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:

- 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* - Add all the values of
together, and let’s call this value*p*.*p(Total)* - Now, for each value of
, divide it by*p*, to obtain*p(Total)*.*w* - Multiplying the resulting values of
by 100 gives us the weight percent values, for each respective element in the starting compound.*w*

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
= 2163.60,*p(Nd)*= 4300.07 and*p(Fe)*= 86.49.*p(B)* - Following Step 2,
has a value of 6550.15.*p(Total)* - Following Step 3, this results in values of
= 0.33,*w(Nd)*= 0.66 and*w(Fe)*= 0.01.*w(B)* - 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:

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

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*= 0.23 and*(Sm)**m*= 1.12.*(Co)* - Following Step 2,
has a value of 1.35.*m(Total)* - Following Step 3, this results in values of
= 0.17 and*a(Sm)*= 0.83.*a(Co)* - 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:

- 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* - Add all the values of
together, and let’s call this value*r*.*r(Total)* - Now, for each value of
, divide it by*r*, to obtain*r(Total)*.*w* - Multiplying the resulting values of
by 100 gives us the weight percent values, for each respective element in the starting compound.*w*

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*= 288.84,*(Nd)**r*= 781.83 and*(Fe)*= 10.81.*r(B)* - Following Step 2,
has a value of 1081.48.*r(Total)* - Following Step 3, this results in values of
= 0.27,*w(Nd)*= 0.72 and*w(Fe)*= 0.01.*w(B)* - 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.

September 2nd, 2009 at 11:30 am

This was the best answer to my problem.

Thanks,

M

November 12th, 2009 at 7:23 am

It is superb…

December 12th, 2009 at 5:25 pm

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

January 21st, 2010 at 9:34 pm

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

January 29th, 2010 at 6:30 pm

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

February 11th, 2010 at 11:16 pm

Excellent work, Thank you

May 7th, 2010 at 6:00 am

Its very useful thank you.

June 6th, 2010 at 12:28 am

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

October 13th, 2010 at 7:25 pm

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?

April 11th, 2011 at 10:46 am

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

July 7th, 2011 at 11:16 am

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

July 8th, 2011 at 7:00 am

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 ?

July 14th, 2011 at 8:52 am

I am very grateful!

August 26th, 2011 at 8:03 am

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

May 12th, 2012 at 1:30 pm

Very Nice!!!!

June 21st, 2012 at 1:35 am

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?

October 1st, 2012 at 4:16 am

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.

January 17th, 2013 at 12:51 pm

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

March 11th, 2013 at 11:53 pm

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

Cheers,

Lim SL

April 30th, 2013 at 4:29 am

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

June 11th, 2013 at 7:38 am

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November 27th, 2013 at 7:56 am

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June 14th, 2014 at 9:38 pm

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.

October 13th, 2014 at 8:11 pm

How can I solve a problem to above ternary alloys?

December 1st, 2014 at 3:51 am

GREAT!!!! Really easy to understand!