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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.

Respuesta :

Erythrosin17
The atomic mass of an element is a result of the weighted average of the masses of its corresponding isotopes. 

We are given: 
Sb-121: mass of 120.9038 amu, x abundance 
Sb-123: mass of 122.9042 amu, 1-x abundance 

To be able to calculate the atomic mass of antimony, we multiply the percent abundances of the isotopes by their respective atomic masses. Then, we add,

Atomic mass = (Atomic Mass of Sb-121) (% Abundance) + (Atomic Mass of Sb-123) (% Abundance )

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

Solving for x,
120.9038x + 122.9042 -122.9042x = 121.760
x = 0.57096 or 57.096% 
1-x = 1-0.57096 = 0.42904 or 42.90% 

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 [tex]\times[/tex] % abundance of isotope 1 ) + (atomic mass [tex]\times[/tex] % 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 [tex]\times[/tex] (x)) + (122.9042 [tex]\times[/tex] (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:

https://brainly.com/question/21536220