How To Convert Atomic Percent To Weight Percent And Vice VersaWed, 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. 3. 4. Add all the values of p together, and let’s call this value p(Total). Now, for each value of p, divide it by p(Total), to obtain w. 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. • • • • 0.01. Following Step 1 above, we first obtain the atomic weights for each element. To two Completing Step 1 results in values of p(Nd) = 2163.60, p(Fe) = 4300.07 and p(B) = 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) = Following Step 4 results in the final values, in weight percent terms, of 33% Nd, 66% significant figures, these are: Nd – 144.24, Fe – 55.85 and B – 10.81. 86.49. • Fe and 1% B. B) Converting from weight percent to atomic percent: 1. 2. 3. 4. 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 m together, and let’s call this value m(Total). Now, for each value of m, divide it by m(Total), to obtain a. Multiplying the resulting values of a by 100 gives us the atomic percent values, for each respective element in the starting compound. 1 we first obtain the atomic weights for each element. Now. r(Fe) = 781. Following Step 3. as being 34% Sm and 66% Co. Fe – 55. these are: Nd – 144. Following Step 2.Example: we encounter a samarium-based permanent magnet material whose composition is listed.35 and Co – 58. related to A) above. these are: Sm – 150. in weight percent terms. • • • • 0. and let’s call this value r(Total). 2 .17 and a(Co) = 0. 72% significant figures. For each element. • Fe and 1% B.99. for each respective element in the starting compound. in atomic percent terms. Add all the values of r together. To two Completing Step 1 results in values of r(Nd) = 288.85 and B – 10. let’s call this value r. Example: we look to evaluate the main hard magnetic phase in neodymium-based permanent magnet material.e.81.81.23 and m(Co) = 1. Following Step 4 results in the final values. divide it by r(Total).72 and w(B) = Following Step 4 results in the final values. Following Step 1 above. m(Total) has a value of 1.48. whose chemical formula consists of 2 atoms of Nd. 2. of 27% Nd. to obtain w.83 and r(B) = Following Step 2.84. Following Step 3.35. in the examples above. 14 atoms of Fe and 1 atom of B [i. of 17% Sm and significant figures. but you’ll probably find that you don’t need to get too much more detailed than I did. For each element listed in the compound. the so-called 2-14-1 stoichiometric composition]. Increasing the number of significant figures in the various values will increase the accuracy of the calculations. multiply the number of atoms of the element by its atomic weight. 3. this results in values of w(Nd) = 0.12.01. feel free to comment or suggest other topics for discussion or review. r(Total) has a value of 1081.24. w(Fe) = 0. Multiplying the resulting values of w by 100 gives us the weight percent values. • • • • • Following Step 1 above. To two Completing Step 1 results in values of m(Sm) = 0. 83% Co. in weight percent terms.83. 10. this results in values of a(Sm) = 0. we first obtain the atomic weights for each element.27. I hope that this is of some use to you. but which involves the chemical formula for a particular metallurgical phase: C) Converting from chemical formula to weight percent: 1. 4. for each value of r. There is one other scenario that we sometimes encounter. 022×1023 water molecules) of approximately 18 grams (the sum of 2 hydrogen atoms. but. Weight is a measurement of gravitational force exerted on matter. Atomic mass is sometimes erroneously confused with atomic weight—the obsolete term for relative atomic mass. The correct values for those weights are closer to 16 and 14.e. The one problem with Dalton's suggestion was that chemists had to know the formulas of chemical compounds before they could determine the weights of atoms. are still listed on many Periodic tables. Mass is an intrinsic property of matter. each with an atomic mass of 1.5 for the atomic weight of monatomic oxygen and 4. a specific isotope of an element) is measured in comparison with the mass of one atom of carbon-12 (12C) that is assigned a mass of 12 atomic mass units (amu). NH for ammonia (actually NH3).022×1023—Avogadro's number—atoms or molecules) weighs its total unit atomic mass (formerly termed atomic weight) in grams. for example usually list the atomic weights of individual elements based upon the natural distribution of isotopes of that element. the natural percentage distribution of isotopes of that element is assumed. 