ATOMS AND MOLECULES-Exercise

Q1. 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 g of oxygen. Calculate the percentage composition of the compound by weight.

Solution:
Total mass of compound: 0.24g
Mass of boron (B): 0.096g
Mass of oxygen (O): 0.144g

Percent by weight of Boron \[\begin{aligned}&=\left(\dfrac{0.096}{0.24}\times 100\right)\\\\&=40\%\end{aligned}\] Percent by weight of Oxygen \[\begin{aligned}&=\left(\dfrac{0.144}{0.24}\times 100\right)\\\\&=60\%\end{aligned}\]

Q2. 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 when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Solution:
When 3.00 g of carbon is burnt in 50.00 g of oxygen, the mass of carbon dioxide formed will still be 11.00 g, governed by the Law of Conservation of Mass and the Law of Definite Proportions.

Step-by-Step Solution

Given Reaction: \[C+O_2\rightarrow CO_2\] Given Data: 3.00 g carbon + 8.00 g oxygen → 11.00 g carbon dioxide

New Scenario:
3.00 g carbon + 50.00 g oxygen

Limiting Reactant Analysis
1 mole carbon = 12 g
1 mole oxygen = 32 g

From stoichiometry: \[12g\,C+32g\,O_2\rightarrow44g\,CO_2\] For 3.00 g carbon:

Oxygen required \[\dfrac{32}{12}\times 3=8g\] We have 50.00 g oxygen, which is much more than needed. So, carbon is limiting.
Mass of CO₂ Formed
Same as before, all 3.00 g carbon reacts with 8.00 g oxygen, producing:
\[3.00g\,C+8.00g\,O2→11.00g\,CO_2\] No more \(CO_2\) will be produced beyond this amount, as carbon is completely consumed.

Law Governing the Answer
This result follows the Law of Definite Proportions (Law of Constant Composition):

"A given compound always contains exactly the same proportion of elements by mass, regardless of the source or amount of reactants."

It also illustrates the Law of Conservation of Mass since the mass of reactants equals the mass of products.

Q3. What are polyatomic ions? Give examples.

A polyatomic ion is a charged group of atoms bonded together by covalent bonds. Unlike simple ions (like Na⁺ or Cl⁻), polyatomic ions are made up of multiple atoms. The entire group behaves as one ion and can carry either a positive or negative charge.

Q4.Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

Solution:

Q5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

Solution:

Q.6 Calculate the molar mass of the following substances.
(a) Ethyne, \(\mathrm{C_2H_2}\)
(b) Sulphur molecule, \(\mathrm{S_8}\)
(c) Phosphorus molecule, \(\mathrm{P_4}\)
(Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, \(\mathrm{HCl}\)
(e) Nitric acid, \(\mathrm{HNO_3}\)

Solution:

Atomic masses used: \(\mathrm{C}\) = 12, \(\mathrm{H}\) = 1, \(\mathrm{S}\) = 32, \(\mathrm{P}\) = 31, \(\mathrm{Cl}\) = 35.5, \(\mathrm{N}\) = 14, \(\mathrm{O}\) = 16.