The Mole Concept 🌌

The mole is a fundamental concept in chemistry for quantifying substances. It represents the amount of a substance containing as many elementary entities as atoms in 12.00 grams of carbon-12. ⚛️

Symbol: Ar

Definition: The relative atomic mass of an element is the mass of one atom compared to... 🧪

Relative Atomic Masses of Some Elements 🌍

Definition: The relative molecular mass of a compound is the mass of one molecule compared with... 🧫

Examples 📝

Calculate the relative molecular mass, Mr, for the following:

• Hydrogen Chloride (HCl): (1 x 1) + (1 x 35.5) = 36.5
• Carbon Dioxide (CO₂): (1 x 12) + (2 x 16) = 44

• Calculate the relative molecular mass of water, H₂O. 💧
• Calculate the mass of 2.5 moles of NaCl. 🧂

Symbol: M_r

Definition: The relative molecular mass of a compound is the mass of one molecule of the compound relative to one-twelfth of the mass of a carbon-12 atom. It can also be calculated as the sum of the relative atomic masses of its elements.

Examples

• Hydrogen chloride (HCl): M_r = (1 × 1) + (1 × 35.5) = 36.5
• Carbon dioxide (CO₂): M_r = (1 × 12) + (2 × 16) = 44
• Sodium sulphate (Na₂SO₄): M_r = (2 × 23) + (1 × 32) + (4 × 16) = 142
• Copper (II) sulphate pentahydrate (CuSO₄·5H₂O): M_r = (1 × 64) + (1 × 32) + (4 × 16) + (5 × 18) = 250

Exercise

• Calculate the relative molecular mass of magnesium oxide (MgO).
• Determine the molecular mass of glucose (C₆H₁₂O₆).

Symbol: MM

Unit: gram per mole, g/mol

Definition: Molar mass is the mass of one mole of a substance. It can also be defined as the relative molecular mass expressed in grams per mole. 📏

Examples 🧪

Symbol: n

Unit: mole (mol)

Definition: The mole is the amount of substance that contains as many entities as atoms in 12.00g of carbon-12. ⚛️

Formulas 📏

Number of Moles: n = m / MM (where n = moles, m = mass, MM = molar mass)

Examples 📝

• Example 1: How many moles of potassium are in 3.9g of potassium? 🔍

  Given: MM of potassium = 39 g/mol

  Solution: n = m / MM = 3.9g / 39g/mol = 0.1 mol

• Example 2: Find the mass of 0.2 moles of ammonia (NH₃). 💧

  Given: MM of NH₃ = 17 g/mol

  Solution: m = n x MM = 0.2 mol x 17 g/mol = 3.4 g

Symbol: N_A (mole^-1)

Definition: Avogadro’s number is the number of particles in exactly one mole of a pure substance, which is approximately 6.02 x 10²³. 🌌

Examples 📝

• Example 1: How many atoms of iron (Fe) are there in 25g of iron? 🔍

  Given: Atomic mass of Fe = 56 g/mol

  Solution: 25g Fe → 25 / 56 mol → 0.446 mol → 0.446 x 6.02 x 10²³ atoms

Definition: Concentration is the amount of solute dissolved in a unit volume of solution. 🧪

Formulas 📏

Concentration (g/dm³): C = m / V

Molarity (mol/dm³): M = n / V

Examples 📝

• Example 1: A solution contains 0.45g of glucose in 0.075 dm³. Find the concentration. 🍬

  Solution: C = 0.45g / 0.075 dm³ = 6 g/dm³

In chemistry, various calculations can be made using moles, including determining amounts in mass, volume, and particles for reactions. ⚗️

Mole to Mole Calculations 🔄

Example: How many moles of chlorine are required to react with 2.5 moles of calcium to produce calcium chloride? 🔍

Reaction: Ca(s) + Cl₂(g) → CaCl₂(s)

Solution:

1 mole of Ca reacts with 1 mole of Cl₂. Therefore, for 2.5 moles of Ca, you will need 2.5 moles of Cl₂. ⚗️

Mole to Mass Calculations ⚖️

Example: What mass of hydrogen can be produced by reacting 6 moles of aluminum with hydrochloric acid? 🔍

Reaction: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

Solution:

6 moles of Al produces 3 moles of H₂. Since 1 mole of H₂ weighs 2g, the mass of H₂ produced is 3 x 2g = 6g. ⚗️

Mass to Mass Calculations ⚖️

Example: Calculate the mass of calcium chloride produced when 40g of calcium carbonate reacts with hydrochloric acid. 🔍

Reaction: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

Solution:

100g of CaCO₃ produces 111g of CaCl₂. Therefore, 40g of CaCO₃ will produce (40g / 100g) * 111g = 44.4g of CaCl₂. ⚗️

Definition: The limiting reagent is the reactant that is completely used up first, limiting the extent of the reaction and the amount of product formed. ⚗️

Note: Identifying the limiting reagent is important for calculating the theoretical yield in a chemical reaction. 📊

