Complete Chapter 4: Chemical Formulas & Equations – Chemistry Guide

Complete Chemistry Chapter 4: Chemical Formulas & Equations

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Complete Study Guide & Key Notes

4.1 Chemical Formula

Elements exist in different forms in this world. There are elements which exist in the form of aggregate of atoms. These elements are represented by their symbols alone, for example, Na, Ca, C, Fe, etc. On the other hand, elements like O₂, N₂, H₂ exist as discrete molecules in which their atoms are chemically bonded to each other. In ozone, three atoms of oxygen are bonded to each other, so its chemical formula is O₃.

Similar to elements, chemical compounds also exist in different forms. Common salt i.e. sodium chloride exists in the form of ions which are bonded together in the form of a crystal. Since ratio between its ions is 1:1, sodium chloride is represented by a formula unit NaCl. Similarly, the other ionic compounds are represented by their formula units which show the minimum ratio present between their ions. Examples are CaCl₂, KBr and BaCl₂, etc.

Covalent compounds generally exist in discrete molecules in which atoms are bonded together. For example, water exists as molecules which are represented by the chemical formula H₂O. It shows that in one molecule of water two atoms of hydrogen are bonded to one atom of oxygen. Similarly, chemical compound, ammonia is represented by NH₃, and methane gas is represented by CH₄. A chemical compound is thus, represented by a chemical formula which is called the molecular formula of that compound and which shows all the types of atoms bonded together in one molecule of that compound. Examples of covalent compounds are HCl, HF, H₂S, PH₃, H₂O₂, H₂SO₄, CO₂, CO, C₂H₂, etc.

Key Notes – Chemical Formula
  • Elements as atoms: Na, Ca, C, Fe (single atoms)
  • Elements as molecules: O₂, N₂, H₂, O₃ (bonded atoms)
  • Ionic compounds: Formula units showing ion ratios (NaCl, CaCl₂)
  • Covalent compounds: Molecular formulas showing actual atoms (H₂O, NH₃)
  • Formula unit: For ionic compounds, shows simplest ion ratio
  • Molecular formula: For covalent compounds, shows actual atom count
Tips & Memorization

Quick Identification:

  • Metals + Non-metals → Ionic compounds (NaCl, CaO)
  • Non-metals + Non-metals → Covalent compounds (H₂O, CO₂)
  • Group 1 metals always +1 (Na⁺, K⁺, Li⁺)
  • Group 2 metals always +2 (Mg²⁺, Ca²⁺, Ba²⁺)
  • Group 17 (Halogens) always -1 (Cl⁻, Br⁻, I⁻)
Exercise 4.1

1. How would you differentiate between the chemical formula of an element and that of a compound? Give examples.

2. Write down the names of ionic and covalent compounds whose formulas have been given in this article.

Answer 1: Element formulas represent single elements either as atoms (Na, Fe) or molecules (O₂, H₂). Compound formulas represent combinations of two or more different elements (NaCl, H₂O, CO₂).

Answer 2:

Ionic compounds: Sodium chloride (NaCl), Calcium chloride (CaCl₂), Potassium bromide (KBr), Barium chloride (BaCl₂).

Covalent compounds: Water (H₂O), Ammonia (NH₃), Methane (CH₄), Hydrochloric acid (HCl), Hydrogen fluoride (HF), Hydrogen sulfide (H₂S), Phosphine (PH₃), Hydrogen peroxide (H₂O₂), Sulfuric acid (H₂SO₄), Carbon dioxide (CO₂), Carbon monoxide (CO), Acetylene (C₂H₂).

4.2 Empirical Formula

Empirical Formula of a compound shows the simplest whole number ratio of atoms present in that compound. All the ionic compounds are represented by their empirical formulas. These formulas show the simplest ratio present between their ions. The empirical formula of calcium fluoride is CaF₂ which shows the ratio present between calcium and fluoride ions in its crystal.

For covalent chemical compounds, which exist as molecules, the empirical formulas may be different from their molecular formulas. For example, hydrogen peroxide is represented by its molecular formula H₂O₂; its empirical formula will be HO. Similarly, a benzene molecule has C₆H₆ as its molecular formula; so its empirical formula will be CH. For water, the molecular formula and the empirical formula are both the same i.e. H₂O because there does not exist any minimum ratio between hydrogen and oxygen atoms.

