Atoms And Molecules

Step 1: Take a flask and take a mixture of sodium hydroxide solution and copper sulphate solution.

Step 2: Weight the flask and note it.

Step 3: Now mix the whole mixture thoroughly.

Step 4: Now, weight the flask.

We will observe that after mixing there is no change in the mass. This explains the law of conservation of mass.

Law of conservation of mass states that “Matter is neither created nor destroys in a chemical reaction. More simply, the mass of products is equal to the mass of reactants in a chemical reaction.

Law of conservation of mass is obeyed.

Explanation:

Total mass of reactants = 15.9g + 10.6g = 26.5g

Total mass of products = 14.2g + 12.3g = 26.5g

As there is no change in the mass of reactants and products, hence law of conservation is obeyed.

The reaction taking place is:

Let the mass of CO₂ is xg.

According to the law of conservation of mass, the mass of reactants is equal to the mass of products.

Therefore,

⇒ x + 112g = 220g

⇒ x = 220g – 112g

⇒ x = 108g

Thus, the mass of carbon dioxide is 108g.

Given: Mass of compound = 0.24g

Mass of oxygen = 0.144g

Mass of boron = 0.096g

To calculate the percent of the composition of oxygen, we apply the formula:

⇒

⇒ % composition of oxygen = 0.6g × 100

⇒ % composition of oxygen = 60%

To calculate the percent of the composition of boron, we apply the formula:

⇒

⇒ % composition of boron = 0.4g × 100

⇒ % composition of boron = 40%

Total percentage composition of compound is 60% + 40% = 100%

Shamita wrote the correct formula of oxygen as O₂ because:

i. A molecule of oxygen has two atoms.

ii. It is diatomic in nature (atomicity = 2)

iii. Thus, writing oxygen as O₂ indicates two separate atoms of oxygen.

Note: Atomicity is the number of atoms constituting a molecule.

if we do not have standard symbols for elements, then it will take a lot of time to write the full name of the elements and compounds every time to describe a reaction.

No, CO does not represent element whereas Co represents element because the first letter of the symbol is always the upper case and the second letter is always lower case.

In CO, the second letter is uppercase whereas in Co, the second letter is lowercase. Hence Co represents an element.

To get the information, first, we will see the elements present in a molecule of the compound.

Second, we will count the number of atoms of each element of that molecule.

In H₂O:

i. 2 atoms of hydrogen and one atom of oxygen are present.

ii. There are total three atoms present in a molecule of water.

For oxygen:

⇒ First, write the symbol of oxygen –

⇒ Now write 2 as a subscript after O –

For nitrogen:

⇒ First, write the symbol of nitrogen –

⇒ Now write 5 as a subscript after N –

The formula of its chloride is MOCl₂

Valency of Ca = 2

Valency of PO₄ = 3

Now, apply the criss cross method

So the formula will be Ca₃(PO₄)₂

a) Common salt

Chemical name – sodium chloride

Chemical formula – NaCl

b) baking soda

Chemical name – sodium bicarbonate

Chemical formula – NaHCO₃

c) washing soda

Chemical name – sodium carbonate

Chemical formula – Na₂CO₃

d) vinegar

Chemical name – acetic acid

Chemical formula – CH₃COOH

0.5 mole of N₂ gas

First, we apply the formula:

⇒ Mass of N₂ = No. of moles × Molar mas of N₂

⇒ Mass of N₂ = 0.5 mole × 2(atomic mass of nitrogen)

⇒ Mass of N₂ = 0.5 × 2 × 14

⇒ Mass of N₂ = 14g

Thus, the mass of N₂ gas is 14g.

0.5 mole of Natoms

First, we apply the formula:

⇒ Mass of Natoms = No. of moles × Molar mass of N

⇒ Mass of Natoms = 0.5 mole × (atomic mass of nitrogen)

⇒ Mass of Natoms = 0.5 × 14

⇒ Mass of N₂ = 7g

Thus, the mass of N atoms is 7g.

