Atoms and Molecuels

Question: (1) A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 of oxygen. Calculate the percentage composition of the compound by weight.

Solution:

Given, Total mass of sample = 0.24 g

Mass of boron = 0.096 g and Mass of oxygen = 0.144 g

% composition of boron

∵ 0.24g of sample contains 0.096 g of boron

∴ 1 g of sample contains `(0.096)/(0.24)` of boron

∴ 100 g of sample contains `(0.096)/(0.24)xx100=40` of boron

% composition of oxygen

∵ 0.24g of sample contains 0.144 g of boron

∴ 1 g of sample contains `(0.144)/(0.24)` of boron

∴ 100 g of sample contains `(0.144)/(0.24)xx100=60` of boron

Thus, composition of boron = 40% and composition of oxygen = 60% Answer

Question: (2) When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Solution:

When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

Thus, even when 3.00 g of carbon is burnt in 50.00 of oxygen, only 11.00 gm of carbon dioxide is produced. In this reaction only 8.00 g of oxygen will be used and rest will be left unused.

Law of Constant Proportion which says In a chemical substance the elements are always present in definite proportions by mass, governs this answer.

Question: (3) What are polyatomic ions? Give example.

Solution:

Ions made of different atoms are called polyatomic ions.

Example:

(a) SO₄^–2 ion (Sulphate ion) . Sulphate ion is made of two different atoms, i.e. sulphur and oxygen.

(b) CO₃^–2 ion (Carbonate ion) . Carbonate ion is made of two different atoms, i.e. carbon and oxygen.

Question: (4) Write the chemical formulae of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Solution:

(a) Magnesium chloride

Symbol of Magnesium : Mg

Symbol of Chloride : Cl

Valency of Magnesium (Mg) : 2

Valency of Chloride (Cl): 1

Mg₁Cl₂

(e) For convenience "1" is not written as subscript of an element while writing the formula of a compound.

Thus, formula of the Magnesium chloride is MgCl₂

(b) Calcium oxide

Symbol of Calcium : Ca

Symbol of Oxide : O

Valency of Calcium (Ca) : 2

Valency of Oxide (O): 2

Ca₂O₂

Here since both of the the valencies equal thus, both of the valencies are omitted from subscript.

Thus, formula of the Calcium oxide is CaO

(c) Copper nitrate

Symbol of copper: Cu

Symbol of nitrate: NO₃

Valency of copper (Cu): 2+

Valency of nitrate: –1

Thus, formula of copper nitrate is Cu(NO₃)₂

(d) Aluminium chloride

Symbol of Aluminium : Al

Symbol of chloride: Cl

Valency of aluminium (Al) = 3

Valency of chloride (Cl) = 1

Thus, formula of Aluminium chloride is AlCl₃

(e) Calcium carbonate

Symbol of calcium : Ca

Symbol of carbonate : CO₃

Valency of calcium : 2

Valency of carbonate: –2

Ca₂(CO₃)₂

Since, valency of both of the atoms are same, i.e. equal to 2, thus no valency will be written.

Thus, formula of calcium carbonate is CaCO₃

Question: (5) Give the name of the elements present in the following compounds.

(a) Quick line

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate

Solution:

(a) Quick line

Chemical Name of quick lime is: Calcium Oxide

And chemical formula of quick lime (calcium oxide) is CaO

Thus, elements present in the quick lime are : Calcium (Ca) and Oxygen (O).

(b) Hydrogen bromide

Chemical formula of hydrogen bromide : HBr

Thus, elements present in the hydrogen bromide are : Hydrogen (H) and Bromine (Br)

(c) Baking powder

Chemical name of baking powder is Sodium Hydrogen Carbonate

Chemical formula of baking powder is NaHCO₃

Thus, elements present in baking power are : sodium (Na), Hydrogen (H), carbon (C) and Oxygen (O)

(d) Potassium sulphate

Chemical formula of potassium sulphate : K₂SO₄

Thus, elements present in potassium sulphate are : potassium (K), sulphur (S) and oxygen (O)

Question: (6) Calculate the molar mass of the following substances.

(a) Ethyne, C₂H₂

(b) Sulphur molecule, S₈

(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO₃

Solution:

(a) Ethyne, C₂H₂

Atomic mass of Carbon (C) = 12 u

Atomic mass of hydrogen (H) = 1 u

Thus, molar mass of C₂H₂

= (12 u × 2) + (1 u × 2)

= 24 u + 2 u = 26 u

Thus, molar mass of Ethyne, C₂H₂ = 26 u

(b) Sulphur molecule, S₈

Atomic mass of sulphur (S) = 32 u

Thus, molar mass of Sulphur molecule, S₈

= 32 u × 8 = 256 u

Thus, molar mass of Sulphur molecule, S₈ = 256 u

(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)

Atomic mass of phosphorous (P) = 32 u

Thus, molecular mass of Phosphorus molecule, P₄

= 32 u × 4 = 128 u

(d) Hydrochloric acid, HCl

Atomic mass of hydrogen (H) = 1 u

Atomic mass of chlorine (Cl) = 35.5 u

Molecular mass of Hydrochloric acid, HCl

= 1 u + 35.5 u = 36.5 u

(e) Nitric acid, HNO₃

Atomic mass of Hydrogen (H) = 1 u

Atomic mass of nitrogen (N) = 14 u

Atomic mass oxygen (O) = 16 u

Thus, molecular mass of Nitric acid, HNO₃

= 1 u + 14 u + (16 u × 3)

= 1 u + 14 u + 48 u = 63 u

Question: (7) What is the mass of

(a) 1 mole of nitrogen atoms?

