Mole concept, composition and concentration expressions

Moles and molar mass, mass and mole fractions, average molar mass, choosing a basis, and solution concentrations (molarity, molality, normality, ppm).

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Why it matters

Reactions happen molecule by molecule, so chemical engineers balance reactors in moles, while plants buy, sell and weigh material in kilograms. Moving cleanly between mass, moles and the many concentration measures (mass %, mole %, molarity, molality, ppm) is required in every material balance, vapour-liquid equilibrium and reactor calculation.

Key ideas

The mole. One mole contains exactly 6.022 140 76 × 10²³ specified entities (Avogadro constant N_A, fixed by the 2019 SI redefinition). The molar mass M (g/mol, numerically equal to kg/kmol) converts mass to moles: n = m / M. In process work the kmol is most convenient because kg / (kg/kmol) = kmol.

Average molar mass of a mixture. A mixture has no single formula, but its mean molar mass is the mole-fraction-weighted average of the component molar masses. Dry air (79 % N₂, 21 % O₂) has M_avg ≈ 29 kg/kmol.

Fractions. Mass fraction w_i = m_i / Σm and mole fraction x_i = n_i / Σn are dimensionless and each set sums to 1. For gases, y is used for mole fraction; for an ideal gas, mole fraction = volume fraction = pressure fraction, which is why gas analyses are reported in mole (volume) %. Liquids and solids are usually reported on a mass basis.

Basis of calculation. Before converting, choose a convenient basis: 100 kg if mass % is given, 100 kmol if mole % is given. All results are then scaled to the real flow.

Concentration measures for solutions.

  • Molarity C (mol/L of solution): depends on temperature because volume does.
  • Molality m (mol/kg of solvent): independent of temperature; used for colligative properties.
  • Normality N (gram-equivalents/L of solution) = molarity × number of equivalents per mole (e.g. 2 for H₂SO₄ in full neutralisation).
  • Mass concentration (kg/m³ or g/L).
  • ppm and ppb: by mass for liquids (1 ppm = 1 mg/kg ≈ 1 mg/L for dilute aqueous solutions), by mole or volume for gases (ppmv).

Dry vs wet basis. Gas analyses (Orsat) are often on a dry basis, excluding water vapour; convert using the water mole fraction before combining with wet-basis data.

Density and specific gravity. Specific gravity SG = ρ / ρ_ref (water at 4 °C, 1000 kg/m³, for liquids). Density links mass and volume, and is needed to move between molarity and molality or mass fraction.

Formulas

  • n = m / M – n amount (kmol), m mass (kg), M molar mass (kg/kmol).
  • x_i = n_i / Σ n_j ; w_i = m_i / Σ m_j – dimensionless; Σx = Σw = 1.
  • w_i = x_i·M_i / Σ(x_j·M_j) – mole to mass fraction.
  • x_i = (w_i / M_i) / Σ(w_j / M_j) – mass to mole fraction.
  • M_avg = Σ x_i·M_i = 1 / Σ(w_i / M_i) – kg/kmol.
  • C = n_solute / V_solution – mol/L (mol/dm³) = kmol/m³.
  • m = n_solute / m_solvent – mol/kg.
  • N = C × (equivalents per mole) – eq/L.
  • ppm (mass) = (m_i / m_total) × 10⁶ ; ppmv (gas) = y_i × 10⁶.
  • C = 1000·ρ·w / M – C in mol/L, ρ in g/cm³, w mass fraction, M in g/mol (links molarity and mass fraction).

Worked examples

Example 1 (standard). An aqueous NaOH solution contains 20 % NaOH by mass and has a density of 1.219 g/cm³. Find the mole fraction of NaOH, the molarity and the molality. M(NaOH) = 40.0 g/mol, M(H₂O) = 18.015 g/mol.

  1. Basis: 1000 g of solution → 200 g NaOH and 800 g water.
  2. Moles: n_NaOH = 200 / 40.0 = 5.00 mol; n_water = 800 / 18.015 = 44.41 mol.
  3. Mole fraction: x_NaOH = 5.00 / (5.00 + 44.41) = 0.1012.
  4. Volume of 1000 g of solution: V = 1000 g / 1.219 g/cm³ = 820.3 cm³ = 0.8203 L.
  5. Molarity: C = 5.00 mol / 0.8203 L = 6.10 mol/L.
  6. Molality: m = 5.00 mol / 0.800 kg = 6.25 mol/kg.

Answer: x = 0.101; C = 6.10 mol/L; m = 6.25 mol/kg

Example 2 (GATE level). A natural gas contains 80 % CH₄, 15 % C₂H₆ and 5 % N₂ by mole. Find (a) the average molar mass, (b) the mass fractions and (c) the kmol in 100 kg of gas. M: CH₄ 16.04, C₂H₆ 30.07, N₂ 28.01 kg/kmol.

  1. Basis 1 kmol of gas. Masses: CH₄ 0.80 × 16.04 = 12.83 kg; C₂H₆ 0.15 × 30.07 = 4.51 kg; N₂ 0.05 × 28.01 = 1.40 kg.
  2. M_avg = Σ x_i·M_i = 12.83 + 4.51 + 1.40 = 18.74 kg/kmol.
  3. Mass fractions: CH₄ 12.83 / 18.74 = 0.685; C₂H₆ 4.51 / 18.74 = 0.241; N₂ 1.40 / 18.74 = 0.075 (sum 1.000).
  4. Moles in 100 kg: n = 100 kg / 18.74 kg/kmol = 5.34 kmol.

