Q145CP - Chemistry: Molecular Nature Of Matter And Change

Question

Antimony has many uses, for example, in infrared devices and as part of an alloy in lead storage batteries. The element has two naturally occurring isotopes, one with a mass of 120.904 amu, and the other with mass of 122.904 amu. (a) Write the ^A_ZX notation for each isotope. (b) Use the atomic mass of antimony from the periodic table to calculate the natural abundance of each isotope.

Step-by-Step Solution

Answer

(a) The notation can be found as ${}_{\mathbf{51}}{}^{\mathbf{120}\mathbf{.}\mathbf{904}}\mathbf{Sb}\mathbf{ }\mathbf{and}\mathbf{ }{}_{\mathbf{51}}{}^{\mathbf{122}\mathbf{.}\mathbf{904}}\mathbf{Sb}$

(b) The natural abundance of an isotope with an atomic mass of 120.904 is 0.572 and the natural abundance of an isotope with an atomic mass of 122.904 is 0.428.

 

Step 1: Finding the notation

The atomic number of antimony is 51 and as given it has two isotopes with mass numbers 120.904and 122.904. It can be denoted as,

${}_{\mathbf{51}}{}^{\mathbf{120}\mathbf{.}\mathbf{904}}\mathbf{Sb}\mathbf{ }\mathbf{and}\mathbf{ }{}_{\mathbf{51}}{}^{\mathbf{122}\mathbf{.}\mathbf{904}}\mathbf{Sb}$

Step 2: Finding natural abundance

Abundance can be calculated as,

$$\begin{gathered} \mathbf{Average}\mathbf{ }\mathbf{Mass}\mathbf{=}\mathbf{Natural}\mathbf{ }\mathbf{abundance}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{isotope}\mathbf{1}\mathbf{\times }\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{isotope}\mathbf{1} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{+}\mathbf{Natural}\mathbf{ }\mathbf{abundance}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{isotope}\mathbf{2}\mathbf{\times }\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{isotope}\mathbf{2} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{121}\mathbf{.}\mathbf{760}\mathbf{=}\mathbf{120}\mathbf{.}\mathbf{904}\mathbf{ }\mathbf{x}\mathbf{+}\mathbf{122}\mathbf{.}\mathbf{904}\mathbf{(}\mathbf{1}\mathbf{‐}\mathbf{x}\mathbf{)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{121}\mathbf{.}\mathbf{760}\mathbf{=}\mathbf{120}\mathbf{.}\mathbf{904}\mathbf{x}\mathbf{+}\mathbf{122}\mathbf{.}\mathbf{904}\mathbf{-}\mathbf{122}\mathbf{.}\mathbf{904}\mathbf{x} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{144} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{572} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{428} \end{gathered}$$

So, the natural abundance of an isotope with an atomic mass of 120.904 is 0.572 and the natural abundance of an isotope with an atomic mass of 122.904 is 0.428.