6. Periodic tables. water(H2O) has a molar mass (the mass of 6. Berzelius could use this information to calculate correct atomic weights. however. because his formulas were often wrong. The first reasonably accurate table of atomic weights was produced by Swedish chemist Jöns Jacob Berzelius (1779–1848) in 1814. this assumption allowed him to develop the concept of atomic weights. Dalton himself had assumed that compounds always had the simplest possible formula: HO for water (actually H2O). This table had been preceded by nearly a decade of work on the chemical composition of compounds. his work inevitably resulted in incorrect values for most of the atomic weights.The atomic mass of an atom (i. A mole of any element or compound (i. Although incorrect. In a series of papers published between 1803 and 1805 English physicist and chemist John Dalton (1766–1844) emphasized the importance of knowing the weights of atoms and outlined an experimental method for determining those weights. bonded with one oxygenatom with an atomic mass of 15. he reported 5. For example.0079 amu. For example.. Berzelius was faced with a decision that confronted anyone who tried to construct a table of atomic weights: What element should form the basis of that table and what would be the atomic weight of that standard element? 3 .9994 amu).2 for monatomic nitrogen.e. But they had no way of knowing chemical formulas without a dependable table of atomic weights. Once those compositions had been determined. In this process.. Atomic weights. In general usage if a specific isotope or isotope distribution is specified when using atomic mass. and so on. Richards spent more than 30 years improving methods for the calculation of atomic weights and redetermining those weights. The numbers that we refer to as atomic weights are all ratios. far too small to use in any table. Between 1893 and 1903. or one-half as heavy as another atom whose weight was selected to be 32. Stas recalculated Berzelius' weights. of course. but used other values for its weight. The problem with Dalton's choice was that atomic weights are determined by measuring the way elements combine with each other. With the introduction of the concept of molecules (e. To say that the atomic weight of oxygen is 16. using Dalton's system. For over a decade. The controversy over standards was complicated by the fact that. and so on. Because hydrogen is the lightest element. Dalton had made the logical conclusion to use hydrogen as the standard for his first atomic table and had assigned a value of 1 for its atomic weight. he produced a table with values very close to those accepted today. is only to say that a single oxygen atom is 16 times as heavy as some other atom whose weight is somehow chosen as 1. determining the atomic weight of another element might require a two-or three-step process. recognized 4 . various chemical societies finally agreed on the latter standard. in 1828. and hydrogen combines with relatively few elements. it also became possible to calculate molecular weights.g. Berzelius thought it made more sense to choose oxygen as the standard for an atomic weight table.. that the correct formula for water was H2O) by Stanislao Cannizarro in 1858. producing results that were unchallenged for nearly half a century. over time. this decision assures that all atomic weights will be greater than one. Oxygen forms compounds with most other elements whose atomic weights can. He arbitrarily assigned a value of 100 as the atomic weight of oxygen. Richards was awarded the Nobel Prize in chemistry in 1914 for these efforts. Berzelius continued working on atomic weights until.The actual weights of atoms are. The debate as to which element was to be used as the standard for atomic weights extended into the twentieth century. for example. be determined in a single step. with the most popular positions being hydrogen with a weight of 1 or oxygen with a weight of 16. thus. An even higher level of precision was reached in the work of the American chemist Theodore William Richards (1868–1918). therefore. Other chemists agreed that oxygen should be the atomic weight standard. physicists and chemists began to use different standards for the atomic weight table and. or eight times as heavy as another atom whose weight has been chosen as 2. The most precise work on atomic weights during the nineteenth century was that of the Belgian chemist Jean Servais Stas (1813–1891). The molecular weight of any compound is equal to the sum of the weights of all the atoms in a molecule of that compound. So. 0000 as the standard for all atomic weights. This dilemma was finally resolved in 1961 when chemists and physicists agreed to set the atomic weight of the carbon-12 isotope as 12.slightly different values for the atomic weights of the elements. 5 .
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