Since an empirical formula does not tell us the actual number of atoms present in that compound, rather it represents the simplest ratio between atoms, it is possible that some compounds may have the same empirical formula. For example, both benzene (C₆H₆) and acetylene (C₂H₂) have the same empirical formula CH.

Key Notes – Empirical Formula
  • Empirical formula: Simplest whole number ratio of atoms
  • All ionic compounds use empirical formulas (formula units)
  • Covalent compounds may have different empirical and molecular formulas
  • Example: H₂O₂ (molecular) → HO (empirical)
  • Example: C₆H₆ (molecular) → CH (empirical)
  • Example: H₂O → Same empirical and molecular formula
  • Different compounds can have same empirical formula (C₆H₆ and C₂H₂ both have CH)
Tips & Memorization

Quick Method: To find empirical formula from molecular formula:

  1. Write the molecular formula (e.g., C₆H₁₂O₆)
  2. Find GCD of all subscripts (for C₆H₁₂O₆, GCD = 6)
  3. Divide all subscripts by GCD (C₁H₂O₁)
  4. Remove “1” subscripts → CH₂O (empirical formula)

Memory Trick: “Empirical = SIMPLEST ratio”

Exercise 4.2

Give two examples of the compounds which have same empirical and molecular formulas.

Answer:

1. Water (H₂O) – Both empirical and molecular formulas are H₂O

2. Carbon dioxide (CO₂) – Both empirical and molecular formulas are CO₂

3. Ammonia (NH₃) – Both empirical and molecular formulas are NH₃

4. Methane (CH₄) – Both empirical and molecular formulas are CH₄

4.3 Chemical Formulas of Binary Ionic Compounds

In order to write down the formula of an ionic compound, first identify the cations and anions and the number of charges present on them. Finally combine the two ions together to form an electrically neutral compound.

If you know the name of binary ionic compound, you can write its chemical formula. Start by writing symbol of cation with its charge. Then write the symbol of anion with its charge and find out how many of these ions are needed to give an electrically neutral compound. For example, write down the formula of lithium oxide. The symbol of lithium cation with its single positive charge is Li⁺. The symbol of anion is O²⁻.

Let us now apply crisscross method to write the formula. In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are then dropped.

Example: Lithium Oxide

Cation: Li⁺, Anion: O²⁻

Crisscross: Li₁O₂ → Li₂O (after simplifying)

Example: Aluminium Oxide

Cation: Al³⁺, Anion: O²⁻

Crisscross: Al₂O₃

Example: Magnesium Nitride

Cation: Mg²⁺, Anion: N³⁻

Crisscross: Mg₃N₂

Atoms and their Cations with charges Atoms and their Anions with charges
H → H⁺O → O²⁻
Na → Na⁺N → N³⁻
Li → Li⁺Cl → Cl⁻
K → K⁺Br → Br⁻
Mg → Mg²⁺
Ca → Ca²⁺
Cu → Cu⁺, Cu²⁺
Ba → Ba²⁺
Fe → Fe²⁺, Fe³⁺
Zn → Zn²⁺
Sn → Sn²⁺, Sn⁴⁺
Al → Al³⁺
Ni → Ni²⁺
Interesting Information!

The composition of all the chemical products we use in our lives, such as shampoos, perfumes, soaps and fertilizers are formed using stoichiometric calculations. Without stoichiometry the chemical industry does not exist.

Key Notes – Binary Ionic Compounds
  • Binary ionic compounds: Metal + Non-metal
  • Crisscross method: Swap charge numbers as subscripts
  • Always simplify subscripts if possible (e.g., Li₂O not Li₂O₂)
  • Total positive charge = Total negative charge
  • Common cations: Group 1 (+1), Group 2 (+2), Al³⁺, Zn²⁺
  • Common anions: Group 17 (-1), Group 16 (-2), Group 15 (-3)
Crisscross Method – Step by Step
  1. Write cation and anion with charges (Al³⁺ O²⁻)
  2. Crisscross charge numbers (Al₂O₃)
  3. Drop charge signs
  4. Simplify if needed (e.g., Ca₂O₂ → CaO)
  5. Check: (3+ × 2) + (2- × 3) = 6+ + 6- = 0 ✓

Memory Aid: “Cross the numbers, drop the signs!”

