mole of gas at room temperature and pressure and at standard temperature and pressure."

MCQ answer key

1. B. Avogadro's number is 6.022 × 10²³ per mole.
2. C. 2 × 1 + 32 + 4 × 16 = 2 + 32 + 64 = 98 g mol⁻¹.
3. D. 2 mol × 24 dm³ mol⁻¹ = 48 dm³.
4. A. C₂H₄O₂ divided by 2 gives the simplest ratio CH₂O.
5. B. The limiting reactant is consumed first and limits the product.
6. C. (8 ÷ 10) × 100 = 80%.
7. B. Molar mass of NaOH = 40 g mol⁻¹, so 0.5 × 40 = 20 g dm⁻³.
8. C. Moles per cubic decimetre, mol dm⁻³.

Frequently asked questions