Chimie : notions de stœchiométrie

Comprehensive Guide to Stoichiometry

Introduction to Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions. It uses balanced chemical equations to determine the amounts of substances involved in chemical processes. This discipline bridges the gap between theoretical chemistry and practical calculations, enabling chemists to predict how much product will form from given reactants, or how much reactant is needed to produce a desired amount of product.

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Part 1: Chemical Formulas and Representation

Elements and Compounds

Elements are pure substances made of only one type of atom. They exist as individual atoms (such as sodium, Na, or calcium, Ca) or as molecules made of identical atoms bonded together. Examples include hydrogen (H₂), chlorine (Cl₂), nitrogen (N₂), copper (Cu), and iron (Fe).

Compounds are pure substances made of two or more types of elements chemically bonded together. Common examples include water (H₂O), ammonia (NH₃), sulfuric acid (H₂SO₄), and sodium chloride (NaCl). Compounds have fixed compositions and distinct properties different from their constituent elements.

Diatomic Elements

Seven elements exist naturally as diatomic molecules — molecules composed of two identical atoms bonded together. These are:

• Hydrogen (H₂)
• Nitrogen (N₂)
• Oxygen (O₂)
• Fluorine (F₂)
• Chlorine (Cl₂)
• Bromine (Br₂)
• Iodine (I₂)

When writing chemical equations, these elements must always be written in their diatomic form, not as single atoms. For example, oxygen gas is written as O₂, not O.

Molecular Formulas

A molecular formula shows the exact number and type of atoms in one molecule of a compound. It consists of:

• Chemical symbols indicating the type of atoms present
• Subscript numbers showing how many atoms of each element are in the molecule

Examples:

Compound	Molecular Formula	Composition
Water	H₂O	2 hydrogen atoms, 1 oxygen atom
Ammonia	NH₃	1 nitrogen atom, 3 hydrogen atoms
Methane	CH₄	1 carbon atom, 4 hydrogen atoms
Glucose	C₆H₁₂O₆	6 carbon atoms, 12 hydrogen atoms, 6 oxygen atoms
Sulfuric acid	H₂SO₄	2 hydrogen atoms, 1 sulfur atom, 4 oxygen atoms

Important notes on molecular formulas: When no subscript is written, it means there is one atom of that element. The order of elements in a formula matters and follows conventions (typically nonmetals before metals, or specific groupings for polyatomic ions).

Empirical Formulas

The empirical formula is the simplest whole number ratio of the different atoms or ions in a compound. It shows the relative proportions of elements but not necessarily the actual number of atoms in one molecule.

Key differences: Molecular formulas show the actual composition, while empirical formulas show only the simplest ratio. For example:

Compound	Molecular Formula	Empirical Formula
Carbon dioxide	CO₂	CO₂
Hydrogen peroxide	H₂O₂	HO
Dinitrogen tetroxide	N₂O₄	NO₂
Phosphorus pentoxide	P₄O₁₀	P₂O₅
Ethene	C₂H₄	CH₂
Glucose	C₆H₁₂O₆	CH₂O

Pattern identification: For organic compounds, the empirical formula is often different from the molecular formula. However, for ionic compounds, the molecular formula and empirical formula are usually identical. For example, sodium chloride has both a molecular formula of NaCl and an empirical formula of NaCl.

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Part 2: Ionic Compounds and Charge Balancing

Understanding Ions and Ionic Charges

Ions are charged particles formed when atoms gain or lose electrons. Cations are positively charged ions (formed by losing electrons), and anions are negatively charged ions (formed by gaining electrons).