4.4 Determining Molecular Formula from Empirical Formula

Molecular formula of a compound can be found out if we know its empirical formula. To calculate the empirical formula of a compound, you need to determine the simplest whole-number ratio of atoms in the compound. This can be done by using experimental data on the mass percent composition of the compound. Molecular formula is then calculated by the following relationship:

Molecular formula = n × (Empirical Formula)

Where n = Molar Mass / Empirical Formula mass

Example: Hydrogen Peroxide

Empirical formula = HO

Empirical formula mass = 1 + 16 = 17 g/mol

Molar mass = 34 g/mol

n = 34 / 17 = 2

Molecular formula = (HO)₂ = H₂O₂

Key Notes – Molecular Formula from Empirical
  • Molecular formula = n × (Empirical formula)
  • n = Molar mass / Empirical formula mass
  • If n = 1, molecular formula = empirical formula
  • n must be a whole number (integer)
  • Round n to nearest whole number if calculated value is close
  • Example: Empirical CH₂O, Molar mass 180 → n=6 → C₆H₁₂O₆
Sample Problem

Empirical formula of a compound is CH. Its molecular mass is 78 g/mol. Find out its molecular formula.

Solution:

Empirical formula mass of CH = 12 + 1 = 13 g/mol

n = Molar mass / Empirical formula mass = 78 / 13 = 6

Molecular formula = (CH)₆ = C₆H₆ (This is benzene)

Exercise 4.4

1. The empirical formula of a compound is CH₂O. Its molar mass is 180 g/mol. Determine its molecular formula.

2. The empirical formula of a compound is CH₂O. Its molar mass is 60 g/mol. Determine its molecular formula.

Answer 1:

Empirical formula mass of CH₂O = 12 + 2 + 16 = 30 g/mol

n = 180 / 30 = 6

Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (This is glucose)

Answer 2:

Empirical formula mass of CH₂O = 12 + 2 + 16 = 30 g/mol

n = 60 / 30 = 2

Molecular formula = (CH₂O)₂ = C₂H₄O₂ (This is acetic acid)

Quick Calculation Method

Formula: Molecular formula = (Empirical formula) × (Molar mass / Empirical mass)

Steps:

  1. Calculate empirical formula mass
  2. Divide molar mass by empirical mass to get n
  3. Multiply all subscripts in empirical formula by n
  4. If n is not whole number, multiply empirical formula to get whole numbers

4.5 Deduce Molecular Formula from Structural Formula

In order to deduce the molecular formula from the structural formula the following steps are taken:

  1. Write down the structural formula of the compound.
  2. Count the number of atoms of each type in the structural formula.
  3. Write the symbols of all the elements.
  4. Write the total number of atoms of each kind as a subscript.
  5. Remove the subscript 1.
Sample Problem: Sulphuric Acid

Structural formula: H-O-S(=O)₂-O-H

Count atoms: 2 H, 1 S, 4 O

Molecular formula: H₂SO₄

Sample Problem: Acetic Acid

Structural formula: CH₃-COOH

Count atoms: 2 C, 4 H, 2 O

Molecular formula: C₂H₄O₂

Key Notes – Structural to Molecular Formula
  • Step 1: Write structural formula clearly
  • Step 2: Count ALL atoms of each element
  • Don’t forget hidden atoms (like H in CH₃ groups)
  • Step 3: Write element symbols in order (C then H then others)
  • Step 4: Write counts as subscripts
  • Step 5: Omit subscript “1” (write H₂O not H₂O₁)
Exercise 4.5

1. Find out the molecular formula of phosphoric acid. Its structural formula is: HO-P(=O)(OH)₂

2. Determine the molecular formula of n-propyl alcohol. Its structural formula is CH₃-CH₂-CH₂-OH

3. Write down the formula of calcium carbonate. Its structural formula is Ca²⁺ and CO₃²⁻

Answer 1: Phosphoric acid: Count atoms = 3 H, 1 P, 4 O → Molecular formula = H₃PO₄

Answer 2: n-propyl alcohol: Count atoms = 3 C, 8 H, 1 O → Molecular formula = C₃H₈O

Answer 3: Calcium carbonate: Ca²⁺ and CO₃²⁻ → Formula = CaCO₃

4.6 Avogadro’s Number (Nₐ)

In a chemical reaction, large number of atoms or molecules of reactants react to give the products. We would very much like to know the mass ratio in which the reactants react. For this purpose, we would also like to express these masses of reactants in grams. To achieve this objective, we need to transform the concepts of chemical formula and atomic mass units into such concepts which may lead us to know the masses of reacting elements and compounds in grams. Avogadro, an Italian scientist, helped us to achieve this objective in the following way.

