Question

How can I determine the empirical formula from percent composition?

Answer 1

Hint: The empirical formula of any given molecule is a type of formula which provides us with an idea about the minimum ratio regarding the constituent elements in terms of whole numbers.
The empirical formula does not provide any idea regarding the spatial arrangement corresponding to the constituent atoms of the considered molecule.

Complete step-by-step answer:The procedure which is used for the determination of empirical formula with the use of its percentage composition could be divided into six steps. It is easier to tabulate the values and do the calculation, rather than writing each individual value separately. Now, we will consider the steps, which are as follows.
The first step consists of, assignment of the symbols to each of the atoms, which would help us represent specific atoms without writing the complete names. For an instance $N$ represents the atom nitrogen, or $O$ represents the atom of oxygen.
In the next step, we are supposed to use the percentage compositions of every element which are present in the considered compound which would be given to us in the provided question. For instance, $C-48%$, $O-15%$.
Next , we would write the respective atomic masses of each of the elements, for instance we know that the carbon would have the value of atomic mass as $12$.
Now, in the next step we are supposed to divide all these determined values of percentage compositions with their respective values of atomic masses of these elements, so as to get the relative value of the moles, respectively.
In the next step, we are supposed to divide these obtained ratios, of all these elements with the smallest relative number ratio of number of moles.
Lastly, we will get a whole number ratio after the previous step.
Now, we will be taking an example in order to understand it more clearly.
Consider the percentage composition of a compound after the chemical analysis gave the following results \[Na=14.31%{ }\] \[S=9.97%{ }\] \[H=6.22%{ }\] \[O=69.5%\]. Now we would calculate the empirical formula of this compound using the following data. Molecular mass of the compound is \[322\]. \[Na=23{ }\] \[S=32{ }\] \[O=16{ }\] \[H=1\].

Element	Percentage	Atomic mass	Moles of atom	Atomic ratio	Simplest ratio
$Na$	$14.31$	$23$	$\dfrac{14.31}{23}=0.622$	$\dfrac{0.622}{0.311}=2$	$2$
$S$	$9.97$	$32$	$\dfrac{9.97}{32}=0.311$	\[\dfrac{0.311}{0.311}=1\]	$1$
$H$	$6.22$	$1$	$\dfrac{6.22}{1}=6.22$	$\dfrac{6.22}{0.311}=20$	$20$
$O$	$69.5$	$16$	$\dfrac{69.5}{16}=4.34$	$\dfrac{4.34}{0.311}=13.95$	$14$

So, the empirical formula came out to be $N{{a}_{2}}S{{O}_{14}}{{H}_{20}}$.

Note: Empirical Formula of any given compound is the simplest formula which provides an idea of the original molecular formula of that compound, just by taking the simplest subscripts of the elementary atoms of that molecule.
It could be derived simply with the use of the mass percentage of those constituent atoms which constitutes the molecule.