The Mole, Avogadro Constant, and Stoichiometry Study Guide

The Mole and the Avogadro Constant

• Measuring Chemical Amounts: Chemical amounts are measured in moles. The mole (symbol: mol) is the SI unit for the amount of a substance.

• The Particle Concept: One mole of any substance contains the same number of the stated particles, whether they are atoms, molecules, or ions.

• The Avogadro Constant:     * One mole contains $6.02 \times 10^{23}$ particles.     * This specific number is known as the Avogadro Constant.

• Examples of the Avogadro Constant in Practice:     * One mole of sodium (Na) contains $6.02 \times 10^{23}$ atoms of sodium.     * One mole of hydrogen ($H_{2}$) contains $6.02 \times 10^{23}$ molecules of hydrogen.     * One mole of sodium chloride ($NaCl$) contains $6.02 \times 10^{23}$ formula units of sodium chloride.

• Molar Mass: The mass of 1 mole of a substance is referred to as the molar mass.     * For elements: The molar mass is the same as the relative atomic mass ($A_{r}$) expressed in grams.     * For compounds: The molar mass is the same as the relative molecular ($M_{r}$) or formula mass in grams.

Molar Volumes of Gas

• Avogadro’s Law: This law states that at the same temperature and pressure, equal amounts of gases occupy the same volume of space. For instance, 1 mole of hydrogen gas occupies the identical volume as 1 mole of methane gas.

• Room Temperature and Pressure (RTP):     * At RTP, the volume occupied by one mole of any gas is $24\,dm^{3}$ or $24,000\,cm^{3}$.     * RTP conditions are defined as a temperature of $20^{\circ}C$ and a pressure of $1\,atmosphere\,(atm)$.

• Molar Gas Volume Calculations ($dm^{3}$):     * $\text{Volume} = \text{Moles} \times 24\,dm^{3}$     * $\text{Moles} = \frac{\text{Volume}}{24}$

• Molar Gas Volume Calculations ($cm^{3}$):     * $\text{Volume} = \text{Moles} \times 24,000\,cm^{3}$     * $\text{Moles} = \frac{\text{Volume}}{24,000}$

• Data for Converting Moles to Volume:     * Hydrogen (3 moles): Volume = $3 \times 24 = 72\,dm^{3}$ or $3 \times 24,000 = 72,000\,cm^{3}$.     * Carbon dioxide (0.25 moles): Volume = $0.25 \times 24 = 6\,dm^{3}$ or $0.25 \times 24,000 = 6,000\,cm^{3}$.     * Oxygen (5.4 moles): Volume = $5.4 \times 24 = 129.6\,dm^{3}$ or $5.4 \times 24,000 = 129,600\,cm^{3}$.     * Ammonia (0.02 moles): Volume = $0.02 \times 24 = 0.48\,dm^{3}$ or $0.02 \times 24,000 = 480\,cm^{3}$.

• Data for Converting Volume to Moles:     * Methane ($225.6\,dm^{3}$): $225.6 \div 24 = 9.4\,mol$.     * Sulfur dioxide ($960\,dm^{3}$): $960 \div 24 = 40\,mol$.     * Oxygen ($1200\,cm^{3}$): $1200 \div 24,000 = 0.05\,mol$.     * Carbon monoxide ($7.2\,dm^{3}$): $7.2 \div 24 = 0.3\,mol$.

Linking Moles, Mass, and Mr

• Relative Mass Equivalents: One mole of any element is equal to the relative atomic mass ($A_{r}$) of that element in grams.     * 1 mole of Carbon (C) has a mass of $12\,g$ (since $A_{r}$ of C is 12).     * 1 mole of Helium (He) has a mass of $4\,g$ (since $A_{r}$ of He is 4).     * 1 mole of Lithium (Li) has a mass of $7\,g$ (since $A_{r}$ of Li is 7).

• Compound Molar Mass: To find the mass of one mole of a compound, the relative atomic masses of all atoms in the formula are summed.     * Example Carbon dioxide ($CO_{2}$): $(1 \times C) + (2 \times O) = (1 \times 12) + (2 \times 16) = 44$. Molar mass is $44\,g/mol$.

• Calculations using the Mole-Mass Formula:     * $\text{Moles} = \frac{\text{Mass}}{M_{r}}$     * $\text{Mass} = \text{Moles} \times M_{r}$     * $\text{Molar Mass } (M_{r}) = \frac{\text{Mass}}{\text{Moles}}$

• Worked Example: Mass of Zinc:     * Question: What is the mass of $0.250$ moles of zinc?     * Answer: $A_{r}$ of Zn from the Periodic Table is 65. Mass = $0.250\,mol \times 65\,g/mol = 16.25\,g$.

• Worked Example: Moles in Sucrose:     * Question: How many moles are in $2.64\,g$ of sucrose ($C_{12}H_{22}O_{11}$, $M_{r} = 342$)?     * Answer: Moles = $2.64 \div 342 = 7.72 \times 10^{-3}\,mol$.

