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
To find the mass of 87Rb, we can rearrange the equation:
Avg mass = (fraction of Rb85) * (mass of 85Rb) + (fraction of Rb87) * (mass of 87Rb)
Given that the average mass is 85.4678 amu, the mass of 85Rb is 84.9117 amu, and the ratio of the isotopes 85Rb/87Rb is 2.591, we can substitute these values into the equation:
85.4678 = (0.7215) * (84.9117) + (0.2784) * M87
Now, we can solve for M87:
85.4678 - (0.7215) * (84.9117) = (0.2784) * M87
M87 = (85.4678 - (0.7215) * (84.9117)) / (0.2784)
M87 = 86.9064 amu
Therefore, the mass of 87Rb is approximately 86.9064 amu.
To find the mass of 87Rb (M87), we can use the equation provided:
Average mass = (0.7215)(84.9117) + (0.2784)M87
Substituting the values, we get:
85.4678 = (0.7215)(84.9117) + (0.2784)M87
Now, let's solve for M87:
85.4678 - (0.7215)(84.9117) = (0.2784)M87
76.5647645 = (0.2784)M87
Dividing both sides by 0.2784:
M87 ≈ 76.5647645 / 0.2784
M87 ≈ 275.403 amu
So, the mass of 87Rb is approximately 275.403 amu.