Putting a tiny bit of science into the thread for anyone interested and just for fun . . .
Assume there are two earphones:
EP1 = 16 ohms, 101db/mw
ER2 = 120 ohms, 118db/mw
Ohms law: V = I * R
Power = V * I = I**2 *R = V**2 / R
If an amp can drive 1V rms, for example, and the EP1s impedance is 16 ohms (assumed constant across the frequency spectrum, which it is not), the drivers would dissipate 62.5 mW. If EP2s impedance is 120 ohms, the drivers would dissipate 8.3 mW.
A standard equation for db gives (Wikipedia) (I'm never sure when to pick this power equation or the amplitude equation where 10 is replaced by 20 when making this calculation. Any inputs?)
Ldb = 10 Log (P1/Preference)
where P1 = input power and Preference equals the power which gives the rated efficiency at 1mW input.
Doing a bit of algebra to get Preference for each phone and calculating the final value at full power:
EP1 should give a maximum loudness of 118 db for 1 volt drive
EP2 should give a maximum loudness of 127 db for 1 volt drive
So, for the assumptions made on driver efficiency and impedance, the lower impedance driver has a lower volume at 1V rms output.
If the efficiency of EP1 is increased, it can easily exceed the output power of the EP2, but then, 127 db is plenty loud, IMO.