Don Herbison-Evans

donherbisonevans@outlook.com

26 August 2014, revised 6 October 2017

**Introduction**

No-one has actually weighed electrons, and maybe their gravitational mass is not the same as their inertial mass as Einstein assumed in his Equivalence Principle. The Eotvos Experiment indirectly implies that electrons are subject to gravity, but it would be nice actually to weigh them.

Electrons constitute the highest proportion of the mass of molecules that are composed of atoms of the lowest atomic number, so it is tempting to use the lightest elements for this experiment. However measurements based on chemical reactions can be complicated by changes in isotopic ratios for molecules containing elements with more than one naturally ocurring stable isotope. Thus the only elements in the lightest dozen that have only one stable isope and so avoid this fractionation are Be, F and Na.

Electrons only contribute about 549 micrograms per Grm Dalton, but balances are now commercially available that can measure several grams to an accuracy of 1 microgram, such as the Sartorius Cubis MSE6.6S-0CE-DM balance.

In the following considerations, we shall use

N = Avogadro's number

and

Atomic numbers (number of electrons per atom):

Be 4

F 9

and

Atomic (inertial) masses derived from mass spectrometry:

Be 9.01218315 Daltons

F 18.99840316 Daltons

e 0.00054858 = 548 micro-Daltons

**Be to BeF _{2} with Weighty Electrons**

Consider the simple conversion of metallic Beryllium to Beryllium Fluoride:

1 gm atom of Be containing N atoms:

= 9.01218315 grams

1 gm molecule of BeF_{2}

= 9.01218315 + 2 x 18.99840316 = 47.00898947 grams

Normalising : 1 gram of Beryllium would produce

47.00898947 / 9.01218315 = 5.21616002 grams BeF_{2}

**Be to BeF _{2} with Weightless Electrons**

1 gm atom of Be containing N atoms minus the mass of 4.N electrons:

= 9.01218315 - 4 x 0.00054858 = 9.01218315 - 0.00219432 = 9.00998883 grams

1 gm molecule of BeF_{2} minus the 4.N electrons in the Be,
and 2 x 9.N electrons in the 2F:

= 47.00898947 - 22 x 0.00054858 = 47.00898947 - 0.01206876 = 46.99692071 grams

Normalising : 1 gram of Beryllium would produce

46.99692071 / 9.00998883 = 5.21609090 grams BeF_{2}

The difference in the weight of BeF_{2} produced by
by 1 gram of Beryllium is

5.21616002 - 5.21609090 = 69.12 micrograms BeF_{2}

**Discussion**

Using inertial masses, 1gm Be with HF would produce 5.21616002 gms BeF2 if electrons are subject to gravity, and 5.21616002 if not, a difference of 69.12 micrograms easily measured by good balances. Has anyone done this?

One would of course work with arbitrary but accurately weighed amounts of Be and scale the results accordingly. Be and F are monoisotopic so there are no fractionation problems to be considered in these reactions, but spray, impurity levels, and corrosion are problems that would need solving.

One might alternatively consider other reactions involving the other mono-isotopic elements:

Number | Atomic Mass | |

For the general case, consider the conversion of molecule S with molecular mass 'S' Daltons into a molecule T with molecular mass 'T' Daltons. Let 's' be the number of electrons in S, and 't' be the number electrons in T. Let 'e' be the mass of an electron in Daltons. Then if electrons have gravitational weight: then 1 gram of S will produce T/S grams of T. If electrons have no gravitational mass: (S-se) grams of S will produce (T-te)grams of T. So 1 gram of S will produce (T-te)/(S-se) grams of T. The difference 'x' will be

= e.(Ts - St)/[S.(S-se)] ≈ e.(T/S

So for the example above:
Be -> BeF_{2}: S = 9, s = 4, T = 47, t = 22:

**Question To Quora: 6 June 2017
1gm Be with HF produces 5.216160 gm BeF2 if gravity affects electrons, 5.216090 if not, difference = 70 mugm, easily measured, so why no-one done this?
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