Question

A chemical compound is formed to have the following composition C=19.57%,FE= 15.2% ,N=22.83% ,K=42.39% . Calculate the empirical formula of the compound.what will be the molecular formula if the molecular mass of the compound is 368?

Answer 1

GIVEN :-

• Carbon, C = 19.57 %

• Iron, Fe = 15.2 %

• Nitrogen, N = 22.83 %

• Potassium, K = 42.39 %

• Molecular mass of the compound = 368  

TO FIND :-

• Empirical formula of the compound.

• Molecular formula of the compound.

SOLUTION :-

 We know,

•  Atomic mass of carbon = 12g

•  Atomic mass of iron = 56g

•  Atomic mass of nitrogen = 14g

•  Atomic mass of potassium = 39g

 $\longrightarrow \sf No \ of \ mole \ of \ carbon = \dfrac{19.57}{12}$

                                         $\sf = 1.63$

 $\longrightarrow\sf No \ of \ mole \ of \ iron = \dfrac{15.2}{56}$

                                     $\sf = 0.27$

 $\longrightarrow \sf No \ of \ mole \ of \ nitrogen = \dfrac{22.83}{14}$

                                            $\sf = 1.63$

$\longrightarrow \sf No \ of \ mole \ of \ potassium = \dfrac{42.39}{39}$

                                             $\sf = 1.087$

Simplest ratio = 1.63 : 0.27 : 1.63 : 1.087

                        = $\sf\dfrac{1.63}{0.27}:\dfrac{0.27}{0.27}:\dfrac{1.63}{0.27}:\dfrac{1.087}{0.27}$

                        = 6 : 1 : 6 : 4

 Empirical formula of the compound = $\bf C_6FeN_6K_4$

 Empirical formula mass = ( 12 × 6 ) + ( 56 × 1 ) + ( 14 × 6 ) + ( 39 × 4 )

                                         = 72 + 56 + 84 + 156

                                         = 368

   $\longrightarrow\sf \dfrac{Molecular \ mass }{Empirical \ formula \ mass }=\dfrac{368}{368} = 1$

  Molecular formula of the compound  = $\bf C_6FeN_6K_4$