Grade 10
  1. Matter and its properties  1.1. States of matter  1.2. Physical and chemical properties of matter  1.3. Classification of substances (elements, compounds, mixtures)  1.4. Changes in Matter (Physical and Chemical Changes)  1.5. Law of conservation of mass and law of definite proportions
  2. Atomic Structure  2.1. Historical development of the atomic model  2.2. Subatomic particles (protons, neutrons, electrons)  2.3. Atomic number, mass number and isotopes  2.4. Electronic Configuration and Energy Levels  2.5. Valency, ion formation and oxidation state
  3. Periodic table  3.1. History and Development of the Periodic Table  3.2. Modern Periodic Law and Periodic Trends  3.3. Groups and Periods in the Periodic Table  3.4. Trends in the Periodic Table  3.5. Metals, non-metals, metalloids and their properties  3.6. Noble gases and their chemical inertness
  4. Chemical bond  4.1. Formation of Chemical Bonds (Octet Rule)  4.2. Ionic Bonding and Properties of Ionic Compounds  4.3. Covalent Bond and Properties of Covalent Compounds  4.4. Metallic bond and its properties  4.5. Molecular geometry and VSEPR theory  4.6. Polarity and dipole moment of molecules  4.7. Intermolecular forces
  5. Chemical Reactions and Equations  5.1. Types of Chemical Reactions  5.2. Writing and balancing chemical equations  5.3. Law of conservation of mass in chemical reactions  5.4. Factors Affecting Chemical Reactions  5.5. Catalysts and their role in chemical reactions  5.6. Exothermic and Endothermic Reactions
  6. Stoichiometry and Chemical Calculations  6.1. Mole concept and Avogadro number  6.2. Molar Mass and Gram-Mole Conversions  6.3. Empirical and molecular formula  6.4. Stoichiometric calculations and chemical yield  6.5. Limiting and Excess Reactants
  7. Acids, Bases and Salts  7.1. Properties and Definitions of Acids and Bases (Arrhenius, Bronsted-Lowry, Lewis)  7.2. pH Scale, pOH, and Indicators  7.3. Neutralization reactions and salt formation  7.4. Strength of Acids and Bases (Strong vs. Weak Acids and Bases)  7.5. Common Acids and Bases in Daily Life  7.6. Salts, their solubility and applications
  8. Metals and Nonmetals  8.1. Physical and chemical properties of metals  8.2. Physical and Chemical Properties of Non-Metals  8.3. Reactivity Series of Metals and Displacement Reactions  8.4. Extraction of metals from ores  8.5. Corrosion, its causes and prevention  8.6. Alloys, their composition and uses
  9. Carbon and its compounds  9.1. Allotropes of Carbon  9.2. Hydrocarbons and their classification (alkanes, alkenes, alkynes)  9.3. Functional groups in organic compounds  9.4. Nomenclature of organic compounds (IUPAC system)  9.5. Isomerism in organic chemistry (structural and geometrical)  9.6. Important organic reactions (combustion, addition, substitution, polymerization)  9.7. Applications of organic compounds in industry and daily life
  10. Environmental Chemistry  10.1. Air Pollution (Sources, Effects and Control)  10.2. Water Pollution (Causes, Effects and Remedies)  10.3. Greenhouse effect and global warming  10.4. Ozone layer depletion and its effects  10.5. Green Chemistry and Its Applications  10.6. sustainable chemistry practice
  11. Thermochemistry  11.1. Heat and Energy Changes in Chemical Reactions  11.2. Exothermic and Endothermic Reactions  11.3. Enthalpy, enthalpy change, and heat of reaction  11.4. Specific heat capacity and calorimetry  11.5. Energy Changes in Combustion Reactions
  12. Electrochemistry  12.1. Electrolytes and non-electrolytes  12.2. Electrolysis and its applications (electroplating, extraction of metals)  12.3. Redox reactions (oxidation and reduction)  12.4. Galvanic cell and electrochemical series  12.5. Corrosion and electrochemical prevention methods
  13. Chemical kinetics and equilibrium  13.1. Rate of chemical reactions and its measurement  13.2. Factors affecting the reaction rate (catalyst, temperature, concentration)  13.3. Reversible and irreversible reactions  13.4. Dynamic equilibrium and equilibrium constant  13.5. Le Chatelier's principle and its applications
  14. Gases and Gas Laws  14.1. Properties of gases and kinetic molecular theory  14.2. Boyle's Law (Pressure-Volume Relation)  14.3. Charles's Law  14.4. Gay-Lussac's Law  14.5. Combined Gas Law and Ideal Gas Law  14.6. Real Gases vs Ideal Gases
  15. Nuclear Chemistry  15.1. Introduction to Radioactivity and Nuclear Reactions  15.2. Types of radioactive decay (alpha, beta, gamma)  15.3. Half-life and applications of radioactive isotopes  15.4. Nuclear fission and fusion reactions  15.5. Nuclear power generation and its environmental impact

Grade 10 → Stoichiometry and Chemical Calculations ↓


Empirical and molecular formula


In chemistry, it is fundamental to understand the structure of compounds. Two types of formulas are important for this purpose: the empirical formula and the molecular formula. These notations allow us to express the elements that make up a compound and their ratios. In this lesson, we will explore these concepts in depth using simple language. We will also look at visual examples to help clarify these important ideas.

