Question

An inorganic salt given the following percentage composition: $Na = 29.11,{\text{ }}S = 40.51$ and $O = 30.38$. Determine the empirical formula of the salt.

Answer 1

We need to remember the atomic mass of different elements, molar ratio formula and its conversion into simple ratio in order to solve this question. In order to find the empirical formula of salt percentage composition determined, is divided by its atomic number and the fraction value is converted into the whole number and the ratio is written with the symbols representing the elements.

Complete step by step answer:
Given data:
$Na = 29.11,{\text{ }}S = 40.51$ and $O = 30.38$
Steps to determine the empirical formula:
The Percentage composition is determined by quantitative analysis.
The percentage of each element is then divided by its atomic mass to obtain atomic ratio or molar ratio.
Then the atomic ratio of each element is divided by the least atomic ratio to get the simplest ratio of atom present in the compound.
If the ratio in the above step comes in fraction, then the value of simplest ratio is multiplied by the smallest integer to obtain the whole number from each element.
At last to get the empirical formula, symbols of each element are written side by side with their respective ratio as a subscript to the lower right hand corner of the symbol.
The molecular formula of a substance can be determined by the empirical formula if its molecular mass is known.
Now, according to the given question let us tabulate all the values in a table and find the empirical formula

Element	Composition (%)	Atomic Mass (g)	Atomic Ratio	Simple Ratio
$Na$	29.11	23	(29.11/23)	(1.266/1.266)=1×2=2
$S$	40.51	32	(40.51/32)	(1.266/1.266)=1×2=2
$O$	30.38	16	(30.38/16)	(1.898/1.266)=1.5×2=3

Therefore, empirical formula is $N{a_2}{S_2}{O_3}$

Note: The molecular mass of the elements should be taken correctly in order to get the correct empirical formula and if the atomic ratio is coming in fraction then step 4 needs to be followed to get the whole number ratio.