Question

A compound on analysis gave the following percentage composition.

Na—14.31 %	S—9.97 %
H—6.22 %	O—69.5 %.

Calculate the molecular formula of the compound, if all the hydrogen present in the compound is combined with oxygen to form water of crystallization. Molecular mass of compound is 322.
[Na = 23, S = 32, H = 1, O = 16]

Answer 1

Elements	Percentage	Atomic mass	Relative number of atoms	Simplest whole number ratio
Na	14.31	23	$\left(\frac{14.31}{23}\right)=0.62$	$\left(\frac{0.62}{0.31}\right)=2$
S	9.97	32	$\left(\frac{9.97}{32}\right)=0.31$	$\left(\frac{0.31}{0.31}\right)=1$
O	69.5	16	$\left(\frac{69.5}{16}\right)=4.34$	$\left(\frac{4.34}{0.31}\right)=14$
H	6.22	1	$\left(\frac{6.22}{1}\right)=6.22$	$\left(\frac{6.22}{0.31}\right)=20$

The empirical formula of the compound is Na₂SH₂₀ O₁₄.
Mass of the empirical formula = (23$\times$2) +32 +(20$\times$1)+ (16$\times$14) = 322 u
$$\begin{gathered} n=\frac{\mathrm{Molecular}\mathrm{Mass}}{\mathrm{Mass}\mathrm{of}\mathrm{Empirical}\mathrm{formula}} \\ n=\frac{322}{322}=1 \end{gathered}$$

Therefore, molecular formula of the compound is Na₂SO₄.10H₂O.