Different compounds can have the same empirical formula.

Reason:

Because an empirical formula only represents the simplest whole number ratio of atoms in a compound, not the actual number of atoms in a molecule (molecular formula). Therefore, different compounds can have the same empirical formula but different molecular formula.

Example:

Benzene (C₆H₆) and acetylene (C₂H₂), both have the same empirical formula CH but different molecular formulae.

Avogadro's number is important.

Reason:
Because Avogadro's number acts as a bridge between the microscopic world of atoms and molecules and macroscopic world we can observe, allowing chemists to relate the number of particles in a substance to its measurable mass, making understanding of chemical reactions and stoichiometric calculations easier.

"Making calculations in chemistry practical and manageable by providing a standard unit for counting large quantities of atoms or molecules, known as a "mole."

1) Given data: 
Mass of compound = 8.657g
Mass of C = 5.217g
Mass of H = 0.962g
Mass of O = 2.478g

2) To find: 
Percentage of C, H and O = ?

3) Calculations:
Formula:
Percentage of element = (Mass of element​ / Mass of compound )×100%

Now:
⇒ Percentage of carbon = (5.217 g​ / 8.657 g )×100
= 60.26%

⇒ Percentage of hydrogen = (0.962 g​ / 8.657 g )×100
= 11.1%

⇒ Percentage of oxygen =( 2.478 g​ / 8.657 g )×100
= 28.62%

∴ % age of C = 60.26%
% age of H = 11.1%
% age of O = 28.62% 

Steps to calculate masses of products of a reversible reaction:

• Write a balanced chemical equation.
• Convert given mass of reactants into moles.
• Multiply the moles of reactant by the appropriate mole ratio from the balanced equation to get moles of product.
• Convert those moles back to mass by multiplying with the molar mass of product, that's the mass of product.

This calculation is done, assuming you are considering theoretical maximum yield at equilibrium, which is usually determined by the equilibrium constant (Kc) of the reaction.

Conditions for writing a chemical equation:

1. Law of conservation of mass:
A chemical equation must obey the law of conservation of mass. This means that no atom should be destroyed or produced during a chemical change. The total number and the type of atoms must remain the same during the chemical change. Thus the total number and kind of atoms on both sides of the equation must be equal or the chemical equation must be balanced.

2. Correct formulas:
The formulas of elements and compounds must be written correctly.

3. Correct mole ratio:
A chemical equation must determine the correct mole ratio among the reactants, the products and between the reactants and the products.

4. Direction of reaction:
A chemical equation must also point out the direction in which the change is proceeding.

5. Physical states of reactants and products:
It is a usual practice to show the normal physical states of reactants and products as a subscript. Solid, liquid and gas are symbolized as 's', 'l' and 'g' respectively. Aqueous (aq) represents the solvated ion.

1) Avogadro's number

i. Definition:
"Avogadro's number is the number of units (atoms, molecules and formula units) in one mole of a substance.

ii. Value:
The number 6.02 × 10²³ is called Avogadro's number.

iii. Representation:
It is represented by Nₐ.

iv. Naming:
Avogadro's number is named after the name of an Italian chemist Amadeo Avogadro.

v. Explanation:
a) Objective of 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 by presenting the idea of Avogadro's number.

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

   2C     +   O₂  ---→ 2CO
(Atom)  (Molecule)  (Molecule)

Since it is not possible to account for the masses of individual atoms or molecules because these are very small particles, we increase the number of reacting species as written below:

          2C     +       O₂            -------→ 2CO
2 × 100 atoms    100 molecules      → 2 × 100 molecules
2 × 10000 atoms  10000 molecules  → 2 × 10000 molecules

Increasing the number of reacting atoms or molecules will not change the ratio in which these are reacting or are being formed.

Increasing the number of reacting species, however, has not solved the problem because this number is still very small. We should increase this number to such a value whereby it is convenient for us to calculate their masses.

Thus

         2C             +              O₂            -------→       2CO
  2×6.02 × 10²³             6.02 × 10²³              2×6.02 × 10²³ 

    atoms                       molecules                  molecules

6.02 × 10²³ is a huge number and we have selected this because
1.00 g = 6.02 × 10²³ amu

Now the amounts of reactants and products in the forementioned equations can be written as follows:
2×6.02 × 10²³×(12.0 amu)=24.00 g carbon atoms
6.02 × 10²³ ×(32.0 amu)=32.00 g oxygen molecules
2×6.02 × 10²³ ×(28.0 amu)=2×28.00 g CO molecules

The mass ratio between the reactants and those of products will then become:
C    +   O₂ → 2CO
24g      32g     56g

You must have realized that starting from a simple equation, we have developed such ratio of masses of the reacting species which can conveniently be used in laboratory.

