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Why is the standard atomic weight of chlorine, 35.5, not a whole number? Like for example, it could be exactly 35 or exactly 36. Please show the solution of formulae on how u reach to 35.5) with some English words so I can write the answer in my homework assignment and please make it small. Thanks

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marked as duplicate by Mithoron, Jon Custer, Tyberius, Todd Minehardt, A.K. Jul 13 '18 at 3:18

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The process is explained (for silicon) here:

The calculation is exemplified for silicon, whose relative atomic mass is especially important in metrology. Silicon exists in nature as a mixture of three isotopes: $\ce{^28Si}$, $\ce{^29Si}$ and $\ce{^30Si}$. The atomic masses of these nuclides are known to a precision of one part in 14 billion for $\ce{^28Si}$ and about one part in one billion for the others. However the range of natural abundance for the isotopes is such that the standard abundance can only be given to about ±0.001% (see table). The calculation is

$$A_r(\ce{Si}) = (27.97693 \times 0.922297) + (28.97649 \times 0.046832) + (29.97377 \times 0.030872) = 28.0854$$

For chlorine, the idea is the same, but you only have two isotopes, $\ce{^35Cl}$ (76%) and $\ce{^37Cl}$ (24%).

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