# Naturally occuring element Z consists of Z-24 (mass=23.99), Z-25 (mass=24.99), and Z-26 (mass=25.98). If the natural abundance of Z-25 is 10.0% and the isotope ratio Z-24/Z-26 is 7.174, what is the natural abundance of the lightest isotope?

Since Z-25 makes up 10% of the element Z in nature the remaining 90% must be the sum of other two isotopes, Z-24 and Z-26. We know that the percent abundance of X-24 is 7.174 times that of Z-26, so we can find the abundance of both isotopes as follows:

Let Z-24 = x and Z-26 = 0.90-X

x/(0.90-x) = 7.174

x = (7.174)(0.90-x)

8.174x = 6.454

x = 0.7897

Z-24 is the lightest isotope:

abundance of Z-24 = 0.7897 = 78.97%

The abundance of Z-26 is 0.1103 = 11.03%

The given masses of the isotopes aren't needed to solve this problem. However, you could use them to find the average mass of element Z in nature since you now know the relative amounts of each isotope. The average mass is the sum of the mass of each isotope multiplied by its percent abundance.