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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

A Cool Guy
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    You might have heard of isotopes. There are two kinds of chlorine: one is 35, and another is 37. – Ivan Neretin Jul 11 '18 at 07:59
  • yes I've heard of it but other elements also have 2 or more than 2 isotopes. – A Cool Guy Jul 11 '18 at 08:06
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    Well, if one isotope is clearly dominant (like 99.9%), you will have an integer mass within two-digit precision. Not the case with Cl, though. – Ivan Neretin Jul 11 '18 at 08:09
  • The ratio for chlorine-35 to chlorine-37 is somewhere around 3:1, due to which this value of 35.5 arises. – Abhigyan Jul 11 '18 at 08:12
  • related https://chemistry.stackexchange.com/questions/99226/atomic-weight-expected-weight https://chemistry.stackexchange.com/questions/9016/the-formula-for-finding-the-percentage-of-isotopes-in-an-elements-atomic-weight https://chemistry.stackexchange.com/questions/38082/what-is-the-difference-between-molecular-mass-average-atomic-mass-and-mola – Mithoron Jul 11 '18 at 18:18
  • @Mithoron I would agree with any of those being a dupe. The previous one I wouldn't because it specifically recognizes in the question that isotopes are factored in, but asks why elements with significant isotopes still aren't integers. – Tyberius Jul 12 '18 at 19:48
  • Because it has various forms so-called as isotopes and also it is not required for the weight to be a whole number. Is anyone's weight a whole number? – PV. Feb 26 '21 at 06:39

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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%).

Glorfindel
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