Moles and Related Calculations

Mole is the unit of amount of substance.
One mole contains approximately 6.02 x 10²³ of specified entities: atoms, ions, molecules, or formula units. - This value is known as Avogadro's constant.

The mass in grams of 1 mole of a substance is equal to its relative atomic mass (Ar) or relative molecular mass (Mr) expressed in grams.
Formula: $ext{mass (g)} = ext{number of moles} imes ext{molar mass (g/mol)}$
For example: - Sodium Hydroxide (NaOH) relative formula mass: $ext{Mr (NaOH)} = 23 + 16 + 1 = 40 ext{ g/mol}$ - To find the number of moles in 60 g of NaOH: $ext{Number of moles} = \frac{ ext{mass}}{ ext{molar mass}} = \frac{60 ext{ g}}{40 ext{ g/mol}} = 1.5 ext{ moles}$

Mass (g) of one mole of various substances: - Carbon (C): 12 g, contains 6.02 x 10²³ carbon atoms - Iron (Fe): 56 g, contains 6.02 x 10²³ iron atoms - Hydrogen (H₂): 2 g, contains 6.02 x 10²³ H₂ molecules - Oxygen (O₂): 32 g, contains 6.02 x 10²³ O₂ molecules - Water (H₂O): 18 g, contains 6.02 x 10²³ H₂O molecules - Magnesium Oxide (MgO): 40 g, contains 6.02 x 10²³ MgO formula units - Calcium Carbonate (CaCO₃): 100 g, contains 6.02 x 10²³ CaCO₃ formula units - Silicon(IV) Oxide (SiO₂): 60 g, contains 6.02 x 10²³ SiO₂ formula units

There exists a constant molar volume for all gases at standard temperature and pressure (STP).
Molar volume = 24 dm³/mol.
Conversion involving various gases: - For every mole of gas, it occupies: 24 dm³ - Examples: - Hydrogen (H₂): 2 g, occupies 24 dm³ - Oxygen (O₂): 32 g, occupies 24 dm³ - Carbon Dioxide (CO₂): 44 g, occupies 24 dm³ - Ethane (C₂H₆): 30 g, occupies 24 dm³

Concentration can be measured in g/dm³ or mol/dm³.
Practical application: Often, volumes are calculated in cm³.
Relationship: 1 dm³ = 1000 cm³
Calculation Example: - Sodium Chloride (NaCl) in 1 dm³ has a concentration of 1 mol/dm³. - For pure sodium chloride, 58 g in 1 dm³ gives: $ext{Conc.} = 58 ext{ g/dm³}$

To find concentration of sodium hydroxide ( ext{NaOH}) in a solution containing 10 g in 250 cm³: 1. Molar mass of NaOH = 40 g/mol. 2. Moles of NaOH: $ext{Moles of NaOH} = \frac{10 ext{ g}}{40 ext{ g/mol}} = 0.25 ext{ mol}$ 3. Volume in dm³: $\frac{250 ext{ cm³}}{1000} = 0.25 ext{ dm³}$ 4. Calculate concentration: $ext{Concentration} = \frac{0.25 ext{ moles}}{0.25 ext{ dm³}} = 1 ext{ mol/dm³}$

Use of ratios from balanced equations: - General approach: mass of reactant → use ratio → moles of reactant → moles of product → mass of product. - Example Reaction: 4Al + 3O₂ ightarrow 2Al₂O₃ - Practice Example: Given 9.2 g of Aluminum, find mass of Aluminum Oxide produced step-by-step using the given molar ratio.

Concept: In a reaction, the limiting reactant determines the extent of reaction completion.
Practice Example: - Sodium ( ext{Na}) and Sulfur ( ext{S}) react: 2Na + S ightarrow Na₂S - Given 9.2 g Na and 8.0 g S, calculate moles of each reactant: - Sodium: 9.2 g / 23 g/mol = 0.40 moles - Sulfur: 8.0 g / 32 g/mol = 0.25 moles - Molar ratio of Na : S = 2 : 1, - Determine limiting reagent based on calculated moles.

Formula for percentage purity of a reactant: $ext{Percentage Purity} = \frac{ ext{Mass of Pure Reactant}}{ ext{Mass of Impure Reactant}} imes 100$
Practical Example: Calculate impurity of chalk reacted with HCl producing CO₂ at given volume.

Formula for percentage yield: $ext{Percentage Yield} = \frac{ ext{Mass of Actual Product}}{ ext{Mass of Theoretical Product}} imes 100$
Example reaction: - Zinc reacts with copper sulfate: ext{Zn (s) + CuSO₄(aq) ightarrow ZnSO₄(aq) + Cu(s)} - Find actual yield (4.8 g copper) vs. theoretical yield.

Steps to find empirical formula: 1. From percentage by mass, calculate masses of each element in 100 g of compound. 2. Calculate moles of each element and find simplest ratio.

To find molecular formula: 1. Determine the molar mass of empirical formula. 2. Find whole number ratio by dividing relative molecular mass (Mr) by empirical formula mass.

Water of crystallization involves the ratio of anhydrous salt to water in a hydrated compound.
Example procedure to compute water lost from hydrates: 1. Weigh the hydrated salt. 2. Dehydrate it. 3. Weigh the anhydrous salt. 4. Calculate molar masses and find water ratio.