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When we look at atomic mass, for example, $\ce{^134Ba}$ (barium) has a mass of $133.90 \,\mathrm{Da}$. Why is this rounded to $134 \,\mathrm{Da}$? This is shown in my book.

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    $\begingroup$ More to the point, that 134 is the sum of the number of protons plus neutrons in the nucleus, so it must be an integer. $\endgroup$ – Jon Custer Sep 14 '16 at 13:52
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The way my teacher first explained this is due to the definition of atomic mass. According to Google:

The mass of an atom of a chemical element expressed in atomic mass units. It is approximately equivalent to the number of protons and neutrons in the atom (the mass number) or to the average number allowing for the relative abundances of different isotopes.

Now, the key words are

...the average number allowing for the relative abundances of different isotopes.

What does that mean? We go back to the definition of an isotope. An isotope of an element is a species that has the same number of protons in the nucleus, yet a different number of neutrons. For example, $\ce{C^12}$ and $\ce{C^14}$ are isotopes of each other. Sidenote, isotopes are not to be confused with allotropes, which are different forms of the same element. For example, graphite and diamond are allotropes of carbon, since both, either, or neither might be made up of $\ce{C^12}$ ($6p$, $6n$) or $\ce{C^14}$ ($6p$, $8n$). Now how do we represent this with the periodic table? Simply, we take the weighted average. Now, I do not have enough information or data to calculate the relative abundances of each isotope of carbon, and the periodic table does not either. We have an average. The atomic mass of $\ce{C}$ is given as $12.011$. That means a few different things. Firstly, there are more isotopes that are closer to having a sum of $p+n=12$, and less that have $p+n=14$ or $15$. Secondly, this means that although this number is not exact, it represents the most common $p+n$. Averaging all of those different $p+n$'s up, from each isotope, multiplying they by their relative abundances, and averaging should give our periodic table value of $12.011$.

Now it is important to remember that this number is not rounded. You might be asking, "Then why, in our calculations, do we sometimes round this number?" Well it depends on the problem, because for many problems, you do not round, but for the ones that you do, it is because it is more convenient to assume that the number that is given rounded is the most common isotope. Was that sentence confusing? Essentially, it is easy and generally correct to say that $12.011$ for carbon can be rounded to $12$, and $12$, meaning $\{p,n\}\in\{6,6\}$. We can do this by assuming that carbon in its most common form has $6p$, and $6n$.

In orthocresol's comment above, he is also correct. He has a perfectly valid point regarding common word usage in speaking. However, I do not think that this is the main reason.

Regarding Jon Custer's comment above, yes, that would be true, but only if we had a single isotope, which we do not.

Now, make sure you can differentiate between atomic mass and atomic weight! :)

Hope this helps!

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When a specific isotope like $\ce{^134Ba}$ is given, then there is no rounding or averaging necessary. The atomic mass is simply taken as the mass number, i.e. the the sum of the protons and neutrons in the nucleus. Thus the atomic mass of $\ce{^134Ba}$ is simply $\pu{134u}$.

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