Monatomic ions from main group elements:

• Group I elements form ions with a +1 charge (Li⁺, Na⁺, K⁺)
• Group II elements form ions with a +2 charge (Mg²⁺, Ca²⁺, Ba²⁺)
• Group VII elements form ions with a -1 charge (F⁻, Cl⁻, Br⁻, I⁻)
• Transition metals can form multiple ions: Cu²⁺, Fe²⁺, Fe³⁺, Zn²⁺

Polyatomic Ions

Polyatomic ions are ions containing more than one atom bonded together. Common polyatomic ions include:

Ion	Formula	Charge
Hydrogen ion	H⁺	+1
Ammonium ion	NH₄⁺	+1
Nitrate ion	NO₃⁻	-1
Hydroxide ion	OH⁻	-1
Carbonate ion	CO₃²⁻	-2
Sulfate ion	SO₄²⁻	-2

Deducing Ionic Compound Formulas

Ionic compounds have no overall charge — the positive charge from cations must be balanced by the negative charge from anions. To write the formula of an ionic compound:

Method 1: Charge Balance Method

1. Identify the cation and its charge
2. Identify the anion and its charge
3. Determine how many of each ion are needed so the total positive charge equals the total negative charge
4. Write the cation first, then the anion

Example: Iron(II) sulfate

• Fe²⁺ has a +2 charge
• SO₄²⁻ has a -2 charge
• One Fe²⁺ balances one SO₄²⁻: (+2) + (-2) = 0
• Formula: FeSO₄

Example: Zinc chloride

• Zn²⁺ has a +2 charge
• Cl⁻ has a -1 charge
• Two Cl⁻ ions are needed to balance one Zn²⁺: (+2) + 2×(-1) = 0
• Formula: ZnCl₂

Method 2: Swap-and-Drop Method

When ions have different charges, swap their charges and use them as subscripts, then simplify if needed:

• Cu²⁺ and Cl⁻: swap to get CuCl₂ (no simplification needed)
• Mg²⁺ and O²⁻: swap to get MgO₂, which simplifies to MgO
• Al³⁺ and S²⁻: swap to get AlS₃... wait, this needs rebalancing: Al₂S₃ (2 Al give +6, 3 S give -6)

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Part 3: Writing and Balancing Chemical Equations

Word Equations

A word equation describes a chemical reaction using the names of reactants and products instead of chemical symbols. The general format is:

Reactants → Products

Examples:

• Sodium hydroxide + Hydrochloric acid → Sodium chloride + Water
• Carbon + Oxygen → Carbon dioxide
• Ammonia + Nitric acid → Ammonium nitrate

Symbol Equations

A symbol equation (or chemical equation) uses chemical formulas instead of names. When writing symbol equations, follow these guidelines:

• Place reactants on the left of an arrow, products on the right
• Write nonmetals that exist as diatomic molecules (H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂) correctly
• Include state symbols in parentheses:
  • (s) = solid
  • (l) = liquid
  • (g) = gas
  • (aq) = aqueous (dissolved in water)

Example unbalanced equation:

H₂ + O₂ → H₂O

Balancing Chemical Equations

Chemical equations must be balanced to obey the Law of Conservation of Mass: mass cannot be created or destroyed in a chemical reaction. A balanced equation has the same number of atoms of each element on both sides.

Steps to balance an equation:

1. Write the unbalanced equation with correct formulas
2. Count atoms of each element on both sides
3. Add coefficients (numbers before formulas) to balance atoms
4. Start with the most complex compound
5. Balance one element at a time
6. Use fractions if necessary, then multiply through to get whole numbers
7. Never change subscripts — only adjust coefficients

Important tip: When a group of atoms (like NO₃⁻) appears unchanged on both sides, count the entire group as one entity rather than individual atoms.

Balancing examples:

Example 1: Hydrogen combustion

Unbalanced: H₂ + O₂ → H₂O

Balanced: 2H₂ + O₂ → 2H₂O

Example 2: Iron oxidation

Unbalanced: Fe + O₂ → Fe₂O₃

Balanced: 4Fe + 3O₂ → 2Fe₂O₃

Example 3: Methane combustion

Unbalanced: CH₄ + O₂ → CO₂ + H₂O

Balanced: CH₄ + 2O₂ → CO₂ + 2H₂O

Example 4: Magnesium oxide with nitric acid

MgO(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂O(l)

Check: 1 Mg, 1 O + 2 H, 2 N, 6 O → 1 Mg, 2 N, 6 O + 2 H, 1 O. Balanced ✓

Ionic Equations

Ionic equations show the actual particles involved in a reaction. In aqueous solution, ionic compounds dissociate (separate) into their component ions.

