Moles

A mole is defined as the amount of a substance that contains Avogadro's number of particles. It refers to the quantity that includes exactly 6.02214076 × 10²³ elementary entities of the substance.

The particles could be atoms, molecules, ions, electrons, protons, neutrons, or other types. Therefore, it is important to always specify the type of particle involved.

The number 6.02214076 × 10²³ (commonly approximated as 6.02 × 10²³) is called the Avogadro constant or Avogadro's number, represented by the symbol N_A.

The mole is a unit of measurement. Just as 1 dozen pens equals 12 pens, 1 mole of a substance contains 6.02 × 10²³ of its representative particles.

Molar Mass

The molar mass of a molecular compound is the mass in grams that equals the sum of the atomic masses of all atoms in its molecular formula.

Example : Molar Mass of Glucose (C₆H₁₂O₆)

Solution:

Carbon (C): 12 g/mol × 6 = 72 g/mol
Hydrogen (H): 1 g/mol × 12 = 12 g/mol
Oxygen (O): 16 g/mol × 6 = 96 g/mol

Total molar mass of C₆H₁₂O₆ = 72 + 12 + 96 = 180 g/mol

Mole in Terms of Mass

A mole of a substance (element or compound) has a mass equal to its gram atomic mass or gram molecular mass.

Examples:

1 mole of oxygen atoms = 16 g (gram atomic mass)
1 mole of oxygen molecules (O₂) = 32 g (gram molecular mass)

The unit for molar mass is g/mol.

More Examples:

Chlorine gas (Cl₂): 35.5 × 2 = 71 g/mol
Carbon dioxide gas (CO₂): 12 + (16 × 2) = 44 g/mol

Mole in Terms of R.A.M. or R.M.M.

Relative Atomic Mass (R.A.M.) is the ratio of the actual mass of an atom to 1/12th the mass of a carbon-12 atom. R.A.M. values are often decimals because they represent an average based on the isotopes of the element.

Examples of Relative Atomic Mass:

Hydrogen = 1.00794
Carbon = 12.017
Nitrogen = 14.0067
Oxygen = 15.9994

Thus, 1 mole of a substance equals its R.A.M. (for elements) or R.M.M. (for compounds).

Calculation of Mole in Terms of R.M.M. or R.A.M.

Example 1

Problem: Calculate the number of moles in 40 g of calcium carbonate (CaCO₃).

Solution:

Mass of CaCO₃ = 40 g
R.M.M. of CaCO₃ = 40 + 12 + (16 × 3) = 100 g/mol
Mole (n) = Mass / R.M.M. = 40 g / 100 g/mol = 0.4 mol

Mole in Terms of Number

The term number refers to the count of particles such as atoms, ions, molecules, electrons, protons, or neutrons in a given amount of a substance.

The number of particles involved in a chemical reaction can be determined. Avogadro found that 12.00 g of the carbon-12 isotope (¹²₆C) contains approximately 6.02 × 10²³ atoms.

From his experiments, he concluded that:

The gram atomic mass of any element contains the same number of atoms.
The gram molar mass of any compound contains the same number of molecules.
The gram formula mass of any ionic compound contains the same number of ions.

Thus, 1 gram formula mass of any substance contains 6.02 × 10²³ particles (atoms, ions, molecules, etc.), known as Avogadro's number.

Calculation of Mole in Terms of Number

Example : Number of Particles in 8g of Iron (II) Sulphide (FeS)

Solution:

Molar mass of Fe = 56 g/mol
Molar mass of S = 32 g/mol
G.M.M. of FeS = 56 + 32 = 88 g/mol
Mole of FeS = 8 g / 88 g/mol = 0.09 mol

1 mole of FeS = 6.023 × 10²³ molecules
Number of particles = 0.09 × 6.023 × 10²³ = 0.54 × 10²³ molecules of FeS

Percentage Composition of Compounds

Percentage composition is the mass percentage of each element in a compound. The sum of the mass percentages of all elements in a compound is always equal to 100%.

Formulas:

Percentage composition = (Mass of the element / Molar mass of compound) × 100
Percentage composition = (Number of moles of element × Atomic mass of element) / Molar mass of compound × 100

Using the molar masses of the elements and the compound, we can calculate the mass percentage of each element.

For example, carbon dioxide (CO₂) has a molar mass of 44.01 g, made up of 12.01 g for carbon and 32.00 g for two oxygen atoms.

