We have given,

Mass of boron = 0.096g

Mass of oxygen = 0.144g

Mass of sample =0.24g

Thus, percentage of boron by weight in the compound = $\frac{0.096 \mathrm{x} 100}{0.24} =40\%$

Thus, percentage of oxygen by weight in the compound =$\frac{0.144 \mathrm{x} 100}{0.24} =60\%$

When 3.0 g carbon is burnt in 8.00 g oxygen, 11.00 g carbon dioxide is formed. It means all of carbon and oxygen are used up and carbon and oxygen are combined in the ratio of 3 : 8 to form carbon dioxide. Thus when there is 3 g carbon and 50.0 g oxygen, then also only 8 g oxygen will be used and 11.0 g carbon dioxide will be formed. The remaining oxygen is not used us. This indicates law of definite proportions which says that in compounds, the combining elements are present in definite proportions by mass.

A group of atoms carrying a charge is known as a
polyatomic ion. For example,  hydroxide ion (OH^–), carbonate ion (CO₃^–2).

Chemical formulae of the following is:

(a) Magnesium chloride – MgCl₂

(b) Calcium oxide – CaO

(c) Copper nitrate – Cu(NO₃)₂

(d) Aluminium chloride – AlCl₃

(e) Calcium carbonate – CaCO₃