On the disk an individual bit weighs nothing, it's just a change in magnetic polarity; see TheTXI's answer for a more elaborate explanation of this.

In RAM, however, bits are comprised of electrons (or lack thereof) and they *do* have a mass which is about 9.10938215 Ã— 10^{âˆ’31} kg. So for a GiB of memory, assuming equal distribution for zero and one bits, we get around

4294967296 *n* Ã— 9.10938215 Ã— 10^{âˆ’31} kg

4294967296 would be the number of one bits in memory (assumed to be 50Â %) and *n* would be the number of electrons that are on average in one bit. I have found one source^{1} that specified this number at around 10^{5}.

So we can give an estimate of how much mass 1 GiB (or 1 GB) of memory would have:

1 GiB, half filled with ones â‰ˆ 3.91 Ã— 10^{âˆ’16} kg = 391 femtograms

1 GiB, completely filled with ones â‰ˆ 7.82 Ã— 10^{-16} kg = 782 femtograms

1 GB, half filled with ones â‰ˆ 3.64 Ã— 10^{âˆ’16} kg = 364 femtograms

1 GB, completely filled with ones â‰ˆ 7.29 Ã— 10^{âˆ’16} kg = 729 femtograms

So in general you can assume that weight to be pretty unnoticeable (or, with hard disks to be downright nonexistant).

^{1} These lecture slides, but they are in German.

`7x`

. I'll leave determining the value of`x`

as an exercise for the reader. – Pesto Jul 23 '09 at 13:54