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A compound on analysis,was found to have the following composition :(i) Na=14.31% (ii) S=9.97% (iii) O=69.50% (iv) H=6.22%.Calculate the molecular formula of the compound assuming that whole of hydrogen in the compound is present as water of crystallization .Molecular mass of the compound is 322.

$\begin{array}{1 1}(a)\;Na_2SO_4.10H_2 O&(b)\;Na_2SO_4\\(c)\;NaSO_49H_2O&(d)\;Na_2SO_3 10H_2O\end{array}$

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A)
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The empirical formula $=Na_2SH_{20}O_{14}$
Empirical formula mass =$(2\times 23)+32+(20\times 1)$
$\qquad\qquad\qquad\qquad\;=322$
Molecular mass =322
$n=\large\frac{Mol.mass}{Emp.mass}$
$\;\;\;=\large\frac{322}{322}$
$\;\;\;=1$
$\therefore$ Molecular formula =$Na_2SH_{20}O_{14}$
Whole of the hydrogen is present in the form of water.Thus,10 water molecules are present in the molecule.
So molecular formula =$Na_2SO_4.10H_2 O$
Hence (a) is the correct answer.
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