Question

2.) calculate the average atomic mass of bromine. one isotope of bromine has an atomic mass of 78.92 amu and a relative abundance of 50.69%. the other major isotope of bromine has an atomic mass of 80.92 amu and a relative abundance of 49.31%.

Explanation:

Step1: Convert percentages to decimals

For the first isotope, \( 50.69\% = \frac{50.69}{100}=0.5069 \)
For the second isotope, \( 49.31\% = \frac{49.31}{100}=0.4931 \)

Step2: Calculate contribution of each isotope

Contribution of first isotope: \( 78.92 \, \text{amu} \times 0.5069 \)
\( 78.92\times0.5069 \approx 40.00 \, \text{amu} \) (approximate calculation: \( 78.92\times0.5 = 39.46, 78.92\times0.0069\approx0.544, 39.46 + 0.544 = 40.004 \))

Contribution of second isotope: \( 80.92 \, \text{amu} \times 0.4931 \)
\( 80.92\times0.4931\approx 80.92\times0.5 - 80.92\times0.0069\approx40.46 - 0.558 = 39.902 \, \text{amu} \)

Step3: Sum the contributions

Average atomic mass \( = 40.004 + 39.902 = 79.906 \, \text{amu} \) (more accurately, using exact multiplication:

First isotope: \( 78.92\times0.5069 = 78.92\times(0.5 + 0.0069)=78.92\times0.5+78.92\times0.0069 = 39.46+0.544548 = 40.004548 \)

Second isotope: \( 80.92\times0.4931 = 80.92\times(0.5 - 0.0069)=80.92\times0.5 - 80.92\times0.0069 = 40.46 - 0.558348 = 39.901652 \)

Sum: \( 40.004548+39.901652 = 79.9062 \, \text{amu} \))

Answer:

The average atomic mass of bromine is approximately \( 79.91 \, \text{amu} \) (or more precisely \( 79.906 \, \text{amu} \))