Question

From the above data what is the molecular mass of $C{{H}_{4}}$ containing all isotopes of carbon but hydrogen on $_{1}^{1}H$. (Given that atomic mass of hydrogen$=1.008$)

Isotope	Relative abundance ($%$ )	Atomic mass (a)
${{12}_{c}}$	98.8	12
${{13}_{c}}$	1.18	13.1
${{14}_{c}}$	0.02	14.1

A. $16.240u$
B. $16.004u$
C. $16.125u$
D. $16.42u$

Answer 1

Hint: The mass of a given molecule is the molecular mass (m)- it is calculated in Daltons (Da or u). Different molecules of the same compound may have different molecular masses, since they contain different element isotopes.

Complete answer:
There are steps to find out the molecular mass of the compound which are as follows:
Step 1. We have to dictate the molecular mass of the molecule.
Step 2. By the help of a periodic table find the atomic mass of each element in the molecule.
Step 3. Now multiply each element’s atomic mass with the number of atoms of the element present in the molecule. This number is present on the subscript next to the element symbol in the following molecular formula.
Step 4. Now the last step is adding these values together for each different atom in the molecule you will find the molecular mass.
Here we have the following data that isotope of carbon ${{12}_{c}}$ has relative abundance in percent which is $98.8$ and the atomic mass is $12$ going further we get ${{13}_{c}}$ has relative abundance in percent which is $1.18$ and the atomic mass is $13.1$and we have ${{14}_{c}}$ which has relative abundance in percent given $0.02$ and the atomic mass is $14.1$. But hydrogen has $_{1}^{1}H$ given its atomic mass as $1.008$.
Now by using the steps given to us we get:
We here divide it by \[100\] because the relative abundance is given in percentage.
\[=\dfrac{1583.9616+20.21576+0.36264}{100}\]
 Now by calculating it we get
\[=16.0454u\]

So, the molecular mass of $C{{H}_{4}}$ is \[16.0454u\] nearly equal to option D.

Note:
We should know that the formula unit mass of a substance is equal to the sum of the atomic masses of all the atoms present in the formula unit of a compound so we can take an example of calcium oxide which has a formula unit of \[CaO\]. One more point is that the formula unit mass is calculated in the same way as molecular mass as we see in above.