What information is necessary to determine the atomic mass of the element chlorine?
(1) the atomic mass of each artificially produced isotope of chlorine, only
(2) the relative abundance of each naturally occurring isotope of chlorine, only
(3) the atomic mass and the relative abundance of each naturally occurring isotope of chlorine
(4) the atomic mass and the relative abundance of each naturally occurring and artificially produced isotope of chlorine

Answer is: (3) the atomic mass and the relative abundance of each naturally occurring isotope of chlorine.

m(³⁵Cl) = 34.97 amu; the average atomic mass of chlorine-35.  

m(³⁷Cl) = 36.97 amu; the average atomic massof chlorine-37.

ω(³⁵Cl) = 75.76% ÷ 100% = 0.7576; fractional abudance of chlorine-35.

ω(³⁷Cl) = 24.24% ÷ 100% = 0.2424; fractional abudance of chlorine-37.

m(Cl) = m(³⁵Cl) · ω(³⁵Cl) + m(³⁷Cl) · ω(³⁷Cl).  

m(Cl) = 34.97 amu · 0.7576 + 36.97 amu · 0.2424.  

m(Cl) = 35.4548 amu; average atomic mass of chlorine.

the answer is true.

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