Calculate the average atomic mass of iron. Its composition is 5.90\(\%\) with a mass of 53.94 amu, 91.72\(\%\) with a mass of 55.93 amu, 2.10\(\%\) with a mass of 56.94 amu, and 0.280\(\%\) with a mass of 57.93 amu.

Short Answer

Expert verified
The average atomic mass of iron is \(55.85\) amu.

Step by step solution

01

Convert the relative abundance from percentages to fractions

First, convert the given percentages to decimal form. This can be done by dividing each percentage by 100. So, for iron: 5.90% becomes 0.059, 91.72% becomes 0.9172, 2.10% becomes 0.021, and 0.280% becomes 0.0028.
02

Multiply each isotope's mass by its relative abundance

Next, take the mass of each isotope and multiply it by its relative abundance (now in decimal form). This gives the weighted contribution of each isotope to the overall atomic mass. The results are: \(0.059 * 53.94 = 3.1826\) amu, \(0.9172 * 55.93 = 51.304776\) amu, \(0.021 * 56.94 = 1.19574\) amu, \(0.0028 * 57.93 = 0.162204\) amu.
03

Sum up the weighted masses

Add all the results obtained in step 2 to get the average atomic mass. Therefore, the average atomic mass of iron is \(3.1826 + 51.304776 + 1.19574 + 0.162204 = 55.84532\) amu.

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