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There is a discrepancy among different sources on the molecular weight of bilirubin, $\ce{C33H36N4O6}$:

What is the correct value and why?

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    $\begingroup$ This may be above my pay grade, but try 33*M_C + 36*M_H + 4*M_N + 6*M_O ;) $\endgroup$ – jezzo Jul 29 '20 at 22:04
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    $\begingroup$ And that's supposed to be a different value? 0.01 difference? They could be both right! Isotopic content isn't constant. $\endgroup$ – Mithoron Jul 29 '20 at 22:08
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    $\begingroup$ I would get 584.66 with this formula. lenntech.com/calculators/molecular/… But wiki and fisher scientific both say 584.673 $\endgroup$ – stem-econ Jul 29 '20 at 22:09
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    $\begingroup$ I understand that this difference of 0.013 is trivial. I just would like to know why some would get 584.673, and some would get 584.66. How come nobody gets 584.669 or 584.671, etc.? $\endgroup$ – stem-econ Jul 29 '20 at 22:11
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    $\begingroup$ Huh, I guess it's valid question,but you should elaborate more in this vein in the body of question. $\endgroup$ – Mithoron Jul 29 '20 at 22:23
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The discrepancy is simply due to round-off error. The atomic masses of these elements are usually given as follows. These are known as the standard conventional form, or standard formal short form, since they are given without an interval indicating the uncertainty in their values (see https://en.wikipedia.org/wiki/Standard_atomic_weight):

C = 12.011

H = 1.008

N = 14.007

O = 15.999

Hence 33 C + 36 H + 4 N + 6 O = 584.673

However, if someone were to round off all the masses to four significant figures, but then report the result to five significant figures, we would have:

C = 12.01

H = 1.008

N = 14.01

O = 16.00

And now 33 C + 36 H + 4 N + 6 O = 584.658 = 584.66

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    $\begingroup$ Which shows that it makes sense to leave a "guard digit" on the numbers in the calculation, and wait to round until you have the answer (This would give the molar mass as 584.67 g/mol at two decimal points). $\endgroup$ – Karsten Theis Jul 30 '20 at 18:06
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    $\begingroup$ Yes, it's best practice to leave the rounding to the end (including not rounding the results of intermediate calculations). Though note that, for some calculations, a single guard digit isn't always sufficient. $\endgroup$ – theorist Jul 30 '20 at 22:40
  • $\begingroup$ Here's a toy example: Suppose your data is {1.214, 1.224, 1.214}, and you want to express the sum of these numbers to the nearest tenth. Then, with one guard digit, you would have: 1.21 + 1.22 + 1.21 = 3.64 = 3.6. Without any rounding, you would have: 1.214 + 1.224 + 1.214 = 3.652 = 3.7. $\endgroup$ – theorist Jul 31 '20 at 2:58

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