# 4.1 The Mole Concept 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 $6.023 \times 10^{23}$ 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 = $6.023 \times 10^{23}$ carbon atoms * 1 mole of water molecules = $6.023 \times 10^{23}$ $\mathrm{H_2O}$ 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) = $16 \, \mathrm{g}$ * One mole of Oxygen molecules ($\mathrm{O_2}$) = $32 \, \mathrm{g}$ * One mole of Water molecules ($\mathrm{H_2O}$) = $18 \, \mathrm{g}$ * One mole of Sodium ions ($\mathrm{Na^+}$) = $23 \, \mathrm{g}$ * One mole of Sodium Chloride (NaCl) formula units = $58.5 \, \mathrm{g}$ --- ## 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 $0^\circ \mathrm{C}$ (273.15 K) and 1 atm pressure. * At STP, one mole of any ideal gas occupies a volume of **$22.414 \, \mathrm{dm^3}$**. This relationship allows for direct conversions between the volume, moles, and mass of a gas at STP. **Key Relationships at STP:** * $22.414 \, \mathrm{dm^3}$ of any gas = 1 mole = $6.02 \times 10^{23}$ molecules. * $22.414 \, \mathrm{dm^3}$ of $\mathrm{H_2}$ gas = $2 \, \mathrm{g}$ = 1 mole of $\mathrm{H_2}$. * $22.414 \, \mathrm{dm^3}$ of $\mathrm{NH_3}$ gas = $17 \, \mathrm{g}$ = 1 mole of $\mathrm{NH_3}$. ### Worked Example 4.1: Volume of a Gas **Problem:** Determine the volume of 2.5 moles of chlorine molecules ($\mathrm{Cl_2}$) at STP. **Solution:** 1. **Recall the molar volume relationship:** 1 mole of any gas at STP occupies $22.414 \, \mathrm{dm^3}$. 2. **Set up the calculation:** $$ \text{Volume} = \text{moles} \times \text{Molar Volume} $$ 3. **Substitute the values and solve:** $$ \text{Volume of } \mathrm{Cl_2} = 2.5 \, \text{mol} \times 22.414 \, \frac{\text{dm}^3}{\text{mol}} $$ $$ \text{Volume of } \mathrm{Cl_2} = 56.035 \, \text{dm}^3 $$ --- ## 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. * A gas with a higher molar mass will have a higher density. * A gas with a lower molar mass will have a lower density. This relationship allows for the calculation of a gas's molar mass if its density at STP is known. ### Worked Example 4.2: Molar Mass from Density **Problem:** Calculate the gram molecular mass of a gas which has a density of $1.43 \, \mathrm{g/dm^3}$ at STP. **Solution:** 1. **Given values:** * Density = $1.43 \, \mathrm{g/dm^3}$ * This means $1 \, \mathrm{dm^3}$ of the gas has a mass of $1.43 \, \mathrm{g}$. 2. **Apply the concept of molar volume:** The molar mass is the mass of one mole, which occupies $22.4 \, \mathrm{dm^3}$ at STP (using the common approximation). 3. **Calculate the mass of $22.4 \, \mathrm{dm^3}$ of the gas:** $$ \text{Molar Mass} = \text{Density} \times \text{Molar Volume} $$ $$ \text{Molar Mass} = 1.43 \, \frac{\text{g}}{\text{dm}^3} \times 22.4 \, \frac{\text{dm}^3}{\text{mol}} $$ $$ \text{Molar Mass} = 32.032 \, \text{g/mol} $$ Therefore, the gram molecular mass of the gas is **$32.032 \, \text{g/mol}$**. --- ## 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: $$ 2\mathrm{H_{2(g)}} + \mathrm{O_{2(g)}} \rightarrow 2\mathrm{H_2O_{(g)}} $$ This equation tells us: * **Mole Ratio:** 2 moles of $\mathrm{H_2}$ react with 1 mole of $\mathrm{O_2}$ to produce 2 moles of $\mathrm{H_2O}$. * **Particle Ratio:** $2 \times (6.02 \times 10^{23})$ molecules of $\mathrm{H_2}$ react with $1 \times (6.02 \times 10^{23})$ molecules of $\mathrm{O_2}$ to produce $2 \times (6.02 \times 10^{23})$ molecules of $\mathrm{H_2O}$. * **Mass Ratio:** $4 \, \mathrm{g}$ of $\mathrm{H_2}$ (2 moles) react with $32 \, \mathrm{g}$ of $\mathrm{O_2}$ (1 mole) to produce $36 \, \mathrm{g}$ of $\mathrm{H_2O}$ (2 moles). * **Volume Ratio (at STP):** $44.828 \, \mathrm{dm^3}$ of $\mathrm{H_2}$ ($2 \times 22.414$) react with $22.414 \, \mathrm{dm^3}$ of $\mathrm{O_2}$ to produce $44.828 \, \mathrm{dm^3}$ of $\mathrm{H_2O}$ vapor. For reactions involving limiting reactants, refer to . For calculations involving percentage yield, see . ### Worked Example 4.3: Mass-to-Volume Stoichiometry **Problem:** When 100 g of magnesium is treated with dilute hydrochloric acid, what volume of hydrogen can be collected at STP? $$ \mathrm{Mg_{(s)}} + 2\mathrm{HCl_{(aq)}} \rightarrow \mathrm{MgCl_{2(aq)}} + \mathrm{H_{2(g)}} $$ **Solution:** 1. **Calculate moles of the given substance (Mg):** * Molar mass of Mg = $24 \, \text{g/mol}$ $$ \text{Moles of Mg} = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, \text{g}}{24 \, \text{g/mol}} = 4.17 \, \text{mol} $$ 2. **Use the mole ratio from the balanced equation:** From the equation, 1 mole of Mg produces 1 mole of $\mathrm{H_2}$. Therefore, $4.17$ moles of Mg will produce $4.17$ moles of $\mathrm{H_2}$. 3. **Convert moles of the desired substance ($\mathrm{H_2}$) to volume at STP:** $$ \text{Volume of } \mathrm{H_2} = \text{moles} \times \text{Molar Volume} $$ $$ \text{Volume of } \mathrm{H_2} = 4.17 \, \text{mol} \times 22.4 \, \frac{\text{dm}^3}{\text{mol}} = 93.408 \, \text{dm}^3 $$ --- ## 4.1.4 Mole-Mole Calculation These calculations use the coefficients from a balanced equation to find the number of moles of one substance based on the moles of another. ### Worked Example 4.5: Mole-Mole Calculation **Problem:** How much MgO is formed when 4 moles of magnesium react with excess $\mathrm{O_2}$? Also, calculate the moles of $\mathrm{O_2}$ required. $$ 2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO} $$ **Solution:** **Part 1: Moles of MgO formed** 1. **Identify the mole ratio:** From the equation, 2 moles of Mg produce 2 moles of MgO. The ratio is $2:2$ or $1:1$. 2. **Calculate moles of MgO:** $$ 4 \, \text{mol Mg} \times \frac{2 \, \text{mol MgO}}{2 \, \text{mol Mg}} = 4 \, \text{mol MgO} $$ * **Result:** 4 moles of MgO are formed. **Part 2: Moles of $\mathrm{O_2}$ required** 1. **Identify the mole ratio:** From the equation, 2 moles of Mg react with 1 mole of $\mathrm{O_2}$. The ratio is $2:1$. 2. **Calculate moles of $\mathrm{O_2}$:** $$ 4 \, \text{mol Mg} \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol Mg}} = 2 \, \text{mol O}_2 $$ * **Result:** 2 moles of $\mathrm{O_2}$ are required. --- ## 4.1.5 Mole-Mass Calculations These calculations involve converting a given mass of a substance to moles, using the mole ratio to find the moles of another substance, and then converting those moles back to mass. ### Worked Example 4.6: Mole-Mass Calculation **Problem:** If 25 g of potassium chlorate ($\mathrm{KClO_3}$) is decomposed, calculate: a) How many moles of $\mathrm{O_2}$ are produced. b) How much mass in grams of $\mathrm{O_2}$ is produced. $$ 2\mathrm{KClO_3} \rightarrow 2\mathrm{KCl} + 