Question

part b
a certain element x has four isotopes.

• 94.93% of x has a mass of 31.97207 amu.
• 0.7640% of x has a mass of 32.97146 amu.
• 4.290% of x has a mass of 33.96786 amu.
• 1.600×10⁻²% of x has a mass of 35.96709 amu.

what is the average atomic mass of element x?
express your answer numerically to four significant figures.
view available hint(s)
input box amu

Explanation:

Step1: Convert percentages to decimals

For each isotope, divide the percentage by 100 to get the decimal fraction.

• First isotope: $94.93\% = \frac{94.93}{100} = 0.9493$
• Second isotope: $0.7640\% = \frac{0.7640}{100} = 0.007640$
• Third isotope: $4.290\% = \frac{4.290}{100} = 0.04290$
• Fourth isotope: $1.600\times 10^{-2}\% = \frac{1.600\times 10^{-2}}{100} = 1.600\times 10^{-4}$

Step2: Calculate the contribution of each isotope to the average atomic mass

Multiply the mass of each isotope by its decimal fraction.

• First isotope: $31.97207\ \text{amu} \times 0.9493 = 31.97207\times0.9493 \approx 30.351$
• Second isotope: $32.97146\ \text{amu} \times 0.007640 = 32.97146\times0.007640 \approx 0.2519$
• Third isotope: $33.96786\ \text{amu} \times 0.04290 = 33.96786\times0.04290 \approx 1.457$
• Fourth isotope: $35.96709\ \text{amu} \times 1.600\times 10^{-4} = 35.96709\times0.00016 \approx 0.005755$

Step3: Sum the contributions

Add up the contributions from all four isotopes.
$30.351 + 0.2519 + 1.457 + 0.005755 \approx 32.065655$

Step4: Round to four significant figures

The sum is approximately $32.07$ (rounded to four significant figures).

Answer:

32.07