Chemical & Physical Changes, Conservation Laws, Formulae, Concentration and Stoichiometry

Physical Change

• Definition: A change that can be seen or felt (e.g., changes in state, texture, shape, or appearance) but does not involve the breakup of the particles or the formation of new chemical substances.
• During a physical change the form of matter may change, its identity stays the same (no new chemical species are formed, only the arrangement or state of the existing substance changes).
• Core idea: The molecular structure of each substance is preserved; chemical bonds within molecules remain intact.
• Typical Examples:
  • Change of state (e.g., melting ice to liquid water, boiling liquid water to steam: $\text{H}<em>2\text{O}</em>{(\text{l})} \rightarrow \text{H}<em>2\text{O}</em>{(\text{g})}$). This involves overcoming intermolecular forces, not breaking covalent bonds.
  • Conduction: transfer of energy through a material without altering its chemical makeup (e.g., a metal spoon getting hot in soup).
  • Dissolving: one substance disperses evenly in another but its particles (molecules or ions) remain intact; it can often be reversed by physical means (e.g., evaporating the solvent).
  • Cutting paper, crushing a can, mixing sand and water.
• Key Characteristics to remember:
  1. Arrangement of particles
  • Molecules may re-arrange spatially (e.g., moving closer together or further apart as state changes) but chemical bonds don’t break. Intermolecular forces are altered, but intramolecular bonds are not.
  • Ex: Boiling water – individual water molecules ($\text{H}<em>2\text{O}$) move apart from each other, but the covalent bonds between hydrogen and oxygen atoms within each $\text{H}</em>2\text{O}$ molecule remain unbroken.
  1. Conservation of mass
  • Total mass, number of atoms, and number of molecules remain constant because no atoms are lost or gained, and no new molecules are formed.
  1. Energy changes
  • Energy exchanges are present but smaller than in chemical change. They typically involve overcoming relatively weak intermolecular forces, not strong chemical bonds.

Chemical Change

• Definition: Formation of new substances in a chemical reaction; one type of matter is transformed into another. This involves the breaking of old chemical bonds and the formation of new ones.
• Result: Products often display very different properties (e.g., color change, gas evolution, precipitate formation, temperature change, light emission) from the original reactants, indicating a fundamental change in composition.
• Canonical Examples:
  1. Decomposition of hydrogen peroxide
  • $2\text{H}<em>2\text{O}</em>2 \rightarrow 2\text{H}<em>2\text{O} + \text{O}</em>2$
  • In this reaction, the $\text{O–H}$ and $\text{O–O}$ bonds in $\text{H}<em>2\text{O}</em>2$ break, and new $\text{H–O}$ bonds (in water) and $\text{O–O}$ bonds (in oxygen gas) form.
  1. Synthesis of water
  • $2\text{H}<em>2 + \text{O}</em>2 \rightarrow 2\text{H}_2\text{O}$
  • Covalent bonds in $\text{H}<em>2$ and $\text{O}</em>2$ molecules break, and new $\text{H–O}$ covalent bonds form to create water molecules.
• Important Points:
  1. Arrangement of particles
  • Particles change identity due to the rearrangement of atoms and formation of new bonds. For example, each $\text{H}<em>2\text{O}</em>2$ molecule initially present breaks down to eventually form 2 $\text{H}<em>2\text{O}$ molecules and 1 $\text{O}</em>2$ molecule. Therefore, the number of molecules changes (e.g., 2 molecules of reactant yield 3 molecules of product).
  1. Energy changes
  • Large compared with physical change: substantial energy is absorbed to break existing bonds (endothermic process) and released when new bonds are formed (exothermic process). The net energy change can be significantly exothermic or endothermic.
  1. Reversibility
  • Generally hard to reverse (e.g., getting $\text{H}<em>2\text{O}</em>2$ back from $\text{H}<em>2\text{O}$ and $\text{O}</em>2$ is almost impossible under normal conditions, requiring complex processes or significant energy input).
  1. Mass conservation
  • Mass is still conserved according to the Law of Conservation of Mass, meaning the total mass of reactants equals the total mass of products. However, as noted, the number of molecules may differ before and after the reaction, but the number of atoms of each element remains constant.

