Question

13g of a hydrocarbon contains 1g of hydrogen. Its formula is
1)\[{C_2}{H_2}\]
2) \[{C_2}{H_3}\]
3) \[{C_3}{H_4}\]
4) \[{C_4}{H_7}\]

Answer 1

Hint: Hydrocarbons are nothing but compounds containing hydrogen and carbon as their chief components. When there is a question of deriving the formula of hydrocarbons, think about the mole concept, get to know the atomic mass of the elements and the concept of empirical formula and molecular formula.

Complete answer: Empirical formula is a simplest whole number ratio of constituent atoms of any compound.
We know that a molecular formula is the total number of atoms present in any compound.
To understand better, let us consider an example.
n-Butane - molecular formula is \[{C_4}{H_{10}}\]
For empirical formula we have to take the constituent elements in ratio,
The elements are C and H so,
Empirical formula – C : H = 4 : 10
On simplification we have 2 : 5 i.e., \[{C_2}{H_5}\]
So, the empirical formula of \[{C_4}{H_{10}}\]is \[{C_2}{H_5}\].
If the molecular formula is given then it is easy to calculate the percentage composition or the amount of constituent elements present in a compound.
For example: water - \[{H_2}O\].
The molecular mass of water is 18(\[\left( {2 \times 1} \right) + \left( {1 \times 16} \right) = 18\])
Percentage composition = \[\dfrac{{no{\text{ }}of{\text{ }}atoms \times atomic{\text{ }}mass}}{{molecular{\text{ }}mass}} \times 100\]
\[
  \% H = \dfrac{{2 \times 1}}{{18}} \times 100 = 11.79\% \\
  \% O = \dfrac{{1 \times 16}}{{18}} \times 100 = 88.21\% \\
 \]
Making it to the whole number we can consider it like 12g of H and 88g of O.
But we have the molecular mass of the hydrocarbon and we have to find the formula of it.
Given: molecular mass of hydrocarbon is 13g.
Mass of hydrogen in this hydrocarbon is 1g, which suggests that there is only one H atom.
So, we can consider that there is one C atom. Hence mass of C atom in this hydrocarbon is 12g.
Mole of H : Mole of C
 H : C = \[\dfrac{1}{1}:\dfrac{{12}}{{12}}\]
H : C = 1 : 1.
Therefore, the empirical formula of the compound is \[CH\].
So the compound is \[{C_2}{H_2}\].
Hence the correct option is 1) \[{C_2}{H_2}\].

Note:
The empirical formula shows the simplest whole number ratio of atoms in a compound. Molecular formulas show the number of each type of atom in a molecule. There is also a structural formula which shows how the atoms in a molecule are bonded to each other.