Question

How many hydrogen atoms are represented in the formula $ {\left( {{{C}}{{{H}}_{{3}}}} \right)_{{2}}}{{C}}{{{H}}_{{2}}} $ ?

Answer 1

Hint: In the above question, we are asked to find out number of hydrogen atoms represented in the formula $ {\left( {{{C}}{{{H}}_{{3}}}} \right)_{{2}}}{{C}}{{{H}}_{{2}}} $ . We have to first expand the molecule and then we can count the number of hydrogen present in it.

Complete step by step solution
The given formula can be represented as $ {{C}}{{{H}}_{{3}}}{{ - C}}{{{H}}_{{2}}}{{ - C}}{{{H}}_{{3}}} $
Here, 3 hydrogen atoms are present in first carbon, 2 hydrogen atoms in second carbon and 3 hydrogen atoms in third carbon. So, the total number of hydrogen present is equal to $ {{3 + 2 + 3 = 8}} $ .
Hence, 8 hydrogen atoms are present in $ {\left( {{{C}}{{{H}}_{{3}}}} \right)_{{2}}}{{C}}{{{H}}_{{2}}} $ .

Additional Information
Alkanes are the major constituents of crude oil and the most important constituents of most petroleum products. Alkanes are also known as saturated or saturated hydrocarbons. The carbon atoms which form the carbon backbone are linked together to form a chain (linear or branched alkanes), a circle (cyclic alkenes), or a mixture of both. The alkanes are distinguished from the other hydrocarbons because they are completely saturated with hydrogen. This means that no additional hydrogen atoms can be added to these hydrocarbons without destroying the carbon backbone. A linear or branched alkane that has four carbon atoms will also have 10 and only 10 hydrogen atoms. Linear alkanes, i.e., alkanes whose carbon atoms are arranged in a straight chain or row, are called normal paraffins, normal alkanes, or n-alkanes. The branched alkanes are often referred to as isoparaffins or isoalkanes.

Note
In the above question, we can see that the compound given is an alkane. So, we are given 3 carbon atoms and hence, number of hydrogen by general formula is $ {{2n + 2 = 2 \times 3 + 2 = 8}} $
The general of –
Alkane is $ {{{C}}_{{n}}}{{{H}}_{{{2n + 2}}}} $
Alkene is $ {{{C}}_{{n}}}{{{H}}_{{{2n}}}} $
Alkyne is $ {{{C}}_{{n}}}{{{H}}_{{{2n - 2}}}} $