Rubidium has two naturally occurring isotopes. The average atomic mass of Rb is 85.4678 amu. If 72.15% of Rb is found as Rb-85 (84.9117 amu), what is the mass of the other isotope?

86.908 amu

It is given that:

Average atomic mass of Rb = 85.4678 amu.

Also the mass of Rb-85 = 84.9117 amu and its Percentage = 72.15% = 0.7215

Let the mass of other isotope (Rb-87) = x

So, Percentage of other isotope (Rb-87) = 100 - 72.15 = 27.85% = 0.2785

Average mass = (Percentage * Mass of Rb-85)  + (Percentage * Mass of Rb-87)

85.4678 = (0.7215 * 84.9117) + (0.2785 * x)

85.4678 = 61.2638 + 0.2785x

0.2785 x = 24.204

x = 24.204 / 0.2785

x = 86.908 amu

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x = 86.908 amu

Explanation:

Average mass of isotope =  85.4678 amu

Rb-85 = 84.9117 amu, Percentage = 72.15% = 0.7215

Other isotope = x, Percentage = 100 - 72.15 = 27.85% = 0.2785

Average mass = (Percentage * Mass of Rb-85)  + (Percentage * Mass of Rb-87)

85.4678 = (0.7215 * 84.9117) + (0.2785 * x)

85.4678 = 61.2638 + 0.2785x

0.2785x = 24.204

x = 24.204 / 0.2785

x = 86.908 amu

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