Question

Beryllium gives a compound $ \text{X} $ with the following percentage composition: Be Molecular weight of $ -61%,\,N-37.8%,\,Cl-48%,\,H-8.1% $ $ \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}}} $