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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Na = 14.31% , moles 14.31/23 = 0.622
S = 9.97% , moles = 9.97/32 = 0.311
H = 6.22 % , moles = 6.22/1 = 6.22
O = 69.5 %, moles = 69.5/16 = 4.34
Divide all moles by 0.311 to get mole ratio
Na:S:H:O = 2:1:20:14
empirical formula is Na2SH20O14
empirical formula wt = 46+32+20+224 = 322.
Moleculat weight = 322 (given)
since molecular wt and empirical formula weight are same,
empirical formula = molecular formula.
Given that all Hydrogen is present as water of crystallisation.
Therefore the formula can be expressed as Na2SO4.10H2O
S = 9.97% , moles = 9.97/32 = 0.311
H = 6.22 % , moles = 6.22/1 = 6.22
O = 69.5 %, moles = 69.5/16 = 4.34
Divide all moles by 0.311 to get mole ratio
Na:S:H:O = 2:1:20:14
empirical formula is Na2SH20O14
empirical formula wt = 46+32+20+224 = 322.
Moleculat weight = 322 (given)
since molecular wt and empirical formula weight are same,
empirical formula = molecular formula.
Given that all Hydrogen is present as water of crystallisation.
Therefore the formula can be expressed as Na2SO4.10H2O
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