Question

Calculate the atomic mass for iron given the following data for its natural isotopes: $$ \begin{array}{llc} { }^{54} \mathrm{Fe} & 53.940 \mathrm{amu} & 5.82 \% \\ { }^{56} \mathrm{Fe} & 55.935 \mathrm{amu} & 91.66 \% \\ { }^{57} \mathrm{Fe} & 56.935 \mathrm{amu} & 2.19 \% \\ { }^{58} \mathrm{Fe} & 57.933 \mathrm{amu} & 0.33 \% \end{array} $$

Short answer

The atomic mass of iron is 55.51 amu.

Step-by-step solution

Understanding the Concept

To calculate the atomic mass of an element with multiple isotopes, we use the weighted average method. This involves multiplying the mass of each isotope by its abundance (in decimal form) and adding up the results.

Converting Percentages to Decimals

Convert each isotope's percentage abundance into a decimal by dividing by 100. - For \({ }^{54} \mathrm{Fe}\), 5.82\% becomes 0.0582.- For \({ }^{56} \mathrm{Fe}\), 91.66\% becomes 0.9166.- For \({ }^{57} \mathrm{Fe}\), 2.19\% becomes 0.0219.- For \({ }^{58} \mathrm{Fe}\), 0.33\% becomes 0.0033.

Calculating the Contribution of Each Isotope

Multiply the mass of each isotope by its decimal abundance:- \({ }^{54} \mathrm{Fe}: 53.940 \mathrm{amu} \times 0.0582 = 3.139548 \mathrm{amu}\)- \({ }^{56} \mathrm{Fe}: 55.935 \mathrm{amu} \times 0.9166 = 51.247971 \mathrm{amu}\)- \({ }^{57} \mathrm{Fe}: 56.935 \mathrm{amu} \times 0.0219 = 1.246877 \mathrm{amu}\)- \({ }^{58} \mathrm{Fe}: 57.933 \mathrm{amu} \times 0.0033 = 0.191179 \mathrm{amu}\)

Summing the Contributions

Add together all the weighted contributions to get the average atomic mass of iron:\[3.139548 \mathrm{amu} + 51.247971 \mathrm{amu} + 1.246877 \mathrm{amu} + 0.191179 \mathrm{amu} = 55.508575 \mathrm{amu}\]

Round off and Conclude

The average atomic mass of iron, rounded to two decimal places, is 55.51 amu. This value represents the atomic mass you would expect to find on the periodic table for iron.

Key concepts

Isotopes

Isotopes are variations of a particular chemical element that share the same number of protons but differ in the number of neutrons. This means that while they belong to the same element, isotopes have different atomic masses. For instance, iron has several naturally occurring isotopes such as

• \(^{54}\text{Fe}\)
• \(^{56}\text{Fe}\)
• \(^{57}\text{Fe}\)
• \(^{58}\text{Fe}\)

All these isotopes of iron have the same atomic number, 26, but different atomic masses due to the variance in the number of neutrons. Understanding isotopes is crucial because they contribute to an element's average atomic mass. Each isotope's individual atomic mass and its abundance directly influence this average value, as seen on the periodic table.
This concept is vital in chemistry, especially when calculating the atomic mass using all available isotopic data of any given element.

Weighted Average Method

The weighted average method is a key concept in determining the atomic mass of an element with multiple isotopes. Each isotope contributes to the total atomic mass based on its mass and abundance. To calculate this, you perform the following steps:

• Multiply the atomic mass of each isotope by its abundance expressed as a decimal.
• Add up each of the products obtained.

This compilation of isotope masses weighted by their abundance gives us what is known as the weighted average. It's a bit like calculating your grades in school, where certain assignments like exams are worth more than daily homework. For the element iron, we would multiply each isotope's atomic mass by its decimal abundance, and then sum these contributions to find the weighted average atomic mass.
By using this method, you ensure that each isotope's contribution to the element's average atomic mass is justly represented according to its natural occurrence.

Percentage Abundance

Percentage abundance refers to how common or rare a particular isotope is within a sample of an element. This percentage is crucial when calculating the atomic mass using multiple isotopes. In the calculation process, you convert the percentage abundance of each isotope into a decimal:

• Simply divide each percentage by 100.

For example:

• The isotope \(^{56}\text{Fe}\) has a percentage abundance of 91.66%, which converts to 0.9166 in decimal form.
• \(^{54}\text{Fe}\) with 5.82% becomes 0.0582
Percentage abundance signifies the significant weight of common isotopes in atomic mass calculations, as seen with \(^{56}\text{Fe}\), which is the most abundant iron isotope, heavily influencing the average atomic mass result.
Recognizing the importance of percentage abundance helps in comprehending how each isotope affects the overall atomic weight and composition of a given element when seen on the periodic table.