Chlorine has two stable isotopes of mass 34.97 u and 36.97 u.Suppose the percent

Question

Chlorine has two stable isotopes of mass 34.97 u and 36.97 u.Suppose the percent abundance of the lighter isotope were 69.54%.What would be the atomic mass of Cl? (The actual abundance is75.77%.)


The mass spectrum of a hypothetical element shows that 69.03% ofthe atoms have a mass of 27.977 u, 2.059% have a mass of 28.976 u,and the remaining have a mass of 29.974 u. Calculate the averageatomic mass of this element (in units of u).

Bromine occurs naturally as two isotopes, one with a mass of 78.918u and the other with a mass of 80.916 u. Suppose the average atomicmass from the periodic table were 80.77 amu. What would be thepercent abundance of the lighter isotope?

Explanation / Answer

for part a: You would multiply molar mass of the lighter isotope by the percentabundance given: (34.97 u)(69.54%) = 24.32 amu Next, multiply the molar mass of the heavier isotope by the percentabundance left over (36.97 u)(100%-69.54%) = 11.26 amu Finally, add the 2 answers: 24.32 amu + 11.26 amu = 35.58 amu For part b follow the same steps: part B: (27.977 u * 69.03%) + (28.976 u * 2.059%) + (29.974 u *(100%-69.03%-2.059%)) = 28.57 amu for part C: Solve it algebraicly, in this case the percentages areunknown, let them be x and y respectively. We know (78.918 u * x%) + (80.916 u * y%) = 80.77 amu and also that (x% + y%) =100% x% = (100% - y%) so substitute for x in terms of y, and solve for y: (78.918 u * (100% - y%)) + (80.916 u * y%) = 80.77 amu 78.918 u - 78.918y% + 80.916y% = 80.77 amu 78.918 u + 10.998y% = 80.77 amu y% = 16.83% x% = 100% - 16.83% = 83.16% so the isotope with mass 78.918 u has a natural abundance of16.83% and the isotope with mass 80.916 u has a natural abundance of83.16% PLEASE RATE and COMMENT Hope this helps!

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