The main issue is depending on the option used, the swap command is about two quite different concepts.
In your example you have roughly 16 GB of virtual memory, from which:
This 14.5 GB of available memory is partially RAM, partially disk (swap area) unless there is no swap area at all, not your case..

One can conclude that you have 4.74 GB of available RAM. (14.5G - 10G + 240M)

Free RAM being wasted RAM, a substantial part of this so called available RAM is containing data anyway, essentially disk cache. The difference between it and the allocated RAM is the former can be stolen instantaneously without harm, the previously cached data being still available on disk.

If we sum up the available RAM (4.74 GB) and the used RAM (1.77 GB), we get 6.5 GB of RAM. This number looks odd and there is no doubt you have more RAM installed on this machine.

My guess is you have 8 GB or RAM. The 1.5 GB difference is not part of the virtual memory. It is held mainly by the kernel and possibly by hardware components as non pageable memory, always stored in RAM.

Note also that despite being unused, your swap area has still a positive effect on your system as 250 MB of RAM would have been made unusable (being reserved) without it.

Finally, note that a portion of the processes virtual memory space is not accounted in the "swap -s" statistics, this is the memory that correspond memory mapped files.