# How to calculate the equivalent weight of a metal in a metal sulphate?

The question is:

$$\pu{2g}$$ of a metal is present in$$\pu{4.6g}$$ of its metal sulphate. Find the equivalent weight of the metal.

I understand the concept that equivalent weight of an element is the molecular weight divided by the n-factor (in this case it's the valency), but I couldn't arrive at a conclusion.

Could you also help me understand how equivalent weight of a metal is related to the equivalent weight of the sulphate? In my textbook, it's written like this:

$$\textrm{eq. wt. of metal} = \textrm{eq. wt. of metal sulphate(2-)}$$

and then they carried on with the calculations.

PS I know the answer, I need to know how one should conceptually arrive at it!

## 1 Answer

I assume that if you are dealing in equivalent weights then you must be well associated with the concepts of equivalents.

Now, $$\textrm{equivalent} = \textrm{weight} / \textrm{eq. weight}$$

and since compounds combine in equal amounts of equivalents we have

$$\frac{\textrm{weight of metal}}{ \textrm{eq weight}} = \frac{\textrm{weight of sulphate}}{ \textrm{its eq weight}}$$

$$\rightarrow 2/x = 2.6/48$$

Hence $$x= 39.92$$.