Answer: M_avg = 18.74 kg/kmol; w = 0.685, 0.241, 0.075; 5.34 kmol per 100 kg

Common mistakes

  • Averaging molar masses with mass fractions as weights (M_avg = Σ w_i·M_i is wrong; use 1 / Σ(w_i / M_i)).
  • Dividing by litres of solvent instead of litres of solution for molarity, or by kg of solution instead of kg of solvent for molality.
  • Assuming 1 L of solution weighs 1 kg for concentrated solutions – use the given density.
  • Mixing dry-basis and wet-basis gas analyses.
  • Treating ppm in a gas as mass-based when it is reported by volume.
  • Forgetting that a mole % for gases is also the volume % only for ideal gases.

For GATE CH

Composition conversions rarely appear alone; they are the first step of material-balance, VLE and combustion numericals. Expect NAT questions asking for average molar mass, mass or mole fractions from a given analysis, and conversions involving density (molarity ↔ mass %). Practise always fixing a basis first and checking that fractions sum to one.

Quick check

  1. What is the mole fraction of ethanol in a mixture of 46 g ethanol (46 g/mol) and 54 g water (18 g/mol)?
  2. What is the average molar mass of dry air with 79 mol % N₂ (28) and 21 mol % O₂ (32)?
  3. Which concentration measure does not change with temperature: molarity or molality?
  4. A 0.5 mol/L H₂SO₄ solution has what normality (full neutralisation)?
  5. How many kmol are in 88 kg of CO₂ (M = 44)?

Answers: 1. 0.25; 2. 28.84 kg/kmol; 3. molality; 4. 1.0 N; 5. 2 kmol.

Interview questions

All Process Calculations interview questions

Try answering each one aloud before you open it.

  1. 1.What is the mole concept in chemistry?Concept

    A mole is the amount of substance containing exactly 6.022 140 76 × 10²³ specified entities (the Avogadro constant, fixed by the 2019 SI redefinition; earlier it was defined via 12 g of carbon-12). The molar mass M links mass to amount, n = m / M, and is numerically the same in g/mol and kg/kmol. Engineers use moles because reaction stoichiometry is written in moles, not masses.

  2. 2.Explain the difference between molarity and molality.Concept

    Molarity is the concentration of a solution expressed as the number of moles of solute per liter of solution. Molality, on the other hand, is the concentration expressed as the number of moles of solute per kilogram of solvent. Molarity depends on the volume of the solution, which can change with temperature, while molality depends on the mass of the solvent, which remains constant with temperature.

  3. 3.Why is the mole concept important in chemical engineering?Application

    The mole concept is crucial in chemical engineering because it allows engineers to quantify and balance chemical reactions. It helps in calculating the amounts of reactants and products involved in a reaction, which is essential for designing reactors, optimizing processes, and ensuring safety and efficiency in chemical production.

  4. 4.What happens to the molarity of a solution if the temperature increases?Application

    If the temperature increases, the volume of the solution typically expands, leading to a decrease in molarity. This is because molarity is defined as moles of solute per liter of solution, and an increase in volume results in a lower concentration of solute per unit volume.

  5. 5.How do you calculate the mole fraction of a component in a mixture?Concept

    The mole fraction of a component in a mixture is calculated by dividing the number of moles of that component by the total number of moles of all components in the mixture. It is a dimensionless quantity and is often used in thermodynamic calculations.

  6. 6.Explain why molality is preferred over molarity in certain calculations.Application

    Molality is preferred over molarity in calculations involving temperature changes because it is independent of temperature. Since molality is based on the mass of the solvent, which does not change with temperature, it provides a more consistent measure of concentration in scenarios where temperature fluctuations occur.

  7. 7.What is the significance of Avogadro's number in the mole concept?Concept

    Avogadro's number, approximately 6.022 × 10²³, is significant because it provides a link between the macroscopic scale of substances we can measure and the microscopic scale of atoms and molecules. It allows chemists and engineers to count particles by weighing them, facilitating the calculation of reactants and products in chemical reactions.

  8. 8.Calculate the molarity of a solution containing 5 moles of solute in 2 liters of solution.Numerical

    Molarity (M) is calculated using the formula: M = moles of solute / liters of solution. Here, M = 5 moles / 2 liters = 2.5 M. Therefore, the molarity of the solution is 2.5 moles per liter.

  9. 9.If 10 grams of NaCl is dissolved in 500 grams of water, calculate the molality of the solution. (Molar mass of NaCl = 58.44 g/mol)Numerical

    First, calculate the moles of NaCl: moles = mass / molar mass = 10 g / 58.44 g/mol = 0.171 moles. Molality (m) is moles of solute per kilogram of solvent: m = 0.171 moles / 0.5 kg = 0.342 mol/kg. Therefore, the molality of the solution is 0.342 mol/kg.

  10. 10.What is the role of mole fraction in Raoult's Law?Application

    In Raoult's Law, the mole fraction of a component in a liquid mixture is used to determine the partial vapor pressure of that component. According to Raoult's Law, the partial vapor pressure of a component is equal to the mole fraction of the component in the liquid phase multiplied by the vapor pressure of the pure component. This relationship is crucial for understanding and predicting the behavior of solutions in equilibrium.

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