According to above-mentioned equation, 24g of C contains 2×6.02 × 10²³ atoms of carbon, 32g of oxygen contains 6.02 × 10²³ molecules of oxygen and 56g of carbon monoxide contains 2×6.02 × 10²³ of its molecules.

2) Mole

i. Definition:
"The quantity of a substance containing Avogadro's number of particles (Nₐ) is called a mole."

ii. Explanation:
A mole of a substance means its 6.02 × 10²³ particles which can be atoms, molecules or ions. When we use the term mole of a substance, we must also refer to what type of particles are present in this substance.

iii. Examples:
1 mole of C atoms = 12g = 6.02 × 10²³ atoms
1 mole of O₂ molecules = 32g = 6.02 × 10²³ molecules
1 mole of NaCl = 58.5g = 6.02 × 10²³ formula units

iv. Importance:
Mole is important because atoms and molecules are so small that they cannot be counted. The mole concept allows us to count atoms and molecules by weighing microscopically small amounts of matter.

Given data:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Mass of CH₄ = 10g

To find:
Mass of CO₂ produced = ?

Calculations:
According to balanced chemical equation;
1 mole of CH₄ produced CO₂ = 1 mole
No. of moles of CH₄ = Given mass​ / Molar mass

Molar mass of CH₄ = 12 + (4 × 1) g mol⁻¹
" " " "                       = 12 + 4 g mol⁻¹
" " " "                       = 16 g mol⁻¹  

No. of moles of CH₄ = 10 g​ / 16 g mol⁻¹
" " " "                         = 0.625 moles  

Now;
1 mole of CH₄ produced CO₂ = 1 mole
0.625 moles of CH₄ will produce CO₂ = 1 × 0.625 moles
" " " " " "                                              = 0.625 mole  

→ Mass of CO₂ produced = No. of moles of CO₂ × Molar mass of CO₂  

Molar mass of CO₂ = 12 + (16 × 2) g/mol
" " " "                       = 12 + 32 g/mol
" " " "                       = 44 g/mol  

Now;
Mass of CO₂ produced = 0.625 mol × 44 g mol⁻¹
" " " "                              = 27.5 g  

∴ 27.5g of CO₂ will be produced when we react 10g of CH₄ with excess of O₂.

Given data:
No. of moles of CO = 10 moles
3C(s) + O₂(g) + H₂O(l) → H₂(g) + 3CO(g)

To find:
No. of Moles of coal (C) = ?

Calculations:
According to balanced chemical equation:
3 moles of coal (C) are needed to produce 3 moles of CO.

So,
3 moles of CO need coal(C) = 3 moles
1 mole of CO will need coal = 3 / 3​ moles
" " " " "                                   = 1 mole
10 moles of CO will need coal = 1 × 10 moles
" " " " " "                                   = 10 moles

∴ 10 moles of coal are needed to produce 10 moles of CO. 

Given data:
No. of moles of sulphur = 10 moles

 2H₂S + SO₂ → 2H₂O + 3S

To find:

Mass of SO₂​ = ?

Calculations:
According to balanced chemical equation;
1 moles of sulphur produced. SO₂​ = 3 mole  i.e;
3 moles of sulphur needed SO₂= 1 mole
1 mole of sulphur will need SO₂ =( 1/ 3)​ mole
10 moles of sulphur will need SO₂ = ( 1 / 3 )​×10 mole
" " " " " "                                   SO₂ = 3.33 moles.

So, we have to calculate mass of SO₂, so:
Mass of SO₂​ = No. of moles of SO₂​ × Molar mass of SO₂​  

Molar mass of SO₂​ = 32 + (16 × 2) g mol⁻¹
" "                           = 32 + 32 g mol⁻¹
" "                          = 64 g mol⁻¹

 Now;
Mass of SO₂​ = 3.33 mol × 64 g mol⁻¹
" " " "           = 213.12 g

∴ 213.12 g of SO₂​ is needed to produce 10 moles of sulphur.

Given data:
1 kg of urea fertilizer is to be prepared.
2 NH₃ ​(aq)+CO₂​(aq) → CO(NH₂)₂(aq) +H₂O(l)

To find:
Mass of ammonia needed = ?

Calculations:
According to balanced chemical equation;
1 mole of urea (CO(NH₂)₂​) need ammonia ( NH₃ ​) = 2 moles

 Now;
No. of moles of urea in 1 kg = ?
∴ 1 kg = 1000 g
⇒ No. of moles of urea = (Given mass​ / Molar mass) 

 Molar mass of urea (CO(NH₂)₂) = 12 + 16 + 2(14 + 2×1)
" " "                                               = 12 + 16 + 2(14 + 2)
" " "                                               = 28 + 2(16)
" " "                                               = 28 + 32 g/mol
" "                                                 = 60 g/mol  

Now;
No. of moles of urea = 1000 g​ / 60 g mol⁻¹
" " " "                         = 16.67 moles  

1 mole of urea need NH₃​ = 2 moles
16.67 moles of urea need NH₃ = 2 × 16.67
" "                                             = 33.34 moles  

∴ Mass of ammonia = No. of moles × Molar mass
Molar mass of NH₃​ = 14 + (3 × 1) g/mol
" " "                          = 14 + 3 g/mol
" " "                          = 17 g/mol

Now;
Mass of ammonia (NH₃​) = 33.34 mol × 17 g mol⁻¹
" " " "                              = 566.78 g
∴ 566.78 g of ammonia is needed to produce 1 kg of urea fertilizer. 

a. 3 g of H₂

Solution:
Given data:
Mass of H₂​ = 3g

To find:
No. of atoms in 3g of H₂ = ?

