Question

Beryllium gives a compound X with the following percentage composition:$\text{ Be }=\text{ }6.1{\scriptstyle{}^{0}/{}_{0}}\text{ }$ , $\text{ N = 37}\text{.8}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ , $\text{ Cl = 48}{\scriptstyle{}^{\text{0}}/{}_{\text{0}}}\text{ }$ ,$\text{ H = 8}\text{.1}{\scriptstyle{}^{0}/{}_{0}}\text{ }$ . The molecular weight of X is $\text{148 g mo}{{\text{l}}^{\text{-1}}}$ and that of $\text{ Be}$ is$\text{ 9 g mo}{{\text{l}}^{\text{-1}}}$. The molecular formula of the compound is:
(A)$\text{ Be}{{\text{N}}_{\text{4}}}\text{C}{{\text{l}}_{\text{2}}}{{\text{H}}_{\text{12}}}\text{ }$
(B) $\text{ Be}{{\text{N}}_{\text{2}}}\text{Cl}{{\text{H}}_{\text{6}}}\text{ }$
(C) $\text{ Be}{{\text{N}}_{\text{4}}}\text{C}{{\text{l}}_{\text{2}}}{{\text{H}}_{\text{6}}}\text{ }$
(D) $\text{ Be}{{\text{N}}_{\text{4}}}\text{Cl}{{\text{H}}_{\text{8}}}\text{ }$

Answer 1

Hint: Equate the percentage given with the given molecular mass to find the total mass of each element present in the compound. Then divide the mass by the atomic weight to get the number of moles of each element present per mole of the compound. This will give you the number of atoms of each element present in the compound

Complete answer:
First, let us find the total mass of each element present in one mole of the compound using simple cross multiplication. The general formula after rearranging it to our needs will be
Mass composition % = $\dfrac{Total\,mass\,of\,element}{Molar\,mass\,of\,compound}$
Therefore, total mass of element = mass % of each element one by one.
- For$\text{Be}$,
$\begin{align}
  & \text{ Total mass of Be}=\dfrac{6.1}{100}\times 148 \\
 & \text{Total mass of Be}=9.028\cong 9 \\
\end{align}$
- For$\text{N}$,
$\begin{align}
  & \text{ Total mass of N}=\dfrac{37.8}{100}\times 148 \\
 & \text{Total mass of N}=55.944\cong 56 \\
\end{align}$
- For$\text{Cl}$,
$\begin{align}
  & \text{ Total mass of Cl}=\dfrac{48}{100}\times 148 \\
 & \text{Total mass of Cl}=71.04\cong 71 \\
\end{align}$
- For $\text{H}$ ,
$\begin{align}
  & \text{ Total mass of H}=\dfrac{8.1}{100}\times 148 \\
 & \text{Total mass of H}=11.988\cong 12 \\
\end{align}$
Now, we have the total masses of the elements present. If we divide them by the atomic masses of each element, we can find the number of moles of each element present per mole of the compound. This value will be equal to the number of atoms of the elements present per molecule. The general formula will be:
$\text{ Moles of element = }\dfrac{\text{total mass of element}}{\text{atomic mass of element}}$
- For$\text{Be}$,
$\begin{align}
  & \text{ Moles of Be}=\dfrac{9}{9} \\
 & \text{Moles of Be}=1 \\
\end{align}$
- For$\text{N}$,
$\begin{align}
  & \text{ Moles of N}=\dfrac{56}{14} \\
 & \text{Moles of N}=4 \\
\end{align}$
- For$\text{Cl}$,
$\begin{align}
  & \text{ Moles of Cl}=\dfrac{71}{35.5} \\
 & \text{Moles of Cl}=2 \\
\end{align}$
- For $\text{H}$
$\begin{align}
  & \text{ Moles of H}=\dfrac{12}{1} \\
 & \text{Moles of H}=12 \\
\end{align}$
The above data can be tabulated as follows,

Atom	Atomic weight	Mass percentage $({\scriptstyle{}^{0}/{}_{0}})$	Mass of element	Number of moles	Simplest ratio$\left( \dfrac{\text{mole}}{1} \right)$
$\text{ Be }$	$9$	$6.1$	$9$	$1$	$1$
$\text{ N  }$	$14$	$37.8$	$56$	$4$	$4$
$\text{ Cl  }$	$35.5$	$48.0$	$71$	$2$	$2$
$\text{ H  }$	$1$	$8.1$	$12$	$12$	$12$

Thus, we can deduce the empirical formula to be$\text{ Be}{{\text{N}}_{\text{4}}}\text{C}{{\text{l}}_{\text{2}}}{{\text{H}}_{\text{12}}}\text{ }$.
Therefore, $\text{Be}{{\text{N}}_{\text{4}}}\text{C}{{\text{l}}_{\text{2}}}{{\text{H}}_{\text{12}}}$ is the molecular formula.

Hence, (A) is the correct option.

Additional information:
Another way to solve this question is by finding the relative masses and the empirical formula of the given compound. We can find out the relative number of atoms of each element by dividing the percentage composition of the element with the atomic mass of the element. Then, we can find the simplest ratio of the whole by dividing the relative number of atoms with a common term, the common term will be the lowest value of all the relative masses found.

Note:
You should not be confused with the empirical formula and molecular formula. The empirical formula gave us an idea about in what proportion does the elements present in the compound. The molecular formula gave us an exact form of the compound. Let us take acetaldehyde as an example; the molecular formula is$\text{ C}{{\text{H}}_{\text{3}}}\text{CHO }$, but the empirical formula will be$\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\text{O }$. Remember to verify if your answer is right by finding the molecular mass of the empirical formula obtained.