Question

Weight of oxygen in $F{{e}_{2}}{{O}_{3}}$ and $FeO$ in the ratio of:
(A) 3 : 2
(B) 1 : 2
(C) 2 : 1
(D) 3 : 1

Answer 1

Hint: Law of multiple proportions was given by Dalton.
According to the law of multiple proportions if two elements combine to form more than one compound then the different mass of one element which combine with a fixed mass of another element bear a simple ratio to one another. This law can be exemplified with an example.

Complete step by step answer:
- From the classes of chemistry, we are familiar with the calculations of weight of compound, molecular mass and so on. We shall calculate the ratio of weight of oxygen in the given compounds.
- Nitrogen and oxygen combine to form five oxides, which are nitrous oxide ${{N}_{2}}O$, nitric oxide $NO$, nitrogen trioxide ${{N}_{2}}{{O}_{3}}$, nitrogen tetroxide ${{N}_{2}}{{O}_{4}}$ and nitrogen pentoxide ${{N}_{2}}{{O}_{5}}$.
- Weight of oxygen which combines with the fixed weight of nitrogen, so weights of oxygen which combine with 14 parts by weight of nitrogen in their ratio is a simple ratio.
- Mass of two elements that combine with the same mass of other elements in their respective compounds is in the ratio of their atomic masses.
 So, in $F{{e}_{2}}{{O}_{3}}$ 48 g of oxygen combine with 168g of $Fe$, while in 16 g of oxygen combine with 52 g of $Fe$. So the simplest ratio of oxygen in $F{{e}_{2}}{{O}_{3}}$: $FeO$ is 48 : 16 or 3 : 1.
In the mathematical form,
Equivalent weight of element = $\dfrac{Wt.of.element}{Wt.of.Oxygen}\times 8$
If fixed weight of$Fe$reacts with the ‘x’ g of oxygen, so the weight of$O$in$F{{e}_{2}}{{O}_{3}}$ and $FeO$ can be calculated as follows,
Equivalent weight of $F{{e}^{3+}}$ is the ratio of weight of iron in $F{{e}_{2}}{{O}_{3}}$ to that of ${{X}_{F{{e}_{2}}{{O}_{3}}}}$ multiplied by 8 according to the above formula.
Thus,$Eq.wt.=\dfrac{56}{3}=\dfrac{56\times 2}{{{X}_{F{{e}_{2}}{{O}_{3}}}}}\times 8$ ………..(i)
Similarly, equivalent weight of $F{{e}^{2+}}$is the ratio of weight of iron in $F{{e}_{2}}{{O}_{3}}$ to that of${{X}_{F{{e}_{2}}{{O}_{3}}}}$ multiplied by 8 according to the above formula.
Hence, $Eq.wt.=\dfrac{56}{3}=\dfrac{56}{{{X}_{FeO}}}\times 8$ …………………(ii)
By solving equation number (i) and (ii) we have,
\[\dfrac{1}{3}=\dfrac{{{X}_{F{{e}_{2}}{{O}_{3}}}}}{{{X}_{FeO}}}\]
Therefore, the ratio of ${{X}_{F{{e}_{2}}{{O}_{3}}}}$ : ${{X}_{FeO}}$ = 3 : 1
where${{X}_{F{{e}_{2}}{{O}_{3}}}}$ is the weight of oxygen in $F{{e}_{2}}{{O}_{3}}$
${{X}_{FeO}}$ is the weight of oxygen in $FeO$
The correct answer is option “D” .

Note: Equivalent weight of a substance is the number of parts by weight of the substance that combine with or displaced directly or indirectly 1.008 parts by weight of hydrogen or 8 parts by weight of oxygen.
- This weight of oxygen can be also calculated by oxide formation method. In this method a known mass of element changed into oxide directly or indirectly.