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Question: The abundance of isotopes of hydrogen is as follows: \({\text{Isotope}}\)| \(\% {\text{natural a...

The abundance of isotopes of hydrogen is as follows:

Isotope{\text{Isotope}}%natural abundance\% {\text{natural abundance}}Molar mass{\text{Molar mass}}
1H^1H99.98599.98511
2H^2H0.0150.01522

Calculate the mass of the hydrogen atom.
A) 1.00015u1.00015u
B) 1.00025u1.00025u
C) 1.00035u1.00035u
D) 1.00005u1.00005u

Explanation

Solution

We know that the Average atomic mass is defined as the average of the atomic masses of all atoms in their naturally occurring isotope.
We can calculate the atomic mass of the isotope by using given formula as,
Atomic mass of element=Natural abundance %+Molar mass100{\text{Atomic mass of element}} = \dfrac{{{\text{Natural abundance }}\% + {\text{Molar mass}}}}{{100}}
Example: The two naturally occurring isotopes of chlorine (17Cl35,17Cl37)\left( {_{{\text{17}}}{\text{C}}{{\text{l}}^{{\text{35}}}}{{\text{,}}_{{\text{17}}}}{\text{C}}{{\text{l}}^{{\text{37}}}}} \right) in the ratio 77:23.{\text{77:23}}{\text{.}} the average relative atomic mass of chlorine is,
35×77+37×23100 = 35.46amu.\dfrac{{{{35 \times 77 + 37 \times 23}}}}{{{\text{100}}}}{\text{ = 35}}{\text{.46amu}}{\text{.}}

Complete step by step answer:
We must remember that hydrogen has three isotopes which are one hydrogen (protium), two hydrogen (deuterium), and three hydrogen (tritium). The half-life of tritium is 12.32 years12.32{\text{ }}years five hydrogens containing isotope is the most stable, and therefore the least stable isotope is seven hydrogens.
We can calculate the atomic mass of the isotope by using given formula as,
Atomic mass of hydrogen=Natural abundance %+Molar mass100{\text{Atomic mass of hydrogen}} = \dfrac{{{\text{Natural abundance }}\% + {\text{Molar mass}}}}{{100}}
Given data,
% of natural abundance of H1=99.985 = 99.985
% of natural abundance of H2=0.015 = 0.015
Molar mass of H1=1 = 1
Molar mass of H2=2 = 2
By substituting the value of given data we can calculate the average atomic mass of hydrogen as,
The average atomic mass =1×99.985+2×0.015100 = \dfrac{{1 \times 99.985 + 2 \times 0.015}}{{100}}
On simplifying we get,
The average atomic mass =1.00015u = 1.00015u

Therefore, the correct option is A. .

Additional information:
Now we can discuss about the other isotopes of hydrogen as,
1.Protium: It is one among the common isotopes of hydrogen. It’s plenty in nature with an abundance of 99.98%.99.98\% . Mass of protium is 1.007825amu1.007825amu. Hydrogen generally combines with other atoms in compounds and is usually found in H2{H_2}.
2.Deuterium (2 Hydrogens): It contains one proton and one neutron in its nucleus. The nucleus of hydrogen two is termed as deuteron. It’s not radioactive. Its compounds are utilized in qualitative analysis and solvents for hydrogen one. Deuterium oxide is enriched with molecules consisting of deuterium rather than protium.
3.Tritium (3 hydrogens): It contains two neutrons and one proton in its nucleus. Small traces of hydrogen 3 or tritium occur in nature thanks to the synergy of cosmic rays with atmospheric gases. They’re also released a bit at the time of nuclear weapons tests. It’s radioactive; it decays into helium 3 through decay. Hydrogen 3 as a mass of three 0160492u0160492u.

Note: We must remember that each isotope comprehends unique properties. These isotopes are in common want to date. 4 hydrogens to 7 hydrogens are nuclei isotopes that are incorporated within the laboratory. One among the smallest stable isotopes of hydrogen is 7H and therefore the most stable isotope is 5 hydrogens. The foremost stable radioisotope of hydrogen is tritium.