Dear Ivan,
at first you take the atomic masses from the periodic system of elements. We take 1mol from every component - that is the atomic weight in g:
1 mol Mg: 24,3 g
1 mol Sb: 121,8 g
1 mol Na: 23 g
To fabricate 1 mol Mg3Sb2 we need:
3*24,3 g Mg + 2*121,8 g Sb = 72,9 g Mg + 243,6 g Sb.
The mass ratio can be changed of course by a constant factor, if you need more or less as 316,5 g Mg3Sb2.
Now the Na doping. Therefore we need the expected concentration of Na. Lets assume, you want to have a concentration of 1017 atoms Na per mol. A mol contains 6*1023 particles (molecules Mg3Sb2). That means you need
23 g Na*1017 /6*1023 = 3,8 µg Na. You can calculate the volume concentration if you have the material densities.
With Regard
R. Mitdank
Herr Professor Rüdiger & Ivan Opara
Obviously the only issue here would be trying to add 3.8 µg Na to anything. One could perhaps make up a more concentrated solution and dilute accordingly. Otherwise doping levels would have to be increased at least a thousand-fold (10000 X would be better) to limit errors in weighing.
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