Chemistry Ch.15 – Molar Volume, Concentration & Stoichiometry | EverExams.com
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Chemistry · Chapter 15

Molar Volume, Concentration & Stoichiometry

Complete exercise solutions — MCQ Quiz, Short Questions, Constructed Response & Numericals with full working.

Molar Volume at RTP Empirical Formula Limiting Reactant Percentage Yield Molar Concentration
Before You Start
Student Study Guidelines

📌 How to Study This Chapter Effectively

1
Memorise the molar volume at RTP (24 dm³/mol). Every gas volume question uses this single constant — it is your starting point.
2
Always convert volumes to dm³ and volumes in cm³ to dm³ by dividing by 1000 before substituting into formulas.
3
For empirical and molecular formulas: divide each element’s percentage by its atomic mass, find the smallest ratio, then multiply by n = Molar Mass / Empirical Mass.
4
To find the limiting reactant: compute moles of each reactant, divide by their stoichiometric coefficients, and compare — the smallest value points to the limiting reactant.
5
Percentage yield = (Actual Yield / Theoretical Yield) × 100. Theoretical yield is always calculated from the limiting reactant only.
6
In titration calculations, always write the balanced equation first and use the mole ratio to bridge the two reactants.
7
Attempt the MCQ quiz below with the answer hidden, then check your score. Study every explanation for questions you get wrong before moving to the numerical problems.
Section A

Multiple Choice Quiz

Choose the correct answer. Click your choice, then press Submit. Results and answer key appear below.

