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.