A compound, on analysis, gave the following percentage composition: Na = 14.31 %, S = 9.97 %, H = 6.22 %, O = 69.5 % What would be the molecular formula of the compound assuming that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation. Molecular weight of the compound is 322

Question

A compound, on analysis, gave the following percentage composition: Na = 14.31 %, S = 9.97 %, H = 6.22 %, O = 69.5 % What would be the molecular formula of the compound assuming that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation. Molecular weight of the compound is 322 (A) Na2SO4 (B) Na2SO4⋅ 10H2O (C) Na2SH10O12 (D) Na2SO4 ⋅ 7H2O

Tardigrade Admin · Accepted Answer

Element	%	Atomic ratio	Least ratio	Empirical formula
Na	14.31	$\frac{14.31}{23} = 0.622$	$\frac{0.622}{0.311} = 2$
S	9.97	$\frac{9.97}{32} = 0.311$	$\frac{0.311}{0.311} = 1$	$N a_{2} S H_{20} O_{14}$
H	6.22	$\frac{6.22}{1} = 6.22$	$\frac{6.22}{0.311} = 20$
O	69.5	$\frac{69.5}{16} = 4.34$	$\frac{4.34}{0.311} = 14$

Empirical Formula mass

= 46 + 32 + 20 + 224 = 322

n = \frac{Molecular mass}{Empirical formula mass} = \frac{322}{322} = 1

Hence, molecular formula

= N a_{2} S H_{20} O_{14}

Since, all the hydrogen is in the form of water, thus there are

10 H_{2} O

molecules.
Hence, the molecular formula is

N a_{2} S O_{4} \cdot 10 H_{2} O

$N a_{2} S O_{4}$

$N a_{2} S O_{4} \cdot 10 H_{2} O$

$N a_{2} S H_{10} O_{12}$

$N a_{2} S O_{4} \cdot 7 H_{2} O$

Na2​SO4​

Na2​SO4​⋅10H2​O

Na2​SH10​O12​

Na2​SO4​⋅7H2​O

$N a_{2} S O_{4}$

$N a_{2} S O_{4} \cdot 10 H_{2} O$

$N a_{2} S H_{10} O_{12}$

$N a_{2} S O_{4} \cdot 7 H_{2} O$