Equivalent weight of the anhydrous acid [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 5 years ago.








[Improve this question](/posts/87972/edit) 




> 
> The cupric salt of a monobasic acid contains $3$ molecules of water of hydration per atom of $\ce{Cu}$. $\pu{1 g}$ of hydrated salt yielding on strong heating $\pu{0.3306 g}$ of $\ce{CuO}$. What is the equivalent weight of anhydrous acid?
> 
> 
> 


My approach:
Let the formula of the acid be $\ce{HE}$ (where $\ce{E}$ is some element). So, the formula of the salt is $\ce{CuE2\*3H2O}$


Don't know how to proceed from here. Any help would be appreciated.




Let the anhydrous acid be $\ce{HA}$ and assume its Molar Mass to be $X$. There is no a priori reason to assume $\ce{A}$ is a mere element.


The copper salt in question is $\ce{CuA2.3H2O}$. 


Its Molar Mass is:


$$63.55+2\times(X-1)+3\times 18=115.55+2X$$


Assume reasonably and without losses, that:


$$\ce{CuA2.3H2O(s) -> CuO(s)}$$


So that $1\ \mathrm{mol}$ of the hydrated salt decomposes to $1\ \mathrm{mol}$ of cupric oxide.


The Molar Mass of $\ce{CuO}$ is $79.545$, so $0.3306\ \mathrm{g}$ of it is:


$$\frac{0.3306}{79.545}=4.156\times 10^{-3}\ \mathrm {mol}$$


So now we know that $1\ \mathrm{g}$ of the hydrated salt is $4.156\times 10^{-3}\ \mathrm {mol}$, so its Molar Mass is:


$$\frac{1}{4.156\times 10^{-3}}=240.6\ \mathrm{g/mol}$$


And with the above:


$$115.55+2X=240.6$$


Solve for $X$.


As the acid is monobasic, its Molar Mass is also its Equivalent Mass.


 3