Calculating Particles, Atoms, and Ions

• Worked Example: Magnesium Chloride ($MgCl_{2}$):     * Molecules in 1 mole: $1 \times 6.02 \times 10^{23} = 6.02 \times 10^{23}$.     * Atoms in 1 mole: Periodic formula has 3 atoms (1 Mg, 2 Cl). Total atoms = $3 \times 6.02 \times 10^{23} = 18.06 \times 10^{23}$.     * Chloride ions in 1 mole: 2 ions per unit. Total ions = $2 \times 6.02 \times 10^{23} = 12.04 \times 10^{23}$.     * Magnesium ions in 2 moles: 1 ion per unit. Total ions in 2 moles = $2 \times (1 \times 6.02 \times 10^{23}) = 12.04 \times 10^{23}$.

• Worked Example: Water ($H_{2}O$):     * Question: In $15.7\,g$ of water ($M_{r} = 18$), how many molecules and atoms are there?     * Molecules: Moles = $15.7 \div 18 = 0.872\,mol$. Molecules = $0.872 \times 6.02 \times 10^{23} = 5.25 \times 10^{23}$.     * Atoms: Each molecule has 3 atoms (2 H, 1 O). Total atoms = $3 \times 5.25 \times 10^{23} = 1.58 \times 10^{24}$.

Reacting Masses and Limiting Reactants

• Methodology for Reacting Mass Calculations:     1. Find the amount in moles of the known substance using its mass.     2. Use the balanced chemical equation to identify the molar ratio between reactants and products.     3. Convert the calculated moles back into grams using relative masses.

• Worked Example: Combustion of Magnesium:     * Equation: $2Mg(s) + O_{2}(g) \rightarrow 2MgO(s)$.     * Question: Calculate the mass of $MgO$ produced from $6.0\,g$ of $Mg$.     * Calculation: Moles of $Mg = 6 \div 24 = 0.25$. Ratio of $Mg:MgO$ is $2:2$ ($1:1$). Produced moles of $MgO = 0.25$. Mass of $MgO = 0.25 \times (24 + 16) = 10\,g$.

• Worked Example: Decomposition of Aluminium Oxide:     * Equation: $2Al_{2}O_{3} \rightarrow 4Al + 3O_{2}$.     * Question: Find mass of Al produced from 51 tonnes of $Al_{2}O_{3}$.     * Mass in grams: $51 \times 10^{6}\,g$. $M_{r}$ of $Al_{2}O_{3} = 102$. Moles = $51,000,000 \div 102 = 500,000$.     * Ratio: $2:4$ ($1:2$). Moles Al = $1,000,000$. Mass Al = $1,000,000 \times 27 = 27,000,000\,g$ ($27\,tonnes$).

• Limiting Reactants:     * A reaction stops when one reactant is consumed; this is the limiting reactant.     * The remaining reactant is the excess reactant.     * The amount of product corresponds directly to the amount of limiting reactant.     * Worked Example: Sodium and Sulfur:         * Question: $9.2\,g$ Sodium reacts with $8.0\,g$ Sulfur to produce $Na_{2}S$. Identify the limiting reactant.         * Step 1: Moles $Na = 9.2 \div 23 = 0.40$. Moles $S = 8.0 \div 32 = 0.25$.         * Step 2: Balanced equation: $2Na + S \rightarrow Na_{2}S$. Molar ratio of $Na:S$ is $2:1$.         * Step 3: $0.40\,mol$ of Na requires $0.20\,mol$ of S. Since $0.25\,mol$ of S is present, S is in excess and Na is the limiting reactant.

Calculating Concentration

• Definitions:     * Solute: Solid substance that dissolves into a liquid.     * Solvent: The liquid in which the solute dissolves.     * Solution: The resulting mixture of solute and solvent.     * Concentration: The amount of solute per specific volume of solvent. Units are $g/dm^{3}$ or $mol/dm^{3}$.

• Volume Conversions:     * $1\,dm^{3} = 1000\,cm^{3}$.     * $1\,dm^{3} = 1\,litre$.     * Convert $cm^{3}$ to $dm^{3}$: Divide by 1000.     * Convert $dm^{3}$ to $cm^{3}$: Multiply by 1000.

• Concentration Formulas:     * $\text{Concentration } (g/dm^{3}) = \frac{\text{Mass of solute } (g)}{\text{Volume of solution } (dm^{3})}$     * $\text{Concentration } (mol/dm^{3}) = \frac{\text{Moles of solute } (mol)}{\text{Volume of solution } (dm^{3})}$

• Worked Example: NaOH Mass Concentration:     * Question: $10\,g$ NaOH dissolved in $2\,dm^{3}$. Calculate concentration in $g/dm^{3}$.     * Answer: $10 \div 2 = 5\,g/dm^{3}$.

