4.1 The Mole Concept | Stoichiometry | Chemistry 11

The mole is the SI unit used by chemists to measure the amount of a substance. It provides a consistent method to weigh and count atoms, molecules, and ions.

A mole is defined as the amount of substance that contains $6.022 \times 1 0^{23}$ representative particles (atoms, molecules, or ions). This specific number is known as Avogadro's number ( $N_{A}$ ).

Think of it like a "chemist's dozen":

1 dozen eggs = 12 eggs
1 mole of carbon atoms = $6.022 \times 1 0^{23}$ carbon atoms
1 mole of water molecules = $6.022 \times 1 0^{23}$ $H_{2} O$ molecules

The mass of one mole of a substance (in grams) is numerically equal to its atomic mass, formula mass, or molecular mass in atomic mass units (amu). This crucial link allows for conversions between mass and moles.

Examples:

One mole of Oxygen atoms (O) = $16 g$
One mole of Oxygen molecules ( $O_{2}$ ) = $32 g$
One mole of Water molecules ( $H_{2} O$ ) = $18 g$
One mole of Sodium ions ( $N a^{+}$ ) = $23 g$
One mole of Sodium Chloride (NaCl) formula units = $58.5 g$

4.1.1 Molar Volume ( $V_{m}$ )

The molar volume is the volume occupied by one mole of any gas at Standard Temperature and Pressure (STP), which is $0^{\circ} C$ (273.15 K) and 1 atm pressure.

At STP, one mole of any ideal gas occupies a volume of $22.414 d m^{3}$ .

This relationship allows for direct conversions between the volume, moles, and mass of a gas at STP.

Key Relationships at STP:

$22.414 d m^{3}$ of any gas = 1 mole = $6.022 \times 1 0^{23}$ molecules.
$22.414 d m^{3}$ of $H_{2}$ gas = $2 g$ = 1 mole of $H_{2}$ .
$22.414 d m^{3}$ of $N H_{3}$ gas = $17 g$ = 1 mole of $N H_{3}$ .

Worked Example 4.1: Volume of a Gas

Problem: Determine the volume of 2.5 moles of chlorine molecules ( $C l_{2}$ ) at STP.

Solution:

Recall the molar volume relationship: 1 mole of any gas at STP occupies $22.414 d m^{3}$ .
Set up the calculation: $Volume = moles \times Molar Volume$
Substitute the values and solve: $Volume of C l_{2} = 2.5 mol \times 22.414 \frac{dm ^{3}}{mol}$ $Volume of C l_{2} = 56.035 dm^{3}$

4.1.2 Molar Mass and Density of Gases

Density is the mass per unit volume of a substance. Since one mole of any gas occupies the same volume at STP, the density of a gas is directly proportional to its molar mass.

$M = ρ \times V_{m}$

where $M$ = molar mass (g/mol), $ρ$ = density (g/dm³) at STP, and $V_{m} = 22.414 d m^{3} /mol$ .

A gas with a higher molar mass will have a higher density.
A gas with a lower molar mass will have a lower density.

Worked Example 4.2: Molar Mass from Density

Problem: Calculate the gram molecular mass of a gas which has a density of $1.43 g/d m^{3}$ at STP.

Solution:

Given values:
- Density = $1.43 g/d m^{3}$
- Molar volume at STP = $22.4 d m^{3} /mol$
Apply the formula: $Molar Mass = Density \times Molar Volume$ $Molar Mass = 1.43 \frac{g}{dm ^{3}} \times 22.4 \frac{dm ^{3}}{mol}$ $Molar Mass = 32.032 g/mol$

Therefore, the gram molecular mass of the gas is $32.032 g/mol$ (this corresponds to $O_{2}$ or $S$ ).

4.1.3 Stoichiometric Calculations and Mole Ratio

A balanced chemical equation provides quantitative information about a reaction, known as stoichiometry. The coefficients in the equation represent the mole ratio of reactants and products.

Consider the formation of water: $2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(g)}$

This equation tells us:

Mole Ratio: 2 moles of $H_{2}$ react with 1 mole of $O_{2}$ to produce 2 moles of $H_{2} O$ .
Particle Ratio: $2 \times (6.022 \times 1 0^{23})$ molecules of $H_{2}$ react with $1 \times (6.022 \times 1 0^{23})$ molecules of $O_{2}$ to produce $2 \times (6.022 \times 1 0^{23})$ molecules of $H_{2} O$ .
Mass Ratio: $4 g$ of $H_{2}$ (2 moles) react with $32 g$ of $O_{2}$ (1 mole) to produce $36 g$ of $H_{2} O$ (2 moles).
Volume Ratio (at STP): $44.828 d m^{3}$ of $H_{2}$ ( $2 \times 22.414$ ) react with $22.414 d m^{3}$ of $O_{2}$ to produce $44.828 d m^{3}$ of $H_{2} O$ vapor.

Worked Example 4.3: Mass-to-Volume Stoichiometry

Problem: When $100 g$ of magnesium is treated with dilute hydrochloric acid, calculate the volume of hydrogen gas produced at STP.

$M g_{(s)} + 2 HC l_{(aq)} \to MgC l_{2 (aq)} + H_{2 (g)}$

Solution:

Calculate moles of Mg: $n (Mg) = \frac{m}{M} = \frac{100 g}{24 g/mol} = 4.167 mol$
Use mole ratio: From the equation, 1 mol Mg produces 1 mol $H_{2}$ . $n (H_{2}) = 4.167 mol$
Calculate volume at STP: $V = n \times 22.414 = 4.167 \times 22.414 = 93.4 d m^{3}$

For reactions involving limiting reactants, refer to . For calculations involving percentage yield, see .

