Problem 52 - Introductory Chemistry Atoms First

Question

Bromine exists as only two isotopes in nature, \({ }_{35}^{79} \mathrm{Br}\) (atomic mass \(78.918336\) amu, \% natural abundance \(=50.69 \%\) and \({ }_{35}^{81}\) Br (atomic mass \(80.916289\) amu). (a) What is the percent natural abundance for \({ }_{35}^{81} \mathrm{Br}\) ? (b) Calculate the atomic mass of the naturally occurring mixture of isotopes.

Step-by-Step Solution

Answer

(a) The percent natural abundance for \({ }_{35}^{81} \mathrm{Br} = 100 \% - 50.69 \% = 49.31 \% \). (b) The atomic mass of the naturally occurring mixture of isotopes = \((78.918336 \text{ amu}) (0.5069) + (80.916289 \text{ amu}) (0.4931) = 79.91646 \text{ amu}\).

Step 1: Calculate the percent natural abundance of \({ }_{35}^{81} \mathrm{Br}\)

Since there are only two isotopes, and their percent natural abundances must sum up to 100%, we have: Percent natural abundance of \({ }_{35}^{81} \mathrm{Br} = 100 \% - 50.69 \%\) Now, calculate the result.

Step 2: Calculate the atomic mass of the naturally occurring mixture of isotopes

To find the atomic mass of the naturally occurring mixture of isotopes, we need to multiply the atomic mass of each isotope by its percent natural abundance and sum them up: Atomic mass of mixture = (atomic mass of \({ }_{35}^{79} \mathrm{Br}\))(percent abundance of \({ }_{35}^{79} \mathrm{Br}\)) + (atomic mass of \({ }_{35}^{81} \mathrm{Br}\))(percent abundance of \({ }_{35}^{81} \mathrm{Br}\)) Use the given information and the calculated percent abundance of \({ }_{35}^{81} \mathrm{Br}\) to find the atomic mass of the mixture. Remember to divide the percentage by 100 to convert it to a fraction.

Step 3: Present the results

After completing the calculations in Steps 1 and 2, present the results for the percent natural abundance of \({ }_{35}^{81} \mathrm{Br}\) and the atomic mass of the naturally occurring mixture of isotopes.

Key Concepts

Percent Natural Abundance

In nature, elements can exist as a mixture of different isotopes. Each isotope has a specific percent natural abundance, representing the proportion of that isotope relative to the total amount of the element found in nature. This is expressed as a percentage, indicating how common a particular isotope is. When you have two isotopes, like in the case of bromine, their individual percent abundances must add up to 100%. For example, if \(^{79}\mathrm{Br}\) has a 50.69% natural abundance, then \(^{81}\mathrm{Br}\) must account for the remaining percentage. So, you simply subtract the known percent from 100% to find the unknown abundance:

• Percent abundance of \(^{81}\mathrm{Br} = 100\% - 50.69\% = 49.31\%\)

This result shows how likely you are to find \(^{81}\mathrm{Br}\) compared to \(^{79}\mathrm{Br}\) in a natural sample.

Atomic Mass

The atomic mass of an element is an average mass of its isotopes, weighted according to their natural abundances. It's not simply the mass of one atom; instead, it's a reflection of the different isotopes that make up the element as found in nature. To calculate it, take the atomic mass of each isotope, multiply by its percent natural abundance (converted to a decimal), and then sum these values.

• For bromine, the calculation is:
  \[\text{Atomic mass of bromine} = (78.918336\, \text{amu} \times 0.5069) + (80.916289\, \text{amu} \times 0.4931)\]

This gives us the weighted average reflecting the isotopes' contribution to the element's atomic mass, making it a more accurate representation than any single isotope's mass alone.

Bromine Isotopes

Bromine, a chemical element with the symbol Br, exists in nature primarily as two isotopes: \(^{79}\mathrm{Br}\) and \(^{81}\mathrm{Br}\). Each isotope has a slightly different atomic mass due to the difference in the number of neutrons. However, they share the same number of protons, maintaining the identity of bromine.

• \(^{79}\mathrm{Br}\) has an atomic mass of 78.918336 amu and a 50.69% natural abundance.
• \(^{81}\mathrm{Br}\) has an atomic mass of 80.916289 amu with a 49.31% abundance.

These isotopes' masses reflect differences due to the different numbers of neutrons in the nucleus, affecting the atomic mass but not the chemical behavior. Understanding their characteristics helps scientists and students predict the element's behavior in various reactions and its role in compounds.