• Worked Example: NaOH Mole Concentration:     * Question: Concentration of NaOH when $80\,g$ is dissolved in $500\,cm^{3}$ of water.     * Moles: $M_{r}$ of $NaOH = 40$. $Moles = 80 \div 40 = 2$.     * Volume: $500 \div 1000 = 0.5\,dm^{3}$.     * Concentration: $2 \div 0.5 = 4\,mol/dm^{3}$.

Titration Calculations

• Purpose: Titrations analyze exact concentrations by determining volumes needed for neutralization (e.g., acid-base reactions).

• Worked Example: HCl vs. NaOH:     * Question: $25.0\,cm^{3}$ of HCl titrated against $12.1\,cm^{3}$ of $0.100\,mol/dm^{3}$ NaOH.     * Moles NaOH: $(12.1 \div 1000) \times 0.100 = 1.21 \times 10^{-3}\,mol$.     * Moles HCl: Ratio is $1:1$, so $1.21 \times 10^{-3}\,mol$.     * Concentration HCl: $1.21 \times 10^{-3} \div 0.025\,dm^{3} = 0.0484\,mol/dm^{3}$.

• Worked Example: Volume required for neutralization:     * Question: Volume of $0.50\,mol/dm^{3}$ $HNO_{3}$ to neutralize $25.00\,cm^{3}$ of $0.80\,mol/dm^{3}$ KOH.     * Moles KOH: $0.80 \times 0.025 = 0.02\,mol$.     * Moles $HNO_{3}$: Ratio $1:1$, so $0.02\,mol$.     * Volume $HNO_{3}$: $0.02 \div 0.50 = 0.040\,dm^{3} = 40\,cm^{3}$.

Empirical and Molecular Formulae

• Empirical Formula: The simplest whole-number ratio of atoms in a compound.     * Calculation Steps:         1. List elements/components.         2. Record masses or percentages.         3. Divide by relative atomic mass ($A_{r}$) to find moles.         4. Divide all mole values by the smallest mole value to find the ratio.         5. Convert to whole numbers if necessary.

• Worked Example: Carbohydrate X:     * Data: Carbon: $31.58\%$, Hydrogen: $5.26\%$. Oxygen calculation: $100 - 31.58 - 5.26 = 63.16\%$.     * Moles: C: $2.63$, H: $5.26$, O: $3.95$.     * Ratio (dividing by $2.63$): C: 1, H: 2, O: 1.5.     * Whole number ratio (multiply by 2): C: 2, H: 4, O: 3. Empirical formula: $C_{2}H_{4}O_{3}$.

• Molecular Formula: Indicates the actual number of atoms.     * Calculation: $Multiplier = \frac{\text{Relative formula mass of molecular formula}}{\text{Relative formula mass of empirical formula}}$.     * Worked Example: Empirical formula $C_{4}H_{10}S_{1}$ ($M_{r} = 90$). True $M_{r} = 180$. Multiplier = $180 \div 90 = 2$. Molecular formula: $C_{8}H_{20}S_{2}$.

Hydrated Salts and Water of Crystallisation

• Hydrated Salts: Salts containing water molecules within their crystalline structure (e.g., $CuSO_{4} \cdot 2H_{2}O$).

• Method: Heat a sample until a constant mass is reached to drive off water, then compare the mass of the anhydrous salt vs. mass lost (water).

• Worked Example: Hydrated Copper Sulfate:     * Data: $11.25\,g$ of hydrated salt becomes $7.19\,g$ of anhydrous salt.     * Mass of Water: $11.25 - 7.19 = 4.06\,g$.     * Moles: $CuSO_{4}$ ($159.5\,g/mol$): $7.19 \div 159.5 = 0.045$. Water ($18\,g/mol$): $4.06 \div 18 = 0.226$.     * Ratio: $0.045:0.226$ converts to approximately $1:5$. Formula: $CuSO_{4} \cdot 5H_{2}O$.

Percentage Yield, Purity, and Composition

• Percentage Yield:     * $\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$     * Yield is never 100% due to factors like: reactants left in equipment, reversible reactions, purification losses, side reactions, and transfer losses.

• Percentage Composition by Mass:     * $\text{Percentage Composition} = \frac{\text{Total mass of the element in the compound}}{\text{Relative formula mass of the compound}} \times 100$     * Example Iron in Iron(III) oxide ($Fe_{2}O_{3}$): Fe mass = $2 \times 56 = 112$. Total mass = 160. % Fe = $112 \div 160 \times 100 = 70\%$.

• Percentage Purity:     * $\text{Percentage Purity} = \frac{\text{Mass of pure substance}}{\text{Total mass of substance}} \times 100$     * Worked Example: $15\,g$ sample contains $13.5\,g$ lead(II) bromide. Purity = $13.5 \div 15 \times 100 = 90\%$.