Question

an element has the following isotopes: abundance mass/amu 1.40% 203.973 24.1% 205.974 22.1% 206.976 52.4% 207.977 calculate the average atomic mass. amu identify the element (give the symbol). chempad help x₀ xⁿ → ⇌ ← greek

Explanation:

Step1: Convert percentages to decimals

$1.40\% = 0.0140$, $24.1\%=0.241$, $22.1\% = 0.221$, $52.4\%=0.524$

Step2: Calculate the contribution of each isotope

For the first isotope: $0.0140\times203.973 = 2.855622$
For the second isotope: $0.241\times205.974=49.649734$
For the third isotope: $0.221\times206.976 = 45.741696$
For the fourth isotope: $0.524\times207.977=109.970948$

Step3: Calculate the average atomic mass

$2.855622 + 49.649734+45.741696 + 109.970948=208.217998\approx208.22$ amu

Step4: Identify the element

The element with an average atomic mass close to 208.22 amu is bismuth (Bi).

Answer:

208.22 amu
Bi