Example 🔍

In a reaction where 19.5g of zinc reacts with 9.4g of sulfur:

• Zinc (Zn) has 0.3 mol, and sulfur (S) has 0.29 mol.
• Since sulfur has fewer moles, it is the limiting reagent. ⚠️

Avogadro’s Law states that the volume of a gas is directly proportional to the number of moles of gas molecules present, at constant temperature and pressure. 🌡️

Formula 📏

Volume of Gas at RTP (room temperature and pressure): V = n × 24dm³/mol

Volume of Gas at STP (standard temperature and pressure): V = n × 22.4dm³/mol

Example 🔍

Calculate the number of moles of carbon dioxide in 240cm³ of gas at RTP:

Solution:

n = V / Vm = 240cm³ / 24000cm³/mol = 0.01 mol

Definition: Percentage yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage. 📈

Formula: Percentage Yield = (Actual Yield / Theoretical Yield) × 100%

Example 🔍

When 36.8g of benzene (C₆H₆) reacts with chlorine to form chlorobenzene (C₆H₅Cl), the actual yield is 38.8g. Calculate the percentage yield. 📊

Solution:

The theoretical yield is 53.08g of C₆H₅Cl. Therefore:

Percentage Yield = (38.8g / 53.08g) × 100% = 73.1%

Definition: Concentration is the amount of solute dissolved in a unit volume of the solution. 🧪

Formula: Concentration (g/dm³) = mass of solute (g) / volume of solution (dm³)

Molarity: Molarity (mol/dm³) = moles of solute (mol) / volume of solution (dm³)

Example 🔍

A solution of sodium hydroxide contains 3.5g of NaOH in 100 cm³ of solution. Calculate the molarity. 🧪

Solution:

Convert volume to dm³: 100 cm³ = 0.1 dm³

Molar mass of NaOH = 40 g/mol

Moles of NaOH = mass / molar mass = 3.5g / 40g/mol = 0.0875 mol

Molarity = moles / volume = 0.0875 mol / 0.1 dm³ = 0.875 mol/dm³

Exercises 📚

• What is the concentration of 0.5g of sodium hydroxide in 250 cm³ of solution in:
  • g/dm³
  • mol/dm³

Definition: Dilution is the process of adding more solvent to a solution to decrease its concentration. 🧪

Formula: M₁V₁ = M₂V₂

where:

• M₁ = initial concentration
• V₁ = initial volume
• M₂ = final concentration
• V₂ = final volume

Example 🔍

If 0.25 dm³ of concentrated sulfuric acid with a concentration of 18M is diluted to 2.0M, what is the final volume? 💧

Solution:

M₁ = 18M, V₁ = 0.25 dm³, M₂ = 2.0M, V₂ = ?

V₂ = (M₁V₁) / M₂ = (18 × 0.25) / 2 = 2.25 dm³

Exercise 📝

• Calculate the volume of 4M hydrochloric acid needed to prepare 2.5 dm³ of a 0.2M solution. 🧪

Formula: % by mass = (mass of element / total molar mass of compound) × 100%

Example 🔍

Calculate the percentage by mass of each element in sodium carbonate, Na₂CO₃. 🧪

Solution:

Molar mass of Na₂CO₃ = (2 × 23) + 12 + (3 × 16) = 106 g/mol

% Na = (46 / 106) × 100% = 43.3%

% C = (12 / 106) × 100% = 11.3%

% O = (48 / 106) × 100% = 45.3%

Definition: The empirical formula is the simplest whole-number ratio of atoms in a compound. ⚗️

Example 🔍

Determine the empirical formula for a compound containing 80% copper and 20% sulfur. 🧪

Solution:

Cu: 80 / 63.5 = 1.26 mol

S: 20 / 32 = 0.625 mol

Divide by the smallest number of moles: Cu : S = 2 : 1

Empirical formula = Cu₂S

Exercise 📝

• Calculate the empirical formula of a compound with 85.7% carbon and 14.3% hydrogen. 🔍

Definition: The molecular formula shows the actual number of each type of atom in a compound. It is derived from the empirical formula and the molecular mass. ⚗️

Formula: Molecular Formula = (Empirical Formula) × n, where n = (molar mass of compound / molar mass of empirical formula)

Example 🔍

If the empirical formula of a compound is C₂H₄O and the molar mass is 88 g/mol, find the molecular formula. 🧪

Solution:

Molar mass of C₂H₄O = 44 g/mol

n = 88 / 44 = 2

Molecular formula = (C₂H₄O)₂ = C₄H₈O₂

Exercise 📝

• A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Find its empirical and molecular formulas if the molar mass is 180 g/mol. 🔍

Definition: Percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%. 📈

Formula: Percentage Yield = (Actual Yield / Theoretical Yield) × 100%

Example 🔍

If 75g of carbon monoxide reacts to produce 68.4g of methanol, calculate the percentage yield. 🧪

Solution:

Theoretical yield = 80g

Percentage yield = (68.4 / 80) × 100% = 85.5%