3\mathrm{O_2} $$ **Solution:** 1. **Calculate moles of the given substance ($\mathrm{KClO_3}$):** * Molar mass of $\mathrm{KClO_3} = 39 + 35.5 + 3(16) = 122.5 \, \text{g/mol}$ $$ \text{Moles of } \mathrm{KClO_3} = \frac{25 \, \text{g}}{122.5 \, \text{g/mol}} = 0.204 \, \text{mol} $$ 2. **a) Calculate moles of $\mathrm{O_2}$ produced:** * **Mole Ratio:** 2 moles of $\mathrm{KClO_3}$ produce 3 moles of $\mathrm{O_2}$. $$ \text{Moles of } \mathrm{O_2} = 0.204 \, \text{mol } \mathrm{KClO_3} \times \frac{3 \, \text{mol } \mathrm{O_2}}{2 \, \text{mol } \mathrm{KClO_3}} = 0.306 \, \text{mol } \mathrm{O_2} $$ 3. **b) Calculate the mass of $\mathrm{O_2}$ produced:** * Molar mass of $\mathrm{O_2} = 2(16) = 32 \, \text{g/mol}$ $$ \text{Mass of } \mathrm{O_2} = \text{moles} \times \text{molar mass} $$ $$ \text{Mass of } \mathrm{O_2} = 0.306 \, \text{mol} \times 32 \, \text{g/mol} = 9.796 \, \text{g} $$ --- ## 4.1.6 Solution Stoichiometry Stoichiometric calculations for reactions in solutions often use **molarity** as the unit of concentration. * **Molarity (M)** is defined as the number of moles of solute dissolved per cubic decimeter ($\mathrm{dm^3}$) of solution. $$ M = \frac{\text{moles of solute}}{\text{dm}^3 \text{ of solution}} $$ ### Worked Example 4.8: Calculating Molarity **Problem:** A solution contains 40 g of urea ($\mathrm{NH_2CONH_2}$) dissolved in $500 \, \text{cm}^3$ of solution. Calculate the molarity. **Solution:** | Parameter | Calculation | Value | | :--- | :--- | :--- | | **Mass of urea** | Given | $40 \, \text{g}$ | | **Molar Mass of urea** | $14+2(1)+12+16+14+2(1)$ | $60 \, \text{g/mol}$ | | **Moles of urea** | $\frac{40 \, \text{g}}{60 \, \text{g/mol}}$ | $0.667 \, \text{mol}$ | | **Volume of solution** | $\frac{500 \, \text{cm}^3}{1000 \, \text{cm}^3/\text{dm}^3}$ | $0.5 \, \text{dm}^3$ | **Calculate Molarity:** $$ \text{Molarity} = \frac{\text{moles of solute}}{\text{dm}^3 \text{ of solution}} = \frac{0.667 \, \text{mol}}{0.5 \, \text{dm}^3} = 1.334 \, M $$ ### Stoichiometry Formula for Solutions When two solutions react completely, their mole ratio is determined by the balanced equation. This leads to a useful formula: $$ \frac{M_1 V_1}{n_1} = \frac{M_2 V_2}{n_2} $$ Where: * $M_1, V_1$ = Molarity and Volume of reactant 1 * $M_2, V_2$ = Molarity and Volume of reactant 2 * $n_1, n_2$ = Stoichiometric coefficients from the balanced equation for reactants 1 and 2. ### Worked Example 4.10: Using the Stoichiometry Formula **Problem:** What volume of $0.5 \, M$ sodium sulfate will react with $275 \, \text{cm}^3$ of $0.25 \, M$ barium chloride solution to completely precipitate $\mathrm{Ba^{2+}}$ in solution? $$ \mathrm{Na_2SO_{4(aq)}} + \mathrm{BaCl_{2(aq)}} \rightarrow \mathrm{BaSO_{4(s)}} + 2\mathrm{NaCl_{(aq)}} $$ **Solution:** 1. **Identify the variables:** * **Reactant 1 ($\mathrm{Na_2SO_4}$):** $M_1 = 0.5 \, M$, $V_1 = ?$, $n_1 = 1$ * **Reactant 2 ($\mathrm{BaCl_2}$):** $M_2 = 0.25 \, M$, $V_2 = 275 \, \text{cm}^3$, $n_2 = 1$ 2. **Apply the formula:** $$ \frac{M_1 V_1}{n_1} = \frac{M_2 V_2}{n_2} $$ $$ \frac{0.5 \times V_1}{1} = \frac{0.25 \times 275}{1} $$ 3. **Solve for $V_1$:** $$ V_1 = \frac{0.25 \times 275}{0.5} $$ $$ V_1 = 137.5 \, \text{cm}^3 $$ * **Result:** $137.5 \, \text{cm}^3$ of $0.5 \, M$ sodium sulfate is required.