Natural rubidium has an average mass of 85.4678 amu and is composed of isotopes
85Rb (mass = 84.9117 amu) and 87Rb. The ratio of the isotopes 85Rb/87Rb in natural rubidium is 2.591. What is the mass of 87Rb?
Let x be the fraction of the Rb atoms that are Rb85. x/2.591 is the fraction that are Rb87
x + x/2.591 = 1
1.38595 x = 1
x = 0.7215 = fraction that is Rb85
1-x = 0.2784 = fraction that is Rb87
Avg mass = 85.4678 = (0.7215)(84.9117) + (0.2784)M87
Solve for M87
2 answers
M87 = 86.94 amu
not a very good answer... I'm sure the person asking the question is looking for an answer that applies to every question, and not only to this one.
2 answers