How To Convert Atomic Percent To Weight Percent And Vice Versa

Wed, Aug 12, 2009

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How To Convert Atomic Percent To Weight Percent And Vice Versa

In the course of reviewing information on metals and minerals, I often come across chemical composition information that is written in terms of atomic percent, when I am actually more interested in the weight percent values of the elements involved.  A little less frequently I want to do things the other way around, and do a conversion from weight percent to atomic percent.

After searching online, I’ve noticed that what little conversion information is out there, is unnecessarily complicated.  So, I thought I’d share the simple but trusty formulae that I have had pinned on one wall or another for the past couple of decades…

A) Converting from atomic percent to weight percent:

  1. For each element listed in the compound, multiply the atomic percent of the element by its atomic weight [the larger of the two principal numbers listed for each element in the standard periodic table].  For each element, let’s call this value p.
  2. Add all the values of p together, and let’s call this value p(Total).
  3. Now, for each value of p, divide it by p(Total), to obtain w.
  4. Multiplying the resulting values of w by 100 gives us the weight percent values, for each respective element in the starting compound.

Example: we encounter a neodymium-based permanent magnet material whose composition is listed, in atomic percent terms, as being 15% Nd, 77% Fe and 8% B.

  • Following Step 1 above, we first obtain the atomic weights for each element. To two significant figures, these are: Nd – 144.24, Fe – 55.85 and B – 10.81.
  • Completing Step 1 results in values of p(Nd) = 2163.60, p(Fe) = 4300.07 and p(B) = 86.49.
  • Following Step 2, p(Total) has a value of 6550.15.
  • Following Step 3, this results in values of w(Nd) = 0.33, w(Fe) = 0.66 and w(B) = 0.01.
  • Following Step 4 results in the final values, in weight percent terms, of 33% Nd, 66% Fe and 1% B.

B) Converting from weight percent to atomic percent:

  1. For each element listed in the compound, divide the weight percent of the element by its atomic weight.  For each element, let’s call this value m.
  2. Add all the values of m together, and let’s call this value m(Total).
  3. Now, for each value of m, divide it by m(Total), to obtain a.
  4. Multiplying the resulting values of a by 100 gives us the atomic percent values, for each respective element in the starting compound.

Example: we encounter a samarium-based permanent magnet material whose composition is listed, in weight percent terms, as being 34% Sm and 66% Co.

  • Following Step 1 above, we first obtain the atomic weights for each element. To two significant figures, these are: Sm – 150.35 and Co – 58.99.
  • Completing Step 1 results in values of m(Sm) = 0.23 and m(Co) = 1.12.
  • Following Step 2, m(Total) has a value of 1.35.
  • Following Step 3, this results in values of a(Sm) = 0.17 and a(Co) = 0.83.
  • Following Step 4 results in the final values, in atomic percent terms, of 17% Sm and 83% Co.

There is one other scenario that we sometimes encounter, related to A) above, but which involves the chemical formula for a particular metallurgical phase:

C) Converting from chemical formula to weight percent:

  1. For each element listed in the compound, multiply the number of atoms of the element by its atomic weight.  For each element, let’s call this value r.
  2. Add all the values of r together, and let’s call this value r(Total).
  3. Now, for each value of r, divide it by r(Total), to obtain w.
  4. Multiplying the resulting values of w by 100 gives us the weight percent values, for each respective element in the starting compound.

Example: we look to evaluate the main hard magnetic phase in neodymium-based permanent magnet material, whose chemical formula consists of 2 atoms of Nd, 14 atoms of Fe and 1 atom of B [i.e. the so-called 2-14-1 stoichiometric composition].

  • Following Step 1 above, we first obtain the atomic weights for each element. To two significant figures, these are: Nd – 144.24, Fe – 55.85 and B – 10.81.
  • Completing Step 1 results in values of r(Nd) = 288.84, r(Fe) = 781.83 and r(B) = 10.81.
  • Following Step 2, r(Total) has a value of 1081.48.
  • Following Step 3, this results in values of w(Nd) = 0.27, w(Fe) = 0.72 and w(B) = 0.01.
  • Following Step 4 results in the final values, in weight percent terms, of 27% Nd, 72% Fe and 1% B.

Increasing the number of significant figures in the various values will increase the accuracy of the calculations, but you’ll probably find that you don’t need to get too much more detailed than I did, in the examples above.

I hope that this is of some use to you; feel free to comment or suggest other topics for discussion or review.

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This post was written by:

Gareth Hatch - who has written 64 posts on Terra Magnetica.

Gareth is a Founding Principal at Technology Metals Research, LLC. He has expertise in a variety of magnetic materials, devices and applications, and their associated trends and challenges, particularly for renewable energy production. For more information check out his biography page. Don't forget to check out Terra Magnetica at Twitter too.

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13 Responses to “How To Convert Atomic Percent To Weight Percent And Vice Versa”

  1. MB Says:

    This was the best answer to my problem.

    Thanks,
    M

  2. JKR Says:

    It is superb…

  3. Aww Says:

    This was very useful! Thank You very much!! :)

  4. maria Says:

    This is the best solution to my homework problems that I am facing in my civil and materials engineering class. I have been looking online and through chemistry books for last three hours. thanks so much

  5. alex Says:

    This is the best and easiest solution EVER presented to these problems.

  6. Pierre Says:

    Excellent work, Thank you

  7. sen Says:

    Its very useful thank you.

  8. hector Says:

    muchas gracias esto es algo que me ba a ayudar mas en mi trabajo por que yo soy metal sorter por mas de 25anos , conosco todo tipo de metales pero me especializo en refractory metals. GRACIAS DE NUEVO . I was looking for this information for many years,so thanyou very much sincerely


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