Test - February 2025

Calculations Involving Masses (1.43 – 1.53)

• Relative Formula Mass (Mr): Sum of the relative atomic masses (Ar) of all atoms in a formula.

• Moles:

  • Mole = Mass ÷ Mr

  • One mole contains 6.02 × 10²³ particles (Avogadro’s constant).

• Empirical & Molecular Formulae:

  • Empirical formula: Simplest whole number ratio of atoms in a compound.

  • Molecular formula: Actual number of atoms in a molecule (multiple of empirical formula).

• Reacting Masses & Stoichiometry:

  • Use balanced equations to calculate mass of reactants or products.

  • Limiting reactants: The reactant that is completely used up in a reaction.

• Percentage Yield:

  • % Yield = (Actual mass / Theoretical mass) × 100

• Atom Economy:

  • Atom Economy = (Mass of desired product / Total mass of reactants) × 100

Quantitative Analysis (5.8C – 5.14C)

• Titrations:

  • Used to determine concentration of an unknown solution.

  • Key formulas:

    • Moles = Concentration × Volume (dm³)

    • Use balanced equations to calculate unknown concentrations.

• Concentration:

  • Concentration (g/dm³) = Mass of solute (g) ÷ Volume of solution (dm³)

  • Concentration (mol/dm³) = Moles ÷ Volume (dm³)

• Gas Volumes:

  • Molar gas volume at RTP (Room Temperature & Pressure) = 24 dm³/mol

  • Volume of gas (dm³) = Moles × 24

Fuels (8.1 – 8.17)

• Hydrocarbons: Compounds made of hydrogen and carbon atoms only.

• Fossil Fuels: Crude oil, coal, and natural gas (non-renewable).

• Complete Combustion: Hydrocarbon + Oxygen → Carbon dioxide + Water

• Incomplete Combustion: Produces carbon monoxide (CO) and/or soot (carbon).

• Cracking: Thermal decomposition of long-chain alkanes into shorter, more useful hydrocarbons.

• Alternative Fuels: Biofuels (ethanol, biodiesel), hydrogen fuel cells.

Hydrocarbons (9.10C – 9.16C)

• Alkanes (CₙH₂ₙ₊₂): Saturated hydrocarbons (single bonds only).

• Alkenes (CₙH₂ₙ): Unsaturated hydrocarbons (contain at least one C=C double bond).

• Testing for Alkenes:

  • Bromine water: Orange → Colourless (if an alkene is present).

• Reactions of Alkenes:

  • Addition reactions (e.g., with hydrogen, halogens, steam).

• Polymerisation:

  • Alkenes undergo addition polymerisation to form polymers like poly(ethene).

Covalent Bonding (1.28 – 1.31)

• Covalent Bonds:

  • Formed when non-metal atoms share electrons to achieve a full outer shell.

• Simple Covalent Molecules:

  • Low melting/boiling points (weak intermolecular forces).

  • Do not conduct electricity (no free electrons/ions).

• Giant Covalent Structures:

  • Diamond: Hard, each carbon bonded to 4 others, high melting point.

  • Graphite: Layers, soft, conducts electricity (delocalised electrons).

Methods of Separating and Purifying Substances (2.5 – 2.8)

• Filtration: Separates insoluble solids from liquids.

• Crystallisation: Separates a solute from a solution by evaporating the solvent.

• Distillation:

  • Simple distillation: Separates solvent from a solution (e.g., water from salt water).

  • Fractional distillation: Separates a mixture of liquids with different boiling points.

• Chromatography:

  • Used to separate mixtures of dyes/inks.

  • Rf value = Distance moved by substance ÷ Distance moved by solvent.

The Periodic Table (1.19)

• Modern Periodic Table: Arranges elements in order of increasing atomic number.

• Groups: Vertical columns (elements have the same number of outer electrons).

• Periods: Horizontal rows (shows increasing energy levels of electrons).

• Trends:

  • Group 1 (Alkali metals): More reactive down the group.

  • Group 7 (Halogens): Less reactive down the group.

Quantitative Analysis (5.16C & 5.18C)

• Gravimetric Analysis:

  • Uses mass measurements to determine composition.

• Volumetric Analysis:

  • Uses solutions of known concentration to determine unknowns (e.g., titrations).

Practice Questions Below:

1. Calculations Involving Masses (1.43 – 1.53)

1. Calculate the relative formula mass (Mr) of calcium carbonate (CaCO₃).

2. How many moles are in 25g of NaCl? (Mr of NaCl = 58.5)

3. A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

4. Calculate the mass of magnesium oxide produced when 4.8g of magnesium reacts with oxygen. (Mg + O₂ → MgO)

5. A reaction has a theoretical yield of 30g but produces only 21g of product. Calculate the percentage yield.

2. Quantitative Analysis (5.8C – 5.14C)

1. A student titrates 25.0 cm³ of NaOH with 0.100 mol/dm³ HCl and finds the volume of acid used is 20.0 cm³. Calculate the concentration of NaOH.

