Avogadro’s law in relation to molar quantities

Avogadro's law

"At the same temperature and pressure, volumes of all gases contain the same number of particles."

Formula

$n = \frac{V}{V_m}$

Where:

• $V$ = volume of gas.
• $V_m$ = molar volume of gas (22.4 dm³/mol).
• $n$ = number of moles.

Example

For 18.8 dm³ of CO₂ at STP:

$n = \frac{18.8 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 0.839 \text{ mol}$

For 48.8 dm³ of O₂ at STP:

$n = \frac{48.8 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 2.18 \text{ mol}$

The mole in stoichiometric calculations

Stoichiometry is the quantitative relationship between reacting substances in a chemical equation.

Example: Consider the following balanced equation:

$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$

• Molecular mass of CaCO₃ = 100 g/mol.
• Molecular mass of CaO = 56 g/mol.

Calculate the weight of CaO and volume of CO₂ produced from 75 g of CaCO₃.

Solution:

From the equation, 100 g of CaCO₃ produces 56 g of CaO.

The weight of CaO produced by 75 g of CaCO₃ is:

$x = \frac{75 \text{ g} \times 56 \text{ g}}{100 \text{ g}} = 42 \text{ g}$

For the volume of CO₂ produced:

$75 \text{ g CaCO}_3 = \frac{75 \text{ g}}{100 \text{ g/mol}} \times 22.4 \text{ dm}^3 = 16.8 \text{ dm}^3$