Mass of C = 12.01 g
Mass of O = 32.00 g

Percentage composition:

Carbon = (12.01 / 44.01) × 100 = 27.30%
Oxygen = (32.00 / 44.01) × 100 = 72.70%

Thus, any pure sample of carbon dioxide contains 27.30% carbon and 72.70% oxygen by mass.

Calculation of Percentage Composition of Compounds

Example 1: Percentage of Oxygen in Aluminium Sulphate Al₂(SO₄)₃

Solution:

Relative atomic masses: Al = 27, S = 32, O = 16
Relative formula mass = (2 × 27) + 3(32 + (4 × 16))
Relative formula mass = 54 + 3 × 96 = 342
Total mass of oxygen = 16 × 12 = 192
Percentage of O = (192 / 342) × 100 = 56.1%

The percentage of oxygen is 56.1%.

Example 2: Percentage of Copper in Copper Sulphate CuSO₄

Solution:

Relative atomic masses: Cu = 64, S = 32, O = 16
Relative formula mass = 64 + 32 + (16 × 4) = 160
Percentage of Cu = (64 / 160) × 100 = 40%

The percentage of copper is 40%.

Empirical & Molecular Formula

Empirical Formula

The empirical formula shows the simplest ratio between the atoms of elements in a compound. It gives the simplest whole-number ratio of atoms present in the compound.

Molecular Formula

The molecular formula shows the actual number of atoms of each element in a molecule of the compound. It reflects the true composition of a single molecule.

The molecular mass (MM) of a compound is a multiple of its empirical formula mass:

MM = n × empirical formula mass

Empirical formula can also be determined from the percentage composition of a compound.

Example:

For ethanoic acid:

Molecular formula: CH₃COOH or C₂H₄O₂
Atoms: 2 carbon, 4 hydrogen, and 2 oxygen atoms

Empirical formula: CH₂O (ratio 1 : 2 : 1)

Finding Empirical Formula from Percentage Composition

Steps:

Create a table with six columns and rows equal to the number of elements.
Write the elements in the first column.
Write the percentage composition in the second column.
Divide each percentage by the atomic mass of the element to find moles.
Divide each mole value by the smallest mole value to simplify the ratio.
Round or multiply to obtain whole numbers for the subscripts.

Alternative Method:

Write the mass or percentage composition of each element.
Convert to moles by dividing by the atomic mass.
Divide each mole value by the smallest mole value and approximate to the nearest whole number.

Example

A compound contains 31.91% potassium, 28.93% chlorine, and the rest oxygen. Find the chemical formula.

Oxygen % = 100% - (31.91% + 28.93%) = 39.16%
K = 31.91 / 39 ≈ 0.8182
Cl = 28.93 / 35.5 ≈ 0.8149
O = 39.16 / 16 ≈ 2.4475

Dividing by the smallest value (0.8149):

K = 1
Cl = 1
O = 3

Empirical formula: KClO₃

Finding Empirical Formula from Mass

To find the empirical formula from masses:

First divide each element's mass by the total sample mass to get percentages.
Then follow the steps for percentage composition.

This is the formula for benzene.

Example

If carbon and hydrogen are present in a 1:2 ratio and the molecular mass is 28 g/mol, find the molecular formula.

Empirical formula = CH₂
Empirical formula mass = 12 + (2 × 1) = 14 g/mol
n = 28 / 14 = 2
Molecular formula = C₂H₄

This is the molecular formula for ethene.

Moles

Molar Mass

Example : Molar Mass of Glucose (C6H12O6)

Mole in Terms of Mass

Examples:

More Examples:

Mole in Terms of R.A.M. or R.M.M.

Examples of Relative Atomic Mass:

Calculation of Mole in Terms of R.M.M. or R.A.M.

Example 1

Mole in Terms of Number

Calculation of Mole in Terms of Number

Example : Number of Particles in 8g of Iron (II) Sulphide (FeS)

Percentage Composition of Compounds

Formulas:

Calculation of Percentage Composition of Compounds

Example 1: Percentage of Oxygen in Aluminium Sulphate Al2(SO4)3

Example 2: Percentage of Copper in Copper Sulphate CuSO4

Empirical & Molecular Formula

Empirical Formula

Molecular Formula

Example:

Finding Empirical Formula from Percentage Composition

Steps:

Alternative Method:

Example

Finding Empirical Formula from Mass

Example

Example : Molar Mass of Glucose (C₆H₁₂O₆)

Example 1: Percentage of Oxygen in Aluminium Sulphate Al₂(SO₄)₃

Example 2: Percentage of Copper in Copper Sulphate CuSO₄