Homework Answers
Q. Atomic mass of X = 485.682 amu
Q12. most molecules : 1 g of CH4
Q. empirical formula : P2O5
Explanation
Q. Given : abundance of isotope 1 = 82.482%
abundance of isotope 2 = 100% - abundance of isotope 1
abundance of isotope 2 = 100% - 82.482%
abundance of isotope 2 = 17.518%
atomic mass of X = [(abundance of isotope 1) * (atomic mass of isotope 1) + (abundance of isotope 1) * (atomic mass of isotope 1)] / 100
atomic mass of X = [(82.482) * (486.621 amu) + (17.518) * (481.263 amu)] / 100
atomic mass of X = (40137.473 amu + 8430.765 amu) / 100
atomic mass of X = (48568.238 amu) / 100
atomic mass of X = 485.682 amu
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