Avogadro's Mole & Mole Concept

Avogadro's Mole

• One mole of Carbon-12 atoms weighs exactly 0.012 kg, equivalent to 12 g
• 1 atomic mass unit (amu) is equal to 1 g divided by Avogadro's number (Nᴀᵥ)
• Using Nᴀᵥ showcases combinations by weight, while atoms always combine by number
• Avogadro's number (Nᴀᵥ) is 6.0221367 x 10²³, and is the SI count unit

Mole Concept

• One mole is the amount of matter in one unit of atomic or molecular weight
• One mole can be defined as the unit that represents 6.022 x 10²³ particles of the same matter
• 1 mole = atomic weight in grams
• 1 mole = 6.022 x 10²³ atoms/molecules; this is Avogadro's Number
• 1 mole = 22.4 liters at Standard Temperature and Pressure (STP)
• 1 mole of Hydrogen atoms equals 1 g and contains 6.023 x 10²³ hydrogen atoms
• 1 mole of Hydrogen molecules equals 2 grams and contains 6 x 10²³ hydrogen molecules
• 1 mole of Sodium equals 23 g and contains 6.023 x 10²³ sodium atoms

Relative Atomic Masses (RAM) and Relative Molecular Masses (RMM)

• RAM's standard is a carbon isotope with the symbol ¹²C, also known as C-12
• RMM's standard is also the C-12 isotope
• RAM is a relative scale used to compare the masses of different atoms, both RAM and RMM have no units
• The relative molecular mass of Hydrogen Chloride (HCl) is calculated with Hydrogen and Chloride RAMs of 1 and 35.45, respectively
• RMM of HCl = RAM of H + RAM of Cl = 1 + 35.45 = 36.45
• The formula weight of barium chloride dihydrate, BaCl₂.2H₂O, is calculated as follows:
• RMM = RMM of BaCl₂ + [2 x (RMM of H₂O)] = 137.3 + (2 x 35.45) + [2 x (2 + 16)] = 208.2 + 36 = 244.2

Mole Concept Diagram

• Mass is related to moles through "Molar Mass"
• Moles are related to the number of particles through "Nᴀᵥ = 6.02 x 10²³ "
• Moles are related to volume through "1 mole = 22.4 L"

Additional Mole Concept Information

• The molar mass calculation of the compound C₇H₁₄O₂:
• Molar Mass of C₇H₁₄O₂ = (2 x 12.01 g/mol) + (14 x 1.008 g/mol) + (2 x 16.00 g/mol) = 130.18 g/mol
• 2.4 x 10²⁴ molecules of carbon dioxide contains a certain number of moles
• 3 x 10²³ molecules of water contains a certain number of moles

Exercise Examples

• Calculate the volume for 18 g of O₂ at STP
• Calculate the number of N₂ molecules in 10 L of nitrogen gas
• Determine the number of moles of aluminum in 96.7 g of Al
• Calculate how many moles of NH₃ are in 77.5 g of NH₃
• Calculate the mass of 0.0250 mol of P₂O₅

Percent Composition of Compounds

• The composition of a compound can be described in two ways
• The number of its constituent atoms; for example, Methane contains 1 carbon atom and 4 hydrogen atoms
• The percentages by mass of its elements
• Mass percent can be obtained from a compound's formula by comparing the mass of each element in 1 mole to the total mass of 1 mole of the compound
• A molecule of ethanol (C₂H₅OH) contains:
• 2 mol C
• 6 mol H
• 1 mol O
• It is possible to compute the mass percent of each element in (C₁₀H₁₄O)

Empirical Formula and Molecular Formula

• Molecular Formula: The formula of a compound that describes the types of atoms and numbers of each type of atom
• The formula for Ethane is C₂H₆
• Empirical Formula: The formula of a compound expressing the smallest whole-number ratio of atoms in a compound
• The simplest formula is the formula for Ethane: CH₃
• A compound that contains the molecules C₄H₈O₄ has the same empirical formula as a compound that contains C₆H₁₂O₆ molecules
• CH₂O is the empirical formula for both
• It is possible to represent the molecular formula as a multiple of the empirical formula: C₆H₁₂O₆ = (CH₂O)₆

Calculating Empirical Formulas

• For an oxide of aluminum formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen:
• Convert the masses to numbers of moles, using atomic masses
• Start by dividing both numbers by the smallest of the two; this converts the smallest number to 1
• If no whole numbers are obtained, find a set of whole numbers to express the empirical formula
• If we multiply both 1.0 and 1.5 by 2, we get the integers we need
• If 3 mol O, then the empirical formula is AI₂O₃
• 32 grams of sulfur combine with 32 grams of oxygen to form sulfur dioxide, and the empirical formula is calculated:
• Convert the masses to numbers of moles, using atomic masses
• Start by dividing both numbers by the smallest of the two; this converts the smallest number to 1
• The emperical formula is SO₂

Exercise: Empirical Formulas

• When ethene is analyzed it contains 85.72% carbon and 14.28% hydrogen; the empirical formula is CH₂
• Cisplatin has the composition 65.02% platinum, 9.34% nitrogen, 2.02% hydrogen, and 23.63% chlorine; and the empirical formula is Pt₂NH₈Cl₂

Calculation of Molecular Formulas

• In order to obtain the molecular formula, it must be known that the molar mass, and empirical formula can be the same for different compounds
• CH₂O can be the empirical formula for C₂H₄O₂, C₃H₆O₃, C₄H₈O₄, C₅H₁₀O₅, C₆H₁₂O₆, etc
• To find the molecular formula of a hydrocarbon:
• First find the empirical formula
• Molecular formula = n × empirical formula, which can also be written as
• n = Molecular formula / empirical formula
• Divide the molar mass of M.F by the empirical formula mass to find n
• For example, 24 g of carbon combine with 6 g of hydrogen to form a hydrocarbon with RMM of 30:
• If the empirical formula is CH₃ then its molecular formula can be calculated to be C₂H₆

