solve the following isotope related question\na) the former president and vice president have just…

solve the following isotope related question\na) the former president and vice president have just simultaneously discovered new elements. they decided to have a contest to see which element had the greater atomic mass based on the abundance of their respective isotopes. who wins the contest?\nisotope atomic mass abundance\nbidenium - 36 36.013 amu 8.02%\nbidenium - 37 36.9 amu 13%\nbidenium - 40 39.87 amu 78.98%\nisotope atomic mass abundance\nharrisium - 39 38.8 amu 93.11%\nharrisium - 43 43.46 amu 6.89%

solve the following isotope related question\na) the former president and vice president have just simultaneously discovered new elements. they decided to have a contest to see which element had the greater atomic mass based on the abundance of their respective isotopes. who wins the contest?\nisotope atomic mass abundance\nbidenium - 36 36.013 amu 8.02%\nbidenium - 37 36.9 amu 13%\nbidenium - 40 39.87 amu 78.98%\nisotope atomic mass abundance\nharrisium - 39 38.8 amu 93.11%\nharrisium - 43 43.46 amu 6.89%

Answer

Explanation:

Step1: Calculate average atomic mass of Bidenium

The formula for average atomic mass $A = \sum_{i = 1}^{n}m_i\times p_i$, where $m_i$ is the atomic - mass of isotope $i$ and $p_i$ is the percentage abundance of isotope $i$ (in decimal form). For Bidenium: $p_1=0.0802$, $m_1 = 36.013$ amu; $p_2 = 0.13$, $m_2=36.9$ amu; $p_3 = 0.7898$, $m_3 = 39.87$ amu. $A_{Bidenium}=36.013\times0.0802+36.9\times0.13 + 39.87\times0.7898$ $A_{Bidenium}=36.013\times0.0802+36.9\times0.13+39.87\times0.7898$ $=2.8882426+4.797+31.404626$ $=39.09$ amu

Step2: Calculate average atomic mass of Harrisium

For Harrisium: $p_1 = 0.9311$, $m_1=38.8$ amu; $p_2 = 0.0689$, $m_2 = 43.46$ amu. $A_{Harrisium}=38.8\times0.9311+43.46\times0.0689$ $A_{Harrisium}=38.8\times0.9311 + 43.46\times0.0689$ $=36.12668+2.994494$ $=39.12$ amu

Answer:

Harrisium wins the contest.