Steps to write an ionic equation:

1. Write the full balanced symbol equation
2. Identify which compounds are ionic and dissociate them into ions
3. Remove spectator ions (ions that appear unchanged on both sides)
4. Write the net ionic equation showing only the particles that actually react

Example 1: Chlorine displacing iodide ions

Full balanced equation: 2KI(aq) + Cl₂(aq) → 2KCl(aq) + I₂(aq)

Ionic form: 2K⁺(aq) + 2I⁻(aq) + Cl₂(aq) → 2K⁺(aq) + 2Cl⁻(aq) + I₂(aq)

Net ionic equation: 2I⁻(aq) + Cl₂(aq) → 2Cl⁻(aq) + I₂(aq)

(K⁺ ions are spectator ions and removed)

Example 2: Neutralization of hydrochloric acid and sodium hydroxide

Full balanced equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Ionic form: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l)

Net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l)

Example 3: Precipitation of silver chloride

Full balanced equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Ionic form: Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

Net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

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Part 4: Relative Atomic Mass and Relative Molecular Mass

Relative Atomic Mass (Aᵣ)

Relative atomic mass (Aᵣ) is the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of carbon-12. It is found on the periodic table and is usually a whole number or simple decimal.

Key facts about Aᵣ:

• It is always larger than the atomic number (except for hydrogen)
• It reflects the weighted average of all naturally occurring isotopes
• For single atoms, Aᵣ is a dimensionless number

Examples of relative atomic masses:

Element	Symbol	Aᵣ
Hydrogen	H	1
Carbon	C	12
Nitrogen	N	14
Oxygen	O	16
Sodium	Na	23
Magnesium	Mg	24
Aluminium	Al	27
Sulfur	S	32
Chlorine	Cl	35.5
Iron	Fe	56

Relative Molecular Mass (Mᵣ)

Relative molecular mass (Mᵣ) is the sum of the relative atomic masses of all atoms in a molecule. For ionic compounds, it is called the relative formula mass.

Formula:

Mᵣ = sum of (number of atoms × Aᵣ) for each element

Examples of calculating Mᵣ:

Oxygen gas (O₂):

Mᵣ = 2 × 16 = 32

Water (H₂O):

Mᵣ = (2 × 1) + (1 × 16) = 2 + 16 = 18

Sodium carbonate (Na₂CO₃):

Mᵣ = (2 × 23) + (1 × 12) + (3 × 16) = 46 + 12 + 48 = 106

Calcium hydroxide (Ca(OH)₂):

Mᵣ = (1 × 40) + (2 × 16) + (2 × 1) = 40 + 32 + 2 = 74

Ammonium sulfate ((NH₄)₂SO₄):

Mᵣ = (2 × 14) + (8 × 1) + (1 × 32) + (4 × 16) = 28 + 8 + 32 + 64 = 132

Common error: When calculating Mᵣ for compounds with polyatomic ions in parentheses like Ca(OH)₂, multiply the Aᵣ of each atom within the parentheses by the subscript after the parentheses. For example, in Ca(OH)₂, there are 2 oxygen atoms and 2 hydrogen atoms.

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Part 5: The Law of Conservation of Mass and Reacting Masses

The Law of Conservation of Mass

The Law of Conservation of Mass states that mass cannot be created or destroyed in a chemical reaction. Therefore, the total mass of reactants must equal the total mass of products.

