Question

How would you determine the molecular formula from the following empirical formula and molar mass: ${\text{NPC}}{{\text{l}}_{\text{2}}}$,${\text{348}}\,{\text{g/mol}}$?

Answer 1

Hint:To determine the molecular formula of the compound from the empirical formula and molar mass of the compound we have to determine first empirical formula mass and then determine the value of n.Here, n represents the whole number value which is obtained by taking the ratio of molar mass to empirical formula mass.Then by multiplying all subscripts in the empirical formula by n molecular formula of the compound is obtained.

Complete solution:
Here, the empirical formula of the compound given is ${\text{NPC}}{{\text{l}}_{\text{2}}}$.
The molar mass of nitrogen is $14.0067\,{\text{g/mol}}$, the molar mass of phosphorus is $30.9737\,{\text{g/mol}}$, and the molar mass of chlorine is $35.453\,{\text{g/mol}}$.
Now, to determine empirical formula mass we have to add masses of all atoms involved in the empirical formula.
Here, the empirical formula of the compound given is ${\text{NPC}}{{\text{l}}_{\text{2}}}$.
${{\text{M}}_{{\text{empirical}}\,{\text{formula}}}}{\text{ = }}{{\text{M}}_{\text{N}}}\,{\text{ + }}\,{{\text{M}}_{\text{P}}}{\text{ + }}\left( {{{2 \times }}{{\text{M}}_{{\text{Cl}}}}} \right)$
\[{{\text{M}}_{{\text{empirical}}\,{\text{formula}}}}{\text{ = }}\left( {14.0067\,{\text{g/mol}}} \right)\,{\text{ + }}\,\left( {30.9737\,{\text{g/mol}}} \right){\text{ + }}\left( {{{2 \times }}35.453\,{\text{g/mol}}} \right)\]
\[{{\text{M}}_{{\text{empirical}}\,{\text{formula}}}}{\text{ = }}115.8864\,{\text{g/mol}}\]
Thus, the empirical formula mass of the compound is \[115.8864\,{\text{g/mol}}\].
Now, we have to determine the value of n as follows:
${\text{n}}\,{\text{ = }}\dfrac{{{\text{molar}}\,{\text{mass}}}}{{{\text{empirical}}\,{\text{formula}}\,{\text{mass}}}}$
Here, substitute molar mass as ${\text{348}}\,{\text{g/mol}}$ and empirical formula mass as \[115.8864\,{\text{g/mol}}\].
${\text{n}}\,{\text{ = }}\dfrac{{{\text{348}}\,{\text{g/mol}}}}{{115.8864\,{\text{g/mol}}}}$
${\text{n}}\,{\text{ = }}3.0029$
${\text{n}}\, \cong 3$
Thus, the value of the n is 3.
Now, determine the molecular formula of the compound as follows:
${\text{molecular}}\,{\text{formula = }}{\left( {{\text{empirical}}\,{\text{formula}}} \right)_{\text{n}}}$
The empirical formula given is ${\text{NPC}}{{\text{l}}_{\text{2}}}$ and the value of n is 3.
${\text{molecular}}\,{\text{formula = }}{\left( {{\text{NPC}}{{\text{l}}_{\text{2}}}} \right)_3}$
${\text{molecular}}\,{\text{formula = }}{{\text{N}}_3}{{\text{P}}_3}{\text{C}}{{\text{l}}_6}$
Thus, the molecular formula of the compound is ${{\text{N}}_3}{{\text{P}}_3}{\text{C}}{{\text{l}}_6}$.

Note:The empirical formula of the compound is the simplest formula which represents the whole number ratio between the atoms present in the compounds.
The empirical formula mass is the sum of masses of all atoms present in the empirical formula.
The molar mass of the compound represents the mass of one mole of that compound. The unit of the molar mass is ${\text{g/mol}}$.