## Chemistry 10th Edition

% abundance of $^{107}Ag = 51.835\%$ % abundance of $^{109}Ag = 48.165\%$
Atomic mass of Silver $= 107.8682 amu$ Mass of $^{107}Ag = 106.90509 amu$ Mass of $^{109}Ag =108.9047 amu$ Let the fraction of $^{107}Ag = x$ Then the fraction of $^{109}Ag = (1-x)$ The atomic mass of an element is the weighted average of the masses of its constituent isotopes. Therefore, $107.8682amu = x(106.90509amu) + (1-x)(108.9047amu)$ $107.8682=106.90509x+108.9047-108.9047x$ $-1.0365=-1.99961x$ $x = 0.51835$ The fraction of $^{107}Ag = 0.51835$ % abundance of $^{107}Ag = 51.835\%$ The fraction of $^{109}Ag=0.48165$ % abundance of $^{109}Ag = 48.165\%$