Question

The atomic masses of \({ }_{15}^{74} \mathrm{Br}\left(50.69\right.\) percent) and \({ }_{35}^{81} \mathrm{Br}\) (49.31 percent) are 78.9183361 and 80.916289 amu, respectively. Calculate the average atomic mass of bromine. The percentages in parentheses denote the relative abundances.

Short answer

The average atomic mass of bromine is 79.88 amu.

Step-by-step solution

Understanding the Concept

The average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its isotopes. This involves multiplying the atomic mass of each isotope by its fractional abundance (percent abundance divided by 100) and then summing the results.

Convert Percent Abundances to Fractions

To use the percentages as weights, convert them to decimal fractions. For \(_{15}^{74} \mathrm{Br}\), the fractional abundance is \(0.5069\) and for \(_{35}^{81} \mathrm{Br}\), the fractional abundance is \(0.4931\).

Calculate Contribution of Each Isotope

Multiply the atomic mass of each isotope by its fractional abundance. For \(_{15}^{74} \mathrm{Br}\), the calculation is \(78.9183361 \times 0.5069 = 40.0051\). For \(_{35}^{81} \mathrm{Br}\), the calculation is \(80.916289 \times 0.4931 = 39.8776\).

Sum the Contributions

Add the contributions from each isotope to get the average atomic mass of bromine: \(40.0051 + 39.8776 = 79.8827\).

Round the Result

The average atomic mass, when expressed to an appropriate number of significant figures based on the given data, is \(79.88\;\text{amu}\).

Key concepts

Bromine Isotopes

Bromine is a chemical element with the symbol \(_{35}^{ ext{Br}}\) in the periodic table. It has two common isotopes: \(^{74} \text{Br}\) and \(^{81} \text{Br}\). An isotope is a version of an element that has the same number of protons but a different number of neutrons. This difference in neutron count leads to varying atomic masses between isotopes while maintaining the same chemical behavior.

In the case of bromine, both isotopes occur naturally with specific abundances. The isotope \(^{74} \text{Br}\) has an atomic mass of approximately \(78.9183361\) amu and occurs with a relative abundance of 50.69%, whereas \(^{81} \text{Br}\) has an atomic mass of \(80.916289\) amu with a 49.31% abundance.

• The occurrence of multiple isotopes is common, and understanding their distribution is essential to determining an element's average atomic mass.

Calculating the average atomic mass involves using the individual masses and relative abundances of these isotopes.

Weighted Average

The concept of a weighted average is crucial for calculating the average atomic mass of an element. It reflects not just the individual values but also how significant each value's presence is in the mix. In simpler terms, a weighted average considers the 'weight' or contribution of each value.

To calculate a weighted average, follow these steps:

• Multiply each individual mass by its corresponding weight (fraction representing abundance).
• Add all the resulting values together.

For bromine, the weighted average is computed by taking the atomic masses of its isotopes and multiplying them by their respective fractional abundances.
This method reflects both the mass and prevalence of each isotope within bromine's natural occurrence.

Fractional Abundance

Fractional abundance is a key factor in average atomic mass calculations. It is the decimal form of an isotopic abundance percentage, indicating how much of the isotope is present out of the total amount of the element.

• To find the fractional abundance, divide the given percentage by 100.

For example, the percentage abundances of \(^{74} \text{Br}\) and \(^{81} \text{Br}\) are 50.69% and 49.31%, respectively.
By converting these percentages, we get fractional abundances of 0.5069 for \(^{74} \text{Br}\) and 0.4931 for \(^{81} \text{Br}\).

These fractions serve as weights in the weighted average calculation, allowing us to account for how prevalent each isotope is when calculating the average atomic mass.

Significant Figures

Significant figures are vital in expressing an answer accurately in scientific calculations. They represent the precision of a measurement. When multiple measurements are being added, as in the calculation of an average atomic mass, the final result should reflect the least precise measurement's level of precision.

• In practice, it's important to pay attention to the number of significant figures when performing any operation and adjust the final answer accordingly.

In the bromine example, the various operations lead to an intermediate result of 79.8827 amu for the average atomic mass.

Considering the significant figures provided in the original data (four decimal places), rounding the final result yields \(79.88\) amu.
This ensures that the answer respects the precision of the original measurements while providing a scientifically valid result.