An elemental sample of Iron (Fe) is found to contain three naturally occurring stable isotopes: , , — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
An elemental sample of Iron (Fe) is found to contain three naturally occurring stable isotopes: $^{54}Fe$, $^{56}Fe$, and $^{57}Fe$. If their relative percentage abundances are $5\%$, $90\%$, and $5\%$ respectively, what is the highly precise average atomic mass of Iron?
💡 Solution & Explanation
The average atomic mass is calculated using the weighted average of all naturally occurring isotopes based on their fractional abundances. $M_{\text{avg}} = \frac{\sum (\% \text{ abundance} \times \text{isotopic mass})}{100}$. $M_{\text{avg}} = \frac{(5 \times 54) + (90 \times 56) + (5 \times 57)}{100} = \frac{270 + 5040 + 285}{100} = \frac{5595}{100} = 55.95\text{ u}$.