Calculations:
No. of atoms in 3g of H₂​ = No. of molecules × No. of atoms(2) in each molecule of H₂​  

No. of molecules = No. of moles × Nₐ
No. of moles = Given mass​ / Molar mass  

Molar mass of H₂​ = 2 × 1 g mol⁻¹
" "                         = 2 g mol⁻¹  Now;
⇒ No. of moles =3g​ / 2g mol⁻¹ 

No. of moles = 1.5 moles
 No. of molecules = 1.5 x 6.022 x 10²³
                                = 9.03 x 10²³ molecules

No. of atoms in 3g of H₂ = 9.03 x 10²³ x 2
                            = 18.06 x 10²³
                            = 1.81 x 10²⁴ atoms

:. No. of atoms in 3g of H₂ = 1.81 x 10²⁴ atoms 

b) 3.4 moles of N₂

Solution:
Given data:
   No. of moles of N₂ = 3.4 moles

 To find:
   No. of atoms in 3.4 moles of N₂ = ?

Calculations:
   No. of atoms = No. of molecules x No. of atoms(2) in each molecule of N₂.

   No. of molecules = No. of moles x Nₐ
                    = 3.4 x 6.022 x 10²³
                    = 20.47 x 10²³
                    = 2.05 x 10²⁴ molecules

   No. of atoms = 2.05 x 10²⁴ x 2
                = 4.10 x 10²⁴ atoms

:. No. of atoms in 3.4 moles of N₂ = 4.10 x 10²⁴ atoms 

c) 10g of C₆H₁₂ O₆

Solution:

Given data:
   Mass of C₆H₁₂ O₆ = 10g

To find:
   No. of atoms in 10g of C₆H₁₂ O₆  (glucose) = ?

Calculations:
    No. of atoms = No. of molecules x No. of atoms (6+12+6 = 24) in each molecule of glucose

   No. of molecules = No. of moles x Nₐ

   No. of moles = Given mass / Molar mass

   Molar mass of C₆H₁₂ O₆ = (12x6) + (12x1) + (16x6)
                         = 72 + 12 + 96 g mol⁻¹
                         = 180 g mol⁻¹

   Now;
   No. of moles = 10g / 180g mol⁻¹
                        = 0.06 moles

   No. of molecules = 0.06 x 6.022 x 10²³
                                = 0.36 x 10²³
                               = 3.6 x 10²² molecules

   No. of atoms = 3.6 x 10²² x 24
                         = 86.4 x 10²² atoms
                         = 8.64 x 10²³  atoms

:. No. of atoms in 10g of glucose = 8.64 x 10²³ atoms

Given data:
Volume of a glass = 400 cm³
No. of glasses of water should be drunk everyday = 8

To find:
No. of moles of water needed everyday = ?

Calculations:
 No. of moles of H₂O = Given mass / Molar mass

Molar mass of H₂O = (2x1) + 16 g mol⁻¹
                                = 2 + 16 g mol⁻¹
                                = 18 g mol⁻¹

. Given mass of H₂O = ? 
Density = Mass / Volume               .:Density of water = 1 g cm⁻³

Mass of H₂O = Density x Volume
                       = 1 g cm⁻³x 400 cm³
                       = 400 g

Mass of 8 glasses of H₂O = 8 x 400 g
                                         = 3200 g

Now;
No. of moles of H₂O = 3200 g / 18 g mol⁻¹
                                   = 177.78 moles

:. 177.78 moles of water are needed daily for a single adult.

1- Molecular structure of SiO₂:

Silicon dioxide (SiO₂), commonly known a sand, has a three-dimensional network structure. In this structure, each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral arrangement. Each oxygen atom is shared between two silicon atoms, forming a continuous network of Si-O bonds.

2- Determination of empirical formula:

The empirical formula is determined on the basis of the simplest ratio of silicon to oxygen atoms in the lattice structure. In the case of sand, the ratio is 1 silicon atom to 2 oxygen atoms, hence the formula SiO₂. The formula reflects the stoichiometric proportion of the elements in the continuous crystal lattice rather than discrete molecules.

:. The formula SiO₂ represents the simplest atomic ratio in sand's network structure, not discrete molecules.