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I had this question:

Naturally occurring europium consists of two isotopes with a mass of 151 and 153. Europium-151 has an abundance of 48.030 and, and europium-153 has a natural abundance of 51.970. What is the atomic mass of europium? Do not include units. Be sure to round to the correct number of significant figures.

So I do the math for the calculation:
$(151 \times 0.48030)+(153 \times 0.51970) = 152.0394$
I submit 152.0 because, as I understand it, the products of the math in the parentheses should be considered to have 3 sig figs each - ending in the tenths place. So when I add those two sums up, it should end in the tenths place too given addition/subtraction sig figs rules. This is marked as incorrect, and I worry I am misunderstanding something.

I emailed the professor, and he said it was a word question and I needed only to select the fewest sig figs of the ones provided (151 and 153 have 3), so my answer should have been 152 without the 0 in the tenths place. Is this usually the case? What confuses me is I don't think I was allowed to do this for other word problems, but I could be wrong. Is there a difference between word problems and other problems, and is 3 really the amount of sig figs I should have had?

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  • $\begingroup$ Teachers should not be torturing students with significant figure rules. They should rather teach common sense and critical reasoning in science rather than impose significant figure rules as heavenly rules. No wonder chemistry doesn't attract the brightest rather than repels them. Nobody does research with atomic mass measurements like this. I have Linus Pauling' General Chemistry Textbook (Two Nobel Prizes), and it does not even mention these these type of calculations. $\endgroup$
    – M. Farooq
    Feb 8 at 3:04
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    $\begingroup$ Anyway, use significant figures as "common sense". These are approximate rules and the idea is that do not use unnecessary numbers once you quote the answer. Rounding off should be done at end of all calculations. There is nothing such as sig fig rules for word problems etc. $\endgroup$
    – M. Farooq
    Feb 8 at 3:07
  • $\begingroup$ "Do not include units" in a question with all quantities being dimensionless as well as omitted percent sign at the natural abundance value are troubling signs. As for the sig figs, your professor pretty much explained it all right and I'm not sure what kind of answer you expect on top of that. The thing is, significant figures are supposed to be studied in a math class prior to chemistry and physics (because they belong to numerical analysis) so that the questions like "do sig figs depend on the type of question" don't even arise. $\endgroup$
    – andselisk
    Feb 8 at 5:55
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    $\begingroup$ IMHO, this case is a typical example of a task made up for a sake of the task, with rather weird up to ridiculous formulation. They should use rather real life example with meaningful numbers and approach. E.g. volumetry with uncertainties of mass, volume and concentration. $\endgroup$
    – Poutnik
    Feb 8 at 6:50
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    $\begingroup$ Putting the task ridiculousness aside, IMHO the expected procedure is this: isotope masses have 3 sig figs. Multiplied by numbers with more sig figs keeps these 3 sig figs. So does the summation. The result with 3 sig figs is 152. ( as said, ridiculousness aside ). Round up just the final result not to cummulate rounding errors. $\endgroup$
    – Poutnik
    Feb 8 at 9:29
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First, for reasons I'll point out later this is a poorly made-up problem which was only designed to test significant figures, not to teach anything real about chemistry.

Second there are no special rules for "word problems" using significant figures.

The gist is that significant figures are a simple method to do error propagation in calculations. It is essentially based on the notion that the error is +/- 1/2 of the last significant digit, as if the value had been rounded to that digit. So 151 means that the value should be between 150.5 and 151.5.

The whole point of significant figures is to prevent you from dividing 151 by 3.00 and getting some ridiculous precision like 50.333333333333333333333333333333.

Now on to the problem. 151 and 153 are mass numbers for the europium isotopes. Mass numbers are poor values to use for the isotope masses. If fact the known masses are 150.9198502(26) and 152.9212303(26). The numbers in parenthesis give the standard deviation of the measurement. So for 150.9198502 the standard deviation is 0.0000026.

Likewise according to the Wikipedia page, $\ce{^{151}Eu}$ has an abundance of 0.4781(6) and $\ce{^{153}Eu}$ has an abundance of 0.5219(6). The point is that the isotopic variation in samples taken from all over the earth limits the precision of the atomic mass for any element, not the mass of the isotopes which are know with much greater precision.

The same Wikipedia page linked earlier gives the atomic mass of europium as 151.964(1) which is inconsistent with the given percentages. I do not know the how/why there are so many more significant figures in the atomic mass (and I am too lazy to turn this answer into a research project to figure that part out...).

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    $\begingroup$ ....limits the precision of the atomic mass for any element.... While elements with odd atomic number have maximally 2 stable isotopes, some of them have just 1, like 19F or 23Na. In these cases, atomic mass is known with precision of isotopic mass. $\endgroup$
    – Poutnik
    Mar 13 at 8:21

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