Question

One gram of the silver salt of an organic dibasic acid yields, on strong heating, $0.5934$g of silver. If the weight percentage of carbon in it $8$ times the weight percentage of hydrogen and one-half the weight percentage of oxygen, determine the molecular formula of the acid.

• A. $C_4{H}_6{O}_4$
• B. $C_4{H}_6{O}_6$
• C. $C_2{H}_6{O}_2$
• D. $C_5{H}_{10}{O}_5$

Answer 1

The correct option is B $C_4{H}_6{O}_6$

Correct option is B .

Given,

Mass percentage of C = $\frac{1}{8}$ of mass percentage of H

Mass percentage of C = $\frac{1}{2}$ of mass percentage of O

Let the mass percentage of C be x

According to the reaction:

$2Ag+H_{2}A\to A{g}_{2}A+2H^{+}$ , where $A$ represents the acid $C_x{H}_y{O}_z$ whose molecular mass has to be determined.

Also, on heating the silver salt, we get: $A{g}_{2}A\to 2Ag$ $...\left(i\right)$

Weight of the silver salt = 1 gm

Weight of Silver = 0.5934 gm

Number of moles of $A{g}_{2}A=2\left(\frac{1}{216+A}\right)$

Number of moles of $Ag=\frac{0.594}{108}$ (Since, number of moles = given mass /molar mass of compound)

From equation (i), we get that number of moles of $A{g}_{2}A$ = number of moles of Ag

Therefore, $2\left(\frac{1}{216+A}\right)=\frac{0.594}{108}$

$\Rightarrow A=148$

Since the dibasic acid was$H_{2}A$, therefore molecular weight will be

Molecular weight of Acid = molecular weight of A + molecular weight of $H_2$

$\Rightarrow 148+2=150u$

The molar mass of the acid comes out to be 150u .

Now since we have taken the mass percentage of carbon as $x$ , therefore :

Mass percentage of $C$ $H$ $O$

$x$ $\frac{x}{8}$ $2x$

$÷$ Molecular mass: $\frac{x}{12}$ $\frac{x}{8}$ $\frac{x}{8}$

$÷$by $4$ $\frac{x}{3}$ $\frac{x}{2}$ $\frac{x}{2}$

$\Rightarrow$ Empirical formula is $(C_2{H}_3{O}_3{)}_n$

Value of $n=\frac{M}{EM}$$=\frac{150}{24+3+48}=2$

Molecular formula is $C_4{H}_6{O}_6$ .

Correct option is B