Question

humans started to master the process of smelting iron ores to turn into iron tools and weapons in eurasia during the 2nd millennium bc, replacing the use of copper alloys (a metal made by combining two or more metallic elements). in the modern world, iron alloys, such as steel, stainless steel, cast iron and special steels, are by far the most common industrial metals, due to their mechanical properties and low cost. the iron and steel industry is thus very important economically, and iron is the cheapest metal, with a price of a few dollars per kilogram or pound. iron has four stable isotopes: ( ^{54}fe ) (with an abundance of 5.85%), ( ^{56}fe ) (with an abundance of 91.8%), ( ^{57}fe ) (with an abundance of 2.12%), and ( ^{58}fe ) (with an abundance of 0.28%).
explain, in terms of percent abundance, why the atomic mass of iron is closest to the mass number of iron - 56.
the atomic mass of iron is closest to the mass number of iron - 56 because iron 56 has the highest percent among the iron isotopes.
construct a mathematical representation showing how to calculate the overall atomic mass of iron.
iron - 54 - 0.54
iron - 56 - 0.56
iron - 57 - 0.57
iron - 58 - 0.58
final answer: _____________ amu

Explanation:

Step1: Calculate the contribution of each isotope

The formula for atomic mass \(A=\sum_{i} (m_i\times p_i)\), where \(m_i\) is the mass of isotope \(i\) and \(p_i\) is its percent abundance (in decimal form).
For \(^{54}\text{Fe}\): \(m_1 = 54\space amu\), \(p_1=0.0585\), contribution \(C_1=54\times0.0585 = 3.159\space amu\)

Step2: Calculate the contribution of \(^{56}\text{Fe}\)

For \(^{56}\text{Fe}\): \(m_2 = 56\space amu\), \(p_2 = 0.918\), contribution \(C_2=56\times0.918=51.408\space amu\)

Step3: Calculate the contribution of \(^{57}\text{Fe}\)

For \(^{57}\text{Fe}\): \(m_3 = 57\space amu\), \(p_3=0.0212\), contribution \(C_3=57\times0.0212 = 1.2084\space amu\)

Step4: Calculate the contribution of \(^{58}\text{Fe}\)

For \(^{58}\text{Fe}\): \(m_4 = 58\space amu\), \(p_4=0.0028\), contribution \(C_4=58\times0.0028=0.1624\space amu\)

Step5: Sum up the contributions

Atomic mass of Fe \(A = C_1 + C_2+C_3 + C_4\)
\(A=3.159+51.408 + 1.2084+0.1624\)
\(A=(3.159+51.408)+(1.2084 + 0.1624)\)
\(A = 54.567+1.3708\)
\(A=55.9378\space amu\)

Answer:

\(55.94\space amu\)