Atoms and Molecules

Atoms and Molecules


3.1 Laws of chemical combinations-

Given by -Antoine L. Lavoisier and Joseph L. Proust

1- Law of conservation of mass –“Mass can neither be created nor destroyed in a chemical reaction.”
2- Law of constant proportion- “In a chemical substance the elements are always present in definite proportions by mass”.


3.2 Dalton’s atomic theory-

All matter is composed of small particles called atoms.

Postulates-
• All matter is made up of very tiny particles called atom which take part in a chemical reaction•
• Atoms cannot be created nor destroyed in a chemical reaction.
• Atoms of a given element are identical in mass and chemical properties.
• Atoms of different elements have different mass and chemical properties.
• Atoms combine in the ratio of the small whole number to form compounds.
• The relative number of atoms and kind of atoms are constant in a given compound.


3.3 Atoms –

•Building blocks of all matter are atoms.
•Atoms do not exist independently, they combine together to form molecules.
•Atoms are very smaller in size having an atomic radius in nanometres.
1 nm or 1 x 10-9m
1m =109 nm
•Atomic radius of hydrogen =101.

•SYMBOL OF ELEMENTS- Dalton was the first scientist to use the symbol for elements. Later Berzelius suggested that the symbol of the element should be made from one or two letters of the name of elements. Now, IUPAC ( International union of pure and applied chemistry) approves name, symbols and units of elements.

Example- Hydrogen (H)
Helium     (He)
Nitrogen   (N)

•ATOMIC MASS- One atomic mass unit is a mass unit equal to exactly one -twelfth the mass of one atom of carbon-12.


3.3- Molecules-

•Group of two or more atoms that are chemically combined together by a strong attractive force.

•Molecules of elements – Molecules of elements are made up of the same type of atoms.

Example- H2 ( Molecule of hydrogen is made up of two atoms of hydrogen)

•Molecules of a compound- Molecules of a compound are formed by combining elements of different atoms.

Example-Carbon di-es of which metals and non-metals formed which compose compounds.

Two types of ions-

  • Positively charged ions called cations.
  • Negatively charged ions called as an anion.

Example-      NaCl-
Na (+) (-) cation
Cl (-) (-)  anion


3.4 – Writing chemical formula-

•Valency- Combining the power of an element. This will represent the capacity of combining atoms of one element with that of other elements to form a chemical compound.

.•Rules for writing simple formulae-

-Balancing of charges of ions.
-In the case of combining metal and non- metal, we must write the symbol of metal first.

Example-  sodium chloride-sodium(Na) is metal so written first in its chemical formula is NaCl.
-The compound formed with polyatomic ions, number of Ions present in the compound is shown by enclosing the formula of ion in a bracket and write the number of ion outside the packet.
Example- Mg (OH)2

•Formula of the simple compounds-Simple compound is made up of two elements.


3.5 -Molecular mass-

It is the sum of atomic masses of all the atoms in a molecule.

Example- 

The molecular mass of water H2O

Mass of hydrogen = 1u
Oxygen = 16 u
So, molecular mass =2+16 =18 u


3.6 Formula unit mass –

It is the sum of atomic masses of all the atoms in a formula unit of a compound. The only difference from the molecular masses that substance is taken whose constituent particles are ions.
Example- NaCl – 23+35.5=58.5 u


3.7 Mole concept-

•1 mole is defined as a quantity in number having a mass equal to its atomic a molar mass in grams.
• Number of particles in 1 mole of any substance is equal to 6.022 × 1023 This is Avogadro constant shown given by Italian scientist Amedeo Avogadro.
• The word Mole was introduced first by William Ostwald world who drive the term from the Latin word “moles” having meaning ‘heap or pile’.


NCERT Solution


Page: 32


1. In a reaction, 5.3g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.

Answer:

Equation is
Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water

Now, Law of conservation says that
The total mass of reactant =total mass of product
Here,
Total mass of reactant = Sodium carbonate + acetic acid
=5.3 g +6 g
= 11.3 g
Total mass of product= Sodium acetate + carbon dioxide + water

.=8.2 g +2.2 g+ 0.9 g
=11.3 g
Therefore LHS=RHS=11.3 g
Law of conservation agreed.

2. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Answer:

The ratio of hydrogen and oxygen=1:8

That means,
1g of hydrogen = 8g of oxygen.
3g of hydrogen= 3 x 8 = 24g of oxygen is required to complete reaction.

3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer:

Dalton’s atomic theory says “Atoms can neither be created nor destroyed”.

4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer:

The number of atoms and kind of atoms is equal in given compounds.


Page: 35


1-Define the atomic mass unit?