Let us consider the following reaction in which two atoms of carbon react with a molecule of oxygen to produce two molecules of CO.

Example Reaction: 2C + O₂ → 2CO

2C (atoms) + O₂ (molecule) → 2CO (molecules)

When scaled up by Avogadro’s number:

2 × 6.022 × 10²³ atoms of C + 6.022 × 10²³ molecules of O₂ → 2 × 6.022 × 10²³ molecules of CO

Which corresponds to: 24 g C + 32 g O₂ → 56 g CO

The number 6.022 × 10²³ is called Avogadro’s number after the name of an Italian chemist Amedeo Avogadro. Avogadro’s number is the number of units in one mole of a substance. This number is represented as Nₐ.

Key Notes – Avogadro’s Number
  • Avogadro’s number: 6.022 × 10²³
  • Named after Italian scientist Amedeo Avogadro
  • Number of particles in 1 mole of any substance
  • Connects microscopic (atoms) and macroscopic (grams) worlds
  • 1 mole = 6.022 × 10²³ particles (atoms, molecules, ions)
  • 1 g = 6.022 × 10²³ amu (atomic mass units)
  • Example: 12 g C = 1 mole C = 6.022 × 10²³ C atoms
Memory Tricks for Avogadro’s Number

Remember: 6.022 × 10²³

  • “6.02 times 10 to the 23”
  • Same number of particles in 1 mole of ANY substance
  • 1 mole of sand grains = 6.022 × 10²³ grains
  • 1 mole of water molecules = 6.022 × 10²³ H₂O molecules
  • 1 mole of NaCl formula units = 6.022 × 10²³ NaCl units

4.7 The Mole and Molar Mass

Avogadro’s number has an immense significance in Chemistry and the quantity of a substance containing Avogadro’s number of particles (Nₐ) is called a Mole. Mole is a number like a dozen or a gross. A dozen of oranges means 12 oranges. Similarly, a mole of a substance means its 6.022 × 10²³ particles which can be atoms, molecules or ions.

The mass of one mole of a substance is called Molar mass. The molar mass of hydrogen atoms refers to the mass of one mole of hydrogen atoms and its value is 1.008 g. Similarly the molar mass of hydrogen molecules will be 2.016 g.

Examples:

• A mole of carbon atoms contains 6.022 × 10²³ atoms and weighs 12 g.

• A mole of oxygen molecules (O₂) contains 6.022 × 10²³ molecules and weighs 32 g.

• A mole of sodium chloride (NaCl) consists of 6.022 × 10²³ of its formula units and its mass is 58.5 g.

Interesting Information!

Mole is important because atoms and molecules are so small that they can not be counted. The mole concept allows us to count atoms and molecules by weighing macroscopically small amounts of matter.

Key Notes – Mole & Molar Mass
  • Mole: Amount containing Avogadro’s number of particles
  • Molar mass: Mass of 1 mole of substance (g/mol)
  • Molar mass = atomic/molecular mass in grams
  • 1 mole atoms = atomic mass in grams
  • 1 mole molecules = molecular mass in grams
  • 1 mole ionic compound = formula mass in grams
  • Example: Molar mass of H₂O = 18 g/mol
  • Example: Molar mass of NaCl = 58.5 g/mol
Sample Problems

Calculate the molar masses of the following compounds: H₃PO₄, SiO₂, C₆H₁₂O₆, N₂O, MgCO₃

Answers:

H₃PO₄: (3×1) + 31 + (4×16) = 3 + 31 + 64 = 98 g/mol

SiO₂: 28 + (2×16) = 28 + 32 = 60 g/mol

C₆H₁₂O₆: (6×12) + (12×1) + (6×16) = 72 + 12 + 96 = 180 g/mol

N₂O: (2×14) + 16 = 28 + 16 = 44 g/mol

MgCO₃: 24 + 12 + (3×16) = 24 + 12 + 48 = 84 g/mol

Exercise 4.7

Determine the molar masses of the following compounds in g mol⁻¹:

(a) H₂SO₄ (Sulphuric acid) Atomic mass of H=1, S=32, O=16

(b) C₆H₁₂O₆ (Glucose) Atomic mass of C=12, H=1, O=16

Answer (a): H₂SO₄ = 2(1) + 1(32) + 4(16) = 2 + 32 + 64 = 98 g/mol

Answer (b): C₆H₁₂O₆ = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g/mol

Quick Molar Mass Calculation

Steps:

  1. Write the chemical formula
  2. List all elements with their counts
  3. Multiply count × atomic mass for each element
  4. Add all products together
  5. Unit: g/mol

Example: CaCO₃ = Ca(1×40) + C(1×12) + O(3×16) = 40 + 12 + 48 = 100 g/mol

4.8 Chemical Equations and Chemical Reactions

To understand a chemical change and to study the different factors which control it, has always been a focal point in the efforts of chemists. To achieve these objectives, we should have an appropriate way of representing a chemical change. Fortunately, the chemists have developed a very suitable way of representing a chemical change in terms of symbols of elements and formulas of compounds. Representing a chemical change in this way is called a chemical equation.

A chemical equation tells us the elements or compounds which are reacting and those which are being produced as a result of a chemical change. The reacting substances are called as reactants while those being produced are called products. It is customary to write the reactants on the left-hand side and the products on the right-hand side. An arrow head drawn from reactants to products separates the two.

Example: 4Al + 3O₂ → 2Al₂O₃

Reactants: Al and O₂, Products: Al₂O₃

Points to remember when writing chemical equations:

  1. A chemical equation must obey the law of conservation of mass (balanced).
  2. The formulas of elements and compounds must be written correctly.
  3. A chemical equation must determine the correct mole ratio among reactants and products.
  4. A chemical equation must also point out the direction of the reaction.
  5. Show physical states as subscripts: (s) solid, (l) liquid, (g) gas, (aq) aqueous.
Example with States: Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

According to this chemical equation, zinc reacts with sulphuric acid to give zinc sulphate and hydrogen gas. This chemical equation further tells us the mole ratio not only between reactants or products but also between reactants and products. According to the equation, one mole of zinc reacts with one mole of sulphuric acid to produce one mole of zinc sulphate and one mole of hydrogen gas.

Reversible reactions are indicated by ⇌ sign, e.g.: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Reaction involving ions may also be shown in the form of chemical equation. Both AgNO₃ and NaCl are ionic compounds and are soluble in water. When they are mixed in water, they react to form products.

Ionic Reaction: AgNO₃(aq) + NaCl(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

AgCl being insoluble comes out of the aqueous solution in the form of a precipitate.

Key Notes – Chemical Equations
  • Reactants: Left side of equation
  • Products: Right side of equation
  • Arrow (→): Shows direction, use ⇌ for reversible
  • Must be balanced: Same number of each atom on both sides
  • Physical states: (s), (l), (g), (aq)
  • Shows mole ratios between reactants and products
  • Example: 2H₂ + O₂ → 2H₂O means 2 moles H₂ react with 1 mole O₂ to produce 2 moles H₂O

4.9 Calculations Based on Chemical Equation

A complete and balanced chemical equation tells us the mole ratio or molar mass ratio between the reactants and the products. With the help of this ratio, we can find out the molar masses of the products provided we know the molar masses of the reactants. Similarly the molar masses of the reactants can also be found out if we know the molar masses of the products.

Example Reaction: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

1 mole CaCO₃ (100 g) + 2 moles HCl (73 g) → 1 mole CaCl₂ (111 g) + 1 mole H₂O (18 g) + 1 mole CO₂ (44 g)

The total masses of the reactants (100g + 73g = 173g) are equal to the total masses of the products (111g + 18g + 44g = 173g).

Example 1

25g of limestone (CaCO₃) reacts with an excess of hydrochloric acid according to the above equation. How much calcium chloride (CaCl₂) will be produced?

Solution:

Mass of CaCO₃ = 25 g

Molar mass of CaCO₃ = 100 g/mol

Molar mass of CaCl₂ = 111 g/mol

According to equation: 100 g CaCO₃ produces 111 g CaCl₂

1 g CaCO₃ produces 111/100 g CaCl₂

25 g CaCO₃ produces (111/100) × 25 = 27.75 g CaCl₂

Example 2

1.80 moles of ethyl alcohol (C₂H₅OH), when burnt in air completely, will utilize how many moles of oxygen gas? Also calculate the number of moles of CO₂ produced.