2. How many moles of gas are present in 48 dm³ at room temperature and pressure?

3. Calculate the mass of solute in 500 cm³ of a 0.50 g/dm³ NaCl solution.

4. A gas cylinder contains 72 dm³ of oxygen. How many moles of oxygen are present? (Molar gas volume = 24 dm³/mol)

5. In a reaction, 0.050 moles of H₂ reacts with O₂ to form water. What volume of oxygen is required?

3. Fuels (8.1 – 8.17)

1. Write the balanced equation for the complete combustion of propane (C₃H₈).

2. What harmful products are formed in incomplete combustion?

3. Explain how cracking helps to meet fuel demands.

4. Compare the advantages and disadvantages of using biofuels instead of fossil fuels.

5. Describe how fractional distillation is used to separate crude oil into fractions.

4. Hydrocarbons (9.10C – 9.16C)

1. What is the general formula of alkanes and alkenes?

2. Describe how to test for an alkene using bromine water.

3. Name the products when ethene reacts with bromine.

4. Write the balanced equation for the addition reaction of ethene with hydrogen.

5. What type of polymerisation do alkenes undergo? Give an example.

5. Covalent Bonding (1.28 – 1.31)

1. Draw a dot and cross diagram for a molecule of oxygen (O₂).

2. Why do simple covalent molecules have low melting and boiling points?

3. Explain why diamond and graphite have different properties, despite both being made of carbon.

4. Why can graphite conduct electricity but diamond cannot?

5. Describe the bonding in a molecule of water (H₂O).

6. Methods of Separating and Purifying Substances (2.5 – 2.8)

1. Describe how filtration can be used to separate sand from salt water.

2. What is the purpose of crystallisation?

3. How does fractional distillation work to separate ethanol from water?

4. A student carries out chromatography on two inks. One ink has an Rf value of 0.75 and the solvent front moves 8 cm. How far has the ink traveled?

5. How can distillation be used to obtain pure water from seawater?

7. The Periodic Table (1.19)

1. How are elements arranged in the modern periodic table?

2. What do elements in the same group have in common?

3. Describe the trend in reactivity of Group 1 metals as you go down the group.

4. Why do Group 7 elements become less reactive down the group?

5. Explain why noble gases (Group 0) are unreactive.

8. Quantitative Analysis (5.16C & 5.18C)

1. Explain the difference between gravimetric and volumetric analysis.

2. In a titration, 0.025 moles of sulfuric acid reacts with sodium hydroxide. What volume of 0.500 mol/dm³ NaOH solution is needed for neutralisation?

3. A calcium chloride solution contains 0.15 moles in 250 cm³. Calculate its concentration in mol/dm³.

4. A sample of barium sulfate was found to contain 0.05 moles. What is the mass of barium sulfate in the sample? (Mr = 233)

5. A gas sample contains 1.2 moles of CO₂. What volume does this gas occupy at room temperature and pressure?

answers below

1. Calculations Involving Masses (1.43 – 1.53)

1. Mr of CaCO₃ = (Ca: 40) + (C: 12) + (O₃: 16×3) = 100

2. Moles of NaCl = 25 g ÷ 58.5 = 0.427 moles

3. Empirical Formula Calculation:

   • C: 40 ÷ 12 = 3.33

   • H: 6.7 ÷ 1 = 6.7

   • O: 53.3 ÷ 16 = 3.33

   • Divide by smallest: C₁H₂O₁ → CH₂O

4. MgO Calculation:

   • Moles of Mg = 4.8 ÷ 24 = 0.2 moles

   • Mg + ½O₂ → MgO (1:1 ratio)

   • Mass of MgO = 0.2 × 40 = 8.0 g

5. % Yield = (21 ÷ 30) × 100 = 70%

2. Quantitative Analysis (5.8C – 5.14C)

1. NaOH concentration:

   • Moles of HCl = 0.100 × (20 ÷ 1000) = 0.002

   • NaOH:HCl ratio = 1:1 → Moles of NaOH = 0.002

   • Concentration = 0.002 ÷ (25 ÷ 1000) = 0.080 mol/dm³

2. Moles of gas = 48 ÷ 24 = 2 moles

3. Mass of solute = 0.50 × 0.500 = 0.25 g

4. Moles of oxygen = 72 ÷ 24 = 3 moles

5. Oxygen volume = (0.050 ÷ 2) × 24 = 0.60 dm³

3. Fuels (8.1 – 8.17)

1. Balanced equation for propane combustion:
   C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

2. Incomplete combustion products: Carbon monoxide (CO) and/or carbon (soot).

3. Cracking produces smaller, more useful hydrocarbons (e.g., alkenes and shorter alkanes).

4. Biofuels vs Fossil Fuels: Biofuels are renewable and reduce CO₂ emissions, but require land and water.

5. Fractional distillation:

   • Crude oil is heated → Vapours rise → Different hydrocarbons condense at different temperatures.

4. Hydrocarbons (9.10C – 9.16C)

1. General formulae:

   • Alkanes: CₙH₂ₙ₊₂

   • Alkenes: CₙH₂ₙ

2. Bromine water test: Turns colourless in the presence of an alkene.

3. Ethene + Bromine: C₂H₄ + Br₂ → C₂H₄Br₂ (1,2-dibromoethane).

4. Ethene + Hydrogen: C₂H₄ + H₂ → C₂H₆

5. Alkenes undergo addition polymerisation to form polymers like polyethene.

5. Covalent Bonding (1.28 – 1.31)

1. Dot and cross diagram for O₂: Each oxygen shares two electrons (O=O).

2. Low melting/boiling points: Weak intermolecular forces require little energy to break.

3. Diamond vs Graphite:

   • Diamond: Hard, each C bonded to 4 others.

   • Graphite: Layers, each C bonded to 3 others, free electrons allow conduction.

4. Graphite conducts electricity due to delocalised electrons; diamond does not.

5. Water bonding: Two single covalent bonds (H–O–H), bent shape.

6. Methods of Separating and Purifying Substances (2.5 – 2.8)

1. Filtration: Sand is trapped in filter paper, saltwater passes through.

2. Crystallisation: Used to obtain pure solid from a solution by evaporating the solvent.

3. Fractional distillation:

   • Different boiling points → Vapours rise and condense at different temperatures.

4. Rf value calculation:

   • Distance travelled by ink = 0.75 × 8 = 6 cm

5. Distillation of seawater:

   • Heat seawater → Water evaporates → Vapour condenses → Leaves salt behind.

7. The Periodic Table (1.19)

1. Modern periodic table: Arranged by atomic number (protons).

2. Same group elements have the same number of outer electrons.

3. Group 1 reactivity trend: Increases down the group (electrons further from nucleus).

4. Group 7 reactivity trend: Decreases down the group (outer electrons less strongly attracted).

5. Noble gases (Group 0) are unreactive due to a full outer shell.

8. Quantitative Analysis (5.16C & 5.18C)

1. Gravimetric vs Volumetric:

   • Gravimetric = Based on mass

   • Volumetric = Based on solutions and titrations.

2. Titration Calculation:

   • Moles of H₂SO₄ = 0.025

   • Ratio H₂SO₄:NaOH = 1:2 → NaOH moles = 0.050

   • Volume = 0.050 ÷ 0.500 = 0.10 dm³ (100 cm³)

3. Concentration of CaCl₂ = 0.15 ÷ (250 ÷ 1000) = 0.60 mol/dm³

4. Mass of BaSO₄ = 0.05 × 233 = 11.65 g

5. CO₂ gas volume = 1.2 × 24 = 28.8 dm³

explanations below

1. Calculations Involving Masses (1.43 – 1.53)

1. Relative Formula Mass (Mr) of CaCO₃:

   • Add the relative atomic masses (Ar):