Application in stoichiometry: If we know the masses of reactants in a balanced chemical equation, we can calculate the mass of products (or vice versa) using the principle that:

Total mass of reactants = Total mass of products

Calculating Reacting Masses

To calculate reacting masses without using the mole concept:

1. Write the balanced chemical equation
2. Calculate the relative atomic/molecular masses from the equation
3. Determine the mass ratio of reactants to products
4. Use proportions to calculate unknown masses

Example: Calcium burning in oxygen

Balanced equation: 2Ca(s) + O₂(g) → 2CaO(s)

Ar: Ca = 40, O = 16

Masses in the equation:

• Reactants: (2 × 40) + (2 × 16) = 80 + 32 = 112 mass units
• Products: 2 × (40 + 16) = 2 × 56 = 112 mass units

Mass ratio: 80 g Ca + 32 g O₂ → 112 g CaO

If 40 kg of calcium reacts, then:

• Oxygen needed: (32/80) × 40 = 16 kg
• Calcium oxide formed: (112/80) × 40 = 56 kg

Key insight: The mass ratio is independent of the units used. If the equation shows 80 g of Ca produces 112 g of CaO, then 80 tonnes of Ca also produces 112 tonnes of CaO.

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Part 6: The Mole Concept and Avogadro's Constant

What is a Mole?

The mole (symbol: mol) is the SI unit of amount of substance. It provides a way to count particles at the atomic scale by using measurable macroscopic quantities.

One mole contains 6.02 × 10²³ particles — this number is called the Avogadro constant. The particles can be atoms, molecules, ions, electrons, or any other specified entity.

Examples:

• 1 mole of sodium (Na) = 6.02 × 10²³ sodium atoms
• 1 mole of hydrogen gas (H₂) = 6.02 × 10²³ hydrogen molecules
• 1 mole of sodium chloride (NaCl) = 6.02 × 10²³ formula units of NaCl
• 1 mole of electrons = 6.02 × 10²³ electrons

Molar Mass

The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

Key relationship: The molar mass of a substance (in grams) is numerically equal to its relative atomic mass (for elements) or relative molecular mass (for compounds).

Examples:

• Molar mass of carbon = 12 g/mol (since Aᵣ = 12)
• Molar mass of water (H₂O) = 18 g/mol (since Mᵣ = 18)
• Molar mass of sodium chloride (NaCl) = 58.5 g/mol (since Mᵣ = 58.5)
• Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol (since Mᵣ = 180)

Calculating the Number of Moles

The fundamental equation relating moles, mass, and molar mass is:

Where:

• = number of moles (mol)
• = mass of substance (g)
• = molar mass (g/mol)

Example 1: Moles in a mass of water

How many moles are in 36 g of water? (Molar mass of H₂O = 18 g/mol)

Example 2: Moles in oxygen gas

How many moles are in 64 g of oxygen gas? (Molar mass of O₂ = 32 g/mol)

Example 3: Moles in a complex compound

How many moles are in 2.64 g of sucrose (C₁₂H₂₂O₁₁)? (Molar mass = 342 g/mol)

Calculating Mass from Moles

Rearranging the mole formula gives:

Example: What is the mass of 5 moles of carbon dioxide? (Molar mass of CO₂ = 44 g/mol)

Calculating the Number of Particles

The Avogadro constant allows conversion between moles and number of particles:

Example 1: Atoms in a pure element

How many copper atoms are in 3.2 g of copper? (Molar mass of Cu = 64 g/mol)

• Moles:
• Atoms:

Example 2: Ions in an ionic compound

How many chloride ions (Cl⁻) are in 1 mole of magnesium chloride (MgCl₂)?

• Each MgCl₂ formula unit contains 2 Cl⁻ ions
• 1 mole of MgCl₂ = formula units
• Number of Cl⁻ ions = ions

Example 3: Ions in multiple moles

For sodium sulfate (Na₂SO₄), how many sodium ions are in 2 moles?

• Each Na₂SO₄ formula unit contains 2 Na⁺ ions
• Number of Na⁺ =
• Number of Na⁺ = ions

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Part 7: Molar Volume of Gases and Avogadro's Law

Avogadro's Law

Avogadro's Law states that at the same temperature and pressure, equal amounts (in moles) of different gases occupy the same volume of space.

This means that one mole of any gas occupies the same volume as one mole of any other gas under the same conditions.

Molar Gas Volume

At room temperature and pressure (RTP — defined as 20°C and 1 atmosphere), one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³).

This is called the molar gas volume at RTP.

