In an op-amp circuit, the overall gain is always very slightly less, than the calculated gain, because the op-amp has a finite gain. The higher the op-amp's gain, relative to the circuit's ideal gain, the closer it will be. The simplified op-amp gain equation, assumes the op-amp's open loop gain is so high, the error is small enough not to matter. The more accurate formula is below:
G = G
OL/(1+βG
OL)
Where:
G = the closed loop gain
G
OL = the op-amp's open loop gain
β = the negative feedback through the resistive divider
Take a simple inverting amplifier.

Normally we just say AV = RF/RIN = 100k/1k = 100, but let's work out the exact gain, taking into account the op-amp's open loop gain.
β = RIN/RF = 1k/100k = 0.01
According to the data sheet, the op-amp has a gain of 100dB
G
OL = 100 000
G = G
OL/(1+βG
OL) = 100 000/(1+0.01*100 000) = 100 000/(1+1000) = 100 000/1001 = 99.9
If 1% resistors were used to build the circuit, then they would contribute more error, than the op-amp's open loop gain.
The open loop gain is lower at higher frequencies, say at 1kHz, it's 60dB.
G
OL = 1000
G = 1000/(1+1000*0.01) = 1000/11 = 90.9
That's nearly 10% lower than the target, which could be unacceptable in some applications.
Exercise for the reader: what would G be, at 1kHz, if two amplifiers, each with a gain of 31.62 used in series, rather than one amplifier. Now take it further and try three amplifiers, each with a gain of 10.