An element has an isotope with a mass of 203.973 amu and and abundance of 1.40%. another isotope has a mass of 205.9745 amu with an abundance of 24.10%. a third isotope weighs 206.9745 amu and is 22.10%, and the fourth isotope weighs 207.9766 amu and an abundance of 57.40. calculate the average atomic mass and identify the unknown element.

Answer is: the average atomic mass 217.606 amu.

Ar₁= 203.973 amu; the average atomic mass of isotope.

Ar₂ = 205.9745 amu.

Ar₃ = 206.9745 amu.

Ar₄ = 207.9766 amu.

ω₁ = 1.40% = 0.014; mass percentage of isotope.

ω₂ = 24.10% = 0.241.

ω₃ = 22.10% = 0.221.

ω₄ = 57.40% = 0.574.

Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.  

Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.

Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.

Ar = 217.606 amu.

But abundance of isotopes is greater than 100%.

It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).