Question

An organic compound on analysis gave $C = 54.2\%$, $H = 9.2\%$ by mass. Its empirical formula is:
A.$CH{O_2}$
B.$C{H_2}O$
C.${C_2}{H_8}O$
D.${C_2}{H_4}O$

Answer 1

We can define the empirical formula of a compound is the modest whole number ratio of each type of atom in a compound. It could be the same as the compound’s molecular formula but not always. We could calculate an empirical formula from data about the mass of each element present in a compound or from the composition (in terms of percentage).

Complete step by step answer:
Given data contains,
The percentage composition of hydrogen is 9.2%.
The percentage composition of carbon is 54.2%.
We can calculate the percentage of oxygen is $100 - \left( {54.2 + 9.2} \right)$
The percentage composition of oxygen is 36.6%.
We could give the steps for obtain the empirical formula of a compound as follows:
Calculate the mass of each element present in grams.
Obtain the number of moles of each atom present.
Split the number of moles of each element by the smallest number of moles.
Change the numbers to whole numbers. The set of whole numbers are given as subscripts while writing the empirical formula.
Now, let’s calculate the moles of elements in compounds with their molar mass.
The molar mass of hydrogen is $1g/mol$.
The molar mass of carbon is $12g/mol$.
The molar mass of oxygen is $16g/mol$.
The moles of the element are calculated with the help of the mass divided by their molar mass. We could write the formula to calculate of moles of an element as,
Moles of element=$\dfrac{{Mass}}{\text{Molecular Mass}}$
We could calculate the moles of each element now,
The moles of hydrogen atom=$\dfrac{{9.2g}}{{1g/mol}} = 9.2moles$
The moles of carbon atom=$\dfrac{{54.2g}}{{12g/mol}} = 4.51moles$
The moles of oxygen atom=$\dfrac{{36.6g}}{{16g/mol}} = 2.287moles$
The moles of hydrogen, carbon and oxygen are $9.2moles$, $4.51moles$ and $2.3moles$ respectively.
We have to divide all values with least value we got
$H = \dfrac{{9.2}}{{2.3}} = 4$
$C = \dfrac{{4.52}}{{2.3}} = 2$
$O = \dfrac{{2.3}}{{2.3}} = 1$
We got the empirical formula of the compound as ${C_2}{H_4}O$. Option (D) is correct.

Note:
We must remember that an empirical formula does not represent the organization or number of atoms. It is typical for many ionic compounds like calcium chloride $\left( {CaC{l_2}} \right)$ , and for macromolecules such as silicon dioxide $\left( {Si{O_2}} \right)$. The organization of the molecule is a structural formula. The number of each kind of atom in a molecule is called a molecular formula. We can define the percent of a particular element in the sample with the help of elemental analysis. For compounds such as glucose, acetic acid, formaldehyde contains the same empirical formula $C{H_2}O$ but different molecular formulas.