Question

Naturally occurring iron contains:
5.82% 54Fe (atomic mass = 53.940 amu)
91.66 % 56Fe (atomic mass = 55.935 amu)
2.19% 57Fe (atomic mass = 56.935 amu)
0.33% 58Fe (atomic mass = 57.933 amu)
Calculate the average atomic mass of iron.

Answer 1

\[(\frac{5.82}{100}*53.940)+(\frac{91.66}{100}*55.935)+(\frac{2.19}{100}*56.935)+(\frac{0.33}{100}*57.933)=55.8\]