A Compound on analysis gave Na = 14.31% S = 9.97% H= 6.22% and O= 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization. (molecular mass of the compound is 322).
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mole of Na = 14.31/23 = 0.622
mole of S = 9.97/32 = 0.311
mole of H = 6.22/1 = 6.22
mole of O = 69.5/16 = 4.34
relative number of atom of each atom :
for Na = 0.622/0.311 = 2
for S = 0.311/0.311 = 1
for H = 6.22/0.311 = 20
for O = 4.34/0.311 = 14
empirical formula = Na2SH20O14
empirical formula mass = 322
n*empirical formula mass = molecular formula mass
n*322 = 322
n = 1
it is given that all H is present as a H2O
hence H20O10 = 10H2O
molecular formula = Na2SO4 .10H2O
mole of S = 9.97/32 = 0.311
mole of H = 6.22/1 = 6.22
mole of O = 69.5/16 = 4.34
relative number of atom of each atom :
for Na = 0.622/0.311 = 2
for S = 0.311/0.311 = 1
for H = 6.22/0.311 = 20
for O = 4.34/0.311 = 14
empirical formula = Na2SH20O14
empirical formula mass = 322
n*empirical formula mass = molecular formula mass
n*322 = 322
n = 1
it is given that all H is present as a H2O
hence H20O10 = 10H2O
molecular formula = Na2SO4 .10H2O
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