Solution:

Balanced equation: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

No. of moles of ethyl alcohol = 1.80

According to equation: 1 mole C₂H₅OH needs 3 moles O₂

1.80 moles C₂H₅OH needs 3 × 1.80 = 5.4 moles O₂

1 mole C₂H₅OH produces 2 moles CO₂

1.80 moles C₂H₅OH produces 2 × 1.80 = 3.6 moles CO₂

Example 3

Aluminium metal reacts with oxygen to produce aluminium oxide. How many grams of oxygen will be required to react completely with 0.3 moles of aluminium?

Solution:

Balanced equation: 4Al + 3O₂ → 2Al₂O₃

Moles of Al = 0.3

According to equation: 4 moles Al need 3 moles O₂

1 mole Al needs 3/4 moles O₂

0.3 moles Al need (3/4) × 0.3 = 0.225 moles O₂

Molar mass of O₂ = 32 g/mol

Mass of O₂ = 0.225 × 32 = 7.2 g O₂

Example 4

How many molecules of water will be produced if we react 5 g of hydrogen gas with excess of oxygen gas?

Solution:

Balanced equation: 2H₂ + O₂ → 2H₂O

Mass of H₂ used = 5 g

Molar mass of H₂ = 2 g/mol

According to equation: 4 g H₂ produces 36 g H₂O

5 g H₂ produces (36/4) × 5 = 45 g H₂O

Molar mass of H₂O = 18 g/mol

45 g H₂O = 45/18 = 2.5 moles H₂O

Number of molecules = 2.5 × 6.022 × 10²³ = 1.5055 × 10²⁴ molecules

Key Notes – Stoichiometry Calculations
  • Step 1: Write balanced chemical equation
  • Step 2: Convert given quantity to moles
  • Step 3: Use mole ratio from balanced equation
  • Step 4: Convert moles to required units
  • Mole ratio comes from coefficients in balanced equation
  • Example: 2H₂ + O₂ → 2H₂O means 2:1:2 mole ratio
  • Total mass reactants = Total mass products (law of conservation)
Stoichiometry Flowchart

Given mass/volume → Moles → Use mole ratio → Moles of desired → Desired mass/volume/particles

  1. Convert given to moles (mass ÷ molar mass)
  2. Multiply by (coefficient desired/coefficient given)
  3. Convert moles to required units

Example: Given 10g H₂, find g H₂O produced in 2H₂ + O₂ → 2H₂O

10g H₂ → 5 mol H₂ → 5 mol H₂O (2:2 ratio) → 90g H₂O

Complete Question Bank with Solutions

Multiple Choice Questions (MCQs)

(i) How many atoms are present in one gram of H₂O?
(a) 6.022 × 10²³ atoms
(b) 6.022 × 10²⁴ atoms
(c) 0.334 × 10²³ atoms
(d) 2.004 × 10²³ atoms

Explanation: Molar mass of H₂O = 18 g/mol. 1g H₂O = 1/18 = 0.0556 mol. Number of molecules = 0.0556 × 6.022 × 10²³ = 0.334 × 10²³ molecules. Each H₂O molecule has 3 atoms (2H + 1O), so total atoms = 3 × 0.334 × 10²³ ≈ 1.002 × 10²³ atoms. Option (c) gives number of molecules, which is what’s typically calculated first.

(ii) Which is the correct formula of calcium phosphide?
(a) CaP
(b) CaP₂
(c) Ca₂P
(d) Ca₃P₂

Explanation: Calcium ion = Ca²⁺, Phosphide ion = P³⁻. To balance charges: need 3 Ca²⁺ (total +6) and 2 P³⁻ (total -6). So formula is Ca₃P₂.

(iii) How many atomic mass units (amu) are there in one gram?
(a) 1 amu
(b) 10²³ amu
(c) 6.022 × 10²³ amu
(d) 6.022 × 10²² amu

Explanation: 1 amu = 1.66 × 10⁻²⁴ g. So 1 g = 1/(1.66 × 10⁻²⁴) = 6.022 × 10²³ amu. This is Avogadro’s number, which shows the relationship between grams and atomic mass units.

