Question

Calculate the change in percentage of C (by weight) when all the H-atoms are replaced by ‘T’ in $C{{H}_{4}}$ molecule. (Given atomic weight- C: 12, H: 1, T : 3 )
(a)- 75%
(b)- 50%
(c)- 25%
(d)- 40%

Answer 1

The percentage of the component in the molecule can be calculated by the formula $\dfrac{x}{y}\text{ x 100}$, where x is the weight of the component in the compound and y is the total weight of the component. First, calculate the composition of carbon in $C{{H}_{4}}$ and then calculate the composition of carbon in $C{{T}_{4}}$, and then subtract them.

Complete step by step answer:
In the $C{{H}_{4}}$molecule there are four hydrogen atoms and one carbon atom, and we know that the molecular mass of carbon is 12 and the molecular mass of hydrogen is 1.
So its molecular mass is: $12+1+1+1+1=16$.
So the molecular mass is 16 g / mol.
Now for the percentage of composition of carbon in $C{{H}_{4}}$= $\dfrac{12}{16}\text{ x 100 = 0}\text{.75 x 100}$
$=\text{ 75 }\\!\\!$
Now the question says that all the hydrogen atoms are replaced with tritium (T), so the formula will become $C{{T}_{4}}$ and in this molecule the molecular mass of carbon is 12 and the molecular mass of T is 3 (given).
So its molecular mass will be: $12+3+3+3+3=24$
So its molecular mass is 24 g / mol.
Now for the percentage of composition of carbon in $C{{T}_{4}}$ = $\dfrac{12}{24}\text{ x 100}$ $=\text{ 0}\text{.50 x 100}$
$=\text{ 50 }\\!\\!$
So the change in weight of the carbon atom in both the compounds will be: $75-50=25$
So the change is 25%.

Therefore, the correct answer is an option (c)- 25%.

Note: There are three isotopes of hydrogen i.e., hydrogen (H), deuterium (D), and tritium (T), so if the molecular weight of these is not mentioned in the question then take the mass of H = 1, D = 2, and T = 3.