Question

A compound on analysis gave the following percentage composition:
Na = 14.31%, S = 9.97%, H = 6.22% O = 69.50%
Calculate the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallization. Molecular mass of the compound is 322.

Answer 1

Element: Na

Percentage composition: 14.31

Atomic mass: 23

Relative number of atoms: $\frac { 14.31 }{ 23 } \quad =\quad 0.622$

Simplest ratio: $\frac { 0.622 }{ 0.311 } \quad =\quad 2$

Element: S

Percentage composition: 9.97

Atomic mass: 32

Relative number of atoms:$\frac { 9.97 }{ 32 } \quad =\quad 0.311$

Simplest ratio: $\frac { 0.311 }{ 0.311 } \quad =\quad 1$

Element: H

Percentage composition: 6.22

Atomic mass: 1

Relative number of atoms: $\frac { 6.22 }{ 1 } \quad =\quad 6.22$

Simplest ratio: $\frac { 6.22 }{ 0.311 } \quad =\quad 20$

Element: O

Percentage composition: 69.5

Atomic mass: 16

Relative number of atoms: $\frac { 69.5 }{ 16 } \quad =\quad 4.34$

Simplest ratio: $\frac { 4.34 }{ 0.311 } \quad =\quad 14$

The empirical formula = ${ Na }_{ 2 }S{ H }_{ 20 }{ O }_{ 14 }$

Empirical formula mass = $(2\quad \times \quad 23)\quad +\quad 32\quad +\quad (20\quad \times \quad 1)\quad =\quad 322$

Molecular mass = 322

$n\quad =\quad \frac { Molecular\quad mass }{ Empirical\quad mass } \quad =\quad \frac { 322 }{ 322 } \quad =1$

Whole of the hydrogen is present in the form of water. Therefore, 10 water molecules are present in the molecule.

 Therefore, molecular formula = ${ Na }_{ 2 }S{ O }_{ 4 }.10{ H }_{ 2 }O$