If given that neutral sample of iron has isotopes ${ }_{26}^{54}\text{Fe}{,}_{26}^{56}{\text{Fe\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}}_{26}^{57}\text{Fe}$are available as of 5, 10, and 15 total atoms in the universe respectively. What will be the average atomic mass of iron (Fe)?

To find the average atomic mass of iron (Fe) we must consider atomic mass of each isotope and its availability percentage in the universe.

$\text{Average atomic mass formula}=\frac{{\text{m}}_{\text{1}}\times n_1+{\text{m}}_2\times n_2+{\text{m}}_3\times n_3}{n_1+n_2+n_3}$

Here, $n_1=5;n_2=10;n_3=15$ 

Atomic mass of iron (Fe) are ${\text{m}}_1=54;{\text{\hspace{0.17em}\hspace{0.17em}m}}_2=56;{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}m}}_3=57$

From the above formula, average atomic mass of iron is

${\text{m}}_{\text{avg}}\text{of\hspace{0.17em}iron (Fe)}=\frac{54\times 5+10\times 56+57\times 15}{30}=\frac{270+560+855}{30}=56.1$