QuestionJuly 11, 2025

Calculate the atomic mass of element "X", if it has 2 naturally occurring isotopes with the following masses and natural abundances: X-4544.8776 amu 32.88% X-4746.9443 amu 67.12% 46.26 amu 46.34 amu 46.84 amu 45.91 amu 44.99 amu

Calculate the atomic mass of element "X", if it has 2 naturally occurring isotopes with the following masses and natural abundances: X-4544.8776 amu 32.88% X-4746.9443 amu 67.12% 46.26 amu 46.34 amu 46.84 amu 45.91 amu 44.99 amu
Calculate the atomic mass of element "X", if it has 2 naturally occurring isotopes with the following masses and natural abundances:
X-4544.8776 amu 32.88% 
X-4746.9443 amu 67.12% 
46.26 amu
46.34 amu
46.84 amu
45.91 amu
44.99 amu

Solution
4.3(231 votes)

Answer

46.81 amu Explanation 1. Convert percentages to decimals Convert 32.88\% to 0.3288 and 67.12\% to 0.6712. 2. Calculate weighted average Use the formula for atomic mass: ** \text{Atomic Mass} = (m_1 \times f_1) + (m_2 \times f_2) **, where m is mass and f is fractional abundance. Calculate: (4544.8776 \, \text{amu} \times 0.3288) + (4746.9443 \, \text{amu} \times 0.6712). 3. Perform calculations 4544.8776 \times 0.3288 = 1494.7337 4746.9443 \times 0.6712 = 3185.7929 4. Sum the results Add the results: 1494.7337 + 3185.7929 = 4680.5266 amu

Explanation

1. Convert percentages to decimals<br /> Convert $32.88\%$ to $0.3288$ and $67.12\%$ to $0.6712$.<br /><br />2. Calculate weighted average<br /> Use the formula for atomic mass: **$ \text{Atomic Mass} = (m_1 \times f_1) + (m_2 \times f_2) $**, where $m$ is mass and $f$ is fractional abundance.<br /> Calculate: $(4544.8776 \, \text{amu} \times 0.3288) + (4746.9443 \, \text{amu} \times 0.6712)$.<br /><br />3. Perform calculations<br /> $4544.8776 \times 0.3288 = 1494.7337$<br /> $4746.9443 \times 0.6712 = 3185.7929$<br /><br />4. Sum the results<br /> Add the results: $1494.7337 + 3185.7929 = 4680.5266$ amu
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