3.011 X 10²³ number of N atoms.

As we know that 1 mole = 6.022 X 10²³atoms

⇒

⇒ 1/2 mole = 0.5 mole

For 0.5 mole of Natoms

First, we apply the formula:

⇒ Mass of Natoms = No. of moles × Molar mass of N

⇒ Mass of Natoms = 0.5 mole × (atomic mass of nitrogen)

⇒ Mass of Natoms = 0.5 × 14

⇒ Mass of N atoms = 7g

Thus, mass of3.011 × 10²³ number of N atoms is 7g

6.022 X 10²³ number of N₂ molecules.

As we know that 1 mole = 6.022 X 10²³atoms

For 1 mole of N₂molecules:

First, we apply the formula:

⇒ Mass of N₂ = No. of moles × Molar mass of N₂

⇒ Mass of N₂ = 1 mole × 2(atomic mass of nitrogen)

⇒ Mass of N₂ = 1mole × 28u

⇒ Mass of N₂ = 28g

Thus, mass of6.022 × 10²³ number of N₂ molecules is 28g.

The number of particles present in one mole of any substance is has a fixed value of 6.022 × 10²³. This number is called Avogadro constant(N_A).

Mass of Na = 46g

Molar mass of Na = 23u

Apply the formula given below:

Number of particles = number of moles × 6.022 × 10²³

⇒

⇒

⇒ Number of particles = 2 × 6.022 × 10²³

^⇒ Number of particles = 12.046 × 10²³

Thus, 46g of Na contains 12.046 × 10²³ particles.

The number of particles present in one mole of any substance is has a fixed value of 6.022 × 10²³. This number is called Avogadro constant(N_A).

Mass of O₂ = 8g

Molar mass of O₂ = 32u

Apply the formula given below:

Number of particles = number of moles × 6.022 × 10²³

⇒

⇒

⇒ Number of particles = 0.25 × 6.022 × 10²³

^⇒ Number of particles = 1.5 × 10²³

Thus, 8g of O₂ contains 1.5 × 10²³ particles.

The number of particles present in one mole of any substance is has a fixed value of 6.022 × 10²³. This number is called Avogadro constant(N_A).

Number of mole of hydrogen = 0.1 mole

Molar mass of H₂ = 2u

Apply the formula given below:

Number of particles = number of moles × 6.022 × 10²³

⇒ Number of particles = 0.1 mole × 6.022 × 10²³

^⇒ Number of particles = 6.022 × 10²⁴

Thus, 0.1 mole of hydrogen contains 6.022 × 10²⁴particles.

Mass of O₂ gas = 12g

Molar mass of O₂ = 32u

Apply the formula:

⇒

⇒ No. of moles = 0.37

Thus, the 12g of O₂ gas is 0.37 mole

Mass of water(H₂O) = 20g

Molar mass of H₂O = 18u

Apply the formula:

⇒

⇒ No. of moles = 1.11

Thus, the 20g of water is 1.11 mole

Mass of CO₂ gas = 22g

Molar mass of CO₂ = 44u

Apply the formula:

⇒

⇒ No. of moles = 0.5

Thus, the 22g of CO₂ gas is 0.5 mole

In FeCl₂, the valency of Fe is + 2

In FeCl₃, the valency of Fe is + 3

Molar mass of sulphuric acid (H₂SO₄):

⇒ 2(atomic mass of hydrogen) + (atomic mass of Sulphur) + 4(atomic mass of oxygen)

⇒ 2 × 1 + 32 + 4 × 16

⇒ 2 + 32 + 64

⇒ 98 u

Thus, the molar mass of H₂SO₄ is 98 u.

Molar mass of glucose (C₆H₁₂O₆):

⇒ 6(atomic mass of carbon) + 12(atomic mass of hydrogen) + 6(atomic mass of oxygen)

⇒ 6 × 12 + 12 × 1 + 6 × 16

⇒ 72 + 12 + 96

⇒ 108 u

Thus, the molar mass of C₆H₁₂O₆ is 108 u.

For sodium

Given: Mass of sodium = 100g

Molar mass of sodium = 23u

Number of atoms = number of moles × 6.022 × 10²³

⇒

⇒

⇒ Number of atoms = 4.34 × 6.022 × 10²³

^⇒ Number of atomss = 2.613 × 10²⁴

^⇒Thus, 100g of sodium contains 2.613 × 10²⁴ atoms.

For iron

Given: Mass of iron = 100g

Molar mass of iron = 55.8u

Number of atoms = number of moles × 6.022 × 10²³

⇒

⇒

⇒ Number of atoms = 1.79× 6.022 × 10²³

^⇒ Number of atoms = 1.07× 10²⁴

^⇒Thus, 100g of iron contains 1.07× 10²⁴atoms.

Therefore, 100g of sodium has more number of atoms.