Question
The relative atomic mass (A_r) of atoms is the average mass of all the different isotopes of an element (taking into account the amount of each isotopes) on a scale weere ^12C atoms have mass of exactly 12.
Element Iron isotopes 54 26 Fe 56 26 Fe 58 26 Fe Abundance 5.8% 91.8% 2.1% 0.3% Relative atomic mass (A_r) to (3sf) ___
Element Iron isotopes 54 26 Fe 56 26 Fe 58 26 Fe Abundance 5.8% 91.8% 2.1% 0.3% Relative atomic mass (A_r) to (3sf) ___
Answers
GPT-4o mini
To find the relative atomic mass (A_r) of iron (Fe), we can use the formula for the weighted average based on the abundance and mass of each isotope.
We have the following isotopes and their abundances:
- \(^{54}\text{Fe}\): mass = 54, abundance = 5.8%
- \(^{56}\text{Fe}\): mass = 56, abundance = 91.8%
- \(^{58}\text{Fe}\): mass = 58, abundance = 2.1%
- \(^{60}\text{Fe}\): mass = 60, abundance = 0.3% (though this isotope is often negligible in calculations, we'll include it for completeness)
To find the A_r, we perform the following calculations:
1. Convert percentages to fractions:
- \(5.8\% = 0.058\)
- \(91.8\% = 0.918\)
- \(2.1\% = 0.021\)
- \(0.3\% = 0.003\)
2. Multiply each isotope’s mass by its corresponding fraction:
- \(^{54}\text{Fe}: 54 \times 0.058 = 3.132\)
- \(^{56}\text{Fe}: 56 \times 0.918 = 51.408\)
- \(^{58}\text{Fe}: 58 \times 0.021 = 1.218\)
- \(^{60}\text{Fe}: 60 \times 0.003 = 0.18\)
3. Sum these values to get the total:
\[
A_r = 3.132 + 51.408 + 1.218 + 0.18 = 55.938
\]
4. Round to 3 significant figures:
\[
A_r \approx 55.9
\]
Thus, the relative atomic mass of iron (Fe) to 3 significant figures is **55.9**.
We have the following isotopes and their abundances:
- \(^{54}\text{Fe}\): mass = 54, abundance = 5.8%
- \(^{56}\text{Fe}\): mass = 56, abundance = 91.8%
- \(^{58}\text{Fe}\): mass = 58, abundance = 2.1%
- \(^{60}\text{Fe}\): mass = 60, abundance = 0.3% (though this isotope is often negligible in calculations, we'll include it for completeness)
To find the A_r, we perform the following calculations:
1. Convert percentages to fractions:
- \(5.8\% = 0.058\)
- \(91.8\% = 0.918\)
- \(2.1\% = 0.021\)
- \(0.3\% = 0.003\)
2. Multiply each isotope’s mass by its corresponding fraction:
- \(^{54}\text{Fe}: 54 \times 0.058 = 3.132\)
- \(^{56}\text{Fe}: 56 \times 0.918 = 51.408\)
- \(^{58}\text{Fe}: 58 \times 0.021 = 1.218\)
- \(^{60}\text{Fe}: 60 \times 0.003 = 0.18\)
3. Sum these values to get the total:
\[
A_r = 3.132 + 51.408 + 1.218 + 0.18 = 55.938
\]
4. Round to 3 significant figures:
\[
A_r \approx 55.9
\]
Thus, the relative atomic mass of iron (Fe) to 3 significant figures is **55.9**.