The average atomic masses of some elements may vary, depending upon souces of their ores. Naturally occurring boron consist of two isotopes with accurately known masses (10 B, 10.0129 amu and 11B, 11.0931 amu). The actual atomic mass of boron can vary grom 10.807 to 10819, depending on whether the mineral source is from Turkey or the United States. Calculate the percent abundances leading to the two values of the average atomic masses of boron from these two countries.

If a sources is rich in one kind of isotope then the atomic mass will be nearer to its value and if it is rich in another kind then the atomic mass will be nearer to its value.

Let us calculate the percentage abundance in each case.

a) Average atomic mass = [tex]\frac{[percentageofisotope(1)Xatomicmassofisotope(1)]+[percentageofisotope(2)Xatomicmassofisotope(2)}{100}[/tex]

Average atomic mass = 10.807

Let the % of 10B isotope = x

% of 11B isotope =100-x

putting in the formula

[tex]10.807=\frac{[x(10.0129)]+[(100-x)11.0931]}{100}=\frac{10.0129x+1109.31-11.0931x}{100}[/tex]

[tex]1080.7=(-1.0802x+1109.31)[/tex]

[tex]x=26.48percentage=percentageof10B[/tex]

Percentage of 11B= 100-26.48= 73.52%

a) Average atomic mass = [tex]\frac{[percentageofisotope(1)Xatomicmassofisotope(1)]+[percentageofisotope(2)Xatomicmassofisotope(2)}{100}[/tex]

Average atomic mass = 10.819

Let the % of 10B isotope = x

% of 11B isotope =100-x

putting in the formula

[tex]10.819=\frac{[x(10.0129)]+[(100-x)11.0931]}{100}=\frac{10.0129x+1109.31-11.0931x}{100}[/tex]

[tex]1081.9=(-1.0802x+1109.31)[/tex]

[tex]x=25.37percentage=percentageof10B[/tex]

Percentage of 11B= 100-25.37= 74.63%