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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7$\begingroup$ You might have heard of isotopes. There are two kinds of chlorine: one is 35, and another is 37. $\endgroup$– Ivan NeretinJul 11, 2018 at 7:59
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$\begingroup$ yes I've heard of it but other elements also have 2 or more than 2 isotopes. $\endgroup$– A Cool GuyJul 11, 2018 at 8:06
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4$\begingroup$ 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. $\endgroup$– Ivan NeretinJul 11, 2018 at 8:09
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$\begingroup$ The ratio for chlorine-35 to chlorine-37 is somewhere around 3:1, due to which this value of 35.5 arises. $\endgroup$– AbhigyanJul 11, 2018 at 8:12
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$\begingroup$ related chemistry.stackexchange.com/questions/99226/… chemistry.stackexchange.com/questions/9016/… chemistry.stackexchange.com/questions/38082/… $\endgroup$– MithoronJul 11, 2018 at 18:18
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1 Answer
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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%).
answered Jul 11, 2018 at 9:11