## How To Convert Atomic Percent To Weight Percent And Vice Versa

Wed, Aug 12, 2009

Tools 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.

### This post was written by:

- who has written 67 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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## 33 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

9. alex Says:

What if I have an alloy composite? Say I have an alloy made up of 1.5wt% Al2O3 and 98.5 wt% ZnO. How would I calculate the atomic percent of each element in this composite?

10. he he he Says:

simple way. chemist!!!!
any question about chemistry of NdFeB or SmCo ask uff or HEHEHE, true specialist
and very knowledgeable about these materials.

Nd2Fe14B

wt%
Nd 2*144.24= 288.48 =288.48/1081.121= 0.266834147 26.68341471
Fe 14*55.85= 781.83 =781.83/1081.121= 0.723166047 72.31660471
B 1*10.811= 10.811 =10.811/1081.121= 0.009999806 0.999980576
Sum:
1081.121

11. Murli Gopal Krishnamoorty Says:

It was possible to calculate a whole range of alloy compositions in a spreadsheet and plot ternary systems both in at.% and wt.%. Earlier I used to take the ratio of number of atoms using Avagadro Number and take percentage to be more sytematic

I cross checked with

http://www1.asminternational.org/asmenterprise/apd/WeightConverter.aspx

Another useful link for plotting a ternary system is

http://www.phasediagram.dk/Triangular.XLSX
Thanks and regards
dr_murli

12. Kashmira Tank Says:

I would like to ask one question, as how we can covert at% into wt% for the chemical formula such as Ca10(PO4)6(OH)2, as the example is given for Nd2Fe14B, But what about hydrogen, which can not detected in EDAX ?

13. ?????? Says:

I am very grateful!

14. meena Says:

I would like to ask one question,how to calculate atomic ratio for metal

15. shanthini Says:

Very Nice!!!!

16. Sang Says:

Good!

I have a query. I have a protein based substance which has some amount of calcium carbonate in it. I know the weight percent (3.5 %) and atomic percent (4.4 %) of calcium in it. How will I determine the total calcium carbonate present in my substance?

17. hana Says:

what if I wanted to prepare some nano material that should consist 99 wt.% ZnO and 1% Ag?how should I calculate how much starting material I needed?before this I used to determine the percentage by calculating the number of moles. but I’m not sure if it’s the right way.

18. Daisy Says:

Thanks!! its very usefull information to have on hand.

19. Lim SL Says:

Thank you, Gareth! A very well written article, and very useful!!

Cheers,
Lim SL

20. k.s Says:

This is the best answer to my problem. Thank you so much.

21. wow news Says:

this is a very useful website!

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23. Joe T Says:

Thank you for the post Gareth. Despite how simple it seems I always forget and have to look it up. Makes it easy to just have this page bookmarked.

24. sergio Says:

How can I solve a problem to above ternary alloys?

25. EG Says:

GREAT!!!! Really easy to understand!

26. Mohamed Basith Says:

Its very useful thank you.

27. vidhya Says:

useful

28. Richard Says:

How true that there is very little conversation out there. Much too much prior knowledge tends to be assumed. Anyone who is familiar with energy dispersive spectrometry (EDX or EDS) will know that results of the analysis are usually expressed as both atomic % and weight %. This post is really helpful to interpret these results and to make deductions about the stoichiometry of the samples.

Well done.

29. rajesh Says:

Such a simple explanation even a child can understand.

30. Nisha Says:

Thank you for such a simple solution

31. Mohamed Says:

how i can prepare 5% Al-Ti Alloy

32. metal a??rl?k hesapalam Says:

thank you

33. Prashanth Says:

This is quick solution. Thanks for the information.