Mass-Volume Relationship – SS2 Chemistry Lesson Note (Third Term)

A mole is the amount of a substance that contains 6.02 × 10²³ particles, which may be atoms, ions, molecules or electrons. This lesson shows how to link the mass, volume and number of particles of a substance through the mole. It also shows how to use balanced equations to calculate the masses of reactants and products.

Behavioural objectives

By the end of the lesson, students should be able to:

• define the mole, relative atomic mass and relative molecular mass;
• calculate the relative molecular mass of a compound;
• state the molar volume of a gas at s.t.p.;
• use the relationships between amount, mass, molar mass and number of particles to solve problems;
• calculate the masses of reactants and products from a balanced chemical equation.

Mole and molar quantities

What is a mole?

A mole is a fixed number of particles of a substance, which may be atoms, ions, molecules or electrons. This number is approximately 6.02 × 10²³ and is known as Avogadro’s number of particles.

The mole is defined as the amount of a substance which contains as many elementary units as there are atoms in 12 g of Carbon-12.

What is relative atomic mass?

The relative atomic mass of an element is the number of times the average mass of one atom of that element is heavier than one-twelfth the mass of one atom of Carbon-12. It indicates the mass of an atom of an element. For example, the relative atomic masses of hydrogen, oxygen, carbon, sodium and calcium are 1, 16, 12, 23 and 40 respectively.

The relative atomic mass of an element, expressed in grams, contains the same number of atoms, which is 6.02 × 10²³ atoms. For example, 1 mole of hydrogen atoms, with an atomic mass of 1.0 g, contains 6.02 × 10²³ atoms.

What is relative molecular mass?

The relative molecular mass of an element or compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of Carbon-12.

It is the sum of the relative atomic masses of all atoms in one molecule of that substance. It is also called the formula mass. The formula mass refers not only to the relative mass of a molecule but also to that of an ion or radical.

Calculation: relative molecular mass

Calculate the relative molecular mass of:

1. Magnesium chloride
2. Sodium hydroxide
3. Calcium trioxocarbonate

[Mg = 24, Cl = 35.5, Na = 23, O = 16, H = 1, Ca = 40, C = 12]

Solution:

1. MgCl₂ = 24 + 35.5 × 2 = 24 + 71 = 95 g mol^-1
2. NaOH = 23 + 16 + 1 = 40 g mol^-1
3. CaCO₃ = 40 + 12 + 16 × 3 = 100 g mol^-1

Molar volume of gases

The volume occupied by 1 mole of a gas at standard conditions of temperature and pressure (s.t.p.) is 22.4 dm³. Thus 1 mole of oxygen gas, of molar mass 32.0 g mol^-1, occupies a volume of 22.4 dm³ at s.t.p. Also, 1 mole of helium gas, of molar mass 4.0 g mol^-1, occupies a volume of 22.4 dm³ at s.t.p.

Note: When the conditions of temperature and pressure are altered, the molar volume also changes. Standard temperature = 273 K and standard pressure = 760 mmHg.

Relationship between quantities

• Molar mass = mass (g) ÷ amount (moles), that is, M = m/n, in g mol^-1. (Amount = number of moles.)
• Molar volume of a gas = volume (cm³ or dm³) ÷ amount (mole)
• Amount = reacting mass (g) ÷ molar mass (g mol^-1)
• Amount of substance = number of particles ÷ Avogadro’s constant, where Avogadro’s constant = 6.02 × 10²³

Combining the last two expressions:

Reacting mass ÷ Molar mass = Number of particles ÷ 6.02 × 10²³

Worked examples

1. What is the mass of 2.7 moles of aluminium (Al = 27)?

Solution:

Amount = Reacting mass ÷ Molar mass

Reacting mass = Amount × Molar mass

= 2.7 mole × 27 g mol^-1 = 72.9 g

2. What is the number of oxygen molecules in 32 g of the gas? (O = 16, N_A = 6.02 × 10²³)

Solution:

Reacting mass ÷ Molar mass = Number of molecules ÷ 6.02 × 10²³

Number of molecules = (Reacting mass × 6.02 × 10²³) ÷ Molar mass

Molar mass of O₂ = 16 × 2 = 32 g mol^-1

Number of molecules = (32 g × 6.02 × 10²³) ÷ 32 g mol^-1 = 6.02 × 10²³

The number of oxygen molecules is 6.02 × 10²³.

Stoichiometry of reactions

The calculation of the amounts (generally measured in moles or grams) of reactants and products involved in a chemical reaction is known as the stoichiometry of the reaction. In other words, the mole ratio in which reactants combine and products are formed gives the stoichiometry of the reaction.

From the stoichiometry of a given balanced chemical equation, the mass or volume of the reactant needed for the reaction, or of the products formed, can be calculated.

Calculation of masses of reactants and products

Example 1

Calculate the mass of solid product obtained when 16.8 g of NaHCO₃ was heated strongly until there was no further change.

Solution:

The equation for the reaction is:

2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g)

Molar mass of NaHCO₃ = 23 + 1 + 12 + 16 × 3 = 84 g mol^-1

Molar mass of Na₂CO₃ = 23 × 2 + 12 + 16 × 3 = 106 g mol^-1

From the equation:

2 moles of NaHCO₃ produce 1 mole of Na₂CO₃

2 × 84 g of NaHCO₃ produce 106 g of Na₂CO₃

16.8 g of NaHCO₃ will produce X g of Na₂CO₃

X = (106 g × 16.8 g) ÷ (2 × 84 g) = 10.6 g

Mass of solid product obtained = 10.6 g

Example 2

Calculate the number of moles of CaCl₂ that can be obtained from 25 g of limestone [CaCO₃] in the presence of excess acid.

Solution:

The equation for the reaction is:

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

Number of moles = Reacting mass ÷ Molar mass

Molar mass of CaCO₃ = 40 + 12 + 16 × 3 = 100 g mol^-1

Number of moles of CaCO₃ = 25 g ÷ 100 g mol^-1 = 0.25 mole

From the equation of reaction, 1 mole of CaCO₃ yields 1 mole of CaCl₂.

Therefore, 0.25 mole of CaCO₃ yields 0.25 mole of CaCl₂.

Evaluation

1. Define the mole and state the value of Avogadro’s number.
2. Calculate the relative molecular mass of water, H₂O (H = 1, O = 16).
3. State the volume occupied by 1 mole of any gas at s.t.p.
4. Calculate the mass of 0.5 mole of sodium (Na = 23).
5. Calculate the number of moles in 50 g of CaCO₃ (Ca = 40, C = 12, O = 16).

Assignment

In an experiment, 10 cm³ of ethene [C₂H₄] was burnt in 50 cm³ of oxygen.

• a) Which gas was supplied in excess? Calculate the volume of the excess gas remaining at the end of the reaction.
• b) Calculate the volume of CO₂ gas produced.