Question

Beryllium gives a compound $\text{X}$ with the following percentage composition: Be Molecular weight of $-61$ $\text{X}$ is $\text{148 g mo}{{\text{l}}^{-}}^{\text{1}}$ and that of Be is $\text{9 g mo}{\text{l}}^{\text{-1}}.$ The molecular formula of the compound is

• A. $\text{Be}{\text{N}}_{\text{4}}\text{C}{\text{l}}_{\text{2}}{\text{H}}_{\text{12}}$
• B. $\text{Be}{\text{N}}_{\text{2}}\text{Cl}{\text{H}}_6$
• C. $\text{Be}{\text{N}}_{\text{4}}\text{C}{\text{l}}_{\text{2}}{\text{H}}_6$
• D. $\text{Be}{\text{N}}_{\text{4}}\text{Cl}{\text{H}}_{\text{8}}$

Answer 1

Answer: (A)

Atom
Atomic weight (A)
%
%/A
Simplest ratio
Be
9
6.1
0.68
1
N
14
37.8
2.70
4
$\text{Cl}$
35.5
48.0
1.35
2
H
1
8.1
8.10
12
$\therefore$ The molecular formula is $\text{Be}{\text{N}}_{\text{4}}\text{C}{\text{l}}_{\text{2}}{\text{H}}_{\text{12}}$