Chemical Combination | Jamb Chemistry

Jamb(UTME) key points on Chemical Combination

Law of Definite Proportion

1. Definition: A compound always contains the same elements in the same proportion by mass, regardless of the sample size or source.
2. Example: Water (H₂O) always consists of 88.89% oxygen and 11.11% hydrogen by mass.
3. Significance: Demonstrates that chemical compounds are formed in fixed ratios.
4. Application: Basis for determining molecular formulas in compounds.
5. Experiments: Joseph Proust's experiments on copper carbonate established this law.
6. Consistency: Holds true for pure compounds but not for mixtures.
7. Real-World Relevance: Used in quality control in chemical industries to verify product consistency.
8. Limitations: Does not explain why proportions remain constant, which is addressed by atomic theory.
9. Relation to Atomic Theory: Atoms combine in whole-number ratios to form compounds.
10. Practical Application: Used in calculating empirical formulas of unknown compounds.
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Law of Multiple Proportion

11. Definition: When two elements combine to form multiple compounds, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios.
12. Example: Carbon and oxygen form CO (1:1) and CO₂ (1:2).
13. Significance: Provides evidence for discrete atoms combining in specific ratios.
14. Experiments: John Dalton's work on carbon oxides supported this law.
15. Application: Key in understanding molecular structure and chemical bonding.
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Law of Reciprocal Proportion

16. Definition: The ratio of masses of two elements that combine with a fixed mass of a third element is the same as or a simple multiple of the ratio in which they combine with each other.
17. Example: Hydrogen reacts with oxygen to form water, and with sulfur to form H₂S. Oxygen and sulfur combine in a simple ratio in SO₂.
18. Significance: Explains consistent proportional relationships among elements.
19. Applications: Helps predict possible combinations and ratios in new compounds.
20. Experimental Basis: Established through systematic reaction studies involving multiple elements.
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Conservation of Matter

Law of Conservation of Matter

21. Definition: Matter is neither created nor destroyed in a chemical reaction; the mass of reactants equals the mass of products.
22. Formulator: Antoine Lavoisier, the father of modern chemistry.
23. Key Concept: Atoms rearrange during chemical reactions, but their total mass remains unchanged.
24. Example: In combustion of methane, CH₄ + 2O₂ → CO₂ + 2H₂O, the mass of products equals the mass of reactants.
25. Applications: Used in balancing chemical equations.
26. Relation to Energy Conservation: Ties into the principle that energy and mass are conserved.
27. Industries: Essential in chemical engineering for designing efficient processes.
28. Limitations: Does not account for nuclear reactions where mass can convert to energy.
29. Role in Stoichiometry: Ensures accurate calculations in reactants and products.
30. Practical Uses: Helps verify experimental results in laboratory setups.
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Gas Laws and Avogadro’s Law

Gay-Lussac’s Law of Combining Volumes

31. Definition: Gases react in volumes that are in simple whole-number ratios, under constant temperature and pressure.
32. Example: H₂ + Cl₂ → 2HCl; 1 volume of hydrogen reacts with 1 volume of chlorine to produce 2 volumes of hydrogen chloride.
33. Significance: Simplifies the prediction of gaseous reaction outcomes.
34. Relation to Avogadro’s Law: Reinforced the idea that equal volumes of gases contain equal numbers of particles.
35. Applications: Used in determining molecular formulas of gaseous compounds.
36. Historical Context: Joseph Louis Gay-Lussac formulated this law in 1808.
37. Limitations: Applies only to gaseous reactions, not to solids or liquids.
38. Industries: Crucial in industrial gas production and synthesis.
39. Connection to Stoichiometry: Links gas volumes to mole ratios in reactions.
40. Verification: Empirically confirmed through numerous gas-phase reaction studies.
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Avogadro’s Law

41. Definition: Equal volumes of gases at the same temperature and pressure contain an equal number of molecules.
42. Avogadro’s Number: 6.022 × 10²³, the number of particles in one mole of a substance.
43. Molar Volume: At STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 L.
44. Applications: Used to calculate molar masses and gas densities.
45. Example: Comparing CO₂ and O₂ gases shows equal volumes have equal molecule counts.
46. Significance: Basis for the mole concept in chemistry.
47. Experiments: Verified through advanced gas experiments and molecular studies.
48. Industries: Essential in gas stoichiometry for chemical processes.
49. Molecular Formulas: Helps deduce molecular formulas of gases.
50. Impact: Unified the understanding of gaseous behavior and molecular structure.
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Chemical Symbols, Formulae, and Equations

