I am guessing you are talking about 2M with a bus width of 16 bits from 512K with a bus width of 8 bits? As such there are 1024K in 1M because a computer counts in binary and not decimal. Plus you are doubling the bus width so you need twice as many per bus access. Expect is to run really slow though, and you might need to verify that the system BIOS can run RAM at the access speed of the older chips.
512K = 0.5M * 2 per bus = 8 - 512K x 8 bit bus needed for 2M x 16 bit bus