Question

there are four different isotopes of iron.
fe - 54 5.85% 53.9400 amu
fe - 56 91.75% 55.9349 amu
fe - 57 2.12% 56.9354 amu
fe - 58 0.29% 57.9333 amu
number of protons in fe - 54, fe - 56, fe - 57, fe - 58 = 26,26,26,26
previous answers
correct
part b
determine the number of neutrons in each of the following isotopes of iron.
enter your answers as integers separated by commas.
number of neutrons in fe - 54, fe - 56, fe - 57, fe - 58 = 28,30,31,32
previous answers
correct
part c
use the listed natural abundances and masses to calculate its atomic mass.
express your answer in atomic mass units to two decimal places.
atomic mass = amu

Explanation:

Step1: <Formula for atomic mass>

The formula for the atomic mass \(M\) of an element with isotopes is \(M=\sum_{i} (m_i\times a_i)\), where \(m_i\) is the mass of the \(i -\)th isotope and \(a_i\) is its abundance (in decimal form).

Step2: <Convert abundances to decimals>

For \(Fe - 54\): \(a_1=5.85\%=0.0585\), \(m_1 = 53.9400\) amu.
For \(Fe - 56\): \(a_2=91.75\%=0.9175\), \(m_2 = 55.9349\) amu.
For \(Fe - 57\): \(a_3=2.12\%=0.0212\), \(m_3 = 56.9354\) amu.
For \(Fe - 58\): \(a_4=0.29\%=0.0029\), \(m_4 = 57.9333\) amu.

Step3: <Calculate each term>

\(m_1\times a_1=53.9400\times0.0585 = 53.9400\times\frac{5.85}{100}=3.15549\) amu.
\(m_2\times a_2=55.9349\times0.9175=55.9349\times\frac{91.75}{100}=51.31027075\) amu.
\(m_3\times a_3=56.9354\times0.0212 = 56.9354\times\frac{2.12}{100}=1.20703048\) amu.
\(m_4\times a_4=57.9333\times0.0029=57.9333\times\frac{0.29}{100}=0.16800657\) amu.

Step4: <Sum up the terms>

\(M=(3.15549 + 51.31027075+1.20703048 + 0.16800657)\) amu.
\(M=55.8407978\) amu \(\approx55.84\) amu.

Answer:

\(55.84\) amu