⚠️ Answer all 8 questions before clicking Submit. Your selected answer turns green if correct, red if wrong. The correct option is always highlighted.
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📋 Answer Key
Section B
Short Answer Questions
15.1What is the molar volume of a gas at RTP?
The molar volume of any gas at Room Temperature and Pressure (RTP — 20°C and 1 atm) is 24 dm³/mol. This value applies to all ideal gases regardless of their identity.
15.2How is the concept of molar volume useful?
It allows chemists to convert between the volume of a gas and the number of moles without needing to measure mass directly. This is especially useful in stoichiometric calculations for gaseous reactions, where volume is easier to measure than mass.
15.3How does molar volume relate to Avogadro’s Law?
Avogadro’s Law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. Molar volume is the direct application of this law — it is the volume occupied by exactly one mole (6.02 × 10²³ molecules) of any gas at a given temperature and pressure.
15.4Why may two different compounds have the same empirical formula?
Because empirical formulas represent only the simplest whole-number ratio of atoms. Two compounds can share the same ratio but have different actual numbers of atoms in their molecules. For example, ethene (C₂H₄) and propene (C₃H₆) both have the empirical formula CH₂, yet they are entirely different compounds.
15.5Why is percentage yield important?
Percentage yield measures the efficiency of a chemical reaction. It helps compare actual production with the theoretically possible amount, identify losses in a process, and evaluate cost-effectiveness. In industrial chemistry, even a small improvement in yield can result in enormous economic savings.
15.6What is the concentration of a solution containing 49 g of H₂SO₄ in 1 dm³?
Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 98 g/mol. Moles = 49 ÷ 98 = 0.5 mol. Concentration = 0.5 mol ÷ 1 dm³ = 0.5 mol/dm³.
Section C
Constructed Response Questions
15.1How would you identify the limiting reactant in a chemical reaction?
First, write the balanced chemical equation for the reaction. Then calculate the number of moles of each reactant using the formula: moles = mass ÷ molar mass. Divide the moles of each reactant by its stoichiometric coefficient from the balanced equation. The reactant that gives the smallest quotient is the limiting reactant — it will be completely consumed first and determines the maximum yield of product.
15.2Why is it important to find the purity of a compound used as a medicine?
The purity of a medicinal compound is critical because impurities can cause toxic side effects, trigger allergic reactions, or interact negatively with other drugs. Additionally, the dosage prescribed by a doctor assumes a certain purity — if the active ingredient is diluted by impurities, the patient receives less medicine than intended, reducing effectiveness. Regulatory bodies such as WHO set strict purity standards for pharmaceutical compounds to ensure patient safety.
15.3Will there be a limiting reactant in a reversible reaction?
Yes, the concept of a limiting reactant still applies to reversible reactions at the start of the reaction. However, because the reaction does not go to completion — it reaches a state of dynamic equilibrium — both reactants and products remain present at equilibrium. Therefore, while one reactant is initially “limiting,” neither is entirely consumed. The limiting reactant concept applies strictly in the context of irreversible or complete reactions.
15.4How do you know if a formula is empirical or not?
To check whether a formula is empirical, find the greatest common divisor (GCD) of all the subscripts in the formula. If the GCD is 1 (i.e., no subscript can be divided further), the formula is empirical. If the GCD is greater than 1, the formula is molecular and can be simplified — for example, C₆H₁₂O₆ has GCD 6, so its empirical formula is CH₂O.
15.5Write one method to find the molecular mass of a compound without knowing its molecular formula.
Mass spectrometry is a direct method — the compound is vaporised, ionised, and the mass-to-charge ratio of the molecular ion peak (M⁺) gives the molar mass. Alternatively, colligative property methods such as freezing point depression or boiling point elevation can be used: by dissolving a known mass of the compound in a solvent and measuring the change in freezing or boiling point, the molar mass can be calculated using the formula: M = (mass of solute × K) ÷ (change in temperature × mass of solvent).
Section D
Numerical Problems — Worked Solutions
15.1 Calculate the moles, volume and molecules in 4.0 g of CH₄ at RTP.
Formula:Moles = mass ÷ molar mass  |  Volume = moles × 24  |  Molecules = moles × 6.02 × 10²³
Molar mass:CH₄ = 12 + 4(1) = 16 g/mol
Moles:4.0 ÷ 16 = 0.25 mol
Volume:0.25 × 24 = 6 dm³
Molecules:0.25 × 6.02 × 10²³ = 1.505 × 10²³ molecules
Volume = 6 dm³ | Molecules ≈ 1.505 × 10²³
15.2 Mass of NaCl for 150 cm³ of 0.4 mol/dm³ solution?
Volume:150 cm³ = 150 ÷ 1000 = 0.15 dm³
Moles:concentration × volume = 0.4 × 0.15 = 0.06 mol
Molar mass:NaCl = 23 + 35.5 = 58.5 g/mol
Mass:0.06 × 58.5 = 3.51 g
Mass of NaCl = 3.51 g
15.3 Find the percentage of sodium in NaHCO₃.
Molar mass:Na + H + C + 3O = 23 + 1 + 12 + 48 = 84 g/mol
Mass of Na:23 g
% Na:(23 ÷ 84) × 100 = 27.38%
% Sodium = 27.38%
15.4 Empirical formula of iron sulphide: 1.926 g S and 2.333 g Fe.
Moles Fe:2.333 ÷ 55.85 = 0.04178 mol
Moles S:1.926 ÷ 32.06 = 0.06007 mol
Ratio:Fe: 0.04178 ÷ 0.04178 = 1  |  S: 0.06007 ÷ 0.04178 ≈ 1.44
Multiply × 3:Fe = 3, S = 4 → Empirical formula = Fe₃S₄
Empirical Formula: Fe₃S₄
15.5 Molecular formula: 40% C, 6.71% H, 53.3% O; 0.320 mol weighs 28.8 g.
Molar mass:28.8 ÷ 0.320 = 90 g/mol
Moles (per 100g):C: 40 ÷ 12 = 3.33  |  H: 6.71 ÷ 1 = 6.71  |  O: 53.3 ÷ 16 = 3.33
Simplest ratio:Divide by 3.33 → C:1, H:2, O:1 → Empirical formula = CH₂O (mass = 30)
n:90 ÷ 30 = 3
Molecular:(CH₂O)₃ = C₃H₆O₃
Molecular Formula: C₃H₆O₃
15.6 3H₂ + N₂ → 2NH₃. Start with 12 g H₂ and 64 g N₂. Find limiting reactant.
Moles H₂:12 ÷ 2 = 6 mol
Moles N₂:64 ÷ 28 = 2.286 mol
H₂ needed:To consume all N₂: 2.286 × 3 = 6.857 mol H₂ required, but only 6 mol available.
Conclusion:H₂ runs out first → H₂ is the limiting reactant
Limiting Reactant: H₂
15.7 305 g AgNO₃ + excess MgCl₂ → 23.7 g Mg(NO₃)₂. Find % yield.
Equation:2AgNO₃(aq) + MgCl₂(aq) → Mg(NO₃)₂(aq) + 2AgCl(s)
Molar mass:AgNO₃ = 108 + 14 + 48 = 170 g/mol
Moles AgNO₃:305 ÷ 170 = 1.794 mol
Moles Mg(NO₃)₂:1.794 ÷ 2 = 0.897 mol (from 2:1 ratio)
Molar mass:Mg(NO₃)₂ = 24 + 2(14 + 48) = 148 g/mol
Theoretical:0.897 × 148 = 132.8 g
% Yield:(23.7 ÷ 132.8) × 100 = ≈ 17.85%
Percentage Yield ≈ 17.85%