An oxide of sulphur contains 50% of sulphur in it. Its empirical formula is:

Given: The given compound contains 50% of Sulphur (S) and 50 % Oxygen (O) as the compound is a sulphur oxide.

Atomic mass of Sulphur= 32 u

Atomic mass of Oxygen= 16 u

Step-1: Calculate the empirical formula

• Empirical formula: Empirical formula is the simplest formula of the compound which gives the simplest whole number ratio of the atoms of the various elements present in one molecule of the compound. Example: Molecular formula of glucose is C₆H₁₂O₆ but the simplest whole number ratio of atoms in glucose is CH₂O. Therefore, CH₂O is the empirical formula.

Element	% Of Element	Atomic mass (u)	Atomic ratio $\left(\frac{\%ofelement}{Atomicmass}\right)$	Simplest ratio
Sulphur (S)	50	32	$\frac{50}{32}=1.56$	$\frac{1.56}{1.56}=1$
Oxygen (O)	50	16	$\frac{50}{16}=3.125$	$\frac{3.125}{1.56}=2$

As the simplest ratio of Sulphur is 1 and Oxygen is 2, therefore the empirical formula comes out to be SO_2.