(iv) Structural formula of 2-hexene is CH₃-CH=CH-(CH₂)₂-CH₃. What will be its empirical formula?
(a) C₂H₂
(b) CH
(c) C₆H₁₂
(d) CH₂

Explanation: Molecular formula of 2-hexene from structural formula: C₆H₁₂ (6 C atoms, 12 H atoms). Empirical formula = simplest ratio = divide by 6 → CH₂.

(v) How many moles are there in 25 g of H₂SO₄?
(a) 0.765 moles
(b) 0.51 moles
(c) 0.255 moles
(d) 0.4 moles

Explanation: Molar mass of H₂SO₄ = (2×1) + 32 + (4×16) = 2 + 32 + 64 = 98 g/mol. Moles = mass/molar mass = 25/98 = 0.255 moles. So correct answer is (c) 0.255 moles.

(vi) A necklace has 6g of diamonds in it. What are the number of carbon atoms in it?
(a) 6.02 × 10²³
(b) 12.04 × 10²³
(c) 1.003 × 10²³
(d) 3.01 × 10²³

Explanation: Diamond is pure carbon (C). Molar mass of C = 12 g/mol. 6g C = 6/12 = 0.5 moles. Number of atoms = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ atoms ≈ 3.01 × 10²³ atoms.

(vii) What is the mass of Al in 204g of aluminium oxide, Al₂O₃?
(a) 26g
(b) 27g
(c) 54g
(d) 108g

Explanation: Molar mass of Al₂O₃ = (2×27) + (3×16) = 54 + 48 = 102 g/mol. Mass of Al in 1 mole Al₂O₃ = 54g. So in 204g (which is 2 moles), mass of Al = 2 × 54 = 108g.

Questions for Short Answers

2(i) Write down the chemical formula of barium nitride.

Answer: Ba₃N₂

Explanation: Barium ion = Ba²⁺, Nitride ion = N³⁻. To balance charges: need 3 Ba²⁺ (6+) and 2 N³⁻ (6-).

2(iii) How many molecules are present in 1.5 g H₂O?

Answer: 5.02 × 10²² molecules

Explanation: Molar mass of H₂O = 18 g/mol. Moles = 1.5/18 = 0.0833 mol. Molecules = 0.0833 × 6.022 × 10²³ = 5.02 × 10²² molecules.

2(iv) What is the difference between a mole and Avogadro’s number?

Answer: Avogadro’s number (6.022 × 10²³) is a constant representing the number of particles in one mole of a substance. A mole is a unit of measurement (like a dozen) that represents Avogadro’s number of particles of a substance.

2(v) Write down the chemical equation of the following reaction: Copper + Sulphuric acid → Copper sulphate + Sulphur dioxide + Water

Answer: Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

Balanced equation: Check: Left: Cu=1, H=4, S=2, O=8; Right: Cu=1, H=4, S=2, O=8 ✓

Constructed Response Questions

3(i) Different compounds will never have the same molecular formula but they can have the same empirical formula. Explain

Answer: Molecular formula shows the actual number of atoms in a molecule. If two compounds have the same molecular formula, they are the same compound. Empirical formula shows only the simplest ratio of atoms. Different compounds can have the same empirical formula if they have the same elemental ratio but different actual numbers of atoms.

Examples:

  • Benzene: C₆H₆ → Empirical: CH
  • Acetylene: C₂H₂ → Empirical: CH
  • Both have same empirical formula (CH) but different molecular formulas
3(ii) Write down the chemical formulas of the following compounds: Calcium phosphate, Aluminium nitride, Sodium acetate, Ammonium carbonate and Bismuth sulphate.

Answer:

1. Calcium phosphate: Ca₃(PO₄)₂

2. Aluminium nitride: AlN

3. Sodium acetate: CH₃COONa or NaC₂H₃O₂

4. Ammonium carbonate: (NH₄)₂CO₃

5. Bismuth sulphate: Bi₂(SO₄)₃

3(iii) Why does Avogadro’s number have an immense importance in chemistry?

Answer: Avogadro’s number (6.022 × 10²³) is immensely important because:

  1. It connects the microscopic world of atoms and molecules to the macroscopic world we can measure.
  2. It allows us to count atoms by weighing measurable amounts of substances.
  3. It establishes the mole concept, which is fundamental to chemical calculations.
  4. It enables stoichiometric calculations for chemical reactions.
  5. It provides a bridge between atomic mass units (amu) and grams.
  6. Without Avogadro’s number, we couldn’t relate laboratory-scale masses to numbers of atoms/molecules.

