How to convert magnetic moment in emu/g to Bohr magneton?

We can convert the magnetic moment in emu/g (cgs unit) to Bohr magneton (SI unit) easily as explained below:

As we know that 1 emu = 1*e-3 J/T (* represents the multiplication sign, e-3 represents 10 raised to the power 3)
and 1 Bohr magneton = 9.27402*e-24 J/T

So 1 emu = 1.078*e+20 Bohr magneton

Now we need to find the mass of 1 formula unit (f.u.) of the compound.

e.g. For Ho2Ir2O7, we will write the respective molar masses:

Mass of 1 mole of Ho = 164 g

Mass of 1 mole of Ir = 192.2 g

Mass of 1 mole of O = 16 g

So mass of 1 atom is obtained by dividing by avogadro's number = no. of atoms in one mole of an atom (6.023 *e+23)

Mass of 1 Ho atom= 164.9/6.023*e23 = 27.378* e-23 g

Mass of 1 Ir atom = 192.2/6.023*e23 = 31.911* e-23 g

Mass of 1 O atom = 16/6.023*e23 = 2.656* e-23 g

Mass of 1 f.u. of Ho2Ir2O7 = [2(27.378) + 2(31.911) + 7(2.656)]* e-23 = 137.17*e-23 g

Now Moment of 1 f.u. = moment in emu/g * mass of 1 f.u. in g
                                      = moment in emu
Now we multiply the result by  factor 1.078*e+20 (1 emu = 1.078*e+20 Bohar magneton), we get the the magnetic moment in Bohr Magneton.

So three simple steps:

1) Calculate mass of one f.u. of the respective compound

2) Multiply moment in emu/g by mass of one f.u.

3) Further multiply by a  factor 1.078*e+20

Please leave comment if I missed something.

Thanks.