Question

An oxide of iodine (I = 127) contains 25.4 g for 8 g of oxygen. Its formula will be:
(a)- ${{I}_{2}}{{O}_{3}}$
(b)- ${{I}_{2}}O$
(c)- ${{I}_{2}}{{O}_{5}}$
(d)- ${{I}_{2}}{{O}_{7}}$

Answer 1

Hint: There are many steps that must be followed for finding the formula, i.e., find the percentage of the element, moles of the element, simplest molar ratio, and the simplest whole-number ratio.

Complete step-by-step answer:
To find the formula of the compound, there are some steps that must follow one by one. These are:
First, find the percentage of the element present, or if the gram of the element is present then consider the grams of the element in 100 grams of compound.
Then calculate the moles of the element by dividing the given mass or percentage by the atomic mass of the element.
Now, from all the elements take the smallest number of moles and divide that number with every number of moles to get the simplest molar ratio.
 If all the simplest molar ratios are in integer then the formula can be written accordingly. And if it is not, then multiply it with the smallest integer to get the simplest whole-number ratio.
Now, according to the data given in the question, we can tabulate the formula:

Element	Symbol	Percentage/Mass	Atomic mass	Moles of the element	Simplest molar ratio	The simplest whole-number ratio
Iodine	I	25.4	127	$\dfrac{25.4}{127}=0.2$	$\dfrac{0.2}{0.2}=1$	2
Oxygen	O	8	16	$\dfrac{8}{16}=0.5$	$\dfrac{0.5}{0.2}=2.5$	5

So, according to the data plotted above, in the compound, there are two atoms of iodine and five atoms of oxygen. Therefore, the formula will be ${{I}_{2}}{{O}_{5}}$
Hence, the correct answer is an option (c).

Note: There are two types of formula: Empirical formula and Molecular formula. The Empirical formula is the simplest whereas the Molecular formula is the true formula, and these can be related as:
Molecular Formula = n x Empirical formula.
Where n is an integer.