Question

The percentage of $C,{\text{ }}H{\text{ }}and{\text{ }}N$ in an organic compound is 40%, 13.3% and 46.7% respectively then its empirical formula is:

$${A.{\text{ }}{C_3}{H_{13}}{N_3}} \\\ {B.{\text{ }}C{H_2}N} \\\ {C.{\text{ }}C{H_4}N} \\\ {D.{\text{ }}C{H_5}N}$$

Answer 1

We must know that the empirical formula is the simplest positive integer ration possible for atoms present in any compound.

Complete answer:
Let’s start by discussing a little about the empirical formula; empirical formula is the simplest positive integer ration possible for atoms present in any compound. For calculating the empirical formula follows the following:
Convert the percentages into masses.
Then calculate the moles of the given atom present in the molecule for that particular mass
Divide each mole value by the smallest number of moles calculated and rounded off to an integer. Here we get the empirical ratio.
Let’s formula be ${C_n}{H_m}{N_p}$ and the molecular mass be $100g$. Now we are given that Carbon is 40% so it $40g$, hydrogen is 13.3% so it is $13.3g$ and nitrogen is 46.7% so it is $46.7g$.
Now, the atomic weight of C, H, and N are 12, 1, and 14 respectively.
So, n will be 40 divided by 12 which will be 3.33, m will be 13.3 divided by 1 which will be 13.3 and p will be 46.7 divided by 14 which will be 3.33.
So the ratio of $C:H:N$ will be like $3.33:13.3:3.33$, but we require all this in positive integers, and decimals are not allowed so we will try to divide the whole into 3.33. By dividing we get the ratio of $C:H:N$ as $1:4:1$.
So, the answer to this question is $C{H_4}N$.

Note:
We can use empirical formulas to study about the ration in which the atoms are present in the molecule and not the number of elements present in the molecule. Also, the empirical formula does not tell the true identity of the molecule. Well, even after such drawbacks it is very useful for doing compound analysis of some unknown compound.