Answer:
One atomic mass unit is a mass unit equal to exactly one -twelfth the mass of one atom of carbon-12

2-Why is it not possible to see an atom with naked eyes?

Answer:

Atoms are very small having an atomic radius in nanometer so they are not visible with naked eyes.


Page: 39


1. Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Answer:

(i) sodium oxide – Na2O
(ii) aluminium chloride- AlCl3
(iii) sodium sulphide- Na2S
(iv) magnesium hydroxide Mg (OH)2

2. Write down the names of compounds represented by the following formulae:

(i) Al2(SO4)3

(ii) CaCl2

(iii) K2SO4

(iv) KNO3

(v) CaCO3.

Answer:

(i) Al2(SO4)3– Aluminium sulphate
(ii) CaCl2 -Calcium chloride
(iii) K2SO4 – Potassium sulphate
(iv) KNO3– Potassium nitrate
(v) CaCO3. – Calcium carbonate

3. What is meant by the term chemical formula?

Answer:
The chemical formula of a compound represents its constituent elements and number of atoms of each combining elements.

4. How many atoms are present in a

(i) H2S molecule and

(ii) PO43- ion?

Answer:

(i) H2S molecule – Hydrogen -2 atoms Sulphur-1 atom
(ii) PO43 phosphorus – 1 atom Oxygen-4 atom


Page: 40


1. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Answer:

The molecular mass of

  1. H2 =2 x mass of H
    = 2 x 1u
    = 2u
  2. O2 =2 x mass of O
    = 2 x 16u
    = 32u
  3. Cl2 =2 x mass of Cl
    = 2 x 35.5u
    = 71u
  4. CO2 = mass of C + 2 x mass of O
    = 12 + ( 2×16)u
    = 44u
  5. CH4 = mass of C + 4 x mass of H
    = 12 + ( 4 x 1)u
    = 16u
  6. C2H6= 2 x mass of C + 6 x mass of H
    = (2 x 12) +(6 x 1)u
    =24+6
    =30u
  7. C2H4=2 x mass of C + 4 x mass of H
    = (2x 12) +(4 x 1)u
    =24+4
    =28u
  8. NH3= mass of N + 3 x mass of H
    = (14 +3 x 1)u
    = 17u
  9. CH3OH = mass of C + 3x mass of H + mass of O + mass of H
    = (12 + 3×1+16+1)u
    =(12+3+17)u
    = 32u

2. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u,

Na = 23 u,  K=39u, C = 12u, and O=16u.

 

Answer:

We know that the atomic mass of
Zn = 65u
Na = 23u
K = 39u
C = 12u
O = 16u
The formula unit mass of
1- ZnO = 65u+16u = 81u
2-Na2O = (2 x 23)u +16u=46u+16u=62u
3-K2CO3. = (2 x 39)u+12u+(3 x16)u=138 u


Page: 42


1. If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?

Answer:

1 mole of carbon weighs =12g
Mass of 1 atom =mass of one mole of an atom/Avogadro no.

So
mass of 1 carbon atom = 12/6.022 X 10²³
= 1.99 X 10-23g

2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, the atomic mass of Na = 23u, Fe = 56 u)?

Answer:

For sodium atom-
23g of sodium = 6.022 X 10²³ atoms

1g of Sodium = 6.022 X 10²³ atoms/23
100g of Na contains = 6.022 X 10²³ atoms X 100/23
= 2.6182 X 1024 atoms
For iron atom- 56g of Iron = 6.022 X 10²³ atoms
1g of Iron = 6.022 X 10²³ atoms/56
100g of Iron= 6.022 X 10²³ atoms X 100/56
= 1.075 X 1024atoms
On comparing both, we Find that 100g of Na has more atoms than that of iron.


Exercise


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

Answer:

Mass of the sample compound = 0.24g
mass of boron = 0.096g
mass of oxygen = 0.144g

Percentage composition Of
Boron.= mass of boron x 100/mass of the compound
= 0.096g x 100/0.24g
=0.4 x 100
= 40%
Oxygen = mass of oxygen x 100/mass of the compound
=0.144 x100/0.24

= 0.6 x 100
= 60%

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

Answer:
3g of carbon +8 g of oxygen =11g Carbon dioxide
And Ratio of Carbon to oxygen=3:8
If we burn 3g of carbon in 50g of oxygen,
Then, out of 50 g = 8g of oxygen is used to react with 3 g of carbon dioxide which will form 11g of carbon dioxide
Left oxygen=50-8=42 g Which remain unused.

This law proves that law of constant proportion.

3. What are polyatomic ions? Give examples Polyatomic ions are the set of two or more ions which behaves like a single unit but the net charge is not equal to zero.