Chemical Symbols and Formulae

51. Chemical Symbols: Shorthand notations representing elements (e.g., H for hydrogen, O for oxygen).
52. Chemical Formulae: Represent the composition of compounds (e.g., H₂O for water).
53. Empirical Formula: Simplest whole-number ratio of atoms in a compound (e.g., CH for benzene).
54. Molecular Formula: Exact number of each type of atom in a molecule (e.g., C₆H₆ for benzene).
55. Structural Formula: Shows the arrangement of atoms in a molecule.
56. Importance: Enables clear communication in chemistry.
57. Example: NaCl represents sodium chloride, a common salt.
58. Balancing Equations: Ensures the law of conservation of matter is upheld.
59. Applications: Essential in writing reaction mechanisms and predicting outcomes.
60. Complex Reactions: Formulae simplify even intricate reactions, aiding calculations.
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Relative Atomic Mass and the Mole Concept

61. Relative Atomic Mass: Weighted average of isotopic masses, normalized to ¹²C = 12.
62. Mole Concept: Relates macroscopic quantities to atomic-scale entities.
63. One Mole: Contains Avogadro’s number (6.022 × 10²³) of particles.
64. Applications: Used in stoichiometry to link masses, moles, and particle numbers.
65. Example: One mole of NaCl weighs 58.44 g (relative molecular mass).
66. Avogadro’s Number: Bridges the gap between macroscopic and molecular scales.
67. Stoichiometry: Based on mole ratios derived from balanced chemical equations.
68. Industries: Key in chemical manufacturing and pharmaceutical formulations.
69. Dimensional Consistency: Ensures calculations involving moles are accurate.
70. Practical Utility: Simplifies the scaling of reactions for real-world applications.
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Stoichiometry of Reactions

71. Definition: Quantitative relationship between reactants and products in a chemical reaction.
72. Balanced Equations: Serve as the foundation for stoichiometric calculations.
73. Mole Ratios: Derived from coefficients in balanced equations.
74. Limiting Reactant: Determines the maximum amount of product that can form.
75. Excess Reactant: Present in larger quantity than required.
76. Mass-Mass Calculations: Convert reactant mass to product mass using mole ratios.
77. Gas Stoichiometry: Involves molar volumes for gaseous reactants and products.
78. Concentration Calculations: Used in solutions to find molarity or normality.
79. Yield: Percentage of theoretical yield achieved in a reaction.
80. Efficiency: Stoichiometry assesses process efficiency in industrial reactions.
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Advanced Applications and Connections

81. Percent Composition: Determines the mass percent of each element in a compound.
82. Empirical vs. Molecular Formula: Derived using experimental data and molar mass.
83. Titrations: Apply stoichiometry to determine concentrations in acid-base reactions.
84. Thermochemistry: Relates stoichiometry to energy changes in reactions.
85. Gas Laws: Integrated with stoichiometry for reactions involving gases.
86. Electrolysis: Uses stoichiometry to calculate the amount of substance deposited.
87. Kinetics: Relates reaction rates to stoichiometric coefficients
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88. Equilibrium: Stoichiometry balances reversible reactions.
89. Environmental Science: Applied in pollution control and waste management.
90. Medicine: Ensures precise drug formulations and dosages.
91. Agriculture: Used in fertilizer production and soil chemistry.
92. Food Industry: Calculates additives and preservatives in processed foods.
93. Pharmaceuticals: Ensures accuracy in active ingredient proportions.
94. Nuclear Chemistry: Relates stoichiometry to radioactive decay and energy release.
95. Safety: Predicts hazardous reactions by analyzing stoichiometric ratios.
96. Education: Teaches fundamental problem-solving in chemistry.
97. Industrial Scaling: Converts lab-scale reactions to industrial scales.
98. Research: Facilitates synthesis of new materials and compounds.
99. Sustainability: Optimizes resource use by minimizing waste.
100. Future Applications: Drives innovation in green chemistry and renewable energy.
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Jamb(UTME) Key points on 30 Questions and Answers on Formulae, Equations, and the Mole Concept

Questions and Answers on Calculations Involving Formulae

1. Q: Calculate the molecular mass of H₂O.
   A: H = 1 × 2, O = 16. Total = 18 g/mol.
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2. Q: What is the empirical formula of a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass?
   A: C: 40/12 = 3.33, H: 6.7/1 = 6.7, O: 53.3/16 = 3.33. Ratio = 1:2:1, so the formula is CH₂O.
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3. Q: Determine the molecular formula of a compound with an empirical formula CH₂ and molar mass of 56 g/mol.
   A: CH₂ mass = 12 + 2 = 14. Molecular formula = (56/14) × CH₂ = C₄H₈.
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4. Q: Find the mass percentage of oxygen in CO₂.
   A: CO₂ = 12 + (16 × 2) = 44. Oxygen = (32/44) × 100 = 72.73%.
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5. Q: A compound contains 92% carbon and 8% hydrogen. What is its empirical formula?
   A: C: 92/12 = 7.67, H: 8/1 = 8. Ratio = 1:1, so the formula is CH.
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6. Q: How many atoms are present in 2 moles of NaCl?
   A: 1 mole contains 6.022 × 10²³ molecules. In 2 moles = 2 × (6.022 × 10²³) = 1.204 × 10²⁴ atoms.
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7. Q: Calculate the formula mass of Na₂CO₃.
   A: Na = 23 × 2, C = 12, O = 16 × 3. Total = 106 g/mol.
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8. Q: What is the empirical formula of a compound with 80% C and 20% H?
   A: C: 80/12 = 6.67, H: 20/1 = 20. Ratio = 1:3, so the formula is CH₃.
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9. Q: Calculate the number of moles in 88 g of CO₂.
   A: Molar mass of CO₂ = 44 g/mol. Moles = 88/44 = 2 moles.
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10. Q: What is the mass of 0.5 moles of H₂SO₄?
    A: Molar mass = 98 g/mol. Mass = 0.5 × 98 = 49 g.
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Questions and Answers on Chemicl Equations and Composition