     • Ca = 40, C = 12, O₃ = 16 × 3 = 48

     • Total = 40 + 12 + 48 = 100

2. Moles of NaCl in 25g:

   • Formula: Moles = Mass ÷ Mr

   • Given: Mass = 25 g, Mr of NaCl = 58.5

   • Moles = 25 ÷ 58.5 = 0.427 moles

3. Empirical Formula Calculation (Simplest whole number ratio of atoms):

   • C: 40 ÷ 12 = 3.33

   • H: 6.7 ÷ 1 = 6.7

   • O: 53.3 ÷ 16 = 3.33

   • Divide by the smallest (3.33): C₁H₂O₁ → CH₂O

4. Mass of MgO Produced:

   • Moles of Mg = 4.8 g ÷ 24 (Ar of Mg) = 0.2 moles

   • Balanced equation: Mg + ½O₂ → MgO

   • Mole ratio of Mg to MgO = 1:1

   • Mass of MgO = 0.2 × 40 = 8.0 g

5. Percentage Yield:

   • % Yield = (Actual Yield ÷ Theoretical Yield) × 100

   • (21 ÷ 30) × 100 = 70%

2. Quantitative Analysis (5.8C – 5.14C)

1. Titration Calculation (NaOH concentration):

   • Moles of HCl = Concentration × Volume

   • = 0.100 × (20 ÷ 1000) = 0.002 moles

   • Mole ratio HCl:NaOH = 1:1, so moles of NaOH = 0.002

   • Concentration = Moles ÷ Volume

   • = 0.002 ÷ (25 ÷ 1000) = 0.080 mol/dm³

2. Moles of Gas at RTP (24 dm³ per mole):

   • Moles = Volume ÷ 24

   • = 48 ÷ 24 = 2 moles

3. Mass of Solute in Solution:

   • Mass = Concentration × Volume

   • = 0.50 × (500 ÷ 1000) = 0.25 g

4. Moles of O₂ Gas:

   • Moles = Volume ÷ 24

   • = 72 ÷ 24 = 3 moles

5. Oxygen Volume Needed for H₂ Reaction:

   • Balanced equation: 2H₂ + O₂ → 2H₂O

   • 2 moles of H₂ react with 1 mole of O₂

   • O₂ volume = (0.050 ÷ 2) × 24 = 0.60 dm³

3. Fuels (8.1 – 8.17)

1. Complete Combustion of Propane:

   • General formula: Hydrocarbon + Oxygen → CO₂ + H₂O

   • C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

2. Incomplete Combustion Products:

   • Produces carbon monoxide (CO) and/or carbon (soot) due to lack of oxygen.

3. Why is Cracking Important?

   • Long-chain hydrocarbons aren’t useful (high boiling points, not flammable).

   • Cracking breaks them into shorter, more useful hydrocarbons (alkanes & alkenes).

4. Biofuels vs Fossil Fuels:

   • Biofuels (e.g., ethanol, biodiesel) are renewable and produce less CO₂.

   • However, they require large amounts of land & water.

5. Fractional Distillation of Crude Oil:

   • Heated crude oil vaporises.

   • Vapours rise in a fractionating column.

   • Different hydrocarbons condense at different temperatures.

4. Hydrocarbons (9.10C – 9.16C)

1. General Formulas:

   • Alkanes = CₙH₂ₙ₊₂

   • Alkenes = CₙH₂ₙ

2. Alkene Test (Bromine Water):

   • Alkene: Bromine water goes from orange to colourless.

   • Alkane: No change.

3. Ethene + Bromine Reaction:

   • C₂H₄ + Br₂ → C₂H₄Br₂

4. Ethene + Hydrogen Reaction:

   • C₂H₄ + H₂ → C₂H₆

5. Polymerisation:

   • Alkenes join together to form polymers like poly(ethene).

5. Covalent Bonding (1.28 – 1.31)

1. Dot and Cross Diagram for O₂:

   • Each oxygen shares two electrons (O=O double bond).

2. Low Melting Points of Simple Covalent Molecules:

   • Weak intermolecular forces need little energy to break.

3. Diamond vs Graphite:

   • Diamond: Hard, 4 bonds per carbon.

   • Graphite: Soft, 3 bonds per carbon, layers can slide.

4. Why Graphite Conducts Electricity:

   • It has delocalised electrons that can move.

5. Water Bonding:

   • Two single covalent bonds (H–O–H).

6. Methods of Separation (2.5 – 2.8)

1. Filtration: Traps solid (sand), liquid (saltwater) passes through.

2. Crystallisation: Evaporate solvent, leaving solute behind.

3. Fractional Distillation: Separates liquids by boiling points.

4. Chromatography (Rf Calculation):

   • Rf = Distance moved by ink ÷ Distance moved by solvent

   • 0.75 × 8 = 6 cm

5. Distillation of Seawater:

   • Evaporate water, condense vapour, leaving salt behind.

7. The Periodic Table (1.19)

1. Arranged by atomic number (protons).

2. Elements in the same group have the same outer electrons.

3. Group 1 Reactivity Trend:

   • More reactive down the group (outer electron is lost more easily).

4. Group 7 Reactivity Trend:

   • Less reactive down the group (harder to gain an electron).

5. Noble Gases (Group 0) are Unreactive due to full outer shells.

8. Quantitative Analysis (5.16C & 5.18C)

1. Gravimetric (mass) vs Volumetric (titration) methods.

2. Titration Calculation:

   • Volume = 0.050 ÷ 0.500 = 0.10 dm³ (100 cm³)

3. Concentration of CaCl₂:

   • 0.60 mol/dm³

4. Mass of BaSO₄:

   • 11.65 g

5. CO₂ Gas Volume:

   • 28.8 dm³