Answer:

Those ions that contain more than one atom but they behave as a single unit is known as Polyatomic ions.

Example- NH+4,(HSO)-4

4. Write the chemical formula of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Answer:
(a) Magnesium chloride- MgCl2
(b) Calcium oxide- CaO
(c) Copper nitrate- Cu(NO3)2
(d) Aluminium chloride- AlCl3
(e) Calcium carbonate -CaCO3

5. Give the names of the elements present in the following compounds.

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate

Answer:
(a) Quick lime- Calcium and oxygen
(b) Hydrogen bromide- Hydrogen and bromine
(c) Baking powder- Sodium, Carbon, Hydrogen, Oxygen
(d) Potassium sulphate.- Sulphur, Oxygen, Potassium

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

Answer:
(a) Ethyne, C2H2 =2 x Mass of C+2 x Mass of H
= (2×12)+(2×1)
=24+2
=26g
(b) The sulphur molecule, S8
= 8 x Mass of S
= 8 x 32
= 256g
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31)
=4 x Mass of P
= 4 x 31
= 124g

(d) Hydrochloric acid, HCl
= Mass of H+ Mass of Cl
= 1+35.5
= 36.5g

(e) Nitric acid, HNO3
=Mass of H+ Mass of Nitrogen + 3 x Mass of O
= 1 + 14+3×16
= 63g

7. What is the mass of –

(a) 1 mole of nitrogen atoms?

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

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

Answer:
(a) Atomic Mass of nitrogen= mass of 1 mole of nitrogen =14 u
(b) 4 moles of aluminium atoms((Atomic mass of aluminium =27)?
Mass of 1 mole of aluminium =27 u
Mass of 4 mole of aluminium =4 x 27 =108g
(c) 10 moles of sodium sulphite (Na2SO3)?
Mass of 1 mole of sodium sulphite (Na2SO3) = 46+32+48 = 126g
Mass of 10 moles of sodium sulphite (Na2SO3)=1260 g

8. Convert into mole.

(a) 12g of oxygen gas

(b) 20g of water

(c) 22g of carbon dioxide

Answer:
(FORMULA= Number of moles = Mass given / molar mass of molecule )

(a) 12g of oxygen gas
The molar mass of oxygen gas = 2 x atomic mass of O
= 2 x 16 = 32g
Given the mass of oxygen =12 g
By using the formula of moles
Number of moles = 12/32
= 0.375 moles

(b) 20g of water
The molar mass of water = 2 x Mass of H + Mass of O
= 2 x 1 + 16 = 18g
Given the mass of water = 20g
By using the formula of moles
Number of moles
= 20/18 = 1.11 moles

(c) 22g of carbon dioxide
(c) Given: Mass of carbon dioxide = 22g
The molar mass of carbon dioxide = Mass of C + 2 x Mass of O.
= 12 + 2x 16 = 12+32=44g
Given the mass of carbon dioxide = 22g
By using the formula of moles
Number of moles
= 22/44 = 0.5 moles

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Answer:
(a) Atomic Mass of oxygen atom= 16u,
Atomic mass of 0.2 moles of oxygen atoms = 0.2 x 16
= 3.2u

(b) 0.5 mole of water molecules?
(b)Atomic Mass of 1 mole of water molecules = 1 x2 +16=18u

Atomic Mass of 0.5 moles of water molecules = 0.5 x 18 = 9u

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

Answer:
Number of moles = mass given/ molar mass of sulphur

The molecular mass of Sulphur= 8xMass of S
= 8×32
= 256g
Given mass = 16g
By putting these values in an above-given formula

No. Of moles = 16/256
= 0.0625 moles
Now,
The number of molecules = Number of moles x Avogadro number.
(Avogadro number. = 6.022 x 1023 molecules)
= 0.0625 x 6.022 x 1023 molecules
= 3.763 x 1022 molecules of sulphur.

11. Calculate the number of aluminium ions present in 0.051g ofaluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)

Answer:
1 mole = 6.022 x 1023 molecules of aluminium oxide:
Molecular mass of aluminium oxide
= 2 x Mass of aluminium + 3 x Mass of Oxygen
= (2x 27) + (3 x16)
= 54 +48
= 102g

Now,
102g of aluminium oxide = 6.022 x 1023 molecules
0.051g of aluminium oxide has = 0.051 x 6.022 x 101023/102
= 3.011 x 1020 molecules of aluminium oxide. ———–→ equation 1
In one molecule of aluminium oxide = 2 aluminium ions,
By equation 1,
number of aluminium ions present in 0.051g of aluminium oxide
= 2 x 3.011×1020

molecules of aluminium oxide
= 6.022 x1020