According to here, a serial data frame looks like this:
9600 8N1 - 9600 baud, 8 data bits, no parity, and 1 stop bit
So it implies the total data bits are always a multiple of 8 bits.
What if I have only 15 bits to transfer?
Will the 2 data frames be like this?
Frame 1: 8-N-1
Frame 2: 7-N-1
But what if I want to add parity? The number of bits to calculate the parity are different...
You will need to pad both frames so they have 8 bits of data plus the parity bit (if desired). You also need to figure out how the receiver will be able to tell which frame is supposed to be the first frame.
Both ends have to have the same parameters. You normally would not change parameters during a session.
If you set the parameters to 8 N 1, then you'd send two frames.
Historically, you'd normally be sending ASCII characters which are 7 bits and parity. This was back in the day of Teletypes. Nowadays, 8-N-1 tends to be the default choice.
Yes, your serial line parameters can always be 8N1. There is little point in changing it during communications.
If you use 8N1 like most devices do, you are then free to send 8-bit bytes over the serial line, which usually makes most sense for binary data. If you want parity then choose one and the line parameters are 8P1 for all bytes, P being the selected parity scheme, such as odd or even.
Then you are free to choose any method you want to transmit 15 bits as two bytes over the serial link.
The link you have provided answers your questions:
The data size can be from 5 to 9 bits but it has to be agreed before the start of the communication. It cannot be changed during:
First the parity bit is not part of the calculation for the parity check. Also, the parity bit is not part of the chunk of data agreed to be transmitted, it is added to the string of bits, after the data, like in the picture:
