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Topic: Calculating mass of an isotope in AMU  (Read 374 times)

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Offline calejone

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Calculating mass of an isotope in AMU
« on: February 12, 2020, 10:52:12 AM »
I took an exam today and I got stuck on this question that was "What is the mass (in AMU) of 5 atoms of tungsten-188. I ended up with 939.9874794 amu. There were four answers to chose from (multiple choice): 940. amu, 919 amu, and 1.53e-21 amu and I forget the last one exactly (I think it was 187 amu or 1.87e2 amu, which does seem to be the molecular weight g/mol now that I am looking at W-188 online, but didn't know this during the exam).

I initially tried 5 atoms W-188 * 1 mol W-188/6.022e-23 atoms W-188 * 188 grams/1 mol W-188 = 1.56 e-21 grams

I noticed 1.53e-21 was one of the answers but it was off a bit from 1.56 e-21. Then I realized that 1.56e-21 was in grams and the question asked for it in AMU. So I divided this by 1.6606e-24 and got 939.9874794 amu.

So in the end I ended up selecting 940. as my answer. But I am pretty sure that the "." at the end of 940 means that it is precisely 940 amu and doesn't leave room for any uncertainty.

Anyone have any idea what the correct answer would be and if it is 939.9874794 would the selection of 940. be acceptable? Obviously in about a week I will get the exam back and know what the right answer was. But I am more interested in knowing if I know how to do this correctly or if I was way off. Thanks for any input

Offline mjc123

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Re: Calculating mass of an isotope in AMU
« Reply #1 on: February 12, 2020, 11:16:36 AM »
You over-thought it. One atom of 188W has a mass of 188 amu. 5 atoms have a mass of 5*188 = 940 amu.
Actually, to be more precise, the atomic weight of 188W is 187.96 (to 2 dp), but you didn't use this, so it's not the source of your discrepancy, which must be in the rounding of NA and AMU. You were obviously expected to use 188 as the atomic weight.

Offline MNIO

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Re: Calculating mass of an isotope in AMU
« Reply #2 on: February 12, 2020, 04:45:00 PM »
several things to cover here.  Let's start by taking a look at this image.

The first line is your work.  The next two are mine.  The color coding shows what's canceling.  I think it's outstanding that you're using factor label method (aka dimensional analysis) and I encourage you to continue doing so.  Your problem setup is excellent except for 1 thing.  What you converted "to".  The idea is you start with what you're converting from on the left and convert to the right until you have the units you are after.  You want AMU, you converted to "grams" and stopped there.  Why?.  Check out the second line.  I added another unit factor and finished the conversion to AMU's.

Now look at the third line.  We already had atoms.  We know the mass of 1 W-188 atom is about 188AMU so we don't need to convert to moles.  Can you see the Atoms W / atoms W cancel leaving AMU's?

next, let's cover sig figs.  Your possible answers were
  (1) 940. AMU (with 3 sig figs)
  (2) 919 AMU (with 3 sig figs)
  (3) 1.53x10^-21 AMU (with 3 sig figs)
  (4) 187 AMU (also with 3 sig figs)

so you really only need 3 sig figs in your calculation.  Actually, I encourage all my students to carry 1 extra sig fig along to avoid rounding errors.  Anyway, your calcs were
  5 (exact = infinite sig figs) / 6.022e23 (4 sig figs) * 188 (3 sig figs) --> 1.56x10^-21 g (3 sig figs)

now my calcs (2nd equation)
  5 (exact) / 6.022e23 (4) * 188 (3) / 1.66e-24 (3) ---> 940. with 3 sig figs

now the last equation
  5 (exact) * 188 (3) ---> 940. (3 sig figs)

you can use more sig figs if you wish, but you really don't need more that 4 in your final answer.  It's not worth looking up the isotopic mass of W-188 and doing this calc
  5 (exact) * 187.958489 (9 sig figs)= 939.792445 AMU (9 sig figs)
when you're comparing it to choices with 3 sig figs... right?

you should also know that 187.958489(4) doesn't mean 187.9584894.  Those masses are measured.  The (4) is the standard deviation of the measurments.  187.958489(4) means 187.958489 is the average measurement and 1 standard deviation of measurements is within 4 units of the rightmost digit. i.e 67% of the measurements fell within 187.958485 to 187.958493 AMU

IF the choices were close together, then I would recommend using 1 extra sig fig and generating this answer

with 4 sig figs.  And now we compare 940.0 to the choices. But again, not worth it in this case

one last thing I want to cover because I'm not sure you're crystal clear on this.
  The molar mass of W-188 atoms is 187.958489 grams / mole of atoms
  The atomic mass of W-188 atoms is 187.958489 AMU / 1 atom.
note the #'s are the same
note the units are different... g / 6.022x10^23 atoms... vs... AMU / 1 atom

why are the #'s the same while the units are so different?  it's because of how moles of atoms (6.022x10^23 atoms) and AMU's are defined.  there are 6.022x10^23 AMU's in 1 gram!.  See if you can setup a conversion and from AMU / atom to g / mole of atoms. 

If all else fails, remember this.  you can read those masses of atoms and molecules as g / mole and as AMU / atom or AMU / molecule.

  molar mass of Cu = 63.55 g / mole
  atomic mass of Cu = 63.55 AMU / atom
  molecular mass of Cu     doesn't apply here.

  molar mass of W = 183.4 g / mole of W atoms
  atomic mass of W = 183.4 AMU / atom
  molecular mass of W     doesn't apply here
  isotopic mass of W-188 = 187.958489 AMU/atom
  molar mass of W-188 = 187.958489 g/mol

  molar mass = 16.04 g/mol...  16.04g / 6.022x10^23 molecules
  atomic mass   doesn't apply... methane is a molecular not atomic substance
  molecular mass = 16.04 AMU / molecule

make sure you're using molar mass, atomic mass, molecular mass, isotopic mass, relative atomic mass, etc correctly and that you are rock solid on the meaning of those terms!.

good luck!

Offline calejone

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Re: Calculating mass of an isotope in AMU
« Reply #3 on: February 13, 2020, 11:43:26 AM »
Wow thanks for your input. That helped a lot!

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