Exercise: Molecular formulas

• Determine the empirical & molecular formulas for a compound that gives the following masses upon analysis: Cl: 71.65g, C: 24.27g, H: 4.07g. The molar mass of the molecular formula is known to be 98.96 g/mol. The empirical formula is CH₂Cl, and the molecular formula is C₂H₄Cl₂

Stoichiometry Calculations

• In doing stoichiometry, always start with a balanced chemical equation for the reaction:
• C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
• To calculate these, there must be a conversion between masses and moles of substances
• To calculate what mass of oxygen will react with 96.1 g of propane (C₃H₈):
• 96.1 g C₃H₈ x (1 mol C₃H₈ / 44.1 g C₃H₈) = 2.18 mol C₃H₈
• After calculating the Mole Ratio:
• 5 mol O₂ / 1 mol C₃H₈
• Multiplying the number of moles of C₃H₈ by this factor:
• To calculate the mass of carbon dioxide produced when 96.1 g of propane are combusted when oxygen:
• 2.18 mol-C₃H₅ x (5 mol O₂ / 1 mol-C₃H₃) = 10.9 mol O₂
• Conversion of the 10.9 moles of O₂ to grams:
• 10.9 mol-O₂ x (32.0 g O₂ / 1 mol-O₂) = 349 g O₂

Calculating Amount of Reactant and Products

• It is possible to calculate the amount of magnesium oxide after burning 12 grams magnesium; if 1 mole of magnesium (Mg) is 24 grams and 1 mole of oxygen (O) is 16 grams).
• Magnesium + Oxygen → Magnesium oxide can be written as:
• Mg + O₂ → MgO
• Mg(s) + O₂(g) → MgO(s)
• 2Mg(s) + O₂(g) → 2MgO(s)
• Work out the mass of the moles and then in grams of each chemical:
• (2 x 24.305 g) + (2 × 15.999 g) → (2 x [24.305 g + 15.999 g])
• 48 g → 80 g
• Answer the above question:
• 48 g Mg → 80 g MgO
• 12 g Mg ← X g MgO
• X = (12 g × 80 g) / 48 g = 20 g of MgO
• If you want to make copper(II) sulfate by dissolving copper(II) oxide in sulfuric acid:
• How much copper(II) oxide would you need to make 50 grams of copper(II) sulfate?
• Copper (II) oxide + Sulfuric acid → Copper Sulphate + Water
• CuO + H₂SO₄ → CuSO₄ + H₂O
• CuO(aq) + H₂SO₄ (aq) → CuSO₄ (aq) + H₂O (1)
• CuO(aq) + H₂SO₄ (aq) → CuSO₄ (aq) + H₂O (1)
• Work out the mass of the moles and then in grams of each chemical:
• (16 g + 64 g) + (2 g + 32.06 g + 64 g) → (64 g + 32 g + 64 g) + (2 g +16 g)
• 80 g 98 g → 160 g → 18 g
• For the answer:
• 80 g CuO → 160 g CuSO₄
• X g CuO ← 50 g CuSO₄
• X = (80 g × 50 g) / 160 g = 25 g of CuO

Exercises: Amount of Reactant and Products

• Determine how much carbon dioxide is formed by burning 12 g of carbon in oxygen
• Determine how much iron (III) oxide would be made if 10 g of iron was burnt in oxygen

Composition of Solutions: Molarity

• Molarity (M) is defined as the amount of moles of solute per volume of solution in liters
• Molarity (M) = moles of solute / liters of solution
• Calculation: To calculate the molarity of a solution:
• 11.5 g NaOĤ × (1 mol NaOH/ 40.00 g NaOH) = 0.288 mol NaOH
• Molarity = mol solute / L solution, which can also be written as
• (0.288 mol NaOH / 1.50 L solution) = 0.192 M NaOH

Exercises: Molarity and Other Solution Calculations

• Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution; the answer is 1.6045 mol/L
• Adding more solvent to a solution is dilution
• Moles of solute after dilution = Moles of solute before dilution
• It is possible to calculate the volume of 16 M sulfuric acid is used to prepare 1.5 L of a 0.010 M H₂SO₄ solution
• It is possible to calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198 g of water, with a molar mass of 98.09 g
• It is possible to calculate the new concentration when 50.0 mL of water is added to 725.0 mL of 1.25 M NaCl; the answer is 1.169 M
• It is possible to calculate the molality of 128g of KBr is dissolved in 925g of water; the answer is 1.17m
• Mass percent = (mass of solute / mass of solution) × 100%, so: -Suppose a solution is prepared by dissolving 1.0 g of sodium chloride, and has a solution mass of 49 g, the mass percent is calculated as (1.0 g solute / 49 g solution) × 100% = 0.020 × 100% = 2.0% NaCl
• It is possible to calculate how many L of solute are in 52 L of a 6.3% solution
• The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of one component to the number of moles of all components present
• Mole fraction of component A = Xᴀ = NA/NA + NB -If a solution is prepared by mixing 2.2 moles of CH₂Cl₂, 1.2 moles of CH₂Br₂, 1.8 moles of CHCl₃ and 1.0 mole of CHBr₃ it is possible to find the mole fraction of CHBr₃ A solution can be prepared by mixing 1.00 g of ethanol (C₂H₅OH) with 100.0 g water to give a final volume of 101 mL, and then the molarity, mass percent, mole fraction, and molality of ethanol in this solution can be calculated