What is the empirical formula?

The empirical formula of a compound gives the simplest integer ratio of the elements present in it. It does not tell us the exact number of atoms in the molecule, only the ratio of different types of atoms.

Example of empirical formula:

Consider glucose, a simple sugar. Its molecular formula is C 6 H 12 O 6 The molecular formula gives the actual number of each type of atom in a molecule. However, the empirical formula of glucose is CH 2 O This is because the simplest whole-number ratio of carbon, hydrogen, and oxygen atoms is 1:2:1.

        Molecular formula of glucose: C 6 H 12 O 6
        Empirical formula of glucose: CH2O

How to determine the empirical formula

To determine the empirical formula, follow these steps:

  1. Find the mass (in grams) of each element in the compound.
  2. Convert the mass of each element into moles using the molar mass (atomic weight).
  3. Divide the mole value of each element by the smallest mole number calculated.
  4. Multiply the numbers you get to get whole numbers to use in the ratio.

Step-by-step example:

Let us determine the empirical formula of a compound that contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

  1. Suppose we have 100 grams of the compound, which means it contains 40.0 grams of carbon, 6.7 grams of hydrogen, and 53.3 grams of oxygen.
  2. Convert these masses to moles:
    • Carbon: [frac{40.0 ,g}{12.01 ,g/mol} = 3.33 ,mol]
    • Hydrogen: [frac{6.7 ,g}{1.008 ,g/mol} = 6.65 ,mol]
    • Oxygen: [frac{53.3 ,g}{16.00 ,g/mol} = 3.33 ,mol]
  3. Divide the number of moles of each element by the smallest number of moles, in this case, 3.33 moles:
    • Carbon: (frac{3.33}{3.33} = 1)
    • Hydrogen: (frac{6.65}{3.33} ≈ 2)
    • Oxygen: (frac{3.33}{3.33} = 1)
  4. These ratios become sub-digits in the empirical formula: The empirical formula is CH 2 O

What is the molecular formula?

The molecular formula is more comprehensive than the empirical formula. It expresses the exact number of each type of atom in a molecule. The molecular formula is always a multiple of the empirical formula.

Example of molecular formula:

As discussed earlier, the empirical formula of glucose is CH 2 O , but the molecular formula is C 6 H 12 O 6 This tells us that one molecule of glucose contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.

How to determine molecular formula

Follow these steps to find the molecular formula:

  1. First, find the empirical formula mass by adding up the atomic masses of all the atoms in the empirical formula.
  2. Then, divide the molecular mass (given or experimentally determined) by the empirical formula mass, which gives you the multiplier to apply to the sub-numbers in the empirical formula.
  3. Multiply the sub-digits in the empirical formula by this factor to find the molecular formula.

Step-by-step example:

Determine the molecular formula for a compound with an empirical formula of CH and a molecular mass of 78 g/mol.

  1. Calculate the mass by the empirical formula (CH):
    • Carbon: (12.01 ,g/mol)
    • Hydrogen: (1.008 ,g/mol)
    • Total mass: (12.01 + 1.008 = 13.018 ,g/mol)
  2. Divide the molecular mass by the empirical formula mass: [frac{78 ,g/mol}{13.018 ,g/mol} = 6]
  3. Multiply all sub-numbers in the empirical formula CH by this factor:
    • Result: C 6 H 6

Using visuals to understand formulas

Visual representation can simplify our understanding, so let's look at a simple schematic diagram that shows the difference between the molecular and empirical formula of another compound, water.

Molecular formula: H2O Empirical formula: H 2 O (same as molecular)

In this visualization, blue represents the oxygen atom, while red represents the hydrogen atom. For water, the empirical and molecular formulas are the same.

Why it's important to understand formulas

Knowing empirical and molecular formulas helps in various real-world applications, such as designing chemical reactions, creating new materials, and even in pharmacy for drug manufacturing. The difference between empirical and molecular formulas helps chemists to fully understand the structure of compounds and how they interact in different situations.

As you gain more understanding and practice in calculating these formulas, the process will become more intuitive. Remember, each part of the calculation helps reveal the "story" the atoms are telling about their arrangement in the molecule.

Conclusion

Empirical and molecular formulas are fundamental in chemistry, opening the door to understanding chemical structure and reactions. Mastering these concepts will lay a strong foundation for more advanced topics in chemistry.

Keep practicing with different compounds to strengthen your understanding of empirical and molecular formulas. Handling complex calculations becomes easier with practice, which aids you in your journey in chemistry.


Grade 10 → 6.3


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Empirical and molecular formula

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