Atoms and Molecules

% of boron in sample = (0.096 ÷ 0.24) × 100 = 40%
% of oxygen in sample = (0.144 ÷ 0.24) × 100 = 60%
The sample of compound contains 40% boron and 60% oxygen by weight.

When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is produced. It means all of carbon and oxygen are combined in the ratio of 3 : 8 to form carbon dioxide. Thus when there is 3 gm carbon and 50 gm oxygen, then also only 8 gm of oxygen will be used and 11 gm of carbon dioxide will be formed. The remaining oxygen is not used. This indicates law of definite proportions which says that in compounds, the combining elements are present in definite proportions by mass.

When two or more atoms combine together and behave like one entity with a net charge, then it is called polyatomic ion. For example, oxygen atom and hydrogen atom combine to form hydroxide ion (OH–). One carbon atom and three hydrogen atom combine to form carbonate ion (CO3–2).

1. MgCl2
2. CaO
3. Cu(NO3)2
4. AlCl3
5. CaCO3

1. Calcium and oxygen
2. Hydrogen and bromine
3. Sodium, hydrogen, carbon and oxygen
4. Potassium, sulphur and oxygen.

1. 1 mole of nitrogen atoms = 14u = 14gm
2. 4 moles of aluminium atoms = 4 × 27 = 108u = 108gm
3. 1mole of sodium sulphite, Na2SO3 = 2 × 23 + 1 × 32 + 3 × 16 = 126u = 126gm
   10 moles of sodium sulphite = 126 x 10 = 1260u = 1260gm.

1. 32 gm of oxygen gas = 1 mole
   12 gm of oxygen gas = 12 ÷ 32 = 0.375 mole.
2. 18 gm of water = 1 mole
   20 gm of water = 20 ÷ 18 = 1.1 mole.
3. 44 gm of carbon dioxide = 1 mole
   22 gm carbon dioxide = 22 ÷ 44 = 0.5 mole.

1. 1 mole of oxygen atoms = 16 gm
   0.2 moles of oxygen atoms = 16 x 0.2 = 3.2 gm
2. 1 mole of water molecules = 18 gm
   0.5 mole of water molecules = 18 x 0.5 = 9 gm.

1 mole of S8 = 32 x 8 = 256 gm
1 mole of S8 = 6.023 x 1023 molecules
So, 256 gm of S8 = 6.023 x 1023 S8 molecules
Or, 16 gm S8 = (6.023 x 1023 ÷ 256) x 16 = 3.76 x 1022 molecules.

1 mole of aluminium oxide, Al2O3 = 2 x 27 + 3 x 16 = 102u = 102 gm
That is 102 gm Al2O3 has 6.023 x 1023 Al2O3 molecules
Or, 0.51 gm Al2O3 has = (6.023 x 1023 x 0.51) ÷ 102 = 3.01 x 1021 Al2O3 molecules.
We know 1 molecule of Al2O3 has 2Al+++ ions.
Hence, 0.51 gm Al2O3 has = 2 x 3.01 x 1021 Al+++ ions = 6.023 x 1021 aluminium ions.