Examples:

• 1 mole of O₂ at RTP = 24 dm³
• 1 mole of CO₂ at RTP = 24 dm³
• 1 mole of N₂ at RTP = 24 dm³
• 1 mole of H₂ at RTP = 24 dm³

Calculating Moles from Gas Volume

Using the molar gas volume:

(when volume is in dm³)

Or:

(when volume is in cm³)

Example 1: Moles from volume in dm³

How many moles of gas are in 48 dm³ of oxygen at RTP?

Example 2: Moles from volume in cm³

How many moles of gas are in 1200 cm³ of sulfur dioxide at RTP?

Calculating Gas Volume from Moles

(volume in dm³)

Or:

(volume in cm³)

Example: What volume is occupied by 5.4 moles of oxygen gas at RTP?

or

Unit Conversions for Volume

When working with gas volumes, frequently convert between cm³ and dm³:

To convert:

• cm³ to dm³: divide by 1000
• dm³ to cm³: multiply by 1000

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Part 8: Stoichiometric Calculations Using Moles

Three-Step Mole Calculation Process

To solve stoichiometry problems involving masses and the balanced equation:

1. Convert reactant mass to moles: Use for the known reactant
2. Use the molar ratio from the balanced equation: Identify the coefficient ratio between the known and unknown substances
3. Convert moles of product to mass: Use for the desired product

Example 1: Magnesium combustion

Calculate the mass of magnesium oxide produced when 6.0 g of magnesium burns completely in oxygen.

2Mg(s) + O₂(g) → 2MgO(s)

Step 1: Convert magnesium mass to moles

Step 2: Use molar ratio from balanced equation

From the equation: 2 mol Mg → 2 mol MgO (ratio 1:1)

Therefore: 0.25 mol Mg → 0.25 mol MgO

Step 3: Convert MgO moles to mass

Example 2: Hydrogen and oxygen reaction

Calculate the mass of water produced when 4.0 g of hydrogen reacts completely with oxygen.

2H₂(g) + O₂(g) → 2H₂O(l)

Step 1: Moles of hydrogen

Step 2: Molar ratio

From the equation: 2 mol H₂ → 2 mol H₂O (ratio 1:1)

Therefore: 2 mol H₂ → 2 mol H₂O

Step 3: Convert H₂O moles to mass

Example 3: Aluminium oxide decomposition

Calculate the maximum mass of aluminium that can be produced from 51 g of aluminium oxide.

2Al₂O₃ → 4Al + 3O₂

Step 1: Moles of aluminium oxide

Step 2: Molar ratio

From the equation: 2 mol Al₂O₃ → 4 mol Al (ratio 2:4 or 1:2)

Therefore: 0.5 mol Al₂O₃ → 1 mol Al

Step 3: Convert Al moles to mass

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Part 9: Limiting Reactants and Excess Reactants

Concept of Limiting and Excess Reactants

In most chemical reactions, reactants are not present in exact stoichiometric ratios. One reactant will run out first, while another remains after the reaction stops.

• The limiting reactant (or limiting reagent) is the reactant that is completely consumed first, determining how much product can form
• The excess reactant is the reactant that remains after the reaction is complete

Key principle: The amount of product depends on the limiting reactant, not the excess reactant. A reaction cannot continue once the limiting reactant is used up, even if excess reactant remains.

Identifying the Limiting Reactant

Method:

1. Convert the mass of each reactant to moles
2. Write the balanced equation and identify the molar ratio between reactants
3. Determine which reactant "runs out" first by comparing molar amounts to the stoichiometric ratio

Example: Sodium reacting with sulfur

9.2 g of sodium reacts with 8.0 g of sulfur to produce sodium sulfide (Na₂S). Which is the limiting reactant?

Balanced equation: 2Na + S → Na₂S

Step 1: Convert to moles

Step 2: Identify molar ratio

From the equation: 2 mol Na : 1 mol S

Step 3: Compare to available moles

We have 0.40 mol Na and 0.25 mol S.

To react with 0.40 mol Na, we would need:

We have 0.25 mol S, which is more than 0.20 mol needed.

