Question

Amino-acid organic base gave the following results on analysis. $C=78.6$ $0.4gm$ of the platinic chloride left on ignition $0.125gm$ of $pt$ . The molecular formula of the base is:
(A) ${{C}_{7}}{{H}_{9}}N$
(B) ${{C}_{6}}{{H}_{8}}{{N}_{2}}$
(C) ${{C}_{7}}{{H}_{9}}{{N}_{2}}$
(D) ${{C}_{6}}{{H}_{8}}N$

Answer 1

We know that a stronger base readily donates its pairs of electrons. Chemical formula of benzylamine is $\left( {{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}} \right)or\left( {{C}_{7}}{{H}_{9}}N \right)$ and respectively. In both lone pairs are present on nitrogen atoms whereas in benzylamine, nitrogen is attached to the saturated carbon.

Complete step by step solution:
Let us denote monoacidic base by base by B which is $B=(xHyNz)$ and its molecular weight be W
Thus we can say that ${{B}_{2}}[Pt(C{{l}_{6}})]\to Pt$ [Half reaction to show $1:1$ mole ratio]
Now we have to calculate number of moles of platinic chloride and platinum which are $\dfrac{0.4}{408+W}$ and $\dfrac{0.125}{195}$ respectively,
Now we know that one mole of platinic chloride gives 1 mole of platinum;
$\dfrac{0.4}{408+W\times 2}=\dfrac{0.125}{195}$
$W=\dfrac{78+51}{0.25}$
$\Rightarrow W=108$
Now that we got the value of W we have substitute value in one mole of each compound to get actual number of moles present in molecular formula
$C=78.6$
Here the number of moles of carbons in one mole of compound is $\dfrac{108\times 0.786}{12}=7$
Similarly the number of moles nitrogen in 1 mole of compound is $\dfrac{108\times 0.13}{14}=1$
And the number of moles hydrogen in 1 mole of compound is $\dfrac{108\times 8.4}{100\times 1}=9$
Thus, this corresponds to ${{C}_{7}}{{H}_{9}}N$
Therefore, correct answer is option A, i.e. the molecular formula of the base is ${{C}_{7}}{{H}_{9}}N$

Note:
Not that the benzylamine as well as aniline are bases but relatively differs in their basicity, you must know that the lower is the $p{{K}_{b}}$ value (i.e., numeric measurement of the basicity) , stronger is the base. The $p{{K}_{b}}$ value of benzylamine is $4.70$ .