Mole Concept and Stoichiometry

• The mole concept and stoichiometry involves the use of counting methods to quantify microscopic particles (atoms, molecules) and their relationships in chemical reactions.

Core Concepts

• Mass and volume are used to count microscopic particles.
• Stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.

The Atom

• Atoms are the smallest part of an element.
• Atoms are extremely small, exemplified by the fact that 10^15 atoms are needed to cover the head of a pin.
• Relative masses are more practical than actual masses due to the small size of atoms.

Relative Mass

• Most abundant hydrogen atom has an atomic mass unit of 1.0.
• The most abundant carbon atom has an atomic mass unit of 12.0.
• Carbon-12 is assigned 12 atomic mass units on the 12C scale.

Relative Atomic, Isotopic, Molecular, and Formula Masses

• Relative atomic, isotopic, molecular and formula masses are key terms to be defined

Relative Molecular Mass (Mr)

• Mr applies to molecules and is defined as the ratio of average mass of one molecule of a substance to 1/12 the mass of a carbon-12 isotope, the result is expressed on the 12C scale.
• As an example, the Mr of H2O is 18.0.

Relative Formula Mass

• Mr applies to ionic compounds and is the ratio of average mass of one formula unit of the compound to 1/12 the mass of a carbon-12 isotope, the result is expressed on the 12C scale.
• As an example the Mr of NaCl is 58.5

Relative Isotopic Mass (Ar)

• Ar is the ratio of the mass of one atom of an isotope to 1/12 the mass of a carbon-12 isotope, expressed on the 12C scale.

Relative Atomic Mass (Ar)

• Ar of an element is the ratio of the average mass of one atom of the element to 1/12 the mass of a carbon-12 isotope, the result is expressed on the 12C scale.
• Elements may have isotopes, thus, relative atomic mass considers the weighted average of isotopic masses based on abundance.

Calculating Relative Atomic Mass

• The weighted average requires the sum of (isotope mass * relative abundance) of two or more isotopes, all divided by 100
• Can use the equation: Ar = ((% abundance of isotope 1 * mass of isotope 1) + (% abundance of isotope 2 * mass of isotope 2)) / 100

Molecular and Empirical Formulae

• Molecular formula: actual number of atoms of each element in a molecule
• Empirical formula: simplest whole number ratio of different atoms in the compound
• Experimental data is needed to identify the elements when calculating empirical formula from experimental data
• Percentage composition of elements are often given in experimental results
• Additional information such as Mr is needed to find the molecular formula From the molecular formula, the empirical formula of the compound can be deduced by finding the ratios between the amounts of each element

Mole and Avogadro's Constant

• Moles are easier for chemists to use when counting small particles
• One mole of substance contains 6.02 x 10^23 (Avogadro constant) particles
• A mole of substance contains 6.02 x 10^23 elementary entities
• A mole is equal to 6.02 x 10^23 items

Molar Mass (M)

• Molar mass is the mass of one mole of substance and has a unit of g/mol
• Molar mass of an element is the relative atomic mass in g/mol
• Molar mass of a compound is the relative molecular or formula mass in g/mol
• Molar mass (M) is numerically the same as Mr or Ar, but has the unit g/mol

Molar Volume

• Equal volumes of all gases contain the same number of particles at the same temperature and pressure (Avogadro's Law)
• One mole of any gas occupies 22.7 at standard temperature and pressure (STP) or 24.0 at room temperature and pressure (RTP)
• Molar volume of gas at RTP is 24.0 dm^3
• Molar volume of gas at STP is 22.7 dm^3

Key Equations

• 1 cm³ = 1 x 10-3 dm³ = 1 x 10-6 m³
• Concentration (mol dm−3) = moles of solute / volume of solution (dm3)
• Moles = mass (g) / Mr or Ar
• Moles = Volume (dm3) / 24 (at RTP) or 22.7 (at STP)

Mole Relationships In Calculations

• Reagents react in a certain ratio to form products
• Mole ratios are used in balanced chemical equations to show the stoichiometric ratio of substances in a reaction, where the coefficients in the balanced equation indicate the mole ratio of substances involved

Real Life Examples

• Automobile airbags: Inflate through gas-producing chemical reactions to save lives with precise amounts of produced N2 gas
• CO2 scrubbing: CO2 removed from air in space shuttles through reaction with lithium hydroxide (LiOH).

Important Reactions

• When constructing equations do not alter the formula of compound/ions given, and balance only by changing the numbers (coefficient) in front of compounds/ions/elements
• Total atoms of each element on both sides of the equation must be the same (conservation of mass)
• These are some examples of basic chemical reactions: thermal decomposition, combustion, acid and base reaction, acid-carbonate reaction, ammonium carbonate and base reaction, precipitation of insoluble salts, acid-metal reaction an displacement reaction

Chemical reactions:

• Thermal Decomposition: CaCO3(s) → CaO(s) + CO2(g), heat is needed for reactants
• Combustion: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
• Acid + base reaction (forms salt and water): H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
• Acid + carbonate reaction (forms salt, water and carbon dioxide gas): 2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)
• Ammonium salt + base reaction (forms salt, water and ammonia gas): NH4Cl(aq) + NaOH(aq) → NaCl(aq) + NH3(g) + H2O(l), heat is needed for reactants
• Precipitation of insoluble salts: Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)
• *Acid-metal reaction (forms salt and hydrogen gas): Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
• Displacement reaction: Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s)
• Redox reactions: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

Limiting Reagents

• Excess reagents are not completely used up while one or more reagents are in limited quantities
• The amount of product formed is determined by the limiting reagent
• The reaction is stopped when the limiting reagent is consumed

Percentage Yield and Purity

• Percentage yield reflects actual yield compared to theoretical yield.
• Percentage yield is given by the equation: (actual mass of product / theoretical mass of product) x 100
• Percentage purity is (mass of pure substance / mass of impure substance) x 100

Reacting Volumes of Gases and Combustion Problems

• Equal volumes of gases contain equal number of moles of gases, under the same condition of temperature and pressure (Avogadro’s law)
• Hydrocarbons burn completely in sufficient oxygen to form only carbon dioxide gas and water
• CO2 product will be absorbed (reacted) with the by the base, when an excess amount of base (such as NaOH or KOH) is added the volume of excess base is added = volume of CO2 formed.

Balancing Combustion Equations

• Volume of liquid water is negligible when products of gaseous CO2 and H2O are cooled to room temperature.
• Balance combustion equations by setting the number of moles of hydrocarbon to 1

General Combustion Equation

• General formula for COMPLETE combustion of hydrocarbons:
  • CₓHᵧ(g) + (x + y/4)O₂(g) → xCO₂(g) + (y/2)H₂O(l)
  • CxHy(g) + O2(g) -> xCO2(g) + y/2 H2O(1)

Explore the mole concept and stoichiometry, focusing on quantifying microscopic particles and their relationships in chemical reactions. Learn about atoms, relative mass, and the importance of terms like relative atomic, isotopic, molecular, and formula masses.