Therefore: Sodium is the limiting reactant, and sulfur is in excess.

Example 2: Magnesium and oxygen reaction

12.0 g of magnesium reacts with 14.4 g of oxygen to produce magnesium oxide (MgO). Which is limiting?

Balanced equation: 2Mg + O₂ → 2MgO

Moles:

Molar ratio: 2 mol Mg : 1 mol O₂

To react with 0.5 mol Mg:

We have 0.45 mol O₂, which is more than 0.25 mol needed.

Therefore: Magnesium is the limiting reactant.

Calculating Product from Limiting Reactant

Once the limiting reactant is identified, use it (not the excess reactant) to calculate the amount of product formed.

Example: In the magnesium-oxygen reaction above, how much MgO is produced?

The limiting reactant is Mg (0.5 mol).

From the equation: 2 mol Mg → 2 mol MgO (ratio 1:1)

Therefore: 0.5 mol Mg → 0.5 mol MgO

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Part 10: Solutions, Concentration, and Titrations

Terminology in Solutions

Solute: A solid substance that dissolves into a liquid. Amounts are measured in grams (g) or moles (mol).

Solvent: The liquid that the solute dissolves in. Amounts/volumes are measured in cm³ or dm³. Water is the most common solvent.

Solution: The mixture formed when a solute dissolves completely in a solvent. Amounts/volumes are measured in cm³ or dm³.

Concentration: A measure of the amount of solute in a specific volume of solution. It can be expressed as mass concentration (g/dm³) or molar concentration (mol/dm³).

Mass Concentration (g/dm³)

Mass concentration is calculated using:

Example 1: Sodium hydroxide solution

A student dissolved 10 g of sodium hydroxide in enough water to make 2 dm³ of solution. Calculate the concentration.

Example 2: Copper sulfate solution with unit conversion

2.1 g of copper sulfate is dissolved in 1.5 dm³ of water. Calculate the concentration.

Example 3: Mass concentration with volume in cm³

204 mg of iodine is dissolved in 100 cm³ of ethanol. Calculate the concentration in g/dm³.

Convert units: mass = 204 mg = 0.204 g; volume = 100 cm³ = 0.1 dm³

Molar Concentration (mol/dm³)

Molar concentration (molarity) is calculated using:

Conversion between mass and molar concentration:

• To convert from g/dm³ to mol/dm³: divide by molar mass
• To convert from mol/dm³ to g/dm³: multiply by molar mass

Example 1: Moles from concentration and volume

Calculate the amount of solute in moles present in 2.5 dm³ of a solution with concentration 0.2 mol/dm³.

Example 2: Molar concentration from mass and volume

Calculate the concentration in mol/dm³ when 80 g of sodium hydroxide (NaOH) is dissolved in 500 cm³ of water.

Step 1: Calculate Mᵣ of NaOH

Step 2: Calculate moles

Step 3: Convert volume to dm³

Step 4: Calculate concentration

Titrations and Neutralization Reactions

Titration is a laboratory method used to determine the concentration of a solution by reacting it with a solution of known concentration. The process finds the point where reactants have been completely used up — the end-point.

Acid-base titrations: An acid reacts with an alkali until the acid's hydrogen ions are exactly neutralized by the alkali's hydroxide ions.

Procedure for titrations:

1. Measure a known volume of one solution (usually in a conical flask)
2. Add a known concentration solution from a burette until the end-point is reached (indicated by an indicator color change)
3. Record the volume used
4. Repeat to obtain consistent results
5. Calculate using the volumes and concentrations

Titration calculation steps:

1. Write the balanced equation for the reaction
2. Calculate moles of the known solution using
3. Use the molar ratio from the balanced equation to find moles of the unknown solution
4. Calculate the concentration of the unknown solution

Example 1: Hydrochloric acid titration

25.0 cm³ of hydrochloric acid was titrated against 0.100 mol/dm³ sodium hydroxide. 12.1 cm³ of NaOH was required for complete reaction. Determine the concentration of HCl.

