Question

1. an element has the following natural abundances and isotopic masses: 90.92% abundance with 49.99 amu, 0.26% abundance with 48.99 amu, and 8.82% abundance with 47.99 amu. calculate the average atomic mass of this element. show your work.

Explanation:

Step1: Convert abundances to decimals

• Isotope 1: \( 90.92\% = \frac{90.92}{100} = 0.9092 \)
• Isotope 2: \( 0.26\% = \frac{0.26}{100} = 0.0026 \)
• Isotope 3: \( 8.82\% = \frac{8.82}{100} = 0.0882 \)

Step2: Calculate mass contributions

• Isotope 1: \( 49.99 \, \text{amu} \times 0.9092 = 45.450908 \, \text{amu} \)
• Isotope 2: \( 48.99 \, \text{amu} \times 0.0026 = 0.127374 \, \text{amu} \)
• Isotope 3: \( 47.99 \, \text{amu} \times 0.0882 = 4.232718 \, \text{amu} \)

Step3: Sum the contributions

\( 45.450908 + 0.127374 + 4.232718 = 49.811 \, \text{amu} \) (rounded to appropriate precision)

Answer:

The average atomic mass of the element is approximately \(\boldsymbol{49.81 \, \text{amu}}\) (or \(49.811 \, \text{amu}\) for more precision).