Question

determine the empirical formula of a compound containing K=24.75% ,Mn= 34.77% and rest is oxygen​

Answer 1

★Given :-

• K = 24.75 %
• Mn = 34 .77 %

Rest is oxygen

= 100 - ( K + Mn )

= 100 - ( 24.75+ 34.77)

= 100 - 59.52

= 40.48 %

★Relative no of atoms

Divide the % composition of the elements with it's atomic mass

• Atomic mass of K = 39
• Atomic mass of Mn = 55
• Atomic mass of O = 16

$\implies\mathtt{ K = \dfrac{24.75}{39} = 0.63 }$

$\implies\mathtt{ Mn = \dfrac{34.77}{55} = 0.63 }$

$\implies\mathtt{ Mn = \dfrac{40.48}{16} = 2.53}$

★Simple ratio :-

Divide the least value i.e 0.63 in this case with the other values

$\implies\mathtt{ K = \dfrac{0.63}{0.63} = 1}$

$\implies\mathtt{ Mn = \dfrac{0.63}{0.63} = 1}$

$\implies\mathtt{ O = \dfrac{2.53}{0.63} =4 }$

The empirical formula of this compound is KMnO4