Question

A metal forms two oxides. The higher oxide contains 80% metal. Upon oxidation, 0.72 g of the lower oxide gives 0.80 g of the higher oxide. show that this data supports the law of multiple proportions.

Answer 1

Law of multiple proportions

• The Law of multiple proportions states that ‘When two elements combine to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.’

Let us consider x number of O atoms are present in lower oxide and the z number of O atoms are present in the higher oxide upon the addition of y O atoms. given z=x+y

• ${\mathrm{MO}}_{\mathrm{x}}+\mathrm{yO}\to {\mathrm{MO}}_{\mathrm{z}}\mathrm{where}\mathrm{z}=\mathrm{x}+\mathrm{y}$

Upon oxidation, 0.72 g of the lower oxide gives 0.80 g of the higher oxide.

• $$\begin{gathered} {\mathrm{MO}}_{\mathrm{x}}+\mathrm{yO}\to {\mathrm{MO}}_{\mathrm{z}} \\ 0.72\mathrm{g}0.80\mathrm{g} \end{gathered}$$

According to the law of mass conservation, the total mass of the reactant is equal to the total mass of the product.

• Hence the mass of y atoms of oxygen = 0.80-0.72 = 0.08
• $$\begin{gathered} {\mathrm{MO}}_{\mathrm{x}}+\mathrm{yO}\to {\mathrm{MO}}_{\mathrm{z}} \\ 0.72\mathrm{g}0.08\mathrm{g}0.80\mathrm{g} \end{gathered}$$ g

Hence, the ratio of reactants is 0.72 : 0.08 = 1:9. it is a simple whole-number ratio that agrees with the law of proportion.

Hence, the given data supports the law of proportion.