11. Q: Balance the equation: H₂ + O₂ → H₂O.
    A: 2H₂ + O₂ → 2H₂O.
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12. Q: Calculate the mass of water formed when 4 g of hydrogen reacts with excess oxygen.
    A: H₂ + ½O₂ → H₂O. 2 g of H₂ gives 18 g of H₂O.
    4 g of H₂ gives (4/2) × 18 = 36 g H₂O.
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13. Q: Write the balanced equation for the combustion of methane (CH₄).
    A: CH₄ + 2O₂ → CO₂ + 2H₂O.
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14. Q: How many moles of oxygen are required to completely react with 2 moles of CH₄?
    A: CH₄ + 2O₂ → CO₂ + 2H₂O. 1 mole CH₄ requires 2 moles O₂. So, 2 moles CH₄ require 4 moles O₂.
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15. Q: Find the mass of CO₂ produced from 5 g of CH₄ combustion.
    A: CH₄ + 2O₂ → CO₂ + 2H₂O. Molar mass CH₄ = 16 g/mol. Moles of CH₄ = 5/16 = 0.3125.
    1 mole CH₄ gives 1 mole CO₂ (44 g). CO₂ = 0.3125 × 44 = 13.75 g.
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16. Q: Balance the equation: C₄H₁₀ + O₂ → CO₂ + H₂O.
    A: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.
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17. Q: How many grams of NaCl are produced when 2 moles of HCl react with NaOH?
    A: HCl + NaOH → NaCl + H₂O. 1 mole HCl gives 1 mole NaCl (58.44 g).
    2 moles HCl give 2 × 58.44 = 116.88 g NaCl.
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18. Q: Write the balanced equation for the reaction of iron with chlorine gas.
    A: 2Fe + 3Cl₂ → 2FeCl₃.
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19. Q: How many grams of oxygen are required to burn 4 moles of ethane (C₂H₆)?
    A: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. 2 moles C₂H₆ need 7 moles O₂.
    4 moles need (4/2) × 7 = 14 moles O₂. Mass = 14 × 32 = 448 g.
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20. Q: Calculate the mass of H₂O formed when 10 g of H₂ reacts with excess O₂.
    A: H₂ + ½O₂ → H₂O. 2 g H₂ gives 18 g H₂O.
    10 g H₂ gives (10/2) × 18 = 90 g H₂O.
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Questions and Answers on the Mole Concept

21. Q: Define the mole concept.
    A: A mole is the amount of a substance containing 6.022 × 10²³ particles (atoms, ions, or molecules).
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22. Q: Calculate the number of moles in 12 g of oxygen gas (O₂).
    A: Molar mass of O₂ = 32 g/mol. Moles = 12/32 = 0.375 moles.
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23. Q: How many molecules are there in 2 moles of H₂O?
    A: 1 mole = 6.022 × 10²³ molecules.
    2 moles = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules.
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24. Q: How many moles are in 90 g of H₂O?
    A: Molar mass = 18 g/mol. Moles = 90/18 = 5 moles.
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25. Q: What is Avogadro’s number?
    A: 6.022 × 10²³, the number of particles in one mole.
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26. Q: Calculate the mass of 1 mole of NaCl.
    A: Na = 23, Cl = 35.5. Total = 23 + 35.5 = 58.5 g/mol.
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27. Q: How many atoms are in 3 moles of carbon?
    A: 1 mole = 6.022 × 10²³ atoms.
    3 moles = 3 × 6.022 × 10²³ = 1.807 × 10²⁴ atoms.
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28. Q: How many moles of CO₂ are in 88 g?
    A: Molar mass of CO₂ = 44 g/mol. Moles = 88/44 = 2 moles.
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29. Q: Calculate the number of oxygen atoms in 1 mole of H₂O.
    A: 1 mole H₂O has 1 mole of oxygen atoms = 6.022 × 10²³ atoms.
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30. Q: Find the mass of 0.25 moles of CaCO₃.
    A: Molar mass = 40 + 12 + (16 × 3) = 100 g/mol. Mass = 0.25 × 100 = 25 g.

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