Antimony has 2 stable isotopes 121-sb mass 120.9038 and 123-sb mass 122.9042 amu. calculate the percent abundances of these isotopes. hint: use amu from periodic table and assign x value as decimal abundance for one isotope and 1-x value for other to solve.

The % abundance of 121-Sb is 57.096%, and the % abundance of 123-Sb is 42.904%.

The average atomic mass from the isotopes can be calculated as:

Atomic mass = (atomic mass % abundance of isotope 1 ) + (atomic mass % abundance of isotope 2 )

Let the % abundance of isotope 1 is x

% abundance of isotope 2 will be 1-x

121.760 amu = (120.9038 (x)) + (122.9042 (1-x))

x = 0.57096

Abundance % = 57.096

The abundance % of 121-Sb = x = 57.096 %

The abundance % of 123-Sb = 1-x = 1 - 0.57096

The abundance % of 123-Sb = 0.42904

The abundance % of 123-Sb = 42.904%

For more information about isotopes, refer to the link:

link

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

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