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How should the uncertainty of the atomic masses of elements be quoted?

E.g., Mg has an atomic mass of 24.31, what is that value's uncertainty?

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closed as off-topic by Todd Minehardt, airhuff, getafix, bon, Pritt Balagopal Jul 27 '17 at 8:31

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  • $\begingroup$ In general, if you have a measured value of upto 1 decimal place, then the uncertainty in its measurement is +- 0.1 (plus minus 0.1). Similarly for two decimal places, its +- 0.01 and so on. This is easy to remember, however @Paul has given you a mathematical proof as well. $\endgroup$ – Saksham Jul 27 '17 at 15:30
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First of all it is important to realize that the number you quote is not the atomic mass of magnesium, but the avarage (molar) mass of all naturally occuring isotopes. If you visit the wikipedia page on the isotopes of magnesium you will see that the naturally occuring isotopes have the following masses and abundances

isotope       mass (amu)          abundance
Mg-24         23.985041700(14)    0.7899(4)
Mg-25         24.98583692(3)      0.1000(1)
Mg-26         25.982592929(30)    0.1101(3)

As you can see the last numbers of the mass and abundance are within parenthesis. This is a very common way to display (1$\sigma$) uncertainties and the numbers within the parenthesis represent the uncertainty in the last digit, e.g. 0.7899(4) can also be written as 0.7899$\pm$0.0004 and 23.985041700(14) as 23.985041700(14)$\pm$0.000000014. You'll probably see why people prefer the former notation as it is so much shorter than the latter.

To calculate the average mass ($\bar{m}$) you take the weighted sum over the different isotopic masses ($m_i$) using the abundances ($h_i$)

$$ \bar{m}=\sum_i h_im_i=h_{24}m_{24}+h_{25}m_{25}+h_{26}m_{26}=24.30505 $$

To calculate the uncertainty in the average mass you should apply error propagation to find

$$ \Delta \bar{m}=\sqrt{\left (\frac{\partial\bar{m}}{\partial h_{24}} \right )^2\Delta h_{24}^2 + \left (\frac{\partial\bar{m}}{\partial h_{25}} \right )^2\Delta h_{25}^2 + \left (\frac{\partial\bar{m}}{\partial h_{26}} \right )^2\Delta h_{26}^2 + \left (\frac{\partial\bar{m}}{\partial m_{24}} \right )^2\Delta m_{24}^2 + \left (\frac{\partial\bar{m}}{\partial m_{25}} \right )^2\Delta m_{25}^2 + \left (\frac{\partial\bar{m}}{\partial m_{26}} \right )^2\Delta m_{26}^2}\\ =\sqrt{m_{24}^2\Delta h_{24}^2 + m_{25}^2\Delta h_{25}^2 + m_{26}^2\Delta h_{26}^2 + h_{24}^2\Delta m_{24}^2 + h_{25}^2\Delta m_{25}^2 + h_{26}^2\Delta m_{26}^2}=0.013 $$

The average mass of magnesium is thus 24.305(13) amu. Note that the uncertainty in the final answer is dominated by the uncertainty in the abundances.

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