Balanced equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Step 1: Moles of NaOH

Step 2: Molar ratio

From equation: 1 mol HCl : 1 mol NaOH

Therefore:

Step 3: Concentration of HCl

Example 2: Sulfuric acid with different ratio

25.00 cm³ of 0.15 mol/dm³ barium hydroxide was required to neutralize 12.80 cm³ of nitric acid. Calculate the concentration of HNO₃.

Balanced equation: Ba(OH)₂(aq) + 2HNO₃(aq) → Ba(NO₃)₂(aq) + 2H₂O(l)

Step 1: Moles of Ba(OH)₂

Step 2: Molar ratio and moles of HNO₃

From equation: 1 mol Ba(OH)₂ : 2 mol HNO₃

Therefore:

Step 3: Concentration of HNO₃

Calculating volume required in titration:

Example: Calculate the volume of 0.50 mol/dm³ nitric acid required to neutralize 25.00 cm³ of 0.80 mol/dm³ potassium hydroxide.

Balanced equation: KOH(aq) + HNO₃(aq) → KNO₃(aq) + H₂O(l)

Step 1: Moles of KOH

Step 2: Molar ratio

From equation: 1 mol KOH : 1 mol HNO₃, so

Step 3: Volume of HNO₃

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Part 11: Calculating Empirical and Molecular Formulas

Empirical Formula Calculations

To determine the empirical formula from experimental data showing the mass of each element:

1. Write the element symbols
2. Record the mass of each element in grams
3. Divide each mass by the element's atomic mass to get moles
4. Divide all mole values by the smallest to get the molar ratio
5. If ratios are not whole numbers, multiply by a small integer (2, 3, 4, etc.) until all are whole
6. Write the empirical formula using the whole number ratios as subscripts

Example 1: Simple compound

A sample of a compound contains 10 g of hydrogen and 80 g of oxygen. Determine the empirical formula.

Element	Mass (g)	Aᵣ	Moles	Ratio
H	10	1	10 ÷ 1 = 10	10 ÷ 5 = 2
O	80	16	80 ÷ 16 = 5	5 ÷ 5 = 1

Empirical formula: H₂O

Example 2: Compound requiring whole number adjustment

A compound is found to contain the following mass percentages: C = 40%, H = 6.7%, O = 53.3%. Find the empirical formula.

Assume 100 g sample: C = 40 g, H = 6.7 g, O = 53.3 g

Element	Mass (g)	Aᵣ	Moles	Ratio
C	40	12	40 ÷ 12 = 3.33	3.33 ÷ 1.67 = 2
H	6.7	1	6.7 ÷ 1 = 6.7	6.7 ÷ 1.67 = 4
O	53.3	16	53.3 ÷ 16 = 3.33	3.33 ÷ 1.67 = 2

Empirical formula: C₂H₄O

Molecular Formula from Empirical Formula

To find the molecular formula when the empirical formula and molecular mass are known:

1. Calculate the relative formula mass of the empirical formula
2. Divide the molecular mass by the empirical formula mass:
3. Multiply each subscript in the empirical formula by this value to get the molecular formula

Example: The empirical formula of compound Y is C₂H₄O. The molecular mass is 88 g/mol. Determine the molecular formula.

Step 1: Empirical formula mass

Step 2: Find the factor

Step 3: Molecular formula

(C₂H₄O) × 2 = C₄H₈O₂

Key insight: If the empirical and molecular masses are equal, the molecular formula is the same as the empirical formula. If the molecular mass is a multiple of the empirical mass, multiply all subscripts by that multiple.

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Part 12: Yield and Purity Calculations

Theoretical Yield vs. Actual Yield

Theoretical yield is the maximum amount of product that could form if all reactants were converted to products under ideal conditions. It is calculated using stoichiometric calculations based on the limiting reactant.

Actual yield is the amount of product actually obtained from an experiment. It is always less than the theoretical yield due to practical limitations.

