The chemical formula of a compound is a symbolic representation of its composition.

How many atoms are present in a
(i) H₂S molecule and
(ii) PO₄^3-ion?

Ans

(i) In an H₂S molecule, three atoms are present; two of hydrogen and one of sulphur.

(ii) In a PO₄^3- ion, five atoms are present; one of phosphorus and four of oxygen.

→ Molecular mass of H₂ = 2 × Atomic mass of H
= 2 × 1
= 2 u

→ Molecular mass of O₂ = 2 × Atomic mass of O
= 2 × 16
= 32 u

→ Molecular mass of Cl₂= 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u

→ Molecular mass of CO₂= Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16
= 44 u

→ Molecular mass of CH₄ = Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1
= 16 u

→ Molecular mass of C₂H₆ = 2× Atomic mass of C + 6× Atomic mass of H
= 2 × 12 + 6 × 1
= 30 u

→ Molecular mass of C₂H₄= 2 x Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28 u

→ Molecular mass of NH₃ = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3×1
= 17 u

→ Molecular mass of CH₃OH = Atomic mass of C + 3 × Atomic mass of H + Atomic mass of O + Atomic mass of H
= 12 + 3×1 + 8 + 1
= 24 u

→ Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16
= 81 u

→ Formula unit mass of Na₂O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 + 16
= 62 u

→ Formula unit mass of K₂CO₃= 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 + 12 + 2 × 16 

= 78 + 12 + 32 

= 122 u

One mole of carbon atoms weighs 12 g (Given)
i.e., mass of 1 mole of carbon atoms = 12 g
Then, mass of 6.022 × 10²³ number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon = 12 ÷ (6.022 × 10²³)
= 1.9926 x 10^-23 g

Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains = 6.022 × 10²³ g number of atoms
Thus, 100 g of Na contains = 6.022 × 10²³ / 23×100 number of atoms
= 2.6182 × 10²⁴ number of atoms

Again, atomic mass of Fe = 56 u (Given)
Then, gram atomic mass of Fe = 56 g

Now, 56 g of Fe contains = 6.022 × 10²³ g number of atoms

Thus, 100 g of Fe contains = 6.022 × 10²³ / 56 × 100 number of atoms
= 1.0753 × 10²⁴ number of atoms

Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Total mass of Compound = 0.24 g (Given)
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)

 

Thus, percentage of boron by weight in the compound = 0.096 / 0.24 × 100%

= 40%

And, percentage of oxygen by weight in the compound = 0.144 / 0.24 × 100% = 60%

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
 

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive.
In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.