Question

If the equivalent mass of a metal (M) is $x$ and formula of its oxide is ${M_m},{O_n}$ then show that the atomic mass of M is $\dfrac{{2xn}}{m}$.

Answer 1

Hint: The equivalent mass of any element is related to its molecular mass. The atomic mass is the product of n-factor and equivalent mass of atom. Here, n-factor is also called the valency of the element.

Complete step by step answer:
The equivalent weight of a substance is the mass of a given substance which will combine with a fixed quantity of another substance. It has the relation with the atomic mass. This relation can be given as:
$E = \dfrac{M}{n}$
Where, $E = $ equivalent mass of the element
$M = $ Atomic mass of the element
$n = $ n-factor or valency of the element.
The reaction of the metal to form oxide can be given as :
$mM + nO \to {M_m}{O_n}$ $ - (1)$
Now, we know that valency of oxygen is $ - 2$ and let valency of metal M be $V$. Also, there are $m$ molecules of metal M and $n$ molecules of oxygen in metal oxide.
Now, charge on $m$ molecule of metal M $ = Vm$
And charge on $n$ molecules of oxygen $ = - 2n$
As we know that metal oxide is a neutral molecule so the total charge on it is zero. Thus, we can write it as:
$
  Vm - 2n = 0 \\
  Vm = 2n \\
  V = \dfrac{{2n}}{m} \\
 $
Hence, valency of metal is given as $\dfrac{{2n}}{m}$.
Now, as we know that atomic mass is the product of valency and equivalent weight of atom. So, from equation $ - (1)$ we get:
$M = V \times E$
Here, $E = x$(given)= equivalent weight and $M = $molecular weight
Now, from above equation we get:
$
  M = \dfrac{{2n}}{m} \times x \\
  M = \dfrac{{2nx}}{m} \\
 $
Hence, the atomic mass of the metal $M$ is $\dfrac{{2nx}}{m}$.


Note:
Remember that for acids, the n-factor is defined as the number of ${H^ + }$ ions replaced by one mole of acid in a reaction and for bases, the n-factor is defined as the number $O{H^ - }$ ions replaced by one mole of base in a reaction.