Why Actual Yield is Less Than Theoretical

Several factors prevent 100% yield:

• Reactants may remain on glassware surfaces and not fully react
• Many reactions are reversible — products convert back to reactants, so the reaction never goes to completion
• Products may be lost during separation and purification (e.g., during filtration, evaporation, or distillation)
• Side reactions may occur where reactants form unwanted products instead
• Products can react with air or water, decomposing or dissolving
• Transfer of products between containers causes mechanical loss

Percentage Yield

Percentage yield compares the actual yield to the theoretical yield:

Percentage yield is always between 0% and 100%.

Example: A student prepared 1.6 g of dry copper(II) sulfate crystals. The theoretical yield was 2.0 g. Calculate the percentage yield.

Common error: If you calculate a percentage yield greater than 100%, you have made a calculation error. The most frequent mistake is dividing theoretical by actual instead of actual by theoretical. Simply swap the numbers and recalculate.

Percentage Composition by Mass

Percentage composition expresses the mass of each element as a percentage of the total compound mass:

Example 1: Water

H₂O has Mᵣ = 18

Percentage of H:

Percentage of O:

Example 2: Iron(III) oxide

Fe₂O₃ contains: 2 Fe atoms (mass = 2 × 56 = 112) and 3 O atoms (mass = 3 × 16 = 48)

Mᵣ = 112 + 48 = 160

Percentage of Fe:

Percentage of O:

Example 3: Ammonium nitrate (complex ion)

NH₄NO₃ contains: 2 N atoms (2 × 14 = 28), 4 H atoms (4 × 1 = 4), 3 O atoms (3 × 16 = 48)

Mᵣ = 28 + 4 + 48 = 80

Percentage of N:

Percentage of H:

Percentage of O:

Percentage Purity

Percentage purity indicates how much of a sample is the desired pure substance versus contaminants or impurities:

Example: A 15 g sample of lead(II) bromide was found to contain only 13.5 g of pure lead(II) bromide. Calculate the percentage purity.

Interpretation: The sample is 90% pure lead(II) bromide and contains 10% impurities.

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Summary of Key Concepts and Relationships

Fundamental Equations

Concept	Equation	Units
Number of moles		mol = g ÷ g/mol
Mass from moles		g = mol × g/mol
Concentration (mass)		g/dm³ = g ÷ dm³
Concentration (molar)		mol/dm³ = mol ÷ dm³
Gas volume		dm³ = mol × 24 dm³/mol
Number of particles		particles = mol × 6.02 × 10²³
Percentage yield		%
Percentage composition		%
Percentage purity		%

Important Constants and Conversions

• Avogadro constant: particles/mol
• Molar gas volume at RTP: or
• RTP conditions:
• Volume conversion:
• Relative atomic mass relationship: Aᵣ (numerically) = molar mass (in grams)

Problem-Solving Strategies

For reacting mass calculations:

1. Write balanced equation
2. Calculate molar masses
3. Convert known mass to moles
4. Use molar ratio from equation
5. Convert unknown moles to mass

For limiting reactant problems:

1. Convert all masses to moles
2. Identify the molar ratio from the balanced equation
3. Determine which reactant is completely consumed
4. Use the limiting reactant for product calculations

For concentration and titration problems:

1. Always convert volumes to dm³ if needed
2. Calculate moles of known solution
3. Apply molar ratio from balanced equation
4. Calculate concentration or volume of unknown

For formula determination:

1. Convert masses to moles
2. Find simplest whole number ratio (empirical formula)
3. Use molecular mass to find multiplication factor
4. Multiply empirical formula subscripts by factor

Common Pitfalls to Avoid

• Forgetting state symbols in equations — always include (s), (l), (g), or (aq)
• Not balancing polyatomic ions correctly — treat as a unit if unchanged
• Using wrong molar ratio — always read coefficients directly from balanced equation
• Confusing actual and theoretical yield — actual goes in numerator for percentage yield
• Not converting volume units — always use dm³ for concentration calculations
• Calculating percentage yield > 100% — signals an error; check which value is numerator/denominator
• Forgetting to multiply atoms in parentheses — Ca(OH)₂ has 2 O and 2 H, not 1 of each
• Not identifying the limiting reactant — use it for all subsequent calculations, not the excess