Conservation Laws

• Conservation of mass & atoms
  • In any chemical reaction within a closed system (where no matter can enter or leave), total mass and total number of atoms for each element remain constant. Atoms are merely rearranged, not created or destroyed.
  • Shown numerically for $\text{2H}<em>2\text{O}</em>2 \rightarrow \text{2H}<em>2\text{O}+\text{O}</em>2$
  • Reactant mass (2 $\text{H}<em>2\text{O}</em>2$ molecules) $= 4(1.01) + 4(16.0) = 68.04\,\text{u}$ (atomic mass units)
  • Product mass (2 $\text{H}<em>2\text{O}$ molecules + 1 $\text{O}</em>2$ molecule) $= 2(18.02) + 2(16.0) = 68.04\,\text{u}$ (Mass is conserved)
  • Reactant atoms: 4 H & 4 O; Product atoms: 4 H & 4 O. (Atoms are conserved for each element).
• Law of Constant Composition (also known as the Law of Definite Proportions)
  • In a given pure chemical compound, the elements always combine in the same proportion by mass, regardless of the compound's source or method of preparation. This is because molecules of a specific compound always have the same fixed ratio of atoms.
  • Example: In any sample of pure water ($\text{H}_2\text{O}$), the mass ratio is approximately 94% O : 6% H. This ratio is constant because each water molecule always contains two hydrogen atoms for every one oxygen atom.
  • If two elements can combine to form multiple compounds (e.g., $\text{H}<em>2\text{O}$ vs $\text{H}</em>2\text{O}_2$), then the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. This observation leads to the law of multiple proportions.
• Volume relationships in gases (Gay-Lussac’s Law of Combining Volumes)
  • For gaseous reactants and products measured at the same temperature (T) and pressure (P), their volumes react or combine in small whole-number ratios. This law provided early evidence for the molecular nature of gases.

Representing Chemical Change

• Chemical reactions can be described at three main descriptive levels:
  1. Sentence (plain language): A descriptive narrative of the reaction.
  2. Word equation (names): Uses the names of reactants and products, often separated by an arrow.
  3. Symbolic / balanced equation (chemical symbols): Uses chemical formulas and symbols to represent reactants and products, showing the stoichiometric relationships.
• Examples:
  • For the reaction of Iron with Sulphur to form Iron Sulphide:
  • Sentence: Iron reacts with sulphur to form iron sulphide, often with the release of heat.
  • Word: Iron + sulphur $\rightarrow$ iron sulphide.
  • Symbols: $\text{Fe}<em>{(\text{s})} + \text{S}</em>{(\text{s})} \rightarrow \text{FeS}_{(\text{s})}$ (Balanced, showing states).
  • For the reaction $\text{4NH}<em>3 + \text{5O}</em>2 \rightarrow \text{4NO} + \text{6H}_2\text{O}$, where ammonia reacts with oxygen:
  • Sentence: Ammonia gas reacts with oxygen gas to produce nitrogen monoxide gas and water vapor.
  • Word: Ammonia + Oxygen $\rightarrow$ Nitrogen Monoxide + Water.
  • Symbols: $\text{4NH}<em>{3(\text{g})} + \text{5O}</em>{2(\text{g})} \rightarrow \text{4NO}<em>{(\text{g})} + \text{6H}</em>2\text{O}_{(\text{g})}$
• Terminology:
  • Reactants: Substances shown on the left side of the arrow; these are the starting materials that are consumed during the reaction.
  • Products: Substances shown on the right side of the arrow; these are the new substances that are formed as a result of the chemical reaction.
  • Coefficients: The numbers placed in front of chemical formulas in a balanced equation (e.g., the '4' in $\text{4NH}_3$) that indicate the relative number of moles or molecules of each substance.
  • Subscripts: The numbers within a chemical formula (e.g., the '3' in $\text{NH}_3$) that indicate the number of atoms of each element within a single molecule or formula unit.