Solution:

Atomic mass of nitrogen = 14 u

Thus, molar mass of nitrogen = 14 g

Thus, mass of 1 mole of nitrogen atom = 14 g

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27) ?

Solution:

Given, atomic mass of aluminium = 27 u

Thus, mass of 1 moles of aluminium atom = 27 g

Thus, mass of 4 moles of aluminium atoms = 27 g × 4 = 108 g

(c) 10 moles of sodium sulphite (Na₂SO₃)?

Solution:

Atomic mass of sodium (Na) = 23 u

Atomic mass of Sulphur (S) = 32 u

Atomic mass of oxygen (O) = 16 u

Thus, molecular mass of sodium sulphite (Na₂SO₃)

= (23 u × 2) + 32 u + (16 u × 3)

= 46 u + 32 u + 48 u

= 126 u

Thus, mass of 1 mole of sodium sulphite (Na₂SO₃) = 126 g

Thus, mass of 10 moles of sodium sulphite (Na₂SO₃)

= 126 g × 10 = 1260 g

Question: (8) Convert into mole.

(a) 12 g of oxygen gas

Solution:

Atomic mass of oxygen = 16 u

Thus, molecular mass of oxygen (O₂) = 16 u × 2 = 32 u

Thus mole of oxygen = 32 g

∵ 32 g of oxygen gas = 1 mole

∴ 1 g of oxygen gas `=1/32` mole

∴ 12 g of oxygen gas `=1/32xx12=0.375` mole

Thus, 12 g of oxygen gas = 0.375 mole

(b) 20 g of water

Chemical formula of water is H₂O

Atomic mass of hydrogen (H) = 1 u

Atomic mass of oxygen (O) = 16 u

Thus, molecular mass of H₂O = (1 u × 2)+ 16 u

= 2 u + 16 u = 18 u

Thus, mass of 1 mole of water = 18 g

∵ 18 g of water = 1 mole

∴ 1 g of water `=1/18` mole

∴ 20 g of water `=1/18xx20=1.11` mole

Thus, 20 g of water = 1.11 mole

(c) 22 g of carbon dioxide

Chemical formula of carbon dioxide is CO₂

Atomic mass of carbon = 12 u

Atomic mass of oxygen = 16 u

Thus, molecular mass of CO₂

= 12 u + (16 u × 2)

= 12 u + 32 u = 44 u

Thus, mass of 1 mole of carbon dioxide = 44 g

∵ 44 g of carbon dioxide = 1 mole

∴ 1 g of carbon dioxide `=1/44` mole

&htere4; 22 g of carbon dioxide `=1/44xx22 =0.5` mole

Thus, 22 g of carbon dioxide = 0.5 mole

Question: (9) What is the mass of

(a) 0.2 mole of oxygen atoms?

Solution:

Atomic mass of oxygen = 16 u

Thus, mass of 1 mole of oxygen atoms = 16 g

Thus, mass of 0.2 mole of oxygen atoms

= 0.2 × 16 g = 3.2 g

(b) 0.5 mole of water molecules?

Molecular mass of water = 18 u

Thus, mass of 1 mole of water molecule = 18 g

Them, mass of 0.5 mole of water molecule

= 0.5 × 18 g = 9 g

Question: (10) Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.

Solution:

Atomic mass of sulphur (S) = 32 u

Thus, molecular mass of sulphur (S₈) = 32 u × 8 = 256 u

Thus, molar mass of solid sulphur = 256 g

Thus, 256 g of sulphur = 1 mole = 6.022 × 10²³ molecules

∵ 256 g of sulphur contains 6.022 × 10²³ molecules

∴ 1 g of sulphur contains `(6.022xx10^23)/256` molecules

∴ 16 g of sulphur contains `(6.022xx10^23)/256xx16` molecules

`=3.76 xx 10^22` molecules Answer

Question: (11) Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

(Hint: Mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Solution:

Chemical formula of aluminium oxide is Al₂O₃

Now, atomic mass of aluminium = 27 u

Atomic mass of oxygen = 16 u

Thus, molecular mass of aluminium oxide Al₂O₃

= (27 u × 2) + (16 u × 3)

= 54 u + 48 u = 102 u

Thus, 1 mole of aluminium oxide = 102 g

∵ 102 g of aluminium oxide = 6.022 × 10₂₃ molecules

∴ 1 g of aluminium oxide `=(6.022xx20^23)/102` molecules

∴ 0.051 g of aluminium oxide `= (6.022xx20^23)/102xx0.051` molecules

=3.011 × 10²⁰ molecules

Now, in 1 molecule of aluminium oxide (Al₂O₃) number of aluminium ion (Al³⁺) = 2

Thus, in 3.011 × 10²⁰ of aluminium oxide, number of aluminium ion

3.011 × 10²⁰ × 2

= 6.022 × 10²⁰ Answer