Question

What is the empirical formula of an oxide of element (E) containing 40% element by mass? (Atomic mass of E=32)
$(1)E{O_2}$
$(2)EO$
$(3)E{O_3}$
$(4){E_2}{O_3}$

Answer 1

In order to solve the question we need to know what empirical formula means. It is defined as the simplest whole number ratio of constituent atoms of any compound. Example-Ethane has molecular formula of ${C_2}{H_6}$ hence the empirical formula will be $C{H_3}$ ,Butane has molecular formula of ${C_4}{H_{10}}$ so the empirical formula will be ${C_2}{H_5}$ etc.

Complete answer:
There is a relationship between empirical and molecular formulas. There exists a number n which when multiplied to empirical formula gives molecular formula.
Molecular formula=nEmpirical formula
When you calculate the weight of the molecular formula it is called molecular weight and the weight of the empirical formula is called empirical weight.
Molecular weight=nEmpirical weight
Given in the question we have $40\%$ element by mass so the rest of the $60\%$ will be of oxygen.
We are given an atomic mass of E=32 and we know that the atomic mass of oxygen is 16.Using this information we prepare a table. We already know the percentage and atomic weight so we calculate the atomic ratio and then using atomic ratios we calculate the simplest ratios. The ratios should not be in decimal and must be in whole numbers.

Element	% by mass	Atomic weight	Atomic ratio=$\%$ by mass/Atomic weight	Simplest ratio
$E$	$40$	$32$	$40 \div 32 = 1.25$	$1.25 \div 1.25 = 1$
$O$	$60$	$16$	$60 \div 16 = 3.75$	$3.75 \div 1.25 = 3$

Hence the Empirical formula of the oxide will be $E{O_3}$ .Hence option (3) is correct.

Note:
An Empirical formula has no significance in showing the arrangement of atoms or showing the number of atoms since these are the simplest forms of notations. However on the other hand Molecular formula provides complete information about the arrangement as well as the types of atoms present.