Question

Find empirical formula for following percentagecomposition: Na=14.31% S = 9.97%, H = 6.22%, O = 69.5%, calcualte themolecular formula of the compound on the assumption that all thehydrogen in the compound is present in combination with oxygen as waterof crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32,H = 1, 0 = 16].

Answer 1

element- Na

%- 14.31

atomic mass- 23

moles of atom - 14.31/23= 0.62

mole ratio- 0.62/0.31= 2

simplest whole no.= 2

element= S

% =9.97

atomic mass = 32

mole of atom = 9.97/32=0.31

mole ratio = 0.31/0.31=1

simplest whole no.=1

element= H

%=. = 6.22

atomic mass = 1

moles of atom=6.22/1= 6.22

mole ratio= 6.22/0.31= 20.06

simplest whole no.= 20

element= O

%. =69.5

atomic mass= 16

moles of atom = 69.5/16= 4.34

mole ratio= 4.34/0.31= 14

simplest whole no.= 14

Empirical formula= Na2SH20O14

E.F mass Na2SH20O14= 23×2+32+20+16×14= 322

n= molecular mass/E.F mass

n= 322/322= 1

n= 1

molecular formula= Na2 SH20 O14