Descriptive Questions

4(i) Which conditions must be fulfilled before writing a chemical equation for a reaction?

Answer: The following conditions must be fulfilled:

  1. The reactants and products must be identified experimentally.
  2. The correct formulas of all reactants and products must be known.
  3. The physical states of reactants and products should be determined: (s), (l), (g), (aq).
  4. The reaction conditions (temperature, pressure, catalyst) should be known.
  5. The equation must be balanced to obey the law of conservation of mass.
  6. The direction of the reaction (reversible or irreversible) should be indicated.
  7. The mole ratio between reactants and products must be correct.
  8. All information must be experimentally verified before writing the equation.
4(ii) Explain the concepts of Avogadro’s number and mole.

Answer:

Avogadro’s Number: 6.022 × 10²³ is the number of particles (atoms, molecules, ions) in one mole of any substance. Named after Italian scientist Amedeo Avogadro. It’s a fundamental constant that connects atomic scale to laboratory scale.

Mole: The amount of substance that contains Avogadro’s number of particles. It’s a counting unit like dozen (12) or gross (144), but much larger (6.022 × 10²³). One mole of any substance has a mass in grams equal to its atomic/molecular/formula mass.

Relationship: 1 mole = 6.022 × 10²³ particles = molar mass in grams

Examples: 1 mole of C atoms = 6.022 × 10²³ C atoms = 12g; 1 mole of H₂O molecules = 6.022 × 10²³ H₂O molecules = 18g.

4(iv) How many moles of coal are needed to produce 10 moles of CO according to the following equation? C + H₂O → CO + H₂

Answer: 10 moles of coal (C)

Explanation: From balanced equation: C + H₂O → CO + H₂. The mole ratio is 1:1:1:1. So 1 mole C produces 1 mole CO. Therefore, to produce 10 moles CO, we need 10 moles C.

4(v) How much SO₂ is needed in grams to produce 10 moles of sulphur according to the equation: 2H₂S + SO₂ → 2H₂O + 3S?

Answer: 213.3 g SO₂

Explanation: From equation: 1 mole SO₂ produces 3 moles S. To produce 10 moles S, need 10/3 = 3.333 moles SO₂. Molar mass SO₂ = 32 + 32 = 64 g/mol. Mass = 3.333 × 64 = 213.3 g SO₂.

4(vi) How much ammonia is needed in grams to produce 1 kg of urea fertilizer according to: 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O?

Answer: 566.7 g NH₃

Explanation: Molar mass urea [CO(NH₂)₂] = 12 + 16 + (2×14) + (4×1) = 60 g/mol. 1 kg = 1000 g urea = 1000/60 = 16.667 moles urea. From equation: 2 moles NH₃ produce 1 mole urea. So for 16.667 moles urea, need 2 × 16.667 = 33.333 moles NH₃. Molar mass NH₃ = 17 g/mol. Mass = 33.333 × 17 = 566.7 g NH₃.

4(vii) Calculate the number of atoms in the following: (a) 3g of H₂ (b) 3.4 moles of N₂ (c) 10g of C₆H₁₂O₆

Answer:

(a) 3g of H₂: Molar mass H₂ = 2 g/mol. Moles = 3/2 = 1.5 mol. Molecules = 1.5 × 6.022 × 10²³ = 9.033 × 10²³ molecules. Each H₂ has 2 atoms, so total atoms = 2 × 9.033 × 10²³ = 1.8066 × 10²⁴ atoms.

(b) 3.4 moles of N₂: Molecules = 3.4 × 6.022 × 10²³ = 2.0475 × 10²⁴ molecules. Each N₂ has 2 atoms, so total atoms = 2 × 2.0475 × 10²⁴ = 4.095 × 10²⁴ atoms.

(c) 10g of C₆H₁₂O₆: Molar mass = 180 g/mol. Moles = 10/180 = 0.0556 mol. Molecules = 0.0556 × 6.022 × 10²³ = 3.347 × 10²² molecules. Each C₆H₁₂O₆ has 6+12+6 = 24 atoms, so total atoms = 24 × 3.347 × 10²² = 8.033 × 10²³ atoms.

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