The mole is the SI unit used by chemists to measure the amount of a substance. It provides a consistent method to weigh and count atoms, molecules, and ions.

Think of it like a "chemist's dozen":

1 dozen eggs = 12 eggs
1 mole of carbon atoms = $6.022 \times 1 0^{23}$ carbon atoms
1 mole of water molecules = $6.022 \times 1 0^{23}$ $H_{2} O$ molecules

Examples:

One mole of Oxygen atoms (O) = $16 g$
One mole of Oxygen molecules ( $O_{2}$ ) = $32 g$
One mole of Water molecules ( $H_{2} O$ ) = $18 g$
One mole of Sodium ions ( $N a^{+}$ ) = $23 g$
One mole of Sodium Chloride (NaCl) formula units = $58.5 g$

4.1.1 Molar Volume ( $V_{m}$ )

The molar volume is the volume occupied by one mole of any gas at Standard Temperature and Pressure (STP), which is $0^{\circ} C$ (273.15 K) and 1 atm pressure.

At STP, one mole of any ideal gas occupies a volume of $22.414 d m^{3}$ .

This relationship allows for direct conversions between the volume, moles, and mass of a gas at STP.

Key Relationships at STP:

$22.414 d m^{3}$ of any gas = 1 mole = $6.022 \times 1 0^{23}$ molecules.
$22.414 d m^{3}$ of $H_{2}$ gas = $2 g$ = 1 mole of $H_{2}$ .
$22.414 d m^{3}$ of $N H_{3}$ gas = $17 g$ = 1 mole of $N H_{3}$ .

Worked Example 4.1: Volume of a Gas

Problem: Determine the volume of 2.5 moles of chlorine molecules ( $C l_{2}$ ) at STP.

Solution:

Recall the molar volume relationship: 1 mole of any gas at STP occupies $22.414 d m^{3}$ .
Set up the calculation: $Volume = moles \times Molar Volume$
Substitute the values and solve: $Volume of C l_{2} = 2.5 mol \times 22.414 \frac{dm ^{3}}{mol}$ $Volume of C l_{2} = 56.035 dm^{3}$

4.1.2 Molar Mass and Density of Gases

Density is the mass per unit volume of a substance. Since one mole of any gas occupies the same volume at STP, the density of a gas is directly proportional to its molar mass.

$M = ρ \times V_{m}$

where $M$ = molar mass (g/mol), $ρ$ = density (g/dm³) at STP, and $V_{m} = 22.414 d m^{3} /mol$ .

A gas with a higher molar mass will have a higher density.
A gas with a lower molar mass will have a lower density.

Worked Example 4.2: Molar Mass from Density

Problem: Calculate the gram molecular mass of a gas which has a density of $1.43 g/d m^{3}$ at STP.

Solution:

Given values:
- Density = $1.43 g/d m^{3}$
- Molar volume at STP = $22.4 d m^{3} /mol$
Apply the formula: $Molar Mass = Density \times Molar Volume$ $Molar Mass = 1.43 \frac{g}{dm ^{3}} \times 22.4 \frac{dm ^{3}}{mol}$ $Molar Mass = 32.032 g/mol$

Therefore, the gram molecular mass of the gas is $32.032 g/mol$ (this corresponds to $O_{2}$ or $S$ ).

4.1.3 Stoichiometric Calculations and Mole Ratio

A balanced chemical equation provides quantitative information about a reaction, known as stoichiometry. The coefficients in the equation represent the mole ratio of reactants and products.

Consider the formation of water: $2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(g)}$

This equation tells us:

Mole Ratio: 2 moles of $H_{2}$ react with 1 mole of $O_{2}$ to produce 2 moles of $H_{2} O$ .
Particle Ratio: $2 \times (6.022 \times 1 0^{23})$ molecules of $H_{2}$ react with $1 \times (6.022 \times 1 0^{23})$ molecules of $O_{2}$ to produce $2 \times (6.022 \times 1 0^{23})$ molecules of $H_{2} O$ .
Mass Ratio: $4 g$ of $H_{2}$ (2 moles) react with $32 g$ of $O_{2}$ (1 mole) to produce $36 g$ of $H_{2} O$ (2 moles).
Volume Ratio (at STP): $44.828 d m^{3}$ of $H_{2}$ ( $2 \times 22.414$ ) react with $22.414 d m^{3}$ of $O_{2}$ to produce $44.828 d m^{3}$ of $H_{2} O$ vapor.

Worked Example 4.3: Mass-to-Volume Stoichiometry

Problem: When $100 g$ of magnesium is treated with dilute hydrochloric acid, calculate the volume of hydrogen gas produced at STP.

$M g_{(s)} + 2 HC l_{(aq)} \to MgC l_{2 (aq)} + H_{2 (g)}$

Solution:

Calculate moles of Mg: $n (Mg) = \frac{m}{M} = \frac{100 g}{24 g/mol} = 4.167 mol$
Use mole ratio: From the equation, 1 mol Mg produces 1 mol $H_{2}$ . $n (H_{2}) = 4.167 mol$
Calculate volume at STP: $V = n \times 22.414 = 4.167 \times 22.414 = 93.4 d m^{3}$

For reactions involving limiting reactants, refer to . For calculations involving percentage yield, see .