The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = [tex]\frac{14.4}{100}=0.144[/tex]
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = [tex]\frac{14.3}{100}=0.143[/tex]
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = [tex]\frac{71.3}{100}=0.713[/tex]
Formula used for average atomic mass of an element :
[tex]\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})[/tex]
[tex]A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)][/tex]
[tex]A=84.2amu[/tex]
Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu