21 Chemistry -- Foundation and Fundamentals of Chemistry

i). Naturally occuring neon is 90.9% neon-20, 0.3% neon-21, 8.8% neon-22. Use this figure to calculate relative atomic mass of naturally occuring neon.ii).A natural mixture of hydrogen contains 1H1 and 1H2 . Calculate the percentage abundance of these isotopes in the mixture if the aver...

i). Naturally occuring neon is 90.9% neon-20, 0.3% neon-21, 8.8% neon-22. Use this figure to calculate relative atomic mass of naturally occuring neon.

ii).A natural mixture of hydrogen contains 1H1 and 1H2 . Calculate the percentage abundance of these isotopes in the mixture if the average atomic mass of hydrogen is 1.008 amu. 

Question no i)
Solution:

Given, Naturally occuring neon is 90.9% neon-20, 0.3% neon-21, 8.8% neon-22.

Now, the relative atomic mass of naturally occuring neon = ( mass number of isotope neon-20) x ( abundance%) + ( mass number of isotope neon-21) x (abundance%) + ( mass number of isotope neon-22) x (abundance%)/100%

= [(20x90.9%) + (21x0.3%) + (22x8.8%)] /100%

= (1818% + 6.3% + 193.6% ) /100%

=( 2017.9%) /100%

= 20.179 

Question no ii)
Solution:

Given, A natural mixture of hydrogen contains 1H1 and 1H2 . And the average atomic mass of hydrogen is 1.008 amu. 

In this question, we are given two isotopes of Hydrogen.

Mass of isotope 1H1 = 1amu.

Mass of isotope 1H2 = 2amu.

Fractional abundance of isotope 1H1 = x

Fractional abundance of isotope 1H2= (1- x).

To calculate average atomic mass of an element, we use the formula:

Average Atomic mass of Hydrogen = (Fractional abundance of isotope 1H1) *(mass of isotope 1H1) + (Fractional abundance of isotope 1H2) * (mass number of isotope 1H2)

Putting the given information, we get,

Or, 1.008 amu = (1amu *x) + {2amu*(1-x)}

Or, 1.008 = x+ 2 - 2x

Or, x = 0.992

Then, fractional abundance percentage of isotope 1H1 (x) = 0.992*100% = 99.2%

Also, the fractional abundance percentage of isotope 1H2 (1-x) = (100 - 99.2)% = 0.8%

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