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. $N{a}_{2}S{O}_4$
• B. $N{a}_{2}S{O}_4.10H_{2}O$
• C. $N{a}_{2}S{H}_{10}{O}_{12}$
• D. $N{a}_{2}S{O}_4.7H_{2}O$

Answer 1

Answer: (B)

$N{a}_{2}S{O}_4.10H_{2}O$

Element	mass %	Atomic mass(g)	m%/At. mass	Integral ratio
Na	14.31	23	0.622	2
S	9.97	32	0.311	1
H	6.22	1	6.22	20
O	69.5	16	4.34	14

Empirical formula is therefore $N{a}_{2}S{H}_{20}{O}_{14}$

empirical mass=$2\times 23+32+20\times 1+14\times 16=322$g

Given:

Molecular mass = 322 g

$\therefore$ molecular formula =empirical formula=$N{a}_{2}S{H}_{20}{O}_{14}$

Given that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation

If water of crystallization are $n{H}_{2}O$ then, $2n=20$

or $n=10$

Crystallized water is $10H_{2}O$ remaining atoms are $N{a}_{2}S{O}_4$

therefore, the molecule is $N{a}_{2}S{O}_4.10H_{2}O$