Prior to 1961, the atomic mass unit was defined as 1/16 the mass of the atomic mass of oxygen; that is, the atomic mass of oxygen was defined as exactly $\pu{16 amu}$. What was the mass of a $\ce{^{12}C}$ atom prior to 1961 if the atomic mass of oxygen on today’s scale is $\pu{15.9994 amu}$?

I don't fully understand the logic behind the question. Here's my thinking. Atomic mass of all substances were related to $\pu{16.0000 amu}$ prior to 1961. Therefore, we can make the following expression where $X$ is mass of $\ce{^{12}C}$ $$\dfrac{X}{\pu{16.0000amu}}$$

Today, all atomic mass is related to $\pu{12.0000 amu}$ of $\ce{^{12}C}$. We can model the relationship as $$\dfrac{\pu{12.0000amu}}{\pu{15.9994amu}}$$

What the solution manual does is make the two equations equal to each other and solves for $X$.


How are we justified in doing this?


2 Answers 2


Here is another equivalent way of thinking about the problem. The mass of oxygen on the old scale is 16 amu. The mass of oxygen on the new scale is 15.994 amu. When you divide them:

$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} = 1.0000375 \frac{\textrm{old}}{\textrm{new}}$$

Basically, taking a ratio of any value in the old scale to the equivalent value in the new scale (or vice versa) gives you a conversion factor. So for every new amu, there is 1.0000375 old amus. The question is asking you to convert 12 new amus to old amus. Just by looking at the units, it should be obvious what to do:

$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} * 12\ \textrm{new}=12.00045\ \textrm{old}$$

If you explicitly label old and new, it's easy to see that the new units cancel out.

Incidentally, if you keep track of your significant figures, you can see why this redefinition of the amu didn't cause too many problems for chemists—even to five sig figs, the change is negligible.

  • 2
    $\begingroup$ In the old system the atomic mass unit was abbreviated as $\mathrm{amu}$. This system is now obsolete. To avoid conflict, the new system, the unified atomic mass unit uses the symbol $\mathrm{u}$. $\endgroup$ Sep 4, 2014 at 7:27
  • 1
    $\begingroup$ Yeah I know, but I've seen amu used as the unified atomic mass unit in general chemistry textbooks a lot so I used new and old for this example to make it clearer. Personally, I prefer Da to u, if for no other reason than "daltons" is easier to say than "unified atomic mass units". $\endgroup$ Sep 4, 2014 at 16:20

Think about what happened to the scale. For $\ce{^{16}O}$, it decreased from $16.0000\,\mathrm{amu}$ to $15.9994\,\mathrm{amu}$. So the scale was stretched out (since 0 needs to stay constant). That would mean that $\ce{^{12}C}$ would need to decrease from some number greater than $12.0000\,\mathrm{amu}$. With $0$ being constant, there is no addition factor (like when converting from Fahrenheit to Centigrade); it is just a ratio from one scale to the other. $$\frac {\text{unknown}}{16.0000\,\mathrm{amu}}=\frac {12.0000\,\mathrm{amu}}{15.9994\,\mathrm{amu}}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.