Writing Chemical Formulae

• Chemical formula = a compact expression that represents the elements present in a compound and the ratio of atoms of each element.
• Each element is shown by its chemical symbol; a subscript number indicates the number of atoms of that element in one molecule or formula unit (a '1' is usually omitted).
• Key Skills needed for writing and balancing chemical equations:
  1. Recall common element symbols: Essential for converting names into chemical representations.
  2. Write correct reactant/product formulae: Requires knowledge of valencies, polyatomic ions, and common compound names to ensure the formula represents the correct composition.
  3. Balance using conservation laws: Adjust coefficients to ensure the number of atoms of each element is conserved on both sides of the equation. This reflects the Law of Conservation of Atoms.
  4. Add state symbols (s, l, g, aq): Indicate the physical state of each substance (solid, liquid, gas, or aqueous solution), providing important context for experimental conditions.
• Balancing principle: For a closed system, the total mass of reactants must equal the total mass of products ($\sum m<em>{\text{reactants}} = \sum m</em>{\text{products}}$), and more fundamentally, the number of atoms of each elemental type must match on both sides of the equation.
• Sample unbalanced practice equations (to be balanced):
  • $\text{Al}<em>{(\text{s})} + \text{O}</em>{2(\text{g})} \rightarrow \text{Al}<em>2\text{O}</em>{3(\text{s})}$
  • $\text{Mg}<em>{(\text{s})} + \text{HCl}</em>{(\text{aq})} \rightarrow \text{MgCl}<em>{2(\text{aq})} + \text{H}</em>{2(\text{g})}$
  • $\text{CH}<em>{4(\text{g})} + \text{O}</em>{2(\text{g})} \rightarrow \text{CO}<em>{2(\text{g})} + \text{H}</em>{2}\text{O}_{(\text{l})}$

Composition Calculations

• Four common problem-types encountered in quantitative chemistry:
  1. Given chemical formula $\rightarrow$ find percent by mass of each element.
  2. Given percent composition by mass $\rightarrow$ determine the empirical formula of the compound.
  3. Given combustion products (e.g., $\text{CO}<em>2$ and $\text{H}</em>2\text{O}$) from burning an organic compound $\rightarrow$ deduce the reactant formula (combustion analysis, often used for C, H, O compounds).
  4. Determine waters of crystallisation: Calculate the number of water molecules typically associated with ionic compounds in their hydrated solid form (hydrate problems).

Worked Example — % Composition of $\text{H}<em>2\text{SO}</em>4$ (Sulfuric Acid)

To find the percentage by mass of each element in $\text{H}<em>2\text{SO}</em>4$, we use the molar masses (H $\approx$ 1.01 g/mol, S $\approx$ 32.07 g/mol, O $\approx$ 16.00 g/mol):

• Mass of H: $2 \times 1.01 = 2.02\,\text{g}$
• Mass of S: $1 \times 32.07 = 32.07\,\text{g}$
• Mass of O: $4 \times 16.00 = 64.00\,\text{g}$
• Total molar mass of $\text{H}<em>2\text{SO}</em>4 = 2.02 + 32.07 + 64.00 = 98.09\,\text{g/mol}$
• Percentage by mass of each element:
  • H = $\frac{2.02}{98.09} \times 100\% \approx 2.06\%$
  • S = $\frac{32.07}{98.09} \times 100\% \approx 32.71\%$
  • O = $\frac{64.00}{98.09} \times 100\% \approx 65.23\%$ (Note: The percentages sum to 100% due to rounding).

Empirical Formula Example (Compound containing C, H, O)

Given: A compound contains 52.2% C, 13.0% H, and 34.8% O by mass. To find the empirical formula, assume a 100 g sample:

• Moles of C: $52.2\,\text{g C} \times \frac{1\,\text{mol C}}{12.01\,\text{g C}} \approx 4.35\,\text{mol C}$
• Moles of H: $13.0\,\text{g H} \times \frac{1\,\text{mol H}}{1.01\,\text{g H}} \approx 12.87\,\text{mol H}$
• Moles of O: $34.8\,\text{g O} \times \frac{1\,\text{mol O}}{16.00\,\text{g O}} \approx 2.175\,\text{mol O}$
• Divide by the least number of moles to find the simplest whole-number ratio:
  • C: $\frac{4.35}{2.175} \approx 2$
  • H: $\frac{12.87}{2.175} \approx 6$
  • O: $\frac{2.175}{2.175} \approx 1$
• The empirical formula is therefore $\text{C}<em>2\text{H}</em>6\text{O}$.

Lead Oxide Determination (Example of finding formula from mass data)

If 207 g of lead (Pb) reacts completely to form 239 g of a lead oxide, determine its empirical formula:

• Mass of Pb = $207\,\text{g}$
• Total mass of oxide = $239\,\text{g}$
• Mass of O = Total oxide mass - Mass of Pb = $239\,\text{g} - 207\,\text{g} = 32\,\text{g}$
• Moles of Pb: $207\,\text{g Pb} \times \frac{1\,\text{mol Pb}}{207.2\,\text{g Pb}} \approx 1\,\text{mol Pb}$
• Moles of O: $32\,\text{g O} \times \frac{1\,\text{mol O}}{16.00\,\text{g O}} = 2\,\text{mol O}$
• The whole-number ratio of Pb to O is $1 : 2$, so the formula is $\text{PbO}_2$.

Acetic Acid (Vinegar) - Empirical and Molecular Formulae

1. Empirical formula:
   • Masses in a 100 g sample of acetic acid: 39.9 g C, 6.7 g H, 53.4 g O.
   • Moles:
     • C: $39.9\,\text{g} / 12.01\,\text{g/mol} \approx 3.325\,\text{mol}$
     • H: $6.7\,\text{g} / 1.01\,\text{g/mol} \approx 6.634\,\text{mol}$
     • O: $53.4\,\text{g} / 16.00\,\text{g/mol} \approx 3.338\,\text{mol}$
   • Divide by the smallest (approx. 3.325) to get mole ratios: C $\approx 1$, H $\approx 2$, O $\approx 1$.
   • Empirical formula: $\text{CH}_2\text{O}$.
2. Molecular formula:
   • Empirical molar mass ($\text{CH}_2\text{O}$) = $12.01 + 2(1.01) + 16.00 = 30.03\,\text{g/mol}$.
   • If the actual molar mass of acetic acid is known to be $60.06\,\text{g/mol}$.
   • Multiplier = (Actual molar mass) / (Empirical molar mass) = $60.06 / 30.03 \approx 2$.
   • Multiply the subscripts in the empirical formula by 2 to get the molecular formula: $\text{C}<em>{1 \times 2}\text{H}</em>{2 \times 2}\text{O}<em>{1 \times 2} \rightarrow \text{C}</em>2\text{H}<em>4\text{O}</em>2$.
   • This is often written as $\text{CH}_3\text{COOH}$ (showing its carboxylic acid functional group).

Experimental / Classwork Prompts (for practice)

• Balance equations such as: $\text{Mg} + \text{O}<em>2 \rightarrow \text{MgO}$, $\text{H}</em>2\text{O}<em>2 \rightarrow \text{H}</em>2\text{O} + \text{O}<em>2$, $\text{C}</em>3\text{H}<em>8 + \text{O}</em>2 \rightarrow \text{CO}<em>2 + \text{H}</em>2\text{O}$, etc.
• Calculate percent-composition & perform mole calculations for various compounds like $\text{CaCl}_2$, $\text{ZnS}$, or complex chlorinated hydrocarbons.
• Solve problems involving hydration, such as determining the formula of a hydrated salt like copper(II) sulfate pentahydrate ($\text{CuSO}<em>4 \cdot 5\text{H}</em>2\text{O}$).

Molar Volumes of Gases

• Definition: One mole of any ideal gas occupies a specific volume of $22.4\,\text{dm}^3$ (or 22.4 L) at STP (Standard Temperature and Pressure).
  • STP is defined as $0\,\text{°C}$ (273.15 K) and $1\,\text{atm}$ (101.325 kPa).
  • This concept is crucial for stoichiometric calculations involving gases.

Solution Concentration (Molarity)

• Molarity (C) is a common measure of concentration, defined as the number of moles of solute dissolved per unit volume of solution.
• Concentration formula: $C = \frac{n}{V}$
  • where $n$ is the number of moles of solute (in mol) and $V$ is the volume of solution (in $\text{dm}^3$ or L).
  • Units of molarity are $\text{mol} \cdot \text{dm}^{-3}$ (or mol/L).
• Example 1: Calculate the concentration when 3.5 g of NaOH is dissolved in 2.5 $\text{dm}^3$ of solution.
  • Molar mass of NaOH is approximately 40.01 g/mol.
  • Moles of NaOH ($n$) = $\frac{3.5\,\text{g}}{40.01\,\text{g/mol}} \approx 0.0875\,\text{mol}$
  • Concentration ($C$) = $\frac{0.0875\,\text{mol}}{2.5\,\text{dm}^3} = 0.035\,\text{mol} \cdot \text{dm}^{-3}$
• Example 2: How much potassium permanganate ($\text{KMnO}_4$) is needed to prepare 1 $\text{dm}^3$ of a 0.2 $\text{mol} \cdot \text{dm}^{-3}$ solution?
  • Molar mass of $\text{KMnO}_4$ is approximately 158.04 g/mol.
  • Moles ($n$) = $$C \times V = 0.2\,\text{mol} \cdot \text{dm}^{-3} \times 1\,\text{dm}^3 = 0.2\