Question

Calculate the average atomic mass of ${H_2}$ using the following data.

Isotope	per Natural abundance	Molar mass
$_1H$	$99.985$	$1$
$_2H$	$0.015$	$2$

(Write answer up to two decimal digit)

Answer 1

Hint: The average atomic mass can be found by multiplying the abundance of the isotopes with their respective molar mass and then adding them together. This sum has to then be divided by $100$ . The number thus found is the average atomic mass of a substance.

Formula used:Average atomic mass \[ = \dfrac{{abundance\;{{ }}of\;{{ }}isotop{e_1} \times molar\;mas{s_1} + abundance\;{{ }}of\;{{ }}isotop{e_2} \times molar\;mas{s_2}}}{{100}}\]

Complete step by step answer:
The average atomic mass is the atomic mass that is found when all the isotopes of a single element is considered.
To solve this question, we must first understand what an isotope is. An element is said to have isotopes when it has different species of the same element that differ in the number of neutrons present in the nucleus. This leads to a difference in the number of nucleons as well.
This difference in number can lead to a single element having different variations.
In reality, most elements do have different isotopes that are present either in the atmosphere or lithosphere in varying abundance.
In hydrogen, we have THREE different isotopes. They are: Hydrogen, deuterium and tritium.
Out of these, hydrogen is most common and tritium is the least. These three chemical species differ in the number of neutrons.
Hydrogen contains one neutron; deuterium contains two and tritium contains three.
The average atomic mass includes Hydrogen and deuterium. Hydrogen has a natural abundance of $99.985$ and its molar mass is $1$ .
Deuterium has an abundance of $0.015$ and its molar mass is $2$.
Plugging this information in the above equation we get,
\[ = \dfrac{{abundance\;{{ }}of\;{{ }}isotop{e_1} \times molar\;mas{s_1} + abundance\;{{ }}of\;{{ }}isotop{e_2} \times molar\;mas{s_2}}}{{100}}\]
On substituting the values,

\[ \Rightarrow \dfrac{{\left( {99.985 \times 1} \right) + \left( {0.015 \times 2} \right)}}{{100}}\]
on further solving we get,
\[ \Rightarrow \dfrac{{99.985 + 0.03}}{{100}}\]
\[ \Rightarrow \dfrac{{100.015}}{{100}}\]
\[ \Rightarrow 1\]
Therefore, we can conclude by saying that the average atomic mass of Hydrogen is $1.00$ .

Therefore, we can conclude by saying that the average atomic mass of Hydrogen is $1.00$ .

Note: it is important to remember that average atomic mass means that after considering the masses of all isotopes the average that we obtain is the average atomic mass.
Remember that isotopes are changes in neutron number or atomic mass number but not the atomic number.
Isobars contain changes in atomic number and not in atomic mass number.