Answer to Question #121460 in Chemistry for anav

Question

Lead (Pb) (average atomic mass 207.19 amu) has three naturally-occurring isotopes, Pb206, Pb207and
Pb208 with isotopic masses of 205.98 amu, 206.98 amu and 207.98 amu respectively. If the isotopes Pb207and Pb208 are present in equal amounts, calculate the percent abundance of each isotope of Pb, which are, Pb206, Pb207 and Pb208.

Answer 1

Solution:

If w(Pb-207) = w(Pb-208) = x (equal amounts),

then:

w(Pb-206) = 1 - w(Pb-207) - w(Pb-208) = 1 - x - x = 1-2x.

Write the following equation:

[w(Pb-206) × Isotopic mass of Pb-206] + [w(Pb-207) × Isotopic mass of Pb-207] + [w(Pb-208) × Isotopic mass of Pb-208] = Average atomic mass of Pb

Set up average atomic weight equation:

(1-2x)(205.98) + (x)(206.98) + (x)(207.98) = 207.19

205.98 - 411.96x + 206.98x + 207.98x = 207.19

3x = 1.21

x = 0.4033

w(Pb-207) = w(Pb-208) = x = 0.4033 or 40.33%

w(Pb-206) = 1-2x = 1 - 2×0.4033 = 0.1934 or 19.34%

Answer:

The percent abundances are as follows:

Pb-206: